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d of reliability depends on a variety of continuous random variables. 326 NOTE Chapter 5 \| Continuous Random Variables The values of discrete and continuous random variables can be ambiguous. For example, if X is equal to the number of miles (to the nearest mile) you drive to work, then X is a discrete random variable. You count the miles. If X is the distance you drive to work, then you measure values of X and X is a continuous random variable. For a second example, if X is equal to the number of books in a backpack, then X is a discrete random variable. If X is the weight of a book, then X is a continuous random variable because weights are measured. How the random variable is defined is very important. Properties of Continuous Probability Distributions The graph of a continuous probability distribution is a curve. Probability is represented by the area under the curve. The curve is called the probability density function (abbreviated as pdf). We use the symbol f(x) to represent the curve. f(x) is the function that corresponds to the graph; we use the density function f(x) to draw the graph of the probability distribution. Area under the curve is given by a different function called the cumulative distribution function (abbreviated as cdf). The cumulative distribution function is used to evaluate probability as area. • The outcomes are measured, not counted. • The entire area under the curve and above the x-axis is equal to one. • Probability is found for intervals of x values rather than for individual x values. • P(c < x < d) is the probability that the random variable X is in the interval between the values c and d. P(c < x < d) is the area under the curve, above the x-axis, to the right of c and the left of d. • P(x = c) = 0 The probability that x takes on any single individual value is zero. The area below the curve, above the x-axis, and between x = c and x = c has no width, and therefore no area (area = 0). Since the probability is equal to the area, the probability is also zero. • P(c < x < d) is the same as P(c ≤ x ≤ d) because probability is equal to area. We will find the area that represents probability by using geometry, formulas, technology, or probability tables. In general, calculus is needed to find the area under the curve for many probability density functions. When we use formulas to find the area in this textbook, we are using formulas that were found by using the techniques of integral calculus. However, because most students taking this course have not studied calculus, we will not be using calculus in this textbook. There are many continuous probability distributions. When probability is modeled by use of a continuous probability distribution, the distribution used is selected to model and fit the particular situation in the best way. In this chapter and the next, we will study the uniform distribution, the exponential distribution, and the normal distribution. The following graphs illustrate these distributions: Figure 5.2 The graph shows a uniform distribution with the area between x = 3 and x = 6 shaded to represent the probability that the value of the random variable X is in the interval between three and six. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 327 Figure 5.3 The graph shows an exponential distribution with the area between x = 2 and x = 4 shaded to represent the probability that the value of the random variable X is in the interval between two and four. Figure 5.4 The graph shows the standard normal distribution with the area between x = 1 and x = 2 shaded to represent the probability that the value of the random variable X is in the interval between one and two. 5.1 \| Continuous Probability Functions We begin by defining a continuous probability density function. We use the function notation f(x). Intermediate algebra may have been your first formal introduction to functions. In the study of probability, the functions we study are special. We define the function f(x) so that the area between it and the x-axis is equal to a probability. Since the maximum probability is one, the maximum area is also one. For continuous probability distributions, PROBABILITY = AREA. Example 5.1 Consider the function f(x) = 1 20 for 0 ≤ x ≤ 20. x = a real number. The graph of f(x) = 1 20 is a horizontal line. However, since 0 ≤ x ≤ 20, f(x) is restricted to the portion between x = 0 and x = 20, inclusive. 328 Chapter 5 \| Continuous Random Variables Figure 5.5 f(x) = 1 20 for 0 ≤ x ≤ 20. The graph of f(x) = 1 20 is a horizontal line segment when 0 ≤ x ≤ 20. The area between f(x) = 1 20 = 1 20 . where 0 ≤ x ≤ 20 and the x-axis is the area of a rectangle with base = 20 and height Suppose we want to find the area between f(x) = 1 20 and the x-axis where 0 < x < 2. AREA = 20 ⎛ ⎝ 1 20 ⎞ ⎠ = 1 Figure 5.6 ⎛ AREA = (2 – 0) ⎝ ⎞ ⎠ = 0.1 1 20 (2 – 0) = 2 = base of a rectangle REMINDER area of a rectangle = (base)(height) The area corresponds to a probability. The probability that x is between zero and two is 0.1, which can be written This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 329 mathematically as P(0 < x < 2) = P(x < 2) = 0.1. Suppose we want to find the area between f(x) = 1 20 and the x-axis where 4 < x < 15. Figure 5.7 ⎛ AREA = (15 – 4) ⎝ ⎞ ⎠ = 0.55 1 20 (15 – 4) = 11 = the base of a rectangle The area corresponds to the probability P(4 < x < 15) = 0.55. Suppose we want to find P(x = 15). On an x-y graph, x = 15 is a vertical line. A vertical line has no width (or zero ⎞ width). Therefore, P(x = 15) = (base)(height) = (0) ⎛ ⎠ ⎝ = 0 1 20 Figure 5.8 P(X <= x), which can also be written as P(X < x) for continuous distributions, is called the cumulative distribution function or CDF. Notice the less than or equal to symbol. We can also use the CDF to calculate P(X > x). The CDF gives area to the left and P(X > x) gives area to the right. We calculate P(X > x) for continuous distributions as follows: P(X > x) = 1 – P (X < x). 330 Chapter 5 \| Continuous Random Variables Figure 5.9 Label the graph with f(x) and x. Scale the x and y axes with the maximum x and y values. f(x) = 1 20 , 0 ≤ x ≤ 20. To calculate the probability that x is between two values, look at the following graph. Shade the region between x = 2.3 and x = 12.7. Then calculate the shaded area of a rectangle. Figure 5.10 ⎛ P(2.3 < x < 12.7) = (base)(height) = (12.7 − 2.3) ⎝ ⎞ ⎠ = 0.52 1 20 5.1 Consider the function f(x) = 1 8 for 0 ≤ x ≤ 8. Draw the graph of f(x) and find P(2.5 < x < 7.5). 5.2 \| The Uniform Distribution The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. When working out problems that have a uniform distribution, be careful to note if the data are inclusive or exclusive of endpoints. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 331 Example 5.2 The data in Table 5.1 are 55 smiling times, in seconds, of an eight-week-old baby. 10.4 19.6 18.8 13.9 17.8 16.8 21.6 17.9 12.5 11.1 4.9 12.8 14.8 22.8 20.0 15.9 16.3 13.4 17.1 14.5 19.0 22.8 1.3 5.8 8.9 0.7 6.9 9.4 8.9 2.6 9.4 11.9 10.9 7.3 5.9 5.8 7.6 21.7 11.8 3.4 10.0 3.3 6.7 3.7 2.1 7.8 17.9 19.2 9.8 4.5 6.3 10.7 11.6 13.8 18.6 Table 5.1 The sample mean = 11.49 and the sample standard deviation = 6.23. We will assume that the smiling times, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. This means that any smiling time from zero to and including 23 seconds is equally likely. The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution. Let X = length, in seconds, of an eight-week-old baby's smile. The notation for the uniform distribution is X ~ U(a, b) where a = the lowest value of x and b = the highest value of x. The probability density function is f(x) = For this example, X ~ U(0, 23) and f(x) = 1 b − a for a ≤ x ≤ b. 1 23 − 0 for 0 ≤ X ≤ 23. Formulas for the theoretical mean and standard deviation are For this problem, the theoretical mean and standard deviation are μ = a + b 2 and σ = (b − a)2 12 μ = 0 + 23 2 = 11.50 seconds and σ = (23 − 0)2 12 = 6.64 seconds. Notice that the theoretical mean and standard deviation are close to the sample mean and standard deviation in this example. 332 Chapter 5 \| Continuous Random Variables 5.2 The data that follow are the number of passengers on 35 different charter fishing boats. The sample mean = 7.9 and the sample standard deviation = 4.33. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. State the values of a and b. Write the distribution in proper notation, and calculate the theoretical mean and standard deviation. 1 12 4 10 4 14 11 7 11 4 13 2 4 6 3 10 0 12 6 9 10 5 13 4 10 14 12 11 6 10 11 0 11 13 2 Table 5.2 Example 5.3 a. Refer to Example 5.2. What is the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds? Solution 5.3 P(2 < x < 18) = (base)(height) = (18 – 2) ⎛ ⎝ ⎞ ⎠ 1 23 = 16 23 Figure 5.11 b. Find the 90th percentile for an eight-week-old baby's smiling time. Solution 5.3 b. Ninety percent of the smiling times fall below the 90th percentile, k, so P(x < k) = 0.90. ⎛ ⎛ ⎠ ⎝base⎞ P(x < k) = 0.90 ⎝height⎞ ⎠ = 0.90 ⎞ ⎛ ⎠ = 0.90 (k − 0) ⎝ 1 23 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 333 k = (23)(0.90) = 20.7 Figure 5.12 c. Find the probability that a random eight-week-old baby smiles more than 12 seconds knowing that the baby smiles more than eight seconds. Solution 5.3 c. This probability question is a conditional. You are asked to find the probability that an eight-week-old baby smiles more than 12 seconds when you already know the baby ha
s smiled for more than eight seconds. Find P(x > 12\|x > 8) There are two ways to do the problem. For the first way, use the fact that this is a conditional and changes the sample space. The graph illustrates the new sample space. You already know the baby smiled more than eight seconds. for 8 < x < 23 Write a new f(x): f(x) = 1 23 − 8 = 1 15 for 8 < x < 23. P(x > 12\|x > 8) = (23 − 12) ⎛ ⎝ ⎞ ⎠ 1 15 = 11 15 Figure 5.13 For the second way, use the conditional formula from Probability Topics with the original distribution. P(A\|B) = P(A AND B) P(B) For this problem, A is (x > 12) and B is (x > 8). 334 Chapter 5 \| Continuous Random Variables So, P(x > 12\|x > 8) = (x > 12 AND x > 8) P(x > 8) = P(x > 12) P(x > 8) = 11 23 15 23 = 11 15 Figure 5.14 5.3 A distribution is given as X ~ U(0, 20). What is P(2 < x < 18)? Find the 90th percentile. Example 5.4 The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive. a. What is the probability that a person waits fewer than 12.5 minutes? Solution 5.4 a. Let X = the number of minutes a person must wait for a bus. a = 0 and b = 15. X ~ U(0, 15). Write the probability density function. f (x) = 1 15 − 0 = 1 15 for 0 ≤ x ≤ 15. Find P (x < 12.5). Draw a graph. ⎛ P(x < k) = (base)(height) = (12.5 - 0) ⎝ ⎞ ⎠ = 0.8333 1 15 The probability a person waits fewer than 12.5 minutes is 0.8333. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 335 Figure 5.15 b. On the average, how long must a person wait? Find the mean, μ, and the standard deviation, σ. Solution 5.4 b. μ = a + b 2 = 15 + 0 2 σ = (b - a)2 12 = (15 - 0)2 12 = 7.5. On the average, a person must wait 7.5 minutes. = 4.3. The standard deviation is 4.3 minutes. c. Ninety percent of the time, the minutes a person must wait falls below what value? This question asks for the 90th percentile. Solution 5.4 c. Find the 90th percentile. Draw a graph. Let k = the 90th percentile. P(x < k) = (base)(height) = (k − 0)( 1 15 ) ⎛ 0.90 = (k) ⎝ ⎞ ⎠ 1 15 k = (0.90)(15) = 13.5 k is sometimes called a critical value. The 90th percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes. 336 Chapter 5 \| Continuous Random Variables Figure 5.16 5.4 The total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive. a. Find a and b and describe what they represent. b. Write the distribution. c. Find the mean and the standard deviation. d. What is the probability that the duration of games for a team for the 2011 season is between 480 and 500 hours? e. What is the 65th percentile for the duration of games for a team for the 2011 season? Example 5.5 Suppose the time it takes a nine-year old to eat a donut is between 0.5 and 4 minutes, inclusive. Let X = the time, in minutes, it takes a nine-year-old child to eat a doughnut. Then X ~ U(0.5, 4). a. The probability that a randomly selected nine-year-old child eats a doughnut in at least two minutes is _______. Solution 5.5 a. 0.5714 b. Find the probability that a different nine-year-old child eats a doughnut in more than two minutes given that the child has already been eating the doughnut for more than 1.5 minutes. The second question has a conditional probability. You are asked to find the probability that a nine-year-old child eats a doughnut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes. Solve the problem two different ways (see Example 5.3). You must reduce the sample space. First way: Since you know the child has already been eating the doughnut for more than 1.5 minutes, you are no longer starting at a = 0.5 minutes. Your starting point is 1.5 minutes. Write a new f(x): Find P(x > 2\|x > 1.5). Draw a graph. f (x) = 1 4 − 1.5 = 2 5 for 1.5 ≤ x ≤ 4. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 337 Figure 5.17 Solution 5.5 b. 4 5 P(x > 2\|x > 1.5) = (base)(new height) = (4 – 2 The probability that a nine-year-old child eats a donut in more than two minutes given that the child has already been eating the doughnut for more than 1.5 minutes is 4 5 . Second way: Draw the original graph for X ~ U(0.5, 4). Use the conditional formula P(x > 2\|x > 1.5) = P(x > 2 AND x > 1.5) P(x > 1.5) = P(x > 2) P(x > 1.5) = 2 3.5 2.5 3.5 = 0.8 = 4 5 5.5 Suppose the time it takes a student to finish a quiz is uniformly distributed between six and 15 minutes, inclusive. Let X = the time, in minutes, it takes a student to finish a quiz. Then X ~ U(6, 15). Find the probability that a randomly selected student needs at least eight minutes to complete the quiz. Then find the probability that a different student needs at least eight minutes to finish the quiz given that she has already taken more than seven minutes. Example 5.6 Ace Heating and Air Conditioning Service finds that the amount of time a repairman needs to fix a furnace is uniformly distributed between 1.5 and four hours. Let x = the time needed to fix a furnace. Then x ~ U(1.5, 4). a. Find the probability that a randomly selected furnace repair requires more than two hours. b. Find the probability that a randomly selected furnace repair requires less than three hours. c. Find the 30th percentile of furnace repair times. d. The longest 25 percent of furnace repair times take at least how long? (In other words: find the minimum time for the longest 25 percent of repair times.) What percentile does this represent? e. Find the mean and standard deviation 338 Chapter 5 \| Continuous Random Variables Solution 5.6 a. To find f(x): f (x) = 1 4 − 1.5 = 1 2.5 so f(x) = 0.4 P(x > 2) = (base)(height) = (4 – 2)(0.4) = 0.8 Figure 5.18 Uniform distribution between 1.5 and four with shaded area between two and four representing the probability that the repair time x is greater than two Solution 5.6 b. P(x < 3) = (base)(height) = (3 – 1.5)(0.4) = 0.6 The graph of the rectangle showing the entire distribution would remain the same. However the graph should be shaded between x = 1.5 and x = 3. Note that the shaded area starts at x = 1.5 rather than at x = 0. Because X ~ U(1.5, 4), x cannot be less than 1.5. Figure 5.19 Uniform distribution between 1.5 and four with shaded area between 1.5 and three representing the probability that the repair time x is less than three Solution 5.6 c. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 339 Figure 5.20 Uniform distribution between 1.5 and 4 with an area of 0.30 shaded to the left, representing the shortest 30 percent of repair times. P (x < k) = 0.30 P(x < k) = (base)(height) = (k – 1.5)(0.4) 0.3 = (k – 1.5) (0.4); Solve to find k: 0.75 = k – 1.5, obtained by dividing both sides by 0.4 k = 2.25 , obtained by adding 1.5 to both sides The 30th percentile of repair times is 2.25 hours. 30 percent of repair times are 2.5 hours or less. Solution 5.6 d. Figure 5.21 Uniform distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing the longest 25 percent of repair times. P(x > k) = 0.25 P(x > k) = (base)(height) = (4 – k)(0.4) 0.25 = (4 – k)(0.4); Solve for k: 0.625 = 4 − k, obtained by dividing both sides by 0.4 −3.375 = −k, obtained by subtracting four from both sides: k = 3.375 The longest 25 percent of furnace repairs take at least 3.375 hours (3.375 hours or longer). Note: Since 25 percent of repair times are 3.375 hours or longer, that means that 75 percent of repair times are 3.375 hours or less. 3.375 hours is the 75th percentile of furnace repair times. 340 Chapter 5 \| Continuous Random Variables Solution 5.6 e. μ = a + b 2 and σ = (b − a)2 12 μ = 1.5 + 4 2 = 2.75 hours and σ = (4 – 1.5)2 12 = 0.7217 hours 5.6 The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. Let X = the time needed to change the oil on a car. a. Write the random variable X in words. X = __________________. b. Write the distribution. c. Graph the distribution. d. Find P (x > 19). e. Find the 50th percentile. 5.3 \| The Exponential Distribution (Optional) The exponential distribution is often concerned with the amount of time until some specific event occurs. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length, in minutes, of long-distance business telephone calls, and the amount of time, in months, a car battery lasts. It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution. Values for an exponential random variable occur in the following way. There are fewer large values and more small values. For example, the amount of money customers spend in one trip to the supermarket follows an exponential distribution. There are more people who spend small amounts of money and fewer people who spend large amounts of money. Exponential distributions are commonly used in calculations of product reliability, or the length of time a product lasts. Example 5.7 Let X = amount of time (in minutes) a postal clerk spends with his or her customer. The time is known to have an exponential distribution with the average amount of time equal to four minutes. X is a continuous random variable since time is measured. It is given that μ = 4 minutes. To do any calculations, you must know m, the decay parameter. m = 1 μ . Therefore, m = 1 4 = 0.25. The standard deviation, σ, is the same as the mean. μ = σ The distribution notation is X ~ Exp(m). Therefore, X ~ Exp(0.25). The probability density function is f(x) = me-mx. The number e = 2.71828182846... It is a number that is used often in mathematics. Scientific calculators have the key "ex." If you enter one for x, the calculator will di
splay the value e. The curve is f(x) = 0.25e–0.25x where x is at least zero and m = 0.25. For example, f(5) = 0.25e(−0.25)(5) = 0.072. The probability that the postal clerk spends five minutes with the This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 341 customers is 0.072. The graph is as follows: Figure 5.22 Notice the graph is a declining curve. When x = 0, f(x) = 0.25e(−0.25)(0) = (0.25)(1) = 0.25 = m. The maximum value on the y-axis is m. 5.7 The amount of time spouses shop for anniversary cards can be modeled by an exponential distribution with the average amount of time equal to eight minutes. Write the distribution, state the probability density function, and graph the distribution. Example 5.8 a. Using the information in Example 5.7, find the probability that a clerk spends four to five minutes with a randomly selected customer. Solution 5.8 a. Find P(4 < x < 5). The cumulative distribution function (CDF) gives the area to the left. P(x < x) = 1 – e – mx P(x < 5) = 1 – e ( – 0.25)(5) = 0.7135 and P(x < 4) = 1 – e ( – 0.25)(4) = 0.6321 342 Chapter 5 \| Continuous Random Variables Figure 5.23 NOTE You can do these calculations easily on a calculator. The probability that a postal clerk spends four to five minutes with a randomly selected customer is P(4 < x < 5) = P(x < 5) – P(x < 4) = 0.7135 − 0.6321 = 0.0814. On the home screen, enter (1 – e^(–0.255))–(1–e^(–0.254)) or enter e^(–0.254) – e^(–0.255). b. Half of all customers are finished within how long? (Find the 50th percentile). Solution 5.8 b. Find the 50th percentile. Figure 5.24 P(x < k) = 0.50, k = 2.8 minutes (calculator or computer) Half of all customers are finished within 2.8 minutes. You can also do the calculation as follows: This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 343 P(x < k) = 0.50 and P(x < k) = 1 – e – 0.25k Therefore, 0.50 = 1 − e−0.25k and e−0.25k = 1 − 0.50 = 0.5. Take natural logs: ln(e–0.25k) = ln(0.50). So, –0.25k = ln(0.50). Solve for k: k = ln(0.50) -0.25 following two notes. NOTE = 2.8 minutes. The calculator simplifies the calculation for percentile k. See the A formula for the percentile k is k = ln(1 − AreaToTheLe f t) −m where ln is the natural log. On the home screen, enter ln(1 – 0.50)/–0.25. Press the (–) for the negative. c. Which is larger, the mean or the median? Solution 5.8 c. From Part b, the median or 50th percentile is 2.8 minutes. The theoretical mean is four minutes. The mean is larger. 5.8 The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days. Find the probability that a traveler will purchase a ticket fewer than 10 days in advance. How many days do half of all travelers wait? Have each class member count the change he or she has in his or her pocket or purse. Your instructor will record the amounts in dollars and cents. Construct a histogram of the data taken by the class. Use five intervals. Draw a smooth curve through the bars. The graph should look approximately exponential. Then calculate the mean. Let X = the amount of money a student in your class has in his or her pocket or purse. The distribution for X is approximately exponential with mean, μ = _______ and m = _______. The standard deviation, σ = ________. Draw the appropriate exponential graph. You should label the x– and y–axes, the decay rate, and the mean. Shade the area that represents the probability that one student has less than $0.40 in his or her pocket or purse. (Shade P(x < 0.40)). 344 Chapter 5 \| Continuous Random Variables Example 5.9 On the average, a certain computer part lasts 10 years. The length of time the computer part lasts is exponentially distributed. a. What is the probability that a computer part lasts more than seven years? Solution 5.9 a. Let x = the amount of time (in years) a computer part lasts. Find P(x > 7). Draw the graph. μ = 10, so m = 1 μ = 1 10 = 0.1 P(x > 7) = 1 – P(x < 7). Since P(X < x) = 1 – e–mx then P(X > x) = 1 – (1 – e–mx) = e-mx P(x > 7) = e(–0.1)(7) = 0.4966. The probability that a computer part lasts more than seven years is 0.4966. On the home screen, enter e^(-.17). Figure 5.25 b. On the average, how long would five computer parts last if they are used one after another? Solution 5.9 b. On the average, one computer part lasts 10 years. Therefore, five computer parts, if they are used one right after the other would last, on the average, (5)(10) = 50 years. c. Eighty percent of computer parts last at most how long? Solution 5.9 c. Find the 80th percentile. Draw the graph. Let k = the 80th percentile. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 345 Figure 5.26 Solve for k: Eighty percent of the computer parts last at most 16.1 years. k = ln(1 – 0.80) – 0.1 = 16.1years. On the home screen, enter ln(1 – 0.80) – 0.1 . d. What is the probability that a computer part lasts between nine and 11 years? Solution 5.9 d. Find P(9 < x < 11). Draw the graph. Figure 5.27 P(9 < x < 11) = P(x < 11) – P(x < 9) = (1 – e(–0.1)(11)) – (1 – e(–0.1)(9)) = 0.6671 – 0.5934 = 0.0737. The probability that a computer part lasts between nine and 11 years is 0.0737. 346 Chapter 5 \| Continuous Random Variables On the home screen, enter e^(–0.19) – e^(–0.1*11). 5.9 On average, a pair of running shoes can last 18 months if used every day. The length of time running shoes last is exponentially distributed. What is the probability that a pair of running shoes last more than 15 months? On average, how long would six pairs of running shoes last if they are used one after the other? Eighty percent of running shoes last at most how long if used every day? Example 5.10 Suppose that the length of a phone call, in minutes, is an exponential random variable with decay parameter 1 12 . If another person arrives at a public telephone just before you, find the probability that you will have to wait more than five minutes. Let X = the length of a phone call, in minutes. What is m, μ, and σ? The probability that you must wait more than five minutes is _______ . Solution 5.10 • m = 1 12 • μ = 12 • σ = 12 P(x > 5) = 0.6592 5.10 Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay parameter 1 . Let X = the distance people are willing to commute in miles. What is m, μ, and σ? What 20 is the probability that a person is willing to commute more than 25 miles? Example 5.11 The time spent waiting between events is often modeled using the exponential distribution. For example, suppose that an average of 30 customers per hour arrive at a store and the time between arrivals is exponentially distributed. a. On average, how many minutes elapse between two successive arrivals? b. When the store first opens, how long on average does it take for three customers to arrive? c. After a customer arrives, find the probability that it takes less than one minute for the next customer to arrive. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 347 d. After a customer arrives, find the probability that it takes more than five minutes for the next customer to arrive. e. Seventy percent of the customers arrive within how many minutes of the previous customer? f. Is an exponential distribution reasonable for this situation? Solution 5.11 a. Since we expect 30 customers to arrive per hour (60 minutes), we expect on average one customer to arrive every two minutes on average. b. Since one customer arrives every two minutes on average, it will take six minutes on average for three customers to arrive. c. Let X = the time between arrivals, in minutes. By Part a, μ = 2, so m = 1 2 = 0.5. Therefore, X ~ Exp(0.5). The cumulative distribution function is P(X < x) = 1 – e(–0.5)(x). Therefore P(X < 1) = 1 – e(–0.5)(1) ≈ 0.3935. 1 - e^(–0.5) ≈ 0.3935 Figure 5.28 d. P(X > 5) = 1 – P(X < 5) = 1 – (1 – e(−0.5)(5)) = e–2.5 ≈ 0.0821. Figure 5.29 348 Chapter 5 \| Continuous Random Variables 1 – (1 – e ^ (– 0.50)(5)) or e ^ (– 0.50)(5) e. We want to solve 0.70 = P(X < x) for x. Substituting in the cumulative distribution function gives 0.70 = 1 – e–0.5x, so that e−0.5x = 0.30. Converting this to logarithmic form gives –0.5x = ln(0.30), or x = ln(0.30) – 0.5 ≈ 2.41 minutes. Thus, 70 percent of customers arrive within 2.41 minutes of the previous customer. You are finding the 70th percentile k so you can use the formula k = ln(1 – Area _ To _ The _ Le f t _ O f _ k) ( – m) k = ln(1 – 0.70) ( – 0.5) ≈ 2.41 minutes Figure 5.30 f. This model assumes that a single customer arrives at a time, which may not be reasonable since people might shop in groups, leading to several customers arriving at the same time. It also assumes that the flow of customers does not change throughout the day, which is not valid if some times of the day are busier than others. 5.11 Suppose that on a certain stretch of highway, cars pass at an average rate of five cars per minute. Assume that the duration of time between successive cars follows the exponential distribution. a. On average, how many seconds elapse between two successive cars? b. After a car passes by, how long on average will it take for another seven cars to pass by? c. Find the probability that after a car passes by, the next car will pass within the next 20 seconds. d. Find the probability that after a car passes by, the next car will not pass for at least another 15 seconds. Memorylessness of the Exponential Distribution In Example 5.7 recall that the amount of time between customers is exponentially distributed with a mean of two minutes (X ~ Exp(0.5)). Suppose that five minutes have elapsed since the last customer arrived. Since an unusually long amount of time has now elapsed,
it would seem to be more likely for a customer to arrive within the next minute. With the exponential This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 349 distribution, this is not the case—the additional time spent waiting for the next customer does not depend on how much time has already elapsed since the last customer. This is referred to as the memoryless property. Specifically, the memoryless property says the following P (X > r + t \| X > r) = P (X > t) for all r ≥ 0 and t ≥ 0 For example, if five minutes have elapsed since the last customer arrived, then the probability that more than one minute will elapse before the next customer arrives is computed by using r = 5 and t = 1 in the foregoing equation. P (X > 5 + 1 \| X > 5) = P (X > 1) = e ( – 0.5)(1) ≈ 0.6065. This is the same probability as that of waiting more than one minute for a customer to arrive after the previous arrival. The exponential distribution is often used to model the longevity of an electrical or a mechanical device. In Example 5.9, the lifetime of a certain computer part has the exponential distribution with a mean of ten years (X ~ Exp(0.1)). The memoryless property says that knowledge of what has occurred in the past has no effect on future probabilities. In this case it means that an old part is not any more likely to break down at any particular time than a brand new part. In other words, the part stays as good as new until it suddenly breaks. For example, if the part has already lasted ten years, then the probability that it lasts another seven years is P(X > 17\|X > 10) = P(X > 7) = 0.4966. Example 5.12 Refer to Example 5.7 where the time a postal clerk spends with his or her customer has an exponential distribution with a mean of four minutes. Suppose a customer has spent four minutes with a postal clerk. What is the probability that he or she will spend at least an additional three minutes with the postal clerk? The decay parameter of X is m = 1 4 = 0.25, so X ~ Exp(0.25). The cumulative distribution function is P(X < x) = 1 – e–0.25x. We want to find P(X > 7\|X > 4). The memoryless property says that P(X > 7\|X > 4) = P (X > 3), so we just need to find the probability that a customer spends more than three minutes with a postal clerk. This is P(X > 3) = 1 – P (X < 3) = 1 – (1 – e–0.25⋅3) = e–0.75 ≈ 0.4724. Figure 5.31 1–(1–e^(–0.253)) = e^(–0.253). 350 Chapter 5 \| Continuous Random Variables 5.12 Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. If a bulb has already lasted 12 years, find the probability that it will last a total of more than 19 years. Relationship Between the Poisson and the Exponential Distribution There is an interesting relationship between the exponential distribution and the Poisson distribution. Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of μ units of time. Also assume that these times are independent, meaning that the time between events is not affected by the times between previous events. If these assumptions hold, then the number of events per unit time follows a Poisson distribution with mean λ = 1/μ. Recall from the chapter on Discrete Random Variables that if X has the Poisson distribution with mean λ, then P(X = k) = λk e−λ k ! time between events follows the exponential distribution. (k! = k(k–1)(k–2)(k–3)…32*1) . Conversely, if the number of events per unit time follows a Poisson distribution, then the amount of Suppose X has the Poisson distribution with mean λ. Compute P(X = k) by entering 2nd, VARS(DISTR), C: poissonpdf(λ, k). To compute P(X ≤ k), enter 2nd, VARS (DISTR), D:poissoncdf(λ, k). Example 5.13 At a police station in a large city, calls come in at an average rate of four calls per minute. Assume that the time that elapses from one call to the next has the exponential distribution. Take note that we are concerned only with the rate at which calls come in, and we are ignoring the time spent on the phone. We must also assume that the times spent between calls are independent. This means that a particularly long delay between two calls does not mean that there will be a shorter waiting period for the next call. We may then deduce that the total number of calls received during a time period has the Poisson distribution. a. Find the average time between two successive calls. b. Find the probability that after a call is received, the next call occurs in less than 10 seconds. c. Find the probability that exactly five calls occur within a minute. d. Find the probability that fewer than five calls occur within a minute. e. Find the probability that more than 40 calls occur in an eight-minute period. Solution 5.13 a. On average four calls occur per minute, so 15 seconds, or 15 60 calls on average. b. Let T = time elapsed between calls. From Part a, μ = 0.25, so m = = 0.25 minutes occur between successive 1 0.25 = 4. Thus, T ~ Exp(4). The cumulative distribution function is P(T < t) = 1 – e–4t. The probability that the next call occurs in less than 10 seconds (10 seconds = 1/6 minute) is ⎞ ⎛ ⎝ ⎠ ⎞ ⎠ ⎛ ⎝.4866. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 351 Figure 5.32 c. Let X = the number of calls per minute. As previously stated, the number of calls per minute has a Poisson distribution, with a mean of four calls per minute. Therefore, X ~ Poisson(4), and so P(X = 5) = 45 e−4 5 ! poissonpdf(4, 5) = 0.1563 ≈ 0.1563. (5! = (5)(4)(3)(2)(1)) d. Keep in mind that X must be a whole number, so P(X < 5) = P(X ≤ 4). To compute this, we could take P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4). Using technology, we see that P(X ≤ 4) = 0.6288. poisssoncdf(4, 4) = 0.6288 e. Let Y = the number of calls that occur during an eight-minute period. Since there is an average of four calls per minute, there is an average of (8)(4) = 32 calls during each eight minute period. Hence, Y ~ Poisson(32). Therefore, P(Y > 40) = 1 – P(Y ≤ 40) = 1 – 0.9294 = 0.0706. 1 – poissoncdf(32, 40). = 0.0706 5.13 In a small city, the number of automobile accidents occur with a Poisson distribution at an average of three per week. a. Calculate the probability that at most two accidents occur in any given week. b. What is the probability that there are at least two weeks between any two accidents? 352 Chapter 5 \| Continuous Random Variables 5.4 \| Continuous Distribution This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 353 5.1 Continuous Distribution Student Learning Outcomes • The student will compare and contrast empirical data from a random number generator with the uniform distribution. Collect the Data Use a random number generator to generate 50 values between zero and one (inclusive). List them in Table 5.3. Round the numbers to four decimal places or set the calculator MODE to four places. 1. Complete the table. __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Table 5.3 2. Calculate the following: a. b. c. d. x¯ = _______ s = _______ first quartile = _______ third quartile = _______ e. median = _______ Organize the Data 1. Construct a histogram of the empirical data. Make eight bars. 354 Chapter 5 \| Continuous Random Variables Figure 5.33 2. Construct a histogram of the empirical data. Make five bars. Figure 5.34 Describe the Data 1. In two to three complete sentences, describe the shape of each graph. (Keep it simple. Does the graph go straight across, does it have a V shape, does it have a hump in the middle or at either end (and so on). One way to help you determine a shape is to draw a smooth curve roughly through the top of the bars.) 2. Describe how changing the number of bars might change the shape. Theoretical Distribution 1. In words, X = _____________________________________. 2. The theoretical distribution of X is X ~ U(0,1). 3. In theory, based upon the distribution X ~ U(0,1), complete the following. a. μ = ______ b. σ = ______ c. d. first quartile = ______ third quartile = ______ e. median = __________ 4. Are the empirical values (the data) in the section titled Collect the Data close to the corresponding theoretical values? Why or why not? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 355 Plot the Data 1. Construct a box plot of the data. Be sure to use a ruler to scale accurately and draw straight edges. 2. Do you notice any potential outliers? If so, which values are they? Either way, justify your answer numerically. (Recall that any data that are less than Q1 – 1.5(IQR) or more than Q3 + 1.5(IQR) are potential outliers. IQR means interquartile range.) Compare the Data 1. For each of the following parts, use a complete sentence to comment on how the value obtained from the data compares to the theoretical value you expected from the distribution in the section titled Theoretical Distribution: a. minimum value: _______ b. first quartile: _______ c. median: _______ d. third quartile: _______ e. maximum value: _______ f. width of IQR: _______ g. overall shape: _______ 2. Based on your comments in the section titled Collect the Data, how does the box plot fit or not fit what you would expect of the distribution in the section titled Theoretical Distribution? Discussion Question 1. Suppose that the number of v
alues generated was 500, not 50. How would that affect what you would expect the empirical data to be and the shape of its graph to look like? 356 Chapter 5 \| Continuous Random Variables KEY TERMS conditional probability the likelihood that an event will occur given that another event has already occurred decay parameter The decay parameter describes the rate at which probabilities decay to zero for increasing values of x. It is the value m in the probability density function f(x) = me(–mx) of an exponential random variable. It is also equal to m = 1 μ , where μ is the mean of the random variable. exponential distribution a continuous random variable (RV) that appears when we are interested in the intervals of time between some random events, for example, the length of time between emergency arrivals at a hospital; the notation is X ~ Exp(m). The mean is μ = 1 m . The probability density function is f(x) = me−mx, x ≥ 0 and and the standard deviation is σ = 1 m the cumulative distribution function is P(X ≤ x) = 1 − e−mx. memoryless property for an exponential random variable X, the statement that knowledge of what has occurred in the past has no effect on future probabilities This means that the probability that X exceeds x + k, given that it has exceeded x, is the same as the probability that X would exceed k if we had no knowledge about it. In symbols we say that P(X > x + k\|X > x) = P(X > k). Poisson distribution a distribution function that gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event; if there is a known average of λ events occurring per unit time, and these events are independent of each other, then the number of events X occurring in one unit of time has the Poisson distribution. The probability of k events occurring in one unit time is equal to P(X = k) = λk e−λ k ! . uniform distribution a continuous random variable (RV) that has equally likely outcomes over the domain, a < x < b. Notation—X ~ U(a,b). The mean is μ = a + b 2 and the standard deviation is σ = (b − a)2 12 . The probability density function is f(x) = 1 b − a for a < x < b or a ≤ x ≤ b. The cumulative distribution is P(X ≤ x) = x − a b − a . CHAPTER REVIEW 5.1 Continuous Probability Functions The probability density function (pdf) is used to describe probabilities for continuous random variables. The area under the density curve between two points corresponds to the probability that the variable falls between those two values. In other words, the area under the density curve between points a and b is equal to P(a < x < b). The cumulative distribution function (cdf) gives the probability as an area. If X is a continuous random variable, the probability density function (pdf), f(x), is used to draw the graph of the probability distribution. The total area under the graph of f(x) is one. The area under the graph of f(x) and between values a and b gives the probability P(a < x < b). This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 357 Figure 5.35 The cumulative distribution function (cdf) of X is defined by P (X ≤ x). It is a function of x that gives the probability that the random variable is less than or equal to x. 5.2 The Uniform Distribution If X has a uniform distribution where a < x < b or a ≤ x ≤ b, then X takes on values between a and b (may include a and b). All values x are equally likely. We write X ∼ U(a, b). The mean of X is μ = a + b . The standard deviation of X is 2 σ = (b − a)2 12 of X is P(X ≤ x is continuous. . The probability density function of X is f (x) = 1 for a ≤ x ≤ b. The cumulative distribution function b − a Figure 5.36 The probability P(c < X < d) may be found by computing the area under f(x), between c and d. Since the corresponding area is a rectangle, the area may be found simply by multiplying the width and the height. 5.3 The Exponential Distribution (Optional) If X has an exponential distribution with mean μ, then the decay parameter is m = 1 μ , and we write X ~ Exp(m) where x ≥ 0 and m > 0 . The probability density function of X is f(x) = me-mx (or equivalently f (x) = 1 μe distribution function of X is P(X ≤ x) = 1 – e–mx. − x / μ . The cumulative The exponential distribution has the memoryless property, which says that future probabilities do not depend on any past information. Mathematically, it says that P(X > x + k\|X > x) = P(X > k). If T represents the waiting time between events, and if T ~ Exp(λ), then the number of events X per unit time follows the Poisson distribution with mean λ. The probability density function of X is P(X = k) = λk e−k k ! . This may be computed 358 Chapter 5 \| Continuous Random Variables using a TI-83, 83+, 84, 84+ calculator with the command poissonpdf(λ, k). The cumulative distribution function P(X ≤ k) may be computed using the TI-83, 83+,84, 84+ calculator with the command poissoncdf(λ, k). FORMULA REVIEW 5.1 Continuous Probability Functions Probability density function (pdf) f(x): • f(x) ≥ 0 • The total area under the curve f(x) is one. Cumulative distribution function (cdf): P(X ≤ x) 5.2 The Uniform Distribution • cdf: P(X ≤ x) = x − a b − a • mean µ = a + b 2 • standard deviation σ = (b − a)2 12 • P(c < X < d) = (d – c real number between a and b (in some instances, X can take on the values a and b). a = smallest X, b = largest X 5.3 The Exponential Distribution (Optional) Exponential: X ~ Exp(m) where m = the decay parameter X ~ U(a, b) The mean is μ = a + b 2 . The standard deviation is σ = (b – a)2 12 . • pdf: f(x) = me(–mx) where x ≥ 0 and m > 0 • cdf: P(X ≤ x) = 1 – e(–mx) • mean µ = 1 m • standard deviation σ = µ Probability density function: f (x) = 1 b − a for • percentile k: k = ln(1 − AreaToTheLe f tO f k) ( − m) • Additionally ◦ P(X > x) = e(–mx) ◦ P(a < X < b) = e(–ma) – e(–mb) • Memoryless property: P(X > x + k\|X > x) = P (X > k) • Poisson probability: P(X = k) = λk e−k k ! with mean λ • k! = k(k−1)(k−2)(k−3)…321 a ≤ X ≤ b Area to the left of x: P(X < x) = (x – a) ⎛ ⎝ 1 b − a ⎞ ⎠ Area to the right of x: P(X > x) = (b – x) ⎛ ⎝ 1 b − a ⎞ ⎠ Area between c and d: P(c < x < d) = (base)(height) = (d – c) ⎛ ⎝ ⎞ ⎠ 1 b − a Uniform: X ~ U(a, b) where a < x < b • pdf: f (x) = 1 b − a for a ≤ x ≤ b PRACTICE This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 359 5.1 Continuous Probability Functions 1. Which type of distribution does the graph illustrate? Figure 5.37 2. Which type of distribution does the graph illustrate? Figure 5.38 3. Which type of distribution does the graph illustrate? Figure 5.39 360 Chapter 5 \| Continuous Random Variables 4. What does the shaded area represent? P(___< x < ___) Figure 5.40 5. What does the shaded area represent? P(___< x < ___) Figure 5.41 6. For a continuous probablity distribution, 0 ≤ x ≤ 15. What is P(x > 15)? 7. What is the area under f(x) if the function is a continuous probability density function? 8. For a continuous probability distribution, 0 ≤ x ≤ 10. What is P(x = 7)? 9. A continuous probability function is restricted to the portion between x = 0 and 7. What is P(x = 10)? 10. f(x) for a continuous probability function is 1 5 , and the function is restricted to 0 ≤ x ≤ 5. What is P(x < 0)? 11. f(x), a continuous probability function, is equal to 1 12 12)? , and the function is restricted to 0 ≤ x ≤ 12. What is P (0 < x < This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 361 12. Find the probability that x falls in the shaded area. Figure 5.42 13. Find the probability that x falls in the shaded area. Figure 5.43 14. Find the probability that x falls in the shaded area. Figure 5.44 15. f(x), a continuous probability function, is equal to 1 3 ⎛ ⎝x > 3 and the function is restricted to 1 ≤ x ≤ 4. Describe P 2 ⎞ ⎠. 5.2 The Uniform Distribution Use the following information to answer the next 10 questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes: 362 Chapter 5 \| Continuous Random Variables 1.5 2.4 3.6 2.6 1.6 2.4 2.0 3.5 2.5 1.8 2.4 2.5 3.5 4.0 2.6 1.6 2.2 1.8 3.8 2.5 1.5 2.8 1.8 4.5 1.9 1.9 3.1 1.6 Table 5.4 The sample mean = 2.50 and the sample standard deviation = 0.8302. The distribution can be written as X ~ U(1.5, 4.5). 16. What type of distribution is this? 17. In this distribution, outcomes are equally likely. What does this mean? 18. What is the height of f(x) for the continuous probability distribution? 19. What are the constraints for the values of x? 20. Graph P(2 < x < 3). 21. What is P(2 < x < 3)? 22. What is P(x < 3.5\| x < 4)? 23. What is P(x = 1.5)? 24. What is the 90th percentile of square footage for homes? 25. Find the probability that a randomly selected home has more than 3,000 square feet given that you already know the house has more than 2,000 square feet. Use the following information to answer the next eight exercises. A distribution is given as X ~ U(0, 12). 26. What is a? What does it represent? 27. What is b? What does it represent? 28. What is the probability density function? 29. What is the theoretical mean? 30. What is the theoretical standard deviation? 31. Draw the graph of the distribution for P(x > 9). 32. Find P(x > 9). 33. Find the 40th percentile. Use the following information to answer the next 12 exercises. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years. 34. What is being measured here? 35. In words, define the random variable X. 36. Are the data discrete or continuous? 37. The interval of values for x is ______. 38. The distribution for X is ______. 39. Write the probability density function. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Varia
bles 363 40. Graph the probability distribution. a. Sketch the graph of the probability distribution. Figure 5.45 Identify the following values: b. i. Lowest value for x¯ : _______ ii. Highest value for x¯ : _______ iii. Height of the rectangle: _______ iv. Label for x-axis (words): _______ v. Label for y-axis (words): _______ 41. Find the average age of the cars in the lot. 42. Find the probability that a randomly chosen car in the lot was less than four years old. a. Sketch the graph, and shade the area of interest. Figure 5.46 b. Find the probability. P(x < 4) = _______ 364 Chapter 5 \| Continuous Random Variables 43. Considering only the cars less than 7.5 years old, find the probability that a randomly chosen car in the lot was less than four years old. a. Sketch the graph, shade the area of interest. Figure 5.47 b. Find the probability. P(x < 4\|x < 7.5) = _______ 44. What has changed in the previous two problems that made the solutions different? 45. Find the third quartile of ages of cars in the lot. This means you will have to find the value such that 3 4 , or 75 percent, of the cars are at most (less than or equal to) that age. a. Sketch the graph, and shade the area of interest. Figure 5.48 b. Find the value k such that P(x < k) = 0.75. c. The third quartile is _______ 5.3 The Exponential Distribution (Optional) Use the following information to answer the next 10 exercises. A customer service representative must spend different amounts of time with each customer to resolve various concerns. The amount of time spent with each customer can be modeled by the following distribution: X ~ Exp(0.2) 46. What type of distribution is this? 47. Are outcomes equally likely in this distribution? Why or why not? 48. What is m? What does it represent? 49. What is the mean? 50. What is the standard deviation? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 365 51. State the probability density function. 52. Graph the distribution. 53. Find P(2 < x < 10). 54. Find P(x > 6). 55. Find the 70th percentile. Use the following information to answer the next eight exercises. A distribution is given as X ~ Exp(0.75). 56. What is m? 57. What is the probability density function? 58. What is the cumulative distribution function? 59. Draw the distribution. 60. Find P(x < 4). 61. Find the 30th percentile. 62. Find the median. 63. Which is larger, the mean or the median? Use the following information to answer the next eight exercises. Carbon-14 is a radioactive element with a half-life of about 5,730 years. Carbon-14 is said to decay exponentially. The decay rate is 0.000121. We start with one gram of carbon-14. We are interested in the time (years) it takes to decay carbon-14. 64. What is being measured here? 65. Are the data discrete or continuous? 66. In words, define the random variable X. 67. What is the decay rate (m)? 68. The distribution for X is ______. 69. Find the amount (percent of one gram) of carbon-14 lasting less than 5,730 years. The question means that you need to find P(x < 5,730). a. Sketch the graph, and shade the area of interest. Figure 5.49 b. Find the probability. P(x < 5,730) = __________ 366 Chapter 5 \| Continuous Random Variables 70. Find the percentage of carbon-14 lasting longer than 10,000 years. a. Sketch the graph, and shade the area of interest. Figure 5.50 b. Find the probability. P(x > 10,000) = ________ 71. Thirty percent of carbon-14 will decay within how many years? a. Sketch the graph, and shade the area of interest. Figure 5.51 b. Find the value k such that P(x < k) = 0.30. HOMEWORK 5.1 Continuous Probability Functions For each probability and percentile problem, draw the picture. 72. Consider the following experiment. You are one of 100 people enlisted to take part in a study to determine percentage of nurses in America with an R.N. (registered nurse) degree. You ask nurses if they have an R.N. degree. The nurses answer yes or no. You then calculate the percentage of nurses with an R.N. degree. You give that percentage to your supervisor. a. What part of the experiment will yield discrete data? b. What part of the experiment will yield continuous data? 73. When age is rounded to the nearest year, do the data stay continuous, or do they become discrete? Why? 5.2 The Uniform Distribution For each probability and percentile problem, draw the picture. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 367 74. Births are approximately uniformly distributed between the 52 weeks of the year. They can be said to follow a uniform distribution from one to 53 (spread of 52 weeks). f(x) = _________ a. X ~ _________ b. Graph the probability distribution. c. d. μ = _________ e. σ = _________ f. Find the probability that a person is born at the exact moment week 19 starts. That is, find P(x = 19) = _________ g. P(2 < x < 31) = _________ h. Find the probability that a person is born after week 40. i. P(12 < x\|x < 28) = _________ j. Find the 70th percentile. k. Find the minimum for the upper quarter. 75. A random number generator picks a number from one to nine in a uniform manner. f(x) = _________ a. X ~ _________ b. Graph the probability distribution. c. d. μ = _________ e. σ = _________ f. P(3.5 < x < 7.25) = _________ g. P(x > 5.67) h. P(x > 5\|x > 3) = _________ i. Find the 90th percentile. 76. According to a study by Dr. John McDougall of his live-in weight loss program, the people who follow his program lose between six and 15 pounds a month until they approach trim body weight. Let’s suppose that the weight loss is uniformly distributed. We are interested in the weight loss of a randomly selected individual following the program for one month. f(x) = _________ a. Define the random variable. X = _________ b. X ~ _________ c. Graph the probability distribution. d. e. μ = _________ f. σ = _________ g. Find the probability that the individual lost more than 10 pounds in a month. h. Suppose it is known that the individual lost more than 10 pounds in a month. Find the probability that he lost less than 12 pounds in the month. i. P(7 < x < 13\|x > 9) = __________. State this result in a probability question, similarly to Parts g and h, draw the picture, and find the probability. 77. A subway train arrives every eight minutes during rush hour. We are interested in the length of time a commuter must wait for a train to arrive. The time follows a uniform distribution. f(x) = _______ a. Define the random variable. X = _______ b. X ~ _______ c. Graph the probability distribution. d. e. μ = _______ f. σ = _______ g. Find the probability that the commuter waits less than one minute. h. Find the probability that the commuter waits between three and four minutes. i. Sixty percent of commuters wait more than how long for the train? State this result in a probability question, similarly to Parts g and h, draw the picture, and find the probability. 368 Chapter 5 \| Continuous Random Variables 78. The age of a first grader on September 1 at Garden Elementary School is uniformly distributed from 5.8 to 6.8 years. We randomly select one first grader from the class. f(x) = _________ a. Define the random variable. X = _________ b. X ~ _________ c. Graph the probability distribution. d. e. μ = _________ f. σ = _________ g. Find the probability that she is over 6.5 years old. h. Find the probability that she is between four and six years old. i. Find the 70th percentile for the age of first graders on September 1 at Garden Elementary School. Use the following information to answer the next three exercises. The Sky Train from the terminal to the rental–car and long–term parking center is supposed to arrive every eight minutes. The waiting times for the train are known to follow a uniform distribution. 79. What is the average waiting time (in minutes)? a. zero two b. three c. four d. 80. Find the 30th percentile for the waiting times (in minutes). a. two b. 2.4 c. 2.75 three d. 81. The probability of waiting more than seven minutes given a person has waited more than four minutes is? a. 0.125 b. 0.25 c. 0.5 d. 0.75 82. The time (in minutes) until the next bus departs a major bus depot follows a distribution with f(x) = 1 20 where x goes from 25 to 45 minutes. a. Define the random variable. X = ________ b. X ~ ________ c. Graph the probability distribution. d. The distribution is ______________ (name of distribution). It is _____________ (discrete or continuous). e. μ = ________ f. σ = ________ g. Find the probability that the time is at most 30 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement. h. Find the probability that the time is between 30 and 40 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement. i. P(25 < x < 55) = _________. State this result in a probability statement, similarly to Parts g and h, draw the picture, and find the probability. j. Find the 90th percentile. This means that 90 percent of the time, the time is less than _____ minutes. k. Find the 75th percentile. In a complete sentence, state what this means. (See Part j.) l. Find the probability that the time is more than 40 minutes given (or knowing that) it is at least 30 minutes. 83. Suppose that the value of a stock varies each day from $16 to $25 with a uniform distribution. a. Find the probability that the value of the stock is more than $19. b. Find the probability that the value of the stock between $19 and $22. c. Find the upper quartile — 25 percent of all days the stock is above what value? Draw the graph. d. Given that the stock is greater than $18, find the probability that the stock is more than $21. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 369 84. A fireworks show is designed s
o that the time between fireworks is between one and five seconds, and follows a uniform distribution. a. Find the average time between fireworks. b. Find the probability that the time between fireworks is greater than four seconds. 85. The number of miles driven by a truck driver falls between 300 and 700, and follows a uniform distribution. a. Find the probability that the truck driver goes more than 650 miles in a day. b. Find the probability that the truck driver goes between 400 and 650 miles in a day. c. At least how many miles does the truck driver travel on the 10 percent of days with the highest mileage? 5.3 The Exponential Distribution (Optional) 86. Suppose that the length of long-distance phone calls, measured in minutes, is known to have an exponential distribution with the average length of a call equal to eight minutes. Is X continuous or discrete? a. Define the random variable. X = ________________. b. c. X ~ ________ d. μ = ________ e. σ = ________ f. Draw a graph of the probability distribution. Label the axes. g. Find the probability that a phone call lasts less than nine minutes. h. Find the probability that a phone call lasts more than nine minutes. i. Find the probability that a phone call lasts between seven and nine minutes. j. If 25 phone calls are made one after another, on average, what would you expect the total to be? Why? 87. Suppose that the useful life of a particular car battery, measured in months, decays with parameter 0.025. We are interested in the life of the battery. Is X continuous or discrete? a. Define the random variable. X = _________________________________. b. c. X ~ ________ d. On average, how long would you expect one car battery to last? e. On average, how long would you expect nine car batteries to last, if they are used one after another? f. Find the probability that a car battery lasts more than 36 months. g. Seventy percent of the batteries last at least how long? 88. The percent of persons (ages five and older) in each state who speak a language at home other than English is approximately exponentially distributed with a mean of 9.848. Suppose we randomly pick a state. Is X continuous or discrete? a. Define the random variable. X = _________________________________. b. c. X ~ ________ d. μ = ________ e. σ = ________ f. Draw a graph of the probability distribution. Label the axes. g. Find the probability that percentage is less than 12. h. Find the probability that percentage is between eight and 14. i. The percent of all individuals living in the United States who speak a language at home other than English is 13.8. i. Why is this number different from 9.848 percent? ii. What would make this number higher than 9.848 percent? 370 Chapter 5 \| Continuous Random Variables 89. The time (in years) after reaching age 60 that it takes an individual to retire is approximately exponentially distributed with a mean of about five years. Suppose we randomly pick one retired individual. We are interested in the time after age 60 to retirement. Is X continuous or discrete? a. Define the random variable. X = _________________________________. b. c. X ~ = ________ d. μ = ________ e. σ = ________ f. Draw a graph of the probability distribution. Label the axes. g. Find the probability that the person retired after age 70. h. Do more people retire before age 65 or after age 65? i. In a room of 1,000 people over age 80, how many do you expect will not have retired yet? 90. The cost of all maintenance for a car during its first year is approximately exponentially distributed with a mean of $150. a. Define the random variable. X = _________________________________. b. X ~ = ________ c. μ = ________ d. σ = ________ e. Draw a graph of the probability distribution. Label the axes. f. Find the probability that a car required over $300 for maintenance during its first year. Use the following information to answer the next three exercises. The average lifetime of a certain new cell phone is three years. The manufacturer will replace any cell phone failing within two years of the date of purchase. The lifetime of these cell phones is known to follow an exponential distribution. 91. What is the decay rate? a. 0.3333 b. 0.5000 c. 2 d. 3 92. What is the probability that a phone will fail within two years of the date of purchase? a. 0.8647 b. 0.4866 c. 0.2212 d. 0.9997 93. What is the median lifetime of these phones (in years)? a. 0.1941 b. 1.3863 c. 2.0794 d. 5.5452 94. Let X ~ Exp(0.1). a. decay rate = ________ b. μ = ________ c. Graph the probability distribution function. d. On the graph, shade the area corresponding to P(x < 6), and find the probability. e. Sketch a new graph, shade the area corresponding to P(3 < x < 6), and find the probability. f. Sketch a new graph, shade the area corresponding to P(x < 7), and find the probability. g. Sketch a new graph, shade the area corresponding to the 40th percentile and find the value. h. Find the average value of x. 95. Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. a. Find the probability that a light bulb lasts less than one year. b. Find the probability that a light bulb lasts between six and 10 years. c. Seventy percent of all light bulbs last at least how long? d. A company decides to offer a warranty to give refunds to light bulbs whose lifetime is among the lowest two percent of all bulbs. To the nearest month, what should be the cutoff lifetime for the warranty to take place? If a light bulb has lasted seven years, what is the probability that it fails within the 8th year? e. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 371 96. At a 911 call center, calls come in at an average rate of one call every two minutes. Assume that the time that elapses from one call to the next has the exponential distribution. a. On average, how much time occurs between five consecutive calls? b. Find the probability that after a call is received, it takes more than three minutes for the next call to occur. c. Ninety-percent of all calls occur within how many minutes of the previous call? d. Suppose that two minutes have elapsed since the last call. Find the probability that the next call will occur within the next minute. e. Find the probability that fewer than 20 calls occur within an hour. 97. In major league baseball, a no-hitter is a game in which a pitcher, or pitchers, doesn't give up any hits throughout the game. No-hitters occur at a rate of about three per season. Assume that the duration of time between no-hitters is exponential. a. What is the probability that an entire season elapses with a single no-hitter? b. If an entire season elapses without any no-hitters, what is the probability that there are no no-hitters in the following season? c. What is the probability that there are more than three no-hitters in a single season? 98. During the years 1998–2012, a total of 29 earthquakes of magnitude greater than 6.5 occurred in Papua New Guinea. Assume that the time spent waiting between earthquakes is exponential. Assume that the current year is 2013 a. What is the probability that the next earthquake occurs within the next three months? b. Given that six months has passed without an earthquake in Papua New Guinea, what is the probability that the next three months will be free of earthquakes? c. What is the probability of zero earthquakes occurring in 2014? d. What is the probability that at least two earthquakes will occur in 2014? 99. According to the American Red Cross, about one out of nine people in the United States have type B blood. Suppose the blood types of people arriving at a blood drive are independent. In this case, the number of type B blood types that arrive roughly follows the Poisson distribution. a. If 100 people arrive, how many on average would be expected to have type B blood? b. What is the probability that more than 10 people out of these 100 have type B blood? c. What is the probability that more than 20 people arrive before a person with type B blood is found? 100. A website experiences traffic during normal working hours at a rate of 12 visits per hour. Assume that the duration between visits has the exponential distribution. a. Find the probability that the duration between two successive visits to the website is more than 10 minutes. b. The top 25 percent of durations between visits are at least how long? c. Suppose that 20 minutes have passed since the last visit to the website. What is the probability that the next visit will occur within the next five minutes? d. Find the probability that fewer than seven visits occur within a one-hour period. 101. At an urgent care facility, patients arrive at an average rate of one patient every seven minutes. Assume that the duration between arrivals is exponentially distributed. a. Find the probability that the time between two successive visits to the urgent care facility is less than two minutes. b. Find the probability that the time between two successive visits to the urgent care facility is more than 15 minutes. If 10 minutes have passed since the last arrival, what is the probability that the next person will arrive within the c. next five minutes? d. Find the probability that more than eight patients arrive during a half-hour period. REFERENCES 5.2 The Uniform Distribution McDougall, J. A. (1995). The McDougall program for maximum weight loss. New York: Plume 5.3 The Exponential Distribution (Optional) Baseball-Reference.com. (2013). No-hitter. Retrieved from http://www.baseball-reference.com/bullpen/No-hitter U.S. Census Bureau. (n.d.). Retrieved from https://www.census.gov/ 372 Chapter 5 \| Continuous Random Variables World Earthquakes. (2013). Earthquake data for Papua New Guinea. Retrieved from http://www.world-earthquakes.com/ Zhou, Rick. (2013). Exponential distribution lecture slides. Retrieved from www.public.iastate.edu/~riczw/stat330s11/
lecture/lec13.pdf SOLUTIONS 1 Uniform distribution 3 Normal distribution 5 P(6 < x < 7) 7 one 9 zero 11 one 13 0.625 15 The probability is equal to the area from x = 3 2 to x = 4 above the x-axis and up to f(x) = 1 3 . 17 It means that the value of x is just as likely to be any number between 1.5 and 4.5. 19 1.5 ≤ x ≤ 4.5 21 0.3333 23 zero 25 0.6 27 b is 12, and it represents the highest value of x. 29 six 31 Figure 5.52 33 4.8 35 X = The age (in years) of cars in the staff parking lot 37 0.5 to 9.5 39 f(x) = 1 9 41 μ = 5 where x is between 0.5 and 9.5, inclusive. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 373 43 a. Check student’s solution. b. 3.5 7 45 a. Check student's solution b. k = 7.25 c. 7.25 47 No, outcomes are not equally likely. In this distribution, more people require a little bit of time, and fewer people require a lot of time, so it is more likely that someone will require less time. 49 five 51 f(x) = 0.2e–0.2x 53 0.5350 55 6.02 57 f(x) = 0.75e–0.75x 59 Figure 5.53 61 0.4756 63 The mean is larger. The mean is 1 m = 1 0.75 ≈ 1.33 , which is greater than 0.9242. 65 continuous 67 m = 0.000121 69 a. Check student's solution b. P(x < 5,730) = 0.5001 71 a. Check student's solution b. k = 2947.73 73 Age is a measurement, regardless of the accuracy used. 75 a. X ~ U(1, 9) 374 Chapter 5 \| Continuous Random Variables b. Check student’s solution c. f (x) = 1 8 where 1 ≤ x ≤ 9 d. five e. 2.3 f. g. h. 15 32 333 800 2 3 i. 8.2 77 a. X represents the length of time a commuter must wait for a train to arrive on the Red Line. b. X ~ U(0, 8) c. f (x) = 1 8 where ≤ x ≤ 8 d. four e. 2.31 f. g. 1 8 1 8 h. 3.2 79 d 81 b 83 a. The probability density function of X is 1 25 − 16 = 1 9 . P(X > 19) = (25 – 19 Figure 5.54 b. P(19 < X < 22) = (22 – 19 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 375 Figure 5.55 c. The area must be 0.25, and 0.25 = (width) ⎛ ⎝ ⎞ ⎠ 1 9 , so width = (0.25)(9) = 2.25. Thus, the value is 25 – 2.25 = 22.75. d. This is a conditional probability question. P(x > 21\| x > 18). You can do this two ways: ◦ Draw the graph where a is now 18 and b is still 25. The height is 1 (25 − 18) = 1 7 So, P(x > 21\|x > 18) = (25 – 21) ⎛ ⎝ = 4/7. ⎞ ⎠ 1 7 ◦ Use the formula: P(x > 21\|x > 18) = P(x > 21 AND x > 18) P(x > 18) = P(x > 21) P(x > 18) = (25 − 21) (25 − 18) . = 4 7 85 a. P(X > 650) = 700 − 650 700 − 300 = 50 400 = 1 8 = 0.125. b. P(400 < X < 650) = 650 − 400 700 − 300 = 250 400 = 0.625 c. 0.10 = width 700 − 300 farthest 10 percent of days. , so width = 400(0.10) = 40. Since 700 – 40 = 660, the drivers travel at least 660 miles on the 87 a. X = the useful life of a particular car battery, measured in months. b. X is continuous. c. X ~ Exp(0.025) d. 40 months e. 360 months f. 0.4066 g. 14.27 89 a. X = the time (in years) after reaching age 60 that it takes an individual to retire b. X is continuous. c. X ~ Exp ⎛ ⎝ ⎞ ⎠ 1 5 d. five Chapter 5 \| Continuous Random Variables 376 e. five f. Check student’s solution. g. 0.1353 h. before i. 18.3 91 a 93 c 95 Let T = the life time of a light bulb. The decay parameter is m = 1/8, and T ~ Exp(1/8). The cumulative distribution function is P(T < t) = 1 − e a. Therefore, P(T < 1.1175. b. We want to find P(6 < t < 10). To do this, P(6 < t < 10) – P(t < 6) * 10⎞ – 1 8 – 1 8 * 6⎞ = = ⎛ ⎜1 – e ⎝ ⎛ ⎜.7135 – 0.5276 = 0.1859 ⎠ Figure 5.56 c. We want to find 0.70 = P(T > t) = 1 – ⎛ ⎜. ⎞ ⎟ = e ⎠ Solving for t, e – t 8 = 0.70, so – t 8 = ln(0.70), and t = –8ln(0.70) ≈ 2.85 years Or use t = ln(area_to_the_right) ( – m) = ln(0.70) – 1 8 ≈ 2.85 years . This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 \| Continuous Random Variables 377 Figure 5.57 d. We want to find 0.02 = P(T < t) = 1 – e – t 8 . Solving for t, e – t 8 = 0.98, so – t 8 = ln(0.98), and t = –8ln(0.98) ≈ 0.1616 years, or roughly two months. The warranty should cover light bulbs that last less than 2 months. Or use ln(area_to_the_right) = 0.1616. ( – m) = ln(1 – 0.2) – 1 8 e. We must find P(T < 8\|T > 7). Notice that by the rule of complement events, P(T < 8\|T > 7) = 1 – P(T > 8\|T > 7). By the memoryless property (P(X > r + t\|X > r) = P(X > t)). So P(T > 8\|T > 7) = P(T > 1) = 1 – ⎛ ⎜.8825 Therefore, P(T < 8\|T > 7) = 1 – 0.8825 = 0.1175. 97 Let X = the number of no-hitters throughout a season. Since the duration of time between no-hitters is exponential, the number of no-hitters per season is Poisson with mean λ = 3. Therefore, (X = 0) = 30 e – 3 = e–3 ≈ 0.0498 0! You could let T = duration of time between no-hitters. Since the time is exponential and there are three no-hitters per season, then the time between no-hitters is 1 3 season. For the exponential, µ = 1 3 . Therefore, m = 1 μ = 3 and T ~ Exp(3). a. The desired probability is P(T > 1) = 1 – P(T < 1) = 1 – (1 – e–3) = e–3 ≈ 0.0498. b. Let T = duration of time between no-hitters. We find P(T > 2\|T > 1), and by the memoryless property this is simply P(T > 1), which we found to be 0.0498 in part a. c. Let X = the number of no-hitters is a season. Assume that X is Poisson with mean λ = 3. Then P(X > 3) = 1 – P(X ≤ 3) = 0.3528. 99 a. 100 9 = 11.11 b. P(X > 10) = 1 – P(X ≤ 10) = 1 – Poissoncdf(11.11, 10) ≈ 0.5532. 378 Chapter 5 \| Continuous Random Variables c. The number of people with Type B blood encountered roughly follows the Poisson distribution, so the number of people X who arrive between successive Type B arrivals is roughly exponential with mean μ = 9 and m = 1 9 . The cumulative distribution function of X is P(X < x) = 1 − e − x 9 . Thus hus, P(X > 20) = 1 - P(X ≤ 20) = 1 − − 20 9 ⎛ ⎜1 − e ⎝ ⎞ ⎟ ≈ 0.1084. ⎠ NOTE We could also deduce that each person arriving has a 8 9 chance of not having type B blood. So the probability that none of the first 20 people arrive have type B blood is ⎛ ⎝ 8 9 than the exponential because the number of people between type B people is discrete instead of continuous.) ≈ 0.0948 . (The geometric distribution is more appropriate ⎞ ⎠ 20 101 Let T = duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, μ = 7 and the decay constant is m = 1 7 . The cdf is P(T < t) = 1 − e t 7 a. P(T < 2.2485. b. P(T > 15) = 1 − P(T < 15) = 1 − c. P(T > 15\|T > 10) = P(T > 5) = 1 − − 15 7 ⎛ ⎜1 − e ⎝ ⎞ ⎟ ≈ e ⎠ − 15 7 ≈ 0.1173 . − 5 7 ⎛ ⎜.4895 . d. Let X = # of patients arriving during a half-hour period. Then X has the Poisson distribution with a mean of 30 7 , X ~ Poisson ⎛ ⎝ ⎞ ⎠ 30 7 . Find P(X > 8) = 1 – P(X ≤ 8) ≈ 0.0311. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 379 6 \| THE NORMAL DISTRIBUTION Figure 6.1 If you ask enough people about their shoe size, you will find that your graphed data is shaped like a bell curve and can be described as normally distributed. (credit: Ömer Ünlϋ) Introduction Chapter Objectives By the end of this chapter, the student should be able to do the following: • Recognize the normal probability distribution and apply it appropriately • Recognize the standard normal probability distribution and apply it appropriately • Compare normal probabilities by converting to the standard normal distribution The normal, a continuous distribution, is the most important of all the distributions. It is widely used and even more widely abused. Its graph is bell-shaped. You see the bell curve in almost all disciplines, including psychology, business, economics, the sciences, nursing, and, of course, mathematics. Some of your instructors may use the normal distribution to help determine your grade. Most IQ scores are normally distributed. Often, real-estate prices fit a normal distribution. The normal distribution is extremely important, but it cannot be applied to everything in the real world. In this chapter, you will study the normal distribution, the standard normal distribution, and applications associated with 380 them. Chapter 6 \| The Normal Distribution The normal distribution has two parameters: —the mean (μ) and the standard deviation (σ). If X is a quantity to be measured that has a normal distribution with mean (μ) and standard deviation (σ), we designate this by writing Figure 6.2 The curve is symmetric about a vertical line drawn through the mean, μ. In theory, the mean is the same as the median, because the graph is symmetric about μ. With a normal distribution, the mean, median, and mode all lie at the same point. The normal distribution depends only on the mean and the standard deviation. The location of the mean simply indicates the location of the line of symmetry, in a normal distribution. Since the area under the curve must equal one, a change in the standard deviation, σ, causes a change in the shape of the curve; the curve becomes fatter or skinnier depending on σ. A change in μ causes the graph to shift to the left or right. The location of the mean simply indicates the location of the line of symmetry, in a normal distribution. This means there are an infinite number of normal probability distributions. One distribution of special interest is called the standard normal distribution. Your instructor will record the heights of both men and women in your class, separately. Draw histograms of your data. Then draw a smooth curve through each histogram. Is each curve somewhat bell-shaped? Do you think that if you had recorded 200 data values for men and 200 for women that the curves would look bell-shaped? Calculate the mean for each data set. Write the means on the x-axis of the appropriate graph below the peak. Shade the approximate area that represents the probability that one randomly chosen male is taller than 72 inches. Shade the approximate area that represents the probability that 1 randomly chosen female is shorter than 60 inches. If the total area under each curve is one, does either probability appear
to be more than 0.5? 6.1 \| The Standard Normal Distribution The standardized normal distribution is a type of normal distribution, with a mean of 0 and standard deviation of 1. It represents a distribution of standardized scores, called z-scores, as opposed to raw scores (the actual data values). A z-score indicates the number of standard deviation a score falls above or below the mean. Z-scores allow for comparison of scores, occurring in different data sets, with different means and standard deviations. It would not make sense to compare apples and oranges. Likewise, it does not make sense to compare scores from two different samples that have different means and standard deviations. Z-scores can be looked up in a Z-Table of Standard Normal Distribution, in order to find the area under the standard normal curve, between a score and the mean, between two scores, or above or below a score. The standard normal distribution allows us to interpret standardized scores and provides us with one table that we may use, in order to compute areas under the normal curve, for an infinite number of data sets, no matter what the mean or standard deviation. A z-score is calculated as z = sides of the equation by σ gives: (z)(σ) = x − μ . Adding μ to both sides of the equation gives μ + (z)(σ) = x . . The score itself can be found by using algebra and solving for x. Multiplying both x − μ σ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 381 Suppose we have a data set with a mean of 5 and standard deviation of 2. We want to determine the number of standard deviations the score of 11 falls above the mean. We can find this answer (or z-score) by writing or we can solve for z. z = 11 − 5 2 = 3 5 + (z)(2) = 11, 2z = 6 z = 3 We have determined that the score of 11 falls 3 standard deviations above the mean of 5. With a standard normal distribution, we indicate the distribution by writing Z ~ N(0, 1) which shows the normal distribution has a mean of 0 and standard deviation of 1. This notation simply indicates that a standard normal distribution is being used. Z-Scores As described previously, if X is a normally distributed random variable and X ~ N(μ, σ), then the z-score is z = x – μ σ . The z-score tells you how many standard deviations the value x is above, to the right of, or below, to the left of, the mean, μ. Values of x that are larger than the mean have positive z-scores, and values of x that are smaller than the mean have negative z-scores. If x equals the mean, then x has a z-score of zero. When determining the z-score for an x-value, for a normal distribution, with a given mean and standard deviation, the notation above for a normal distribution, will be given. Example 6.1 Suppose X ~ N(5, 6). This equation says that X is a normally distributed random variable with mean μ = 5 and standard deviation σ = 6. Suppose x = 17. Then, z = x – μ σ = 17 – 5 6 = 2. This means that x = 17 is two standard deviations (2σ) above, or to the right, of the mean μ = 5. Notice that 5 + (2)(6) = 17. The pattern is μ + zσ = x. Now suppose x = 1. Then0.67, rounded to two decimal places. This means that x = 1 is 0.67 standard deviations (–0.67σ) below or to the left of the mean μ = 5. This z-score shows that x = 1 is less than 1 standard deviation below the mean of 5. Therefore, the score doesn't fall very far below the mean. Summarizing, when z is positive, x is above or to the right of μ, and when z is negative, x is to the left of or below μ. Or, when z is positive, x is greater than μ, and when z is negative, x is less than μ. The absolute value of z indicates how far the score is from the mean, in either direction. 6.1 What is the z-score of x, when x = 1 and X ~ N(12, 3)? 382 Chapter 6 \| The Normal Distribution Example 6.2 Some doctors believe that a person can lose five pounds, on average, in a month by reducing his or her fat intake and by consistently exercising. Suppose weight loss has a normal distribution. Let X = the amount of weight lost, in pounds, by a person in a month. Use a standard deviation of two pounds. X ~ N(5, 2). Fill in the blanks. a. Suppose a person lost 10 pounds in a month. The z-score when x = 10 pounds is z = 2.5 (verify). This z-score tells you that x = 10 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?). Solution 6.2 a. This z-score tells you that x = 10 is 2.5 standard deviations to the right of the mean five. b. Suppose a person gained three pounds, a negative weight loss. Then z = __________. This z-score tells you that x = –3 is ________ standard deviations to the __________ (right or left) of the mean. Solution 6.2 b. z = –4. This z-score tells you that x = –3 is four standard deviations to the left of the mean. c. Suppose the random variables X and Y have the following normal distributions: X ~ N(5, 6) and Y ~ N(2, 1). If x = 17, then z = 2. This was previously shown. If y = 4, what is z? Solution 6.2 c, where µ = 2 and σ = 1. The z-score for y = 4 is z = 2. This means that four is z = 2 standard deviations to the right of the mean. Therefore, x = 17 and y = 4 are both two of their own standard deviations to the right of their respective means. The z-score allows us to compare data that are scaled differently. To better understand the concept, suppose X ~ N(5, 6) represents weight gains for one group of people who are trying to gain weight in a six-week period and Y ~ N(2, 1) measures the same weight gain for a second group of people. A negative weight gain would be a weight loss. Since x = 17 and y = 4 are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means. 6.2 Fill in the blanks. Jerome averages 16 points a game with a standard deviation of four points. X ~ N(16, 4). Suppose Jerome scores 10 points in a game. The z-score when x = 10 is –1.5. This score tells you that x = 10 is _____ standard deviations to the ______ (right or left) of the mean______ (What is the mean?). The Empirical Rule If X is a random variable and has a normal distribution with mean µ and standard deviation σ, then the Empirical Rule states the following: • About 68 percent of the x values lie between –1σ and +1σ of the mean µ (within one standard deviation of the mean). • About 95 percent of the x values lie between –2σ and +2σ of the mean µ (within two standard deviations of the mean). • About 99.7 percent of the x values lie between –3σ and +3σ of the mean µ (within three standard deviations of the mean). Notice that almost all the x values lie within three standard deviations of the mean. • The z-scores for +1σ and –1σ are +1 and –1, respectively. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 383 • The z-scores for +2σ and –2σ are +2 and –2, respectively. • The z-scores for +3σ and –3σ are +3 and –3, respectively. So, in other words, this is that about 68 percent of the values lie between z-scores of –1 and 1, about 95% of the values lie between z-scores of –2 and 2, and about 99.7 percent of the values lie between z-scores of -3 and 3. These facts can be checked, by looking up the mean to z area in a z-table for each positive z-score and multiplying by 2. The empirical rule is also known as the 68–95–99.7 rule. Figure 6.3 Example 6.3 The mean height of 15-to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15-to 18-year-old male from Chile in 2009–2010. Then X ~ N(170, 6.28). a. Suppose a 15-to 18-year-old male from Chile was 168 cm tall in 2009–2010. The z-score when x = 168 cm is z = _______. This z-score tells you that x = 168 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?). Solution 6.3 a. –0.32, 0.32, left, 170 b. Suppose that the height of a 15-to 18-year-old male from Chile in 2009–2010 has a z-score of z = 1.27. What is the male’s height? The z-score (z = 1.27) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean. Solution 6.3 b. 177.98 cm, 1.27, right 384 Chapter 6 \| The Normal Distribution 6.3 Use the information in Example 6.3 to answer the following questions: a. Suppose a 15-to 18-year-old male from Chile was 176 cm tall from 2009–2010. The z-score when x = 176 cm is z = _______. This z-score tells you that x = 176 cm is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?). b. Suppose that the height of a 15-to 18-year-old male from Chile in 2009–2010 has a z-score of z = –2. What is the male’s height? The z-score (z = –2) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean. Example 6.4 From 1984 to 1985, the mean height of 15-to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15-to 18-year-old males from 1984–1985, and y = the height of one male from this group. Then Y ~ N(172.36, 6.34). The mean height of 15-to 18-year-old males from Chile in 2009–2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15-to 18-year-old male from Chile in 2009–2010, and x = the height of one male from this group. Then X ~ N(170, 6.28). Find the z-scores for x = 160.58 cm and y = 162.85 cm. Interpret each z-score. What can you say about x = 160.58 cm and y = 162.85 cm as they compare to their respective means and standard deviations? Solution 6.4 The z-score for x = 160.58 cm is z = –1.5. The z-score for y = 162.85 cm is z = –1.5. Both x = 160.58 and y = 162.85 deviate the same number of standard deviations from their respective means and in the same direction. 6.4 In 2012, 1,664,479 stude
nts took the SAT exam. The distribution of scores in the verbal section of the SAT had a mean µ = 496 and a standard deviation σ = 114. Let X = a SAT exam verbal section score in 2012. Then, X ~ N(496, 114). Find the z-scores for x1 = 325 and x2 = 366.21. Interpret each z-score. What can you say about x1 = 325 and x2 = 366.21, as they compare to their respective means and standard deviations? Example 6.5 Suppose x has a normal distribution with mean 50 and standard deviation 6. • About 68 percent of the x values lie within one standard deviation of the mean. Therefore, about 68 percent of the x values lie between –1σ = (–1)(6) = –6 and 1σ = (1)(6) = 6 of the mean 50. The values 50 – 6 = 44 and 50 + 6 = 56 are within one standard deviation from the mean 50. The z-scores are –1 and +1 for 44 and 56, respectively. • About 95 percent of the x values lie within two standard deviations of the mean. Therefore, about 95 percent of the x values lie between –2σ = (–2)(6) = –12 and 2σ = (2)(6) = 12. The values 50 – 12 = 38 and 50 + 12 = 62 are within two standard deviations from the mean 50. The z-scores are –2 and +2 for 38 and 62, respectively. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 385 • About 99.7 percent of the x values lie within three standard deviations of the mean. Therefore, about 95 percent of the x values lie between –3σ = (–3)(6) = –18 and 3σ = (3)(6) = 18 of the mean 50. The values 50 – 18 = 32 and 50 + 18 = 68 are within three standard deviations from the mean 50. The z-scores are –3 and +3 for 32 and 68, respectively. 6.5 Suppose X has a normal distribution with mean 25 and standard deviation five. Between what values of x do 68 percent of the values lie? Example 6.6 From 1984–1985, the mean height of 15-to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15-to 18-year-old males in 1984–1985. Then Y ~ N(172.36, 6.34). a. About 68 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively. b. About 95 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________ respectively. c. About 99.7 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively. Solution 6.6 a. About 68 percent of the values lie between 166.02 cm and 178.7 cm. The z-scores are –1 and 1. b. About 95 percent of the values lie between 159.68 cm and 185.04 cm. The z-scores are –2 and 2. c. About 99.7 percent of the values lie between 1153.34 cm and 191.38 cm. The z-scores are –3 and 3. 6.6 The scores on a college entrance exam have an approximate normal distribution with mean, µ = 52 points and a standard deviation, σ = 11 points. a. About 68 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively. b. About 95 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively. c. About 99.7 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively. 6.2 \| Using the Normal Distribution The shaded area in the following graph indicates the area to the left of x. This area could represent the percentage of students scoring less than a particular grade on a final exam. This area is represented by the probability P(X < x). Normal tables, computers, and calculators are used to provide or calculate the probability P(X < x). 386 Chapter 6 \| The Normal Distribution Figure 6.4 The area to the right is then P(X > x) = 1 – P(X < x). Remember, P(X < x) = Area to the left of the vertical line through x. P(X < x) = 1 – P(X < x) = Area to the right of the vertical line through x. P(X < x) is the same as P(X ≤ x) and P(X > x) is the same as P(X ≥ x) for continuous distributions. Suppose the graph above were to represent the percentage of students scoring less than 75 on a final exam, with this probability equal to 0.39. This would also indicate that the percentage of students scoring higher than 75 was equal to 1 minus 0.39 or 0.61. Calculations of Probabilities Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators. NOTE To calculate the probability, use the probability tables provided in Appendix H without the use of technology. The tables include instructions for how to use them. The probability is represented by the area under the normal curve. To find the probability, calculate the z-score and look up the z-score in the z-table under the z-column. Most z-tables show the area under the normal curve to the left of z. Others show the mean to z area. The method used will be indicated on the table. We will discuss the z-table that represents the area under the normal curve to the left of z. Once you have located the z-score, locate the corresponding area. This will be the area under the normal curve, to the left of the z-score. This area can be used to find the area to the right of the z-score, or by subtracting from 1 or the total area under the normal curve. These areas can also be used to determine the area between two z-scores. Example 6.7 If the area to the left is 0.0228, then the area to the right is 1 – 0.0228 = 0.9772. 6.7 If the area to the left of x is 0.012, then what is the area to the right? Example 6.8 The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 387 a. Find the probability that a randomly selected student scored more than 65 on the exam. Solution 6.8 a. Let X = a score on the final exam. X ~ N(63, 5), where μ = 63 and σ = 5. Draw a graph. Calculate the z-score: z = x − μ σ = 65 − 63 5 = 2 5 = .40 The z-table shows that the area to the left of z is 0.6554. Subtracting this area from 1 gives 0.3446. Then, find P(x > 65). P(x > 65) = 0.3446 Figure 6.5 The probability that any student selected at random scores more than 65 is 0.3446. Go into 2nd DISTR. After pressing 2nd DISTR, press 2:normalcdf. The syntax for the instructions is as follows: normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 1099) by pressing 1, the EE key—a 2nd key—and then 99. Or, you can enter 10^99 instead. The number 1099 is way out in the right tail of the normal curve. We are calculating the area between 65 and 1099. In some instances, the lower number of the area might be –1E99 (= –1099). The number –1099 is way out in the left tail of the normal curve. We chose the exponent of 99 because this produces such a large number that we can reasonably expect all of the values under the curve to fall below it. This is an arbitrary value and one that works well, for our purpose. HISTORICAL NOTE The TI probability program calculates a z-score and then the probability from the z-score. Before technology, the z-score was looked up in a standard normal probability table, also known as a Z-table—the math involved to find probability is cumbersome. In this example, a standard normal table with area to the left of the z-score was used. You calculate the z-score and look up the area to the left. The probability is the area to the right. 388 Chapter 6 \| The Normal Distribution Calculate the z-score Press 2nd Distr Press 3:invNorm( Enter the area to the left of z followed by ) Press ENTER. For this Example, the steps are 2nd Distr 3:invNorm(.6554) ENTER The answer is 0.3999, which rounds to 0.4. b. Find the probability that a randomly selected student scored less than 85. Solution 6.8 b. Draw a graph. Then find P(x < 85), and shade the graph. Using a computer or calculator, find P(x < 85) = 1. normalcdf(0,85,63,5) = 1 (rounds to one) The probability that one student scores less than 85 is approximately one, or 100 percent. c. Find the 90th percentile, —that is, find the score k that has 90 percent of the scores below k and 10 percent of the scores above k. Solution 6.8 c. Find the 90th percentile. For each problem or part of a problem, draw a new graph. Draw the x-axis. Shade the area that corresponds to the 90th percentile. This time, we are looking for a score that corresponds to a given area under the curve. Let k = the 90th percentile. The variable k is located on the x-axis. P(x < k) is the area to the left of k. The 90th percentile k separates the exam scores into those that are the same or lower than k and those that are the same or higher. Ninety percent of the test scores are the same or lower than k, and 10 percent are the same or higher. The variable k is often called a critical value. We know the mean, standard deviation, and area under the normal curve. We need to find the z-score that corresponds to the area of 0.9 and then substitute it with the mean and standard deviation, into our z-score formula. The z-table shows a z-score of approximately 1.28, for an area under the normal curve to the left of z (larger portion) of approximately 0.9. Thus, we can write the following: Multiplying each side of the equation by 5 gives Adding 63 to both sides of the equation gives Thus, our score, k, is 69.4. 1.28 = x − 63 5 6.4 = x − 63 69.4 = x. k = 69.4 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 389 Figure 6.6 The 90th percentile is 69.4. This means that 90 percent of the test scores fall at or below 69.4 and 10 percent fall at or above. To get this answer on the calculator, follow this next step: invNorm in 2nd DISTR. invNorm(area to the left, mean, standard deviation) For this problem, invNorm(0.90,63,5) = 69.
4 d. Find the 70th percentile, —that is, find the score k such that 70 percent of scores are below k and 30 percent of the scores are above k. Solution 6.8 d. Find the 70th percentile. Draw a new graph and label it appropriately. k = 65.6 The 70th percentile is 65.6. This means that 70 percent of the test scores fall at or below 65.5 and 30 percent fall at or above. invNorm(0.70,63,5) = 65.6 6.8 The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a randomly selected golfer scored less than 65. Example 6.9 A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment 390 Chapter 6 \| The Normal Distribution are normally distributed and the standard deviation for the times is half an hour. a. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day. Solution 6.9 a. Let X = the amount of time, in hours, a household personal computer is used for entertainment. X ~ N(2, 0.5) where μ = 2 and σ = 0.5. Find P(1.8 < x < 2.75). First, calculate the z-scores for each x-value. z = 1.8 − 2 0.5 z = 2.75 − 2 0.5 = −0.2 0.5 = 0.75 0.5 = − 0.40 = 1.5 Now, use the Z-table to locate the area under the normal curve to the left of each of these z-scores. The area to the left of the z-score of −0.40 is 0.3446. The area to the left of the z-score of 1.5 is 0.9332. The area between these scores will be the difference in the two areas, or 0.9332 − 0.3446 , which equals 0.5886. Figure 6.7 normalcdf(1.8,2.75,2,0.5) = 0.5886 The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886. b. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment. Solution 6.9 b. To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25th percentile, k, where P(x < k) = 0.25. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 391 Figure 6.8 invNorm(0.25,2,0.5) = 1.66 We use invNorm because we are looking for the k-value. The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours. 6.9 The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70. Example 6.10 In the United States smartphone users between the ages of 13 and 55+ between the ages of 13 and 55+ approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively. a. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old. Solution 6.10 a. normalcdf(23,64.7,36.9,13.9) = 0.8186 The z-scores are calculated as z = 23 − 36.9 13.9 z = 64.7 − 36.9 13.9 = −13.9 13.9 = 27.8 13.9 = − 1 = 2 The Z-table shows the area to the left of a z-score with an absolute value of 1 to be 0.1587. It shows the area to the left of a z-score of 2 to be 0.9772. The difference in the two areas is 0.8185. This is slightly different than the area given by the calculator, due to rounding. b. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old. 392 Chapter 6 \| The Normal Distribution Solution 6.10 b. normalcdf(–1099,50.8,36.9,13.9) = 0.8413 c. Find the 80th percentile of this distribution, and interpret it in a complete sentence. Solution 6.10 c. invNorm(0.80,36.9,13.9) = 48.6 The 80th percentile is 48.6 years. 80 percent of the smartphone users in the age range 13–55+ are 48.6 years old or less. 6.10 Use the information in Example 6.10 to answer the following questions: a. Find the 30th percentile, and interpret it in a complete sentence. b. What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27 years old? Example 6.11 In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively. Using this information, answer the following questions. —Round answers to one decimal place. a. Calculate the interquartile range (IQR). Solution 6.11 a. IQR = Q3 – Q1 Calculate Q3 = 75th percentile and Q1 = 25th percentile. Recall that we can use invNorm to find the k-value. We can use this to find the quartile values. invNorm(0.75,36.9,13.9) = Q3 = 46.2754 invNorm(0.25,36.9,13.9) = Q1 = 27.5246 IQR = Q3 – Q1 = 18.8 b. Forty percent of the ages that range from 13 to 55+ are at least what age? Solution 6.11 b. Find k where P(x ≥ k) = 0.40. At least translates to greater than or equal to. 0.40 = the area to the right The area to the left = 1 – 0.40 = 0.60. The area to the left of k = 0.60 invNorm(0.60,36.9,13.9) = 40.4215 k = 40.4. Forty percent of the ages that range from 13 to 55+ are at least 40.4 years. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 393 6.11 Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean μ = 81 points and standard deviation σ = 15 points. a. Calculate the first- and third-quartile scores for this exam. b. The middle 50 percent of the exam scores are between what two values? Example 6.12 A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm. a. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph. Solution 6.12 a. normalcdf(6,10^99,5.85,0.24) = 0.2660 Figure 6.9 b. The middle 20 percent of mandarin oranges from this farm have diameters between ______ and ______. Solution 6.12 b. 1 – 0.20 = 0.80. Outside of the middle 20 percent will be 80 percent of the values. The tails of the graph of the normal distribution each have an area of 0.40. Find k1, the 40th percentile, and k2, the 60th percentile (0.40 + 0.20 = 0.60). This leaves the middle 20 percent, in the middle of the distribution. k1 = invNorm(0.40,5.85,0.24) = 5.79 cm k2 = invNorm(0.60,5.85,0.24) = 5.91 cm So, the middle 20 percent of mandarin oranges have diameters between 5.79 cm and 5.91 cm. c. Find the 90th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence. 394 Chapter 6 \| The Normal Distribution Solution 6.12 c. 6.16, Ninety percent of the diameter of the mandarin oranges is at most 6.16 cm. 6.12 Using the information from Example 6.12, answer the following: a. The middle 45 percent of mandarin oranges from this farm are between ______ and ______. b. Find the 16th percentile, and interpret it in a complete sentence. 6.3 \| Normal Distribution—Lap Times This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 395 6.1 Normal Distribution (Lap Times) Student Learning Outcome • The student will compare and contrast empirical data and a theoretical distribution to determine if Terry Vogel's lap times fit a continuous distribution. Directions Round the relative frequencies and probabilities to four decimal places. Carry all other decimal answers to two places. Use the data from Appendix C. Use a stratified sampling method by lap— races 1 to 20— and a random number generator to pick six lap times from each stratum. Record the lap times below for laps two to seven. _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ Table 6.1 Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes. Figure 6.10 a. Calculate the following: x¯ = _______ s = _______ b. Draw a smooth curve through the tops of the bars of the histogram. Write one to two complete sentences to describe the general shape of the curve. (Keep it simple. Does the graph go straight across, does it have a V-shape, does it have a hump in the middle or at either end, and so on?) Analyze the Distribution 396 Chapter 6 \| The Normal Distribution Using your sample mean, sample standard deviation, and histogram to help, what is the approximate theoretical distribution of the data? • X ~ _____(_____,_____) • How does the histogram help you arrive at the approximate distribution? Describe the Data Use the data you collected to complete the following statements. • The IQR goes from __________ to __________. IQR = __________. (IQR = Q3 – Q1) • • The 15th percentile is _______. • The 85th percentile is _______. • The median is _______. • The empirical probability that a randomly chosen lap time is more than 130 seconds is _______. • Explain the meaning of the 85th percentile of this data. Theoretical Distribution Using the theoretical distribution, complete the following statements. You should use a normal approximation based on your sample data. • The IQR goes from __________ to __________. IQR = _______. • • The 15th percentile is _______. • The 85th percentile is _______. • The median is _______. • The probability that a randomly chosen lap time is more than 130 seconds is _______. • Explain the meaning of the 85th percentile of this distribution. Discussion Questions Do the data
from the section titled Collect the Data give a close approximation to the theoretical distribution in the section titled Analyze the Distribution? In complete sentences and comparing the result in the sections titled Describe the Data and Theoretical Distribution, explain why or why not. 6.4 \| Normal Distribution—Pinkie Length This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 397 6.2 Normal Distribution (Pinkie Length) Student Learning Outcomes • The student will compare empirical data and a theoretical distribution to determine if data from the experiment follow a continuous distribution. Collect the Data Measure the length of your pinkie finger, in centimeters. 1. Randomly survey 30 adults for their pinkie finger lengths. Round the lengths to the nearest 0.5 cm. _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ Table 6.2 2. Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes. Figure 6.11 3. Calculate the following: a. b. x¯ = _______ s = _______ 4. Draw a smooth curve through the top of the bars of the histogram. Write one to two complete sentences to describe the general shape of the curve. Keep it simple. Does the graph go straight across, does it have a V-shape, does it have a hump in the middle or at either end, and so on? Analyze the Distribution Using your sample mean, sample standard deviation, and histogram, what was the approximate theoretical distribution 398 Chapter 6 \| The Normal Distribution of the data you collected? • X ~ _____ (_____, _____) • How does the histogram help you arrive at the approximate distribution? Describe the Data Using the data you collected complete the following statements. Hint—Order the data. REMEMBER (IQR = Q3 – Q1) IQR = _______ • • The 15th percentile is _______. • The 85th percentile is _______. • Median is _______. • What is the theoretical probability that a randomly chosen pinkie length is more than 6.5 cm? • Explain the meaning of the 85th percentile of these data. Theoretical Distribution Using the theoretical distribution, complete the following statements. Use a normal approximation based on the sample mean and standard deviation. IQR = _______ • • The 15th percentile is _______. • The 85th percentile is _______. • Median is _______. • What is the theoretical probability that a randomly chosen pinkie length is more than 6.5 cm? • Explain the meaning of the 85th percentile of these data. Discussion Questions Do the data you collected give a close approximation to the theoretical distribution? In complete sentences and comparing the results in the sections titled Describe the Data and Theoretical Distribution, explain why or why not. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 399 KEY TERMS normal distribution a continuous random variable (RV) where μ is the mean of the distribution and σ is the standard deviation; notation: X ~ N(μ, σ). If μ = 0 and σ = 1, the RV is called the standard normal distribution. standard normal distribution a continuous random variable (RV) X ~ N(0, 1); when X follows the standard normal distribution, it is often noted as Z ~ N(0, 1). z-score the linear transformation of the form z = x – μ σ ; if this transformation is applied to any normal distribution X ~ N(μ, σ), the result is the standard normal distribution Z ~ N(0, 1); If this transformation is applied to any specific value x of the RV with mean μ and standard deviation σ, the result is called the z-score of x. The z-score allows us to compare data that are normally distributed but scaled differently. CHAPTER REVIEW 6.1 The Standard Normal Distribution A z-score is a standardized value. Its distribution is the standard normal, Z ~ N(0, 1). The mean of the z-scores is zero and the standard deviation is one. If z is the z-score for a value x from the normal distribution N(µ, σ), then z tells you how many standard deviations x is above—greater than—or below—less than—µ. 6.2 Using the Normal Distribution The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bellshaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean µ and the standard deviation σ. A special normal distribution, called the standard normal distribution, is the distribution of z-scores. Its mean is zero, and its standard deviation is one. FORMULA REVIEW 6.0 Introduction X ∼ N(μ, σ) μ = the mean, σ = the standard deviation Z = the random variable for z-scores 6.2 Using the Normal Distribution Normal Distribution: X ~ N(µ, σ), where µ is the mean and σ is the standard deviation 6.1 The Standard Normal Distribution Standard Normal Distribution: Z ~ N(0, 1). Calculator function for probability: normalcdf (lower x value of the area, upper x value of the area, mean, standard deviation) Calculator function for the kth percentile: k = invNorm (area to the left of k, mean, standard deviation) Z ~ N(0, 1) z = a standardized value (z-score) mean = 0, standard deviation = 1 To find the kth percentile of X when the z-score is known, k = μ + (z)σ z-score: z = x – μ σ PRACTICE 6.1 The Standard Normal Distribution 1. A bottle of water contains 12.05 fluid ounces with a standard deviation of 0.01 ounces. Define the random variable X in words. X = ____________. 2. A normal distribution has a mean of 61 and a standard deviation of 15. What is the median? 400 3. X ~ N(1, 2) σ = _______ Chapter 6 \| The Normal Distribution 4. A company manufactures rubber balls. The mean diameter of a ball is 12 cm with a standard deviation of 0.2 cm. Define the random variable X in words. X = ______________. 5. X ~ N(–4, 1) What is the median? 6. X ~ N(3, 5) σ = _______ 7. X ~ N(–2, 1) μ = _______ 8. What does a z-score measure? 9. What does standardizing a normal distribution do to the mean? 10. Is X ~ N(0, 1) a standardized normal distribution? Why or why not? 11. What is the z-score of x = 12, if it is two standard deviations to the right of the mean? 12. What is the z-score of x = 9, if it is 1.5 standard deviations to the left of the mean? 13. What is the z-score of x = –2, if it is 2.78 standard deviations to the right of the mean? 14. What is the z-score of x = 7, if it is 0.133 standard deviations to the left of the mean? 15. Suppose X ~ N(2, 6). What value of x has a z-score of three? 16. Suppose X ~ N(8, 1). What value of x has a z-score of –2.25? 17. Suppose X ~ N(9, 5). What value of x has a z-score of –0.5? 18. Suppose X ~ N(2, 3). What value of x has a z-score of –0.67? 19. Suppose X ~ N(4, 2). What value of x is 1.5 standard deviations to the left of the mean? 20. Suppose X ~ N(4, 2). What value of x is two standard deviations to the right of the mean? 21. Suppose X ~ N(8, 9). What value of x is 0.67 standard deviations to the left of the mean? 22. Suppose X ~ N(–1, 2). What is the z-score of x = 2? 23. Suppose X ~ N(12, 6). What is the z-score of x = 2? 24. Suppose X ~ N(9, 3). What is the z-score of x = 9? 25. Suppose a normal distribution has a mean of six and a standard deviation of 1.5. What is the z-score of x = 5.5? 26. In a normal distribution, x = 5 and z = –1.25. This tells you that x = 5 is ____ standard deviations to the ____ (right or left) of the mean. 27. In a normal distribution, x = 3 and z = 0.67. This tells you that x = 3 is ____ standard deviations to the ____ (right or left) of the mean. 28. In a normal distribution, x = –2 and z = 6. This tells you that x = –2 is ____ standard deviations to the ____ (right or left) of the mean. 29. In a normal distribution, x = –5 and z = –3.14. This tells you that x = –5 is ____ standard deviations to the ____ (right or left) of the mean. 30. In a normal distribution, x = 6 and z = –1.7. This tells you that x = 6 is ____ standard deviations to the ____ (right or left) of the mean. 31. About what percent of x values from a normal distribution lie within one standard deviation, left and right, of the mean of that distribution? 32. About what percent of the x values from a normal distribution lie within two standard deviations, left and right, of the mean of that distribution? 33. About what percent of x values lie between the second and third standard deviations, both sides? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 401 34. Suppose X ~ N(15, 3). Between what x values does 68.27 percent of the data lie? The range of x values is centered at the mean of the distribution (i.e., 15). 35. Suppose X ~ N(–3, 1). Between what x values does 95.45 percent of the data lie? The range of x values is centered at the mean of the distribution (i.e., –3). 36. Suppose X ~ N(–3, 1). Between what x values does 34.14 percent of the data lie? 37. About what percent of x values lie between the mean and three standard deviations? 38. About what percent of x values lie between the mean and one standard deviation? 39. About what percent of x values lie between the first and second standard deviations from the mean, both sides? 40. About what percent of x values lie between the first and third standard deviations, both sides? Use the following information to answer the next two exercises: The life of Sunshine CD players is normally distributed with mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. 41. Define the random variable X in words. X = _______________. 42. X ~ _____(_____, _____) 6.2 Using the Normal Distribution 43. How would you represent the area to t
he left of one in a probability statement? Figure 6.12 44. What is the area to the right of one? Figure 6.13 45. Is P(x < 1) equal to P(x ≤ 1)? Why or why not? 402 Chapter 6 \| The Normal Distribution 46. How would you represent the area to the left of three in a probability statement? Figure 6.14 47. What is the area to the right of three? Figure 6.15 48. If the area to the left of x in a normal distribution is 0.123, what is the area to the right of x? 49. If the area to the right of x in a normal distribution is 0.543, what is the area to the left of x? Use the following information to answer the next four exercises: X ~ N(54, 8) 50. Find the probability that x > 56. 51. Find the probability that x < 30. 52. Find the 80th percentile. 53. Find the 60th percentile. 54. X ~ N(6, 2) Find the probability that x is between three and nine. 55. X ~ N(–3, 4) Find the probability that x is between one and four. 56. X ~ N(4, 5) Find the maximum of x in the bottom quartile. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 403 57. Use the following information to answer the next three exercises: The life of Sunshine CD players is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. Find the probability that a CD player will break down during the guarantee period. a. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability. Figure 6.16 b. P(0 < x < ____________) = ___________. Use zero for the minimum value of x. 58. Find the probability that a CD player will last between 2.8 and 6 years. a. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability. Figure 6.17 b. P(__________ < x < __________) = __________ 59. Find the 70th percentile of the distribution for the time a CD player lasts. a. Sketch the situation. Label and scale the axes. Shade the region corresponding to the lower 70 percent. Figure 6.18 b. P(x < k) = __________. Therefore, k = _________. 404 Chapter 6 \| The Normal Distribution HOMEWORK 6.1 The Standard Normal Distribution Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. 60. What is the median recovery time? a. 2.7 b. 5.3 c. 7.4 d. 2.1 61. What is the z-score for a patient who takes 10 days to recover? a. 1.5 b. 0.2 c. 2.2 d. 7.3 62. The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. If the mean is significantly greater than the standard deviation, which of the following statements is true? I. The data cannot follow the uniform distribution. II. The data cannot follow the exponential distribution. III. The data cannot follow the normal distribution. a. b. c. d. I only II only III only I, II, and III 63. The heights of the 430 basketball players were listed on team rosters at the start of the 2005–2006 season. The heights of basketball players have an approximate normal distribution with a mean, µ = 79 inches, and a standard deviation, σ = 3.89 inches. For each of the following heights, calculate the z-score and interpret it using complete sentences: a. 77 inches b. 85 inches c. If a player reported his height had a z-score of 3.5, would you believe him? Explain your answer. 64. The systolic blood pressure, given in millimeters, of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. Systolic blood pressure for males follows a normal distribution. a. Calculate the z-scores for the male systolic blood pressures 100 and 150 millimeters. b. If a male friend of yours said he thought his systolic blood pressure was 2.5 standard deviations below the mean, and that he believed his blood pressure was between 100 and 150 millimeters, what would you say to him? 65. Kyle’s doctor told him that the z-score for his systolic blood pressure is 1.75. Which of the following is the best interpretation of this standardized score? The systolic blood pressure, given in millimeters, of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. If X = a systolic blood pressure score, then X ~ N (125, 14). a. Which answer(s) is/are correct? i. Kyle’s systolic blood pressure is 175. ii. Kyle’s systolic blood pressure is 1.75 times the average blood pressure of men his age. iii. Kyle’s systolic blood pressure is 1.75 above the average systolic blood pressure of men his age. iv. Kyles’s systolic blood pressure is 1.75 standard deviations above the average systolic blood pressure for men. b. Calculate Kyle’s blood pressure. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 405 66. Height and weight are two measurements used to track a child’s development. The World Health Organization measures child development by comparing the weights of children who are the same height and same gender. In 2009, weights for all 80 cm girls in the reference population had a mean µ = 10.2 kg and standard deviation σ = 0.8 kg. Weights are normally distributed. X ~ N(10.2, 0.8). Calculate the z-scores that correspond to the following weights and interpret them: a. 11 kg b. 7.9 kg c. 12.2 kg 67. In 2005, 1,475,623 students heading to college took the SAT exam. The distribution of scores in the math section of the SAT follows a normal distribution with mean µ = 520 and standard deviation σ = 115. a. Calculate the z-score for an SAT score of 720. Interpret it using a complete sentence. b. What math SAT score is 1.5 standard deviations above the mean? What can you say about this SAT score? c. For 2012, the SAT math test had a mean of 514 and standard deviation 117. The ACT math test is an alternative to the SAT math test, and is approximately normally distributed with mean 21 and standard deviation 5.3. If one person took the SAT math test and scored 700 and a second person took the ACT math test and scored 30, who did better with respect to the test that each person took? 6.2 Using the Normal Distribution Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. 68. What is the probability of spending more than two days in recovery? a. 0.0580 b. 0.8447 c. 0.0553 d. 0.9420 69. The 90th percentile for recovery times is – a. 8.89 b. 7.07 c. 7.99 d. 4.32 Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. 70. Based on the given information and numerically justified, would you be surprised if it took less than one minute to find a parking space? a. Yes b. No c. Unable to determine 71. Find the probability that it takes at least eight minutes to find a parking space. a. 0.0001 b. 0.9270 c. 0.1862 d. 0.0668 72. Seventy percent of the time, it takes more than how many minutes to find a parking space? a. 1.24 b. 2.41 c. 3.95 d. 6.05 406 Chapter 6 \| The Normal Distribution 73. According to a study done by De Anza students, the height for Asian adult males is normally distributed with an average of 66 inches and a standard deviation of 2.5 inches. Suppose one Asian adult male is randomly chosen. Let X = height of the individual. a. X ~ _____ (_____ ,_____) b. Find the probability that the person is between 65 and 69 inches. Include a sketch of the graph, and write a probability statement. c. Would you expect to meet many Asian adult males taller than 72 inches? Explain why or why not, and numerically justify your answer. d. The middle 40 percent of heights fall between what two values? Sketch the graph, and write the probability statement. 74. IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual. a. X ~ _____ (_____, _____) b. Find the probability that the person has an IQ greater than 120. Include a sketch of the graph, and write a probability statement. c. MENSA is an organization whose members have the top 2 percent of all IQs. Find the minimum IQ needed to qualify for the MENSA organization. Sketch the graph, and write the probability statement. d. The middle 50 percent of IQs fall between what two values? Sketch the graph, and write the probability statement. 75. The percent of fat calories that a person in the United States consumes each day is normally distributed with a mean of about 36 and a standard deviation of 10. Suppose that one individual is randomly chosen. Let X = percentage of fat calories. a. X ~ _____ (_____, _____) b. Find the probability that the percentage of fat calories a person consumes is more than 40. Graph the situation. Shade in the area to be determined. c. Find the maximum number for the lower quarter of percent of fat calories. Sketch the graph and write the probability statement. 76. Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. a. b. If X = distance in feet for a fly ball, then X ~ _____ (_____, _____) If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled less than 220 feet? Sketch the graph. Scale the horizontal axis X. Shade the region corresponding to the probability. Find the probability. c. Find the 80th percentile of the distribution of fly balls. Sketch the graph, and write the probability statement. 77. In China, four-year-olds average three hours a day uns
upervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount of time spent alone is normally distributed. We randomly select one Chinese four-year-old living in a rural area. We are interested in the amount of time that child spends alone per day. In words, define the random variable X. a. b. X ~ _____ (_____, _____) c. Find the probability that the child spends less than one hour per day unsupervised. Sketch the graph, and write the probability statement. d. What percentage of the children spend more than 10 hours per day unsupervised? e. Seventy percent of the children spend at least how long per day unsupervised? 78. In the 1992 presidential election, Alaska’s 40 election districts averaged 1,956.8 votes per district for a candidate. The standard deviation was 572.3. There are only 40 election districts in Alaska. The distribution of the votes per district for the candidate was bell-shaped. Let X = number of votes for the candidate for an election district. a. State the approximate distribution of X. b. c. Find the probability that a randomly selected district had fewer than 1,600 votes for the candidate. Sketch the Is 1,956.8 a population mean or a sample mean? How do you know? graph, and write the probability statement. d. Find the probability that a randomly selected district had between 1,800 and 2,000 votes for the candidate. e. Find the third quartile for votes for the candidate. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 407 79. Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 21 days and a standard deviation of seven days. In words, define the random variable X. a. b. X ~ _____ (_____, _____) c. If one of the trials is randomly chosen, find the probability that it lasted at least 24 days. Sketch the graph and write the probability statement. d. Sixty percent of all trials of this type are completed within how many days? 80. Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5-mile lap, in a seven-lap race, with a standard deviation of 2.28 seconds. The distribution of her race times is normally distributed. We are interested in one of her randomly selected laps. In words, define the random variable X. a. b. X ~ _____ (_____, _____) c. Find the percent of her laps that are completed in less than 130 seconds. d. The fastest 3 percent of her laps are under _____. e. The middle 80 percent of her laps are from _______ seconds to _______ seconds. 81. Thuy Dau, Ngoc Bui, Sam Su, and Lan Voung conducted a survey as to how long customers at Lucky claimed to wait in the checkout line until their turn. Let X = time in line. Table 6.3 displays the ordered real data, in minutes. 0.50 4.25 5 6 1.75 4.25 5.25 6 7.25 7.25 2 4.25 5.25 6.25 7.25 2.25 4.25 5.5 6.25 7.75 2.25 4.5 5.5 2.5 4.75 5.5 6.5 6.5 8 8.25 2.75 4.75 5.75 6.5 9.5 3.25 4.75 5.75 6.75 9.5 3.75 5 3.75 5 6 6 6.75 9.75 6.75 10.75 Table 6.3 a. Calculate the sample mean and the sample standard deviation. b. Construct a histogram. c. Draw a smooth curve through the midpoints of the tops of the bars. d. In words, describe the shape of your histogram and smooth curve. e. Let the sample mean approximate μ and the sample standard deviation approximate σ. The distribution of X can then be approximated by X ~ _____ (_____, _____) f. Use the distribution in part e to calculate the probability that a person will wait fewer than 6.1 minutes. g. Determine the cumulative relative frequency for waiting less than 6.1 minutes. h. Why aren’t the answers to part f and part g exactly the same? i. Why are the answers to part f and part g as close as they are? j. If only 10 customers were surveyed rather than 50, do you think the answers to part f and part g would have been closer together or farther apart? Explain your conclusion. 82. Suppose that Ricardo and Anita attend different colleges. Ricardo’s GPA is the same as the average GPA at his school. Anita’s GPA is 0.70 standard deviations above her school average. In complete sentences, explain why each of the following statements may be false: a. Ricardo’s actual GPA is lower than Anita’s actual GPA. b. Ricardo is not passing because his z-score is zero. c. Anita is in the 70th percentile of students at her college. 408 Chapter 6 \| The Normal Distribution 83. Table 6.4 shows a sample of the maximum capacity—maximum number of spectators—of sports stadiums. The table does not include horse-racing or motor-racing stadiums. 40,000 40,000 45,050 45,500 46,249 48,134 49,133 50,071 50,096 50,466 50,832 51,100 51,500 51,900 52,000 52,132 52,200 52,530 52,692 53,864 54,000 55,000 55,000 55,000 55,000 55,000 55,000 55,082 57,000 58,008 59,680 60,000 60,000 60,492 60,580 62,380 62,872 64,035 65,000 65,050 65,647 66,000 66,161 67,428 68,349 68,976 69,372 70,107 70,585 71,594 72,000 72,922 73,379 74,500 75,025 76,212 78,000 80,000 80,000 82,300 Table 6.4 a. Calculate the sample mean and the sample standard deviation for the maximum capacity of sports stadiums. b. Construct a histogram. c. Draw a smooth curve through the midpoints of the tops of the bars of the histogram. d. e. Let the sample mean approximate μ and the sample standard deviation approximate σ. The distribution of X can In words, describe the shape of your histogram and smooth curve. then be approximated by X ~ _____ (_____, _____). f. Use the distribution in part e to calculate the probability that the maximum capacity of sports stadiums is less than 67,000 spectators. g. Determine the cumulative relative frequency that the maximum capacity of sports stadiums is less than 67,000 spectators. Hint—Order the data and count the sports stadiums that have a maximum capacity less than 67,000. Divide by the total number of sports stadiums in the sample. h. Why aren’t the answers to part f and part g exactly the same? 84. The length of a pregnancy of a certain female animal is normally distributed with a mean of 280 days and a standard deviation of 13 days. The father was not present from 240 to 306 days before the birth of the offspring, so the pregnancy would have been less than 240 days or more than 306 days long, if he was the father. What is the probability that he was NOT the father? What is the probability that he could be the father? Calculate the z-scores first, and then use those to calculate the probability. 85. A NUMMI assembly line, which has been operating since 1984, has built an average of 6,000 cars and trucks a week. Generally, 10 percent of the cars were defective coming off the assembly line. Suppose we draw a random sample of n = 100 cars. Let X represent the number of defective cars in the sample. What can we say about X in regard to the 68–95–99.7 empirical rule—one standard deviation, two standard deviations, and three standard deviations from the mean being referred to? Assume a normal distribution for the defective cars in the sample. 86. We flip a coin 100 times (n = 100) and note that it only comes up heads 20 percent (p = 0.20) of the time. The mean and standard deviation for the number of times the coin lands on heads is µ = 20 and σ = 4—verify the mean and standard deviation. Solve the following: a. There is about a 68 percent chance that the number of heads will be somewhere between ___ and ___. b. There is about a ____chance that the number of heads will be somewhere between 12 and 28. c. There is about a ____ chance that the number of heads will be somewhere between eight and 32. 87. A child playing a carnival game will be a winner one out of five times. If 190 games are played, what is the probability that there are a. b. c. more than 64 wins somewhere between 34 and 54 wins somewhere between 54 and 64 wins This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 409 88. A social media site provides a variety of statistics on its website that detail the growth and popularity of the site. On average, 28 percent of 18- to 34-year-olds check their social media profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent. a. Find the probability that the percentage of 18- to 34-year-olds who check the social media website before getting out of bed in the morning is at least 30. b. Find the 95th percentile, and express it in a sentence. REFERENCES 6.1 The Standard Normal Distribution CollegeBoard. group http://media.collegeboard.com/digitalServices/pdf/research/TotalGroup-2012.pdf College-bound seniors: (2012). Total 2012 profile report. Retrieved from Joyce, C. A., Janssen, S., & Liu, M. L. (2010). The world almanac and book of facts, 2010. New York, NY: World Almanac Books. London School http://conflict.lshtm.ac.uk/page_125.htm of Hygiene and Tropical Medicine. (2009). Calculation of z-scores. Retrieved from National Center for Education Statistics. (2009). ACT score averages and standard deviations, by sex and race/ethnicity, and percentage of ACT test takers, by selected composite score ranges and planned fields of study: Selected years, 1995 through 2009. Retrieved from http://nces.ed.gov/programs/digest/d09/tables/dt09_147.asp NBA.com. (2013). NBA Media Ventures. Retrieved from http://www.nba.com StatCrunch. viewreport.php?reportid=11960 (2010). Blood pressure of males and females. Retrieved from http://www.statcrunch.com/5.0/ The Mercury News. (n.d.). Retrieved from http://www.mercurynews.com/ Wikipedia. (2013). List of List_of_stadiums_by_capacity stadiums by capacity - Wikipedia. Retrieved from https://en.wikipedia.org/wiki/ 6.2 Using the Normal Distribution Chicago Public Media & Ira Glass. (2013). 403: NUMMI. Retrieved from http://www.thisamericanlife.org/radioarchives/ episode/403/nummi lauramitchell347. (2012, Dec. 28). Smart phone users, by the numbe
rs. Visually. Retrieved from http://visual.ly/smartphone-users-numbers Statistics Brain Research Institute. http://www.statisticbrain.com/facebook-statistics/ (2013). Facebook company statistics – statistic brain. Retrieved from Wikipedia (2013). Naegele's rule. Retrieved from http://en.wikipedia.org/wiki/Naegele's_rule Win at the Lottery. (2013). Scratch-off lottery ticket playing tips. Retrieved from www.winatthelottery.com/public/ department40.cfm SOLUTIONS 1 ounces of water in a bottle 3 2 5 –4 7 –2 9 The mean becomes zero. 11 z = 2 Chapter 6 \| The Normal Distribution 410 13 z = 2.78 15 x = 20 17 x = 6.5 19 x = 1 21 x = 1.97 23 z = –1.67 25 z ≈ –0.33 27 0.67, right 29 3.14, left 31 about 68 percent 33 about 4 percent 35 between –5 and –1 37 about 50 percent 39 about 27 percent 41 The lifetime of a Sunshine CD player measured in years 43 P(x < 1) 45 Yes, because they are the same in a continuous distribution: P(x = 1) = 0 47 1 – P(x < 3) or P(x > 3) 49 1 – 0.543 = 0.457 51 0.0013 53 56.03 55 0.1186 57 a. Check student’s solution b. 3, 0.1979 59 a. Check student’s solution b. 0.70, 4.78 years 61 c 63 a. Use the z-score formula. z = –0.5141. The height of 77 inches is 0.5141 standard deviations below the mean. An NBA player whose height is 77 inches is shorter than average. b. Use the z-score formula. z = 1.5424. The height 85 inches is 1.5424 standard deviations above the mean. An NBA player whose height is 85 inches is taller than average. c. Height = 79 + 3.5(3.89) = 90.67 inches, which is over 7.7 feet tall. There are very few NBA players this tall; so, the answer is no, not likely. 65 a. iv b. Kyle’s blood pressure is equal to 125 + (1.75)(14) = 149.5. 67 Let X = an SAT math score and Y = an ACT math score. a. X = 720 720 – 520 = 1.74 The exam score of 720 is 1.74 standard deviations above the mean of 520. 15 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 \| The Normal Distribution 411 b. z = 1.5 The math SAT score is 520 + 1.5(115) ≈ 692.5. The exam score of 692.5 is 1.5 standard deviations above the mean of 520. c. X – μ σ = 700 – 514 117 ≈ 1.59, the z-score for the SAT. Y – μ σ = 30 – 21 5.3 ≈ 1.70, the z-scores for the ACT. With respect to the test they took, the person who took the ACT did better—has the higher z-score). 69 c 71 d 73 a. X ~ N(66, 2.5) b. 0.5404 c. No, the probability that an Asian male is over 72 inches tall is 0.0082. 75 a. X ~ N(36, 10) b. The probability that a person consumes more than 40 percent of their calories as fat is 0.3446. c. Approximately 25 percent of people consume less than 29.26 percent of their calories as fat. 77 a. X = number of hours that a Chinese four-year-old in a rural area is unsupervised during the day. b. X ~ N(3, 1.5) c. The probability that the child spends less than one hour a day unsupervised is 0.0918. d. The probability that a child spends over 10 hours a day unsupervised is less than 0.0001. e. 2.21 hours 79 a. X = the distribution of the number of days a particular type of criminal trial will take b. X ~ N(21, 7) c. The probability that a randomly selected trial will last more than 24 days is 0.3336. d. 22.77 81 a. mean = 5.51, s = 2.15 b. Check student's solution. c. Check student's solution. d. Check student's solution. e. X ~ N(5.51, 2.15) f. 0.6029 g. The cumulative frequency for less than 6.1 minutes is 0.64. h. The answers to part f and part g are not exactly the same, because the normal distribution is only an approximation to the real one. i. The answers to part f and part g are close, because a normal distribution is an excellent approximation when the sample size is greater than 30. j. The approximation would have been less accurate, because the smaller sample size means that the data does not fit a normal curve as well. Chapter 6 \| The Normal Distribution 412 83 1. mean = 60,136 s = 10,468 2. Answers will vary 3. Answers will vary 4. Answers will vary 5. X ~ N(60136, 10468) 6. 0.7440 7. The cumulative relative frequency is 43/60 = 0.717. 8. The answers for part f and part g are not the same because the normal distribution is only an approximation. 85 n = 100; p = 0.1; q = 0.9 μ = np = (100)(0.10) = 10 σ = npq = (100)(0.1)(0.9) = 3 i. z = ±1: x1 = µ + zσ = 10 + 1(3) = 13 and x2 = µ – zσ = 10 – 1(3) = 7. 68 percent of the defective cars will fall between seven and 13 ii. iii. z = ±2: x1 = µ + zσ = 10 + 2(3) = 16 and x2 = µ – zσ = 10 – 2(3) = 4. 95 percent of the defective cars will fall between four and 16 z = ±3: x1 = µ + zσ = 10 + 3(3) = 19 and x2 = µ – zσ = 10 – 3(3) = 1. 99.7 percent of the defective cars will fall between one and 19 87 n = 190; p = 1 5 = 0.2; q = 0.8 μ = np = (190)(0.2) = 38 σ = npq = (190)(0.2)(0.8) = 5.5136 a. For this problem: P(34 < x < 54) = normalcdf(34,54,48,5.5136) = 0.7641 b. For this problem: P(54 < x < 64) = normalcdf(54,64,48,5.5136) = 0.0018 c. For this problem: P(x > 64) = normalcdf(64,1099,48,5.5136) = 0.0000012 (approximately 0) This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 413 7 \| THE CENTRAL LIMIT THEOREM Figure 7.1 If you want to figure out the distribution of the change people carry in their pockets, using the central limit theorem and assuming your sample is large enough, you will find that the distribution is normal and bell-shaped. (credit: John Lodder) Introduction Chapter Objectives By the end of this chapter, the student should be able to do the following: • Recognize central limit theorem problems • Classify continuous word problems by their distributions • Apply and interpret the central limit theorem for means • Apply and interpret the central limit theorem for sums Why are we so concerned with means? Two reasons are they give us a middle ground for comparison, and they are easy to calculate. In this chapter, you will study means and the central limit theorem. The central limit theorem (clt) is one of the most powerful and useful ideas in all of statistics. There are two alternative forms of the theorem, and both alternatives are concerned with drawing a finite samples size n from a population with a 414 Chapter 7 \| The Central Limit Theorem known mean, μ, and a known standard deviation, σ. The first alternative says that if we collect samples of size n with a large enough n, calculate each sample's mean, and create a histogram of those means, then the resulting histogram will tend to have an approximate normal bell shape. The second alternative says that if we again collect samples of size n that are large enough, calculate the sum of each sample and create a histogram, then the resulting histogram will again tend to have a normal bell shape. The central limit theorem for sample means is more discussed in the world of statistics, but it is important to note that taking each sample's sum and graphing the sums will also result in a normal histogram. There are instances where one wishes to calculate the sum of a sample, as opposed to its mean. In either case, it does not matter what the distribution of the original population is, or whether you even need to know it. The important fact is that the distributions of sample means and the sums tend to follow the normal distribution. The size of the sample, n, that is required in order to be large enough depends on the original population from which the samples are drawn (the sample size should be at least 30 or the data should come from a normal distribution). If the original population is far from normal, then more observations are needed for the sample means or sums to be normal. Sampling is done with replacement. Suppose eight of you roll one fair die ten times, seven of you roll two fair dice ten times, nine of you roll five fair dice ten times, and 11 of you roll ten fair dice ten times. Each time a person rolls more than one die, he or she calculates the sample mean of the faces showing. For example, one person might roll five fair dice and get 2, 2, 3, 4, and 6 on one roll. The mean is .4. The 3.4 is one mean when five fair dice are rolled. This same person would roll the five dice nine more times and calculate nine more means for a total of ten means. Your instructor will pass out the dice to several people. Roll your dice ten times. For each roll, record the faces, and find the mean. Round to the nearest 0.5. Your instructor (and possibly you) will produce one graph (it might be a histogram) for one die, one graph for two dice, one graph for five dice, and one graph for ten dice. Because the mean when you roll one die is just the face on the die, what distribution do these means appear to be representing? Draw the graph for the means using two dice. Do the sample means show any kind of pattern? Draw the graph for the means using five dice. Do you see any pattern emerging? Finally, draw the graph for the means using ten dice. Do you see any pattern to the graph? What can you conclude as you increase the number of dice? As the number of dice rolled increases from one to two to five to ten, the following is happening: 1. The mean of the sample means remains approximately the same. 2. The spread of the sample means (the standard deviation of the sample means) gets smaller. 3. The graph appears steeper and thinner. You have just demonstrated the central limit theorem (clt). The central limit theorem tells you that as you increase the number of dice, the sample means tend toward a normal distribution (the sampling distribution). 7.1 \| The Central Limit Theorem for Sample Means (Averages) Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution). Using a subscript that matches the random variable, suppose a. μ x = the mean of X This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 415 b. σ x = the standard deviation of X If you draw random samples of size n, then as n increases, the random variable X to be normally d
istributed and ¯ , which consists of sample means, tends ¯ X ∼ N ⎛ ⎝μ x, σ x n ⎞ ⎠ The central limit theorem for sample means says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten dice) and calculating their means, the sample means form their own normal distribution (the sampling distribution). The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by the sample size. The variable n is the number of values that are averaged together, not the number of times the experiment is done. ¯ , which consists To put it more formally, if you draw random samples of size n, the distribution of the random variable X of sample means, is called the sampling distribution of the mean. The sampling distribution of the mean approaches a normal distribution as n, the sample size, increases. The random variable X ¯ has a different z-score associated with it from that of the random variable X. The mean x¯ is the ¯ value of X in one sample. z = x¯ − μ x σ x n ⎞ ⎠ ⎛ ⎝ , ¯ . μX is the average of both X and X σ x¯ = σx n = standard deviation of X ¯ and is called the standard error of the mean. To find probabilities for means on the calculator, follow these steps. 2nd DISTR 2:normalcdf normalcd f ⎛ ⎝lower value o f the area, upper value o f the area, mean, standard deviation sample size ⎞ ⎠ where • mean is the mean of the original distribution • • standard deviation is the standard deviation of the original distribution sample size = n Example 7.1 A distribution has a mean of 90 and a standard deviation of 15. Samples of size n = 25 are drawn randomly from the population. a. Find the probability that the sample mean is between 85 and 92. Solution 7.1 a. Let X = one value from the original unknown population. The probability question asks you to find a probability for the sample mean. 416 Chapter 7 \| The Central Limit Theorem Let X ¯ = the mean of a sample of size 25. Because μx = 90, σx = 15, and n = 25, ¯ X ∼ N ⎛ ⎝μ x, σ x n ⎞ ⎠ Find P(85 < x¯ < 92). Draw a graph. P(85 < x¯ < 92) = 0.6997 The probability that the sample mean is between 85 and 92 is 0.6997. Figure 7.2 Find P(85 < x¯ < 92). Draw a graph. P(85 < x¯ < 92) = 0.6997 normalcdf(lower value, upper value, mean, standard error of the mean) The parameter list is abbreviated (lower value, upper value, μ, σ n ). normalcdf(85,92,90, 15 25 ) = 0.6997 b. Find the value that is two standard deviations above the expected value, 90, of the sample mean. Solution 7.1 b. To find the value that is two standard deviations above the expected value 90, use the following formula ⎛ value = µ x + ( # ofSTDEVs) ⎝ σ x n ⎞ ⎠ value = 90 + 2 ⎛ ⎝ 15 25 ⎞ ⎠ = 96. The value that is two standard deviations above the expected value is 96. The standard error of the mean is σx n = 15 25 = 3. Recall that the standard error of the mean is a description of how far (on average) that the sample mean will be from the population mean in repeated simple random samples of size n. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 417 7.1 An unknown distribution has a mean of 45 and a standard deviation of eight. Samples of size n = 30 are drawn randomly from the population. Find the probability that the sample mean is between 42 and 50. Example 7.2 The length of time, in hours, it takes a group of people, 40 years old and older, to play one soccer match is normally distributed with a mean of 2 hours and a standard deviation of 0.5 hours. A sample of size n = 50 is drawn randomly from the population. Find the probability that the sample mean is between 1.8 hours and 2.3 hours. Solution 7.2 Let X = the time, in hours, it takes to play one soccer match. The probability question asks you to find a probability for the sample mean time, in hours, it takes to play one soccer match. ¯ = the mean time, in hours, it takes to play one soccer match. Let X If μX = _________, σX = __________, and n = ___________, then X ~ N(______, ______) by the central limit theorem for means. μX = 2, σX = 0.5, n = 50, and X ~ N ⎛ ⎝2, 0.5 50 ⎞ ⎠ Find P(1.8 < x¯ < 2.3). Draw a graph. normalcdf P(1.8 < x¯ < 2.3) = 0.9977 ⎛ ⎝1.8,2.3,2, .5 50 ⎞ ⎠= 0.9977 The probability that the mean time is between 1.8 hours and 2.3 hours is 0.9977. 7.2 The length of time taken on the SAT exam for a group of students is normally distributed with a mean of 2.5 hours and a standard deviation of 0.25 hours. A sample size of n = 60 is drawn randomly from the population. Find the probability that the sample mean is between two hours and three hours. To find percentiles for means on the calculator, follow these steps. 2nd DIStR 3:invNorm k = invNorm ⎛ ⎝area to the left of k, mean, standard deviation sample size ⎞ ⎠ 418 Chapter 7 \| The Central Limit Theorem where • k = the kth percentile • mean is the mean of the original distribution • • standard deviation is the standard deviation of the original distribution sample size = n Example 7.3 In a recent study reported Oct. 29, 2012, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. Take a sample of size n = 100. a. What are the mean and standard deviation for the sample mean ages of tablet users? b. What does the distribution look like? c. Find the probability that the sample mean age is more than 30 years (the reported mean age of tablet users in this particular study). d. Find the 95th percentile for the sample mean age (to one decimal place). Solution 7.3 a. Because the sample mean tends to target the population mean, we have μχ = μ = 34. The sample standard = 1.5. deviation is given by σχ = σ n = 15 100 = 15 10 b. The central limit theorem states that for large sample sizes (n), the sampling distribution will be approximately normal. c. The probability that the sample mean age is more than 30 is given by P(Χ > 30) = normalcdf(30,E99,34,1.5) = 0.9962. d. Let k = the 95th percentile. ⎝0.95,34, 15 100 k = invNorm ⎛ ⎞ ⎠ = 36.5 7.3 A gaming marketing gap for men between the ages of 30 to 40 has been identified. You are researching a startup game targeted at the 35-year-old demographic. Your idea is to develop a strategy game that can be played by men from their late 20s through their late 30s. Based on the article’s data, industry research shows that the average strategy player is 28 years old with a standard deviation of 4.8 years. You take a sample of 100 randomly selected gamers. If your target market is 29- to 35-year-olds, should you continue with your development strategy? Example 7.4 The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of 60. a. What are the mean and standard deviation for the sample mean number of app engagement minutes by a tablet user? b. What is the standard error of the mean? c. Find the 90th percentile for the sample mean time for app engagement for a tablet user. Interpret this value This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 419 in a complete sentence. d. Find the probability that the sample mean is between eight minutes and 8.5 minutes. Solution 7.4 a. μ x¯ = μ = 8.2 σ x¯ = σ n = 1 60 = 0.13 b. This allows us to calculate the probability of sample means of a particular distance from the mean, in repeated samples of size 60. k = invNorm ⎛ c. Let k = the 90th percentile. ⎝0.90,8.2, 1 60 time for table users is less than 8.37 minutes. ⎞ ⎠ = 8.37. This values indicates that 90 percent of the average app engagement d. P(8 < x¯ < 8.5) = normalcdf ⎛ ⎝8,8.5,8.2, 1 60 ⎞ ⎠ = 0.9293 7.4 Cans of a cola beverage claim to contain 16 ounces. The amounts in a sample are measured and the statistics are n = 34, x¯ = 16.01 ounces. If the cans are filled so that μ = 16.00 ounces (as labeled) and σ = 0.143 ounces, find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces. Do the results suggest that cans are filled with an amount greater than 16 ounces? 7.2 \| The Central Limit Theorem for Sums (Optional) Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose: a. μX = the mean of Χ b. σΧ = the standard deviation of X If you draw random samples of size n, then as n increases, the random variable ΣX consisting of sums tends to be normally distributed and ΣΧ ~ N[(n)(μΧ), ( n )(σΧ)]. The central limit theorem for sums says that if you keep drawing larger and larger samples and taking their sums, the sums form their own normal distribution (the sampling distribution), which approaches a normal distribution as the sample size increases. The normal distribution has a mean equal to the original mean multiplied by the sample size and a standard deviation equal to the original standard deviation multiplied by the square root of the sample size. The random variable ΣX has the following z-score associated with it: a. Σx is one sum. b. z = Σx – (n)(μ X) ( n)(σ X) i. ii. (n)(μX) = mean of ΣX ( n)(σ X) = standard deviation of ΣX To find probabilities for sums on the calculator, follow these steps: 2nd DISTR 420 Chapter 7 \| The Central Limit Theorem 2:normalcdf normalcdf(lower value of the area, upper value of the area, (n)(mean), ( n )(standard deviation)) where, • mean is the mean of the original distribution, • • standard deviation is the standard deviation of the original distribution, and sample size = n. Example 7.5 An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size 80 is drawn randomly from the population. a. Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7,500. b. Find the sum that is 1.5 standard deviations above the mean of the sums. Solution 7.5 Let X = one value from the original unknown population. The probability question asks you to find a prob
ability for the sum (or total of) 80 values. ΣX = the sum or total of 80 values. Because μX = 90, σX = 15, and n = 80, ΣX ~ N[(80)(90), ( 80 )(15)] • mean of the sums = (n)(μX) = (80)(90) = 7200 • • standard deviation of the sums = ( n)(σ X ) = ( 80) (15) sum of 80 values = Σx = 7500 a. Find P(Σx > 7500) P(Σx > 7500) = 0.0127 Figure 7.3 normalcdf(lower value, upper value, mean of sums, stdev of sums) The parameter list is abbreviated(lower, upper, (n)(μX, ( n) (σX)) normalcdf (7500,1E99,(80)(90), ⎛ ⎝ 80⎞ ⎠ (15)) = 0.0127 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 421 REMINDER 1E99 = 1099. Press the EE key for E. b. Find Σx where z = 1.5. Σx = (n)(μX) + (z) ( n) (σΧ) = (80)(90) + (1.5)( 80 )(15) = 7401.2 7.5 An unknown distribution has a mean of 45 and a standard deviation of 8. A sample size of 50 is drawn randomly from the population. Find the probability that the sum of the 50 values is more than 2,400. To find percentiles for sums on the calculator, follow these steps: 2nd DIStR 3:invNorm k = invNorm (area to the left of k, (n)(mean), ( n) (standard deviation)) where, • k is the kth percentile, • mean is the mean of the original distribution, • • standard deviation is the standard deviation of the original distribution, and sample size = n. Example 7.6 In a recent study reported Oct. 29, 2012, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. The sample size is 50. a. What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution? b. Find the probability that the sum of the ages is between 1,500 and 1,800 years. c. Find the 80th percentile for the sum of the 50 ages. Solution 7.6 a. μΣx = nμx = 50(34) = 1,700 and σΣx = n σx = ( 50 ) (15) = 106.01 The distribution is normal for sums by the central limit theorem. b. P(1500 < Σx < 1800) = normalcdf (1500, 1800, (50)(34), ( 50 ) (15)) = 0.7974 c. Let k = the 80th percentile. k = invNorm(0.80,(50)(34), ( 50 ) (15)) = 1789.3 422 Chapter 7 \| The Central Limit Theorem 7.6 In a recent study reported Oct.29, 2012, the mean age of tablet users is 35 years. Suppose the standard deviation is 10 years. The sample size is 39. a. What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution? b. Find the probability that the sum of the ages is between 1,400 and 1,500 years. c. Find the 90th percentile for the sum of the 39 ages. Example 7.7 The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70. a. What are the mean and standard deviation for the sums? b. Find the 95th percentile for the sum of the sample. Interpret this value in a complete sentence. c. Find the probability that the sum of the sample is at least 10 hours. Solution 7.7 a. μΣx = nμx = 70(8.2) = 574 minutes and σΣx = ( n)(σ x) = ( 70 ) (1) = 8.37 minutes b. Let k = the 95th percentile. k = invNorm (0.95,(70)(8.2), ( 70) (1)) = 587.76 minutes Ninety-five percent of the app engagement times are at most 587.76 minutes. c. 10 hours = 600 minutes P(Σx ≥ 600) = normalcdf(600,E99,(70)(8.2), ( 70) (1)) = 0.0009 7.7 The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70. a. What is the probability that the sum of the sample is between seven hours and 10 hours? What does this mean in context of the problem? b. Find the 84th and 16th percentiles for the sum of the sample. Interpret these values in context. 7.3 \| Using the Central Limit Theorem It is important for you to understand when to use the central limit theorem. If you are being asked to find the probability of the mean, use the clt for the means. If you are being asked to find the probability of a sum or total, use the clt for sums. This also applies to percentiles for means and sums. NOTE If you are being asked to find the probability of an individual value, do not use the clt. Use the distribution of its random variable. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 423 Examples of the Central Limit Theorem Law of Large Numbers The law of large numbers says that if you take samples of larger and larger sizes from any population, then the mean x¯ of the samples tends to get closer and closer to μ. From the central limit theorem, we know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller the standard deviation gets. (Remember that .) This means that the sample mean x¯ must be close to the population mean μ. We can say that μ is the value that the sample means approach as n gets larger. The central limit theorem illustrates the law of large numbers. ¯ the standard deviation for X is σ n Central Limit Theorem for the Mean and Sum Examples Example 7.8 A study involving stress is conducted among the students on a college campus. The stress scores follow a uniform distribution with the lowest stress score equal to one and the highest equal to five. Using a sample of 75 students, find: a. b. c. d. the probability that the mean stress score for the 75 students is less than 2 the 90th percentile for the mean stress score for the 75 students the probability that the total of the 75 stress scores is less than 200 the 90th percentile for the total stress score for the 75 students Let X = one stress score. Problems (a) and (b) ask you to find a probability or a percentile for a mean. Problems (c) and (d) ask you to find a probability or a percentile for a total or sum. The sample size, n, is equal to 75. Because the individual stress scores follow a uniform distribution, X ~ U(1, 5) where a = 1 and b = 5 (see Continuous Random Variables for an explanation of a uniform distribution), µ X =a + b 2 (b – a)2 12 σ X = = 1 + 5 2 (5 – 1)2 12 = = 3 = 1.15 In the formula above, the denominator is understood to be 12, regardless of the endpoints of the uniform distribution. For problems (a) and (b), let X ¯ = the mean stress score for the 75 students. Then, ¯ X ~ N ⎛ ⎝3, 1.15 75 ⎞ ⎠ where n = 75. a. Find P( x¯ < 2). Draw the graph. Solution 7.8 a. P( x¯ < 2) = 0 The probability that the mean stress score is less than 2 is about zero. 424 Chapter 7 \| The Central Limit Theorem Figure 7.4 normalcdf ⎛ ⎝1,2,3,1.15 75 ⎞ ⎠ = 0 REMINDER The smallest stress score is one. b. Find the 90th percentile for the mean of 75 stress scores. Draw a graph. Solution 7.8 b. Let k = the 90th precentile. Find k, where P( x¯ < k) = 0.90. k = 3.2 Figure 7.5 The 90th percentile for the mean of 75 scores is about 3.2. This tells us that 90 percent of all the means of 75 stress scores are at most 3.2, and that 10 percent are at least 3.2. invNorm ⎛ ⎝0.90,3,1.15 75 ⎞ ⎠ = 3.2 For problems (c) and (d), let ΣX = the sum of the 75 stress scores. Then, ΣX ~ N[(75)(3), ( 75) (1.15)]. c. Find P(Σx < 200). Draw the graph. Solution 7.8 c. The mean of the sum of 75 stress scores is (75)(3) = 225. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 425 The standard deviation of the sum of 75 stress scores is ( 75) (1.15) = 9.96. P(Σx < 200) = 0 Figure 7.6 The probability that the total of 75 scores is less than 200 is about zero. normalcdf (75,200,(75)(3), ( 75) (1.15)). REMINDER The smallest total of 75 stress scores is 75, because the smallest single score is one. d. Find the 90th percentile for the total of 75 stress scores. Draw a graph. Solution 7.8 d. Let k = the 90th percentile. Find k where P(Σx < k) = 0.90. k = 237.8 Figure 7.7 The 90th percentile for the sum of 75 scores is about 237.8. This tells us that 90 percent of all the sums of 75 scores are no more than 237.8 and 10 percent are no less than 237.8. invNorm(0.90,(75)(3), ( 75) (1.15)) = 237.8 426 Chapter 7 \| The Central Limit Theorem 7.8 Use the information in Example 7.8, but use a sample size of 55 to answer the following questions. a. Find P( x¯ < 7). b. Find P(Σx > 170). c. Find the 80th percentile for the mean of 55 scores. d. Find the 85th percentile for the sum of 55 scores. Example 7.9 Suppose that a market research analyst for a cell phone company conducts a study of their customers who exceed the time allowance included on their basic cell phone contract. The analyst finds that for those people who exceed the time included in their basic contract, the excess time used follows an exponential distribution with a mean of 22 minutes. Consider a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract. Let X = the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted time allowance. X ∼ Exp ⎛ ⎝ ⎞ ⎠ 1 22 . From previous chapters, we know that μ = 22 and σ = 22. ¯ = the mean excess time used by a sample of n = 80 customers who exceed their contracted time allowance. Let X ¯ ~ N ⎛ X ⎝22, 22 80 ⎞ ⎠ by the central limit theorem for sample means. Using the clt to find probability a. Find the probability that the mean excess time used by the 80 customers in the sample is longer than 20 minutes. This is asking us to find P( x¯ > 20). Draw the graph. b. Suppose that one customer who exceeds the time limit for his cell phone contract is randomly selected. Find the probability that this individual customer's excess time is longer than 20 minutes. This is asking us to find P(x > 20). c. Explain why the probabilities in parts (a) and (b) are different. Solution 7.9 a. Find: P( x¯ > 20) P( x¯ > 20) = 0.79199 using normalcdf ⎛ ⎝20,1E99,22, 22 80 ⎞ ⎠ The probability is 0.7919 that the mean excess time used is more than 20 minutes, for a sample of 80 customers who exceed their contracted time allowance. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem
427 Figure 7.8 REMINDER 1E99 = 1099 and –1E99 = –1099. Press the EE key for E. Or just use 1099 instead of 1E99. ⎛ b. Find P(x > 20). Remember to use the exponential distribution for an individual. X ~ Exp ⎝ ⎞ ⎠ . 1 22 P(x > 20) = e ⎛ ⎝− ⎛ ⎝ 1 22 ⎞ ⎞ ⎠(20) ⎠ ( – 0.04545(20)) or e = 0.4029 c. 1. P(x > 20) = 0.4029, but P( x¯ > 20) = 0.7919 2. The probabilities are not equal because we use different distributions to calculate the probability for individuals and for means. 3. When asked to find the probability of an individual value, use the stated distribution of its random variable; do not use the clt. Use the clt with the normal distribution when you are being asked to find the probability for a mean. Using the clt to find percentiles Find the 95th percentile for the sample mean excess time for a sample of 80 customers who exceed their basic contract time allowances. Draw a graph. Solution 7.9 Let k = the 95th percentile. Find k where P( x¯ < k) = 0.95. k = 26.0 using invNorm ⎛ = 26.0 ⎝0.95,22, 22 80 ⎞ ⎠ 428 Chapter 7 \| The Central Limit Theorem Figure 7.9 The 95th percentile for the sample mean excess time used is about 26.0 minutes for a random sample of 80 customers who exceed their contractual allowed time. 95 percent of such samples would have means under 26 minutes; only five percent of such samples would have means above 26 minutes. 7.9 Use the information in Example 7.9, but change the sample size to 144. a. Find P(20 < x¯ < 30). b. Find P(Σx is at least 3000). c. Find the 75th percentile for the sample mean excess time of 144 customers. d. Find the 85th percentile for the sum of 144 excess times used by customers. Example 7.10 U.S. scientists studying a certain medical condition discovered that a new person is diagnosed every two minutes, on average. Suppose the standard deviation is 0.5 minutes and the sample size is 100. a. Find the median, the first quartile, and the third quartile for the sample mean time of diagnosis in the United States. b. Find the median, the first quartile, and the third quartile for the sum of sample times of diagnosis in the United States. c. Find the probability that a diagnosis occurs on average between 1.75 and 1.85 minutes. d. Find the value that is two standard deviations above the sample mean. e. Find the IQR for the sum of the sample times. Solution 7.10 a. We have μx = μ = 2 and σx = σ n 1. 50th percentile = μx = μ = 2, 2. 25th percentile = invNorm(0.25,2,0.05) = 1.97, and = 0.5 10 = 0.05. Therefore, This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 429 3. 75th percentile = invNorm(0.75,2,0.05) = 2.03. b. We have μΣx = n(μx) = 100(2) = 200 and σμx = n (σx) = 10(0.5) = 5. Therefore, 1. 50th percentile = μΣx = n(μx) = 100(2) = 200, 2. 25th percentile = invNorm(0.25,200,5) = 196.63, and 3. 75th percentile = invNorm(0.75,200,5) = 203.37. c. P(1.75 < x¯ < 1.85) = normalcdf(1.75,1.85,2,0.05) = 0.0013 d. Using the z-score equation, z = x¯ – μ x¯ σ x¯ , and solving for x, we get x = 2(0.05) + 2 = 2.1. e. The IQR is 75th percentile – 25th percentile = 203.37 – 196.63 = 6.74. 7.10 Based on data from the National Health Survey, women between the ages of 18 and 24 have an average systolic blood pressures (in mm Hg) of 114.8 with a standard deviation of 13.1. Systolic blood pressure for women between the ages of 18 to 24 follows a normal distribution. a. b. c. If one woman from this population is randomly selected, find the probability that her systolic blood pressure is greater than 120. If 40 women from this population are randomly selected, find the probability that their mean systolic blood pressure is greater than 120. If the sample was four women between the ages of 18–24 and we did not know the original distribution, could the central limit theorem be used? Example 7.11 A study was done about a medical condition that affects a certain group of people. The age range of the people was 14–61. The mean age was 30.9 years with a standard deviation of nine years. a. b. c. In a sample of 25 people, what is the probability that the mean age of the people is less than 35? Is it likely that the mean age of the sample group could be more than 50 years? Interpret the results. In a sample of 49 people, what is the probability that the sum of the ages is no less than 1,600? Is it likely that the sum of the ages of the 49 people are at most 1,595? Interpret the results. d. e. Find the 95th percentile for the sample mean age of 65 people. Interpret the results. f. Find the 90th percentile for the sum of the ages of 65 people. Interpret the results. Solution 7.11 a. P( x¯ < 35) = normalcdf(-E99,35,30.9,1.8) = 0.9886 b. P( x¯ > 50) = normalcdf(50, E99,30.9,1.8) ≈ 0. For this sample group, it is almost impossible for the group’s average age to be more than 50. However, it is still possible for an individual in this group to have an age greater than 50. c. P(Σx ≥ 1,600) = normalcdf(1600,E99,1514.10,63) = 0.0864 d. P(Σx ≤ 1,595) = normalcdf(-E99,1595,1514.10,63) = 0.9005. This means that there is a 90 percent chance that the sum of the ages for the sample group n = 49 is at most 1,595. 430 Chapter 7 \| The Central Limit Theorem e. The 95th percentile = invNorm(0.95,30.9,1.1) = 32.7. This indicates that 95 percent of the people in the sample of 65 are younger than 32.7 years, on average. f. The 90th percentile = invNorm(0.90,2008.5,72.56) = 2101.5. This indicates that 90 percent of the people in the sample of 65 have a sum of ages less than 2,101.5 years. 7.11 According to data from an aerospace company, the 757 airliner carries 200 passengers and has doors with a mean height of 72 inches. Assume for a certain population of men we have a mean of 69 inches inches and a standard deviation of 2.8 inches. a. What mean doorway height would allow 95 percent of men to enter the aircraft without bending? b. Assume that half of the 200 passengers are men. What mean doorway height satisfies the condition that there is a 0.95 probability that this height is greater than the mean height of 100 men? c. For engineers designing the 757, which result is more relevant: the height from part (a) or part (b)? Why? HISTORICAL NOTE Normal Approximation to the Binomial Historically, being able to compute binomial probabilities was one of the most important applications of the central limit theorem. Binomial probabilities with a small value for n (say, 20) were displayed in a table in a book. To calculate the probabilities with large values of n, you had to use the binomial formula, which could be very complicated. Using the normal approximation to the binomial distribution simplified the process. To compute the normal approximation to the binomial distribution, take a simple random sample from a population. You must meet the following conditions for a binomial distribution: • There are a certain number, n, of independent trials. • The outcomes of any trial are success or failure. • Each trial has the same probability of a success, p. Recall that if X is the binomial random variable, then X ~ B(n, p). The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five (np > 5 and nq > 5; the approximation is better if they are both greater than or equal to 10. The product >5 is more or less accepted as the norm here.). This is another accepted rule. So, for whatever value of x we are looking at (the number of successes). We add 0.5 if we are looking for the probability that is less than or equal to that number. We subtract 0.5 if we are looking for the probability that is greater than or equal to that number. Then the binomial can be approximated by the normal distribution with mean μ = np and standard deviation σ = npq . Remember that q = 1 – p. In order to get the best approximation, add 0.5 to x or subtract 0.5 from x (use x + 0.5 or x – 0.5). This is another accepted rule. So, for whatever value of x we are looking at (the number of successes). We add 0.5 if we are looking for the probability that is less than or equal to that number. We subtract 0.5 if we are looking for the probability that is greater than or equal to that number. The number 0.5 is called the continuity correction factor and is used in the following example. Example 7.12 Suppose in a local kindergarten through 12th grade (K–12) school district, 53 percent of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed. a. Find the probability that at least 150 favor a charter school. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 431 b. Find the probability that at most 160 favor a charter school. c. Find the probability that more than 155 favor a charter school. d. Find the probability that fewer than 147 favor a charter school. e. Find the probability that exactly 175 favor a charter school. Let X = the number that favor a charter school for grades K through 5. X ~ B(n, p) where n = 300 and p = 0.53. Because np > 5 and nq > 5, use the normal approximation to the binomial. The formulas for the mean and standard deviation are μ = np and σ = npq . The mean is 159, and the standard deviation is 8.6447. The random variable for the normal distribution is Y. Y ~ N(159, 8.6447). See The Normal Distribution for help with calculator instructions. For Part (a), you include 150 so P(X ≥ 150) has a normal approximation P(Y ≥ 149.5) = 0.8641. normalcdf(149.5,10^99,159,8.6447) = 0.8641. For Part (b), you include 160 so P(X ≤ 160) has a normal approximation P(Y ≤ 160.5) = 0.5689. normalcdf(0,160.5,159,8.6447) = 0.5689 For Part (c), you exclude 155 so P(X > 155) has normal approximation P(y > 155.5) = 0.6572. normalcdf(155.5,10^99,159,8.6447) = 0.6572. For Part (d), you exclude 147 so P(X < 147) has normal approximation P(Y < 146.5) = 0.0741. normalcdf(0,146.5,159,8.6447) = 0.0
741 For Part (e), P(X = 175) has normal approximation P(174.5 < Y < 175.5) = 0.0083. normalcdf(174.5,175.5,159,8.6447) = 0.0083 Because of calculators and computer software that let you calculate binomial probabilities for large values of n easily, it is not necessary to use the the normal approximation to the binomial distribution, provided that you have access to these technology tools. Most school labs have computer software that calculates binomial probabilities. Many students have access to calculators that calculate probabilities for binomial distribution. If you type in binomial probability distribution calculation in an internet browser, you can find at least one online calculator for the binomial. For Example 7.10, the probabilities are calculated using the following binomial distribution: (n = 300 and p = 0.53). Compare the binomial and normal distribution answers. See Discrete Random Variables for help with calculator instructions for the binomial. P(X ≥ 150) :1 - binomialcdf(300,0.53,149) = 0.8641 P(X ≤ 160) :binomialcdf(300,0.53,160) = 0.5684 P(X > 155) :1 - binomialcdf(300,0.53,155) = 0.6576 P(X < 147) :binomialcdf(300,0.53,146) = 0.0742 P(X = 175) :(You use the binomial pdf.)binomialpdf(300,0.53,175) = 0.0083 7.12 In a city, 46 percent of the population favors the incumbent, Dawn Morgan, for mayor. A simple random sample of 500 is taken. Using the continuity correction factor, find the probability that at least 250 favor Dawn Morgan for mayor. 7.4 \| Central Limit Theorem (Pocket Change) 432 Chapter 7 \| The Central Limit Theorem 7.1 Central Limit Theorem (Pocket Change) Student Learning Outcome • The student will demonstrate and compare properties of the central limit theorem. NOTE This lab works best when sampling from several classes and combining data. Collect the Data 1. Count the change in your pocket. (Do not include bills.) 2. Randomly survey 30 classmates. Record the values of the change in Table 7.1. __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Table 7.1 3. Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes. Figure 7.10 4. Calculate the following (n = 1, surveying one person at a time): a. b. x¯ = _______ s = _______ 5. Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 433 Collecting Averages of Pairs: Repeat steps one through five of the section Collect the Data with one exception. Instead of recording the change of 30 classmates, record the average change of 30 pairs. 1. Randomly survey 30 pairs of classmates. 2. Record the values of the average of their change in Table 7.2. __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Table 7.2 3. Construct a histogram. Scale the axes using the same scaling you used for the section titled Collect the Data. Sketch the graph using a ruler and a pencil. Figure 7.11 4. Calculate the following (n = 2, surveying two people at a time): a. b. x¯ = _______ s = _______ 5. Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. Collecting Averages of Groups of Five: Repeat steps one through five (of the section titled Collect the Data), with one exception. Instead of recording the change of 30 classmates, record the average change of 30 groups of five. 1. Randomly survey 30 groups of five classmates. 2. Record the values of the averages of their change. 434 Chapter 7 \| The Central Limit Theorem __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Table 7.3 3. Construct a histogram. Scale the axes using the same scaling you used for the section titled Collect the Data. Sketch the graph using a ruler and a pencil. Figure 7.12 4. Calculate the following (n = 5, surveying five people at a time): a. b. x¯ = _______ s = _______ 5. Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. Discussion Questions 1. Why did the shape of the distribution of the data change, as n changed? Use one to two complete sentences to explain what happened. 2. In the section titled Collect the Data, what was the approximate distribution of the data? 3. X ~ _____(_____, _____) 4. In the section titled Collecting Averages of Groups of Five, what was the approximate distribution of the averages? X ¯ ~ _____(_____, _____) 5. In one to two complete sentences, explain any differences in your answers to the previous two questions. 7.5 \| Central Limit Theorem (Cookie Recipes) This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 435 7.2 Central Limit Theorem (Cookie Recipes) Student Learning Outcome • The student will demonstrate and compare properties of the central limit theorem. Given X = length of time (in days) that a cookie recipe lasted at the Olmstead Homestead. (Assume that each of the different recipes makes the same quantity of cookies.) Recipe # X Recipe # X Recipe # X Recipe # 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 2 2 11 Table 7.4 Calculate the following: a. μx = _______ b. σx = _______ Collect the Data Use a random number generator to randomly select four samples of size n = 5 from the given population. Record your samples in Table 7.5. Then, for each sample, calculate the mean to the nearest tenth. Record them in the spaces provided. Record the sample means for the rest of the class. 1. Complete the following table: 436 Chapter 7 \| The Central Limit Theorem Sample 1 Sample 2 Sample 3 Sample 4 Sample means from other groups: Means: x¯ = ____ x¯ = ____ x¯ = ____ x¯ = ____ Table 7.5 2. Calculate the following: a. b. x¯ = _______ s x¯ = _______ 3. Again, use a random number generator to randomly select four samples from the population. This time, make the samples of size n = 10. Record the samples in Table 7.6. As before, for each sample, calculate the mean to the nearest tenth. Record them in the spaces provided. Record the sample means for the rest of the class. Sample 1 Sample 2 Sample 3 Sample 4 Sample means from other groups Means: x¯ = ____ x¯ = ____ x¯ = ____ x¯ = ____ Table 7.6 4. Calculate the following: a. b. x¯ = ______ s x¯ = ______ 5. For the original population, construct a histogram. Make intervals with a bar width of one day. Sketch the graph using a ruler and pencil. Scale the axes. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 437 Figure 7.13 6. Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. Repeat the procedure for n = 5. 1. For the sample of n = 5 days averaged together, construct a histogram of the averages (your means together with the means of the other groups). Make intervals with bar widths of 1 2 day. Sketch the graph using a ruler and pencil. Scale the axes. Figure 7.14 2. Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. Repeat the procedure for n = 10. 1. For the sample of n = 10 days averaged together, construct a histogram of the averages (your means together with the means of the other groups). Make intervals with bar widths of 1 2 day. Sketch the graph using a ruler and pencil. Scale the axes. 438 Chapter 7 \| The Central Limit Theorem Figure 7.15 2. Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. Discussion Questions 1. Compare the three histograms you have made, the one for the population and the two for the sample means. In three to five sentences, describe the similarities and differences. 2. State the theoretical (according to the clt) distributions for the sample means. a. n = 5: x¯ ~ _____(_____, _____) b. n = 10: x¯ ~ _____(_____, _____) 3. Are the sample means for n = 5 and n = 10 close to the theoretical mean, μx? Explain why or why not. 4. Which of the two distributions of sample means has the smaller standard deviation? Why? 5. As n changed, why did the shape of the distribution of the data change? Use one to two complete sentences to explain what happened. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 439 KEY TERMS average a number that describes the central tendency of the data; there are a number of specialized averages, including the arithmetic mean, weighted mean, median, mode, and geometric mean central limit theorem given a random variable (RV) with a known mean, μ, and known standard deviation, σ, and sampling with size n, we are interested in two new RVs: the sample mean, X ¯ , and the sample sum, ΣΧ If the size (n) of the sample is sufficiently large, then X
¯ ~ N(μ, σ n ) and ΣΧ ~ N(nμ, ( n )(σ)). If the size (n) of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distribution regardless of the shape of the population. The mean of the sample means will equal the population mean, and the mean of the sample sums will equal n times the population mean. The standard deviation of the distribution of the sample means, σ n , is called the standard error of the mean exponential distribution a continuous random variable (RV) that appears when we are interested in the intervals of time between a random events; for example, the length of time between emergency arrivals at a hospital, notation: X ~ Exp(m) The mean is μ = 1 m . The probability density function is f(x) = me–mx, x ≥ 0, and and the standard deviation is σ = 1 m the cumulative distribution function is P(X ≤ x) = 1 – e–mx mean a number that measures the central tendency; a common name for mean is average; the term mean is a shortened form of arithmetic mean;. by definition, the mean for a sample (denoted by x¯ ) is x¯ = sum of all values in the sample number of values in the sample , and the mean for a population (denoted by μ) is μ = sum of all values in the population number of values in the population . continuous random variable (RV) with probability density function (pdf) normal distribution f (x) = 1 σ 2π e a – (x – μ)2 2σ 2 , where μ is the mean of the distribution and σ is the standard deviation; notation: Χ ~ N(μ, σ). If μ = 0 and σ = 1, the RV is called a standard normal distribution sampling distribution given simple random samples of size n from a given population with a measured characteristic such as mean, proportion, or standard deviation for each sample, the probability distribution of all the measured characteristics is called a sampling distribution. standard error of the mean the standard deviation of the distribution of the sample means, or σ n uniform distribution a continuous random variable (RV) that has equally likely outcomes over the domain a < x < b; often referred as the rectangular distribution because the graph of the pdf has the form of a rectangle Notation: X ~ U(a, b). The mean is μ = a + b 2 and the standard deviation is σ = (b – a)2 12 . The probability density function is f (x) = 1 b – a for a < x < b or a ≤ x ≤ b. The cumulative distribution is P(X ≤ x) = x – a b – a CHAPTER REVIEW 7.1 The Central Limit Theorem for Sample Means (Averages) In a population whose distribution may be known or unknown, if the size (n) of the sample is sufficiently large, the distribution of the sample means will be approximately normal. The mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, called the standard error of the mean, is equal to the population standard deviation divided by the square root of the sample size (n). 440 Chapter 7 \| The Central Limit Theorem 7.2 The Central Limit Theorem for Sums (Optional) The central limit theorem tells us that for a population with any distribution, the distribution of the sums for the sample means approaches a normal distribution as the sample size increases. In other words, if the sample size is large enough, the distribution of the sums can be approximated by a normal distribution, even if the original population is not normally distributed. Additionally, if the original population has a mean of μX and a standard deviation of σx, the mean of the sums is nμx and the standard deviation is ( n) (σx), where n is the sample size. 7.3 Using the Central Limit Theorem The central limit theorem can be used to illustrate the law of large numbers. The law of large numbers states that the larger the sample size you take from a population, the closer the sample mean, x¯ , gets to μ. FORMULA REVIEW 7.1 The Central Limit Theorem for Sample Means (Averages) 7.2 The Central Limit Theorem for Sums (Optional) Central limit theorem for sample means: X ¯ ~ N ⎛ ⎝μ x , σx n ⎞ ⎠ Mean X Central ¯ : μx limit standard error of the mean: z = theorem for sample means z-score and x¯ − μ x σ x n ⎛ ⎝ ⎞ ⎠ Central limit theorem for sums: ∑X ~ N[(n)(μx),( n )(σx)] Mean for sums (∑X): (n)(μx) Central deviation limit theorem for sums z-score and standard sums: for z for the sample mean = Σx – (n)(μ X) ( n)(σ X) Standard deviation for sums (∑X): ( n) (σx) Standard error of the mean (standard deviation ( X ¯ )): σ x n PRACTICE 7.1 The Central Limit Theorem for Sample Means (Averages) Use the following information to answer the next six exercises: Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let Χ be the random variable representing the time ¯ be the random variable representing the mean it takes her to complete one review. Assume Χ is normally distributed. Let X time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews. 1. What is the mean, standard deviation, and sample size? 2. Complete the distributions. a. X ~ _____(_____, _____) ¯ ~ _____(_____, _____) b. X This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 441 3. Find the probability that one review will take Yoonie from 3.5 to 4.25 hours. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability. a. Figure 7.16 b. P(________ < x < ________) = _______ 4. Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability. a. Figure 7.17 b. P(________________) = _______ 5. What causes the probabilities in Exercise 7.3 and Exercise 7.4 to be different? 442 Chapter 7 \| The Central Limit Theorem 6. Find the 95th percentile for the mean time to complete one month's reviews. Sketch the graph. a. Figure 7.18 b. The 95th percentile =____________ 7.2 The Central Limit Theorem for Sums (Optional) Use the following information to answer the next four exercises: An unknown distribution has a mean of 80 and a standard deviation of 12. A sample size of 95 is drawn randomly from the population. 7. Find the probability that the sum of the 95 values is greater than 7,650. 8. Find the probability that the sum of the 95 values is less than 7,400. 9. Find the sum that is two standard deviations above the mean of the sums. 10. Find the sum that is 1.5 standard deviations below the mean of the sums. Use the following information to answer the next five exercises: The distribution of results from a cholesterol test has a mean of 180 and a standard deviation of 20. A sample size of 40 is drawn randomly. 11. Find the probability that the sum of the 40 values is greater than 7,500. 12. Find the probability that the sum of the 40 values is less than 7,000. 13. Find the sum that is one standard deviation above the mean of the sums. 14. Find the sum that is 1.5 standard deviations below the mean of the sums. 15. Find the percentage of sums between 1.5 standard deviations below the mean of the sums and one standard deviation above the mean of the sums. Use the following information to answer the next six exercises: A researcher measures the amount of sugar in several cans of the same type of soda. The mean is 39.01 with a standard deviation of 0.5. The researcher randomly selects a sample of 100. 16. Find the probability that the sum of the 100 values is greater than 3,910. 17. Find the probability that the sum of the 100 values is less than 3,900. 18. Find the probability that the sum of the 100 values falls between the numbers you found in (16) and (17). 19. Find the sum with a z-score of –2.5. 20. Find the sum with a z-score of 0.5. 21. Find the probability that the sums will fall between the z-scores –2 and 1. Use the following information to answer the next four exercises: An unknown distribution has a mean 12 and a standard deviation of one. A sample size of 25 is taken. Let X = the object of interest. 22. What is the mean of ΣX? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 443 23. What is the standard deviation of ΣX? 24. What is P(Σx = 290)? 25. What is P(Σx > 290)? 26. True or False: Only the sums of normal distributions are also normal distributions. 27. In order for the sums of a distribution to approach a normal distribution, what must be true? 28. What three things must you know about a distribution to find the probability of sums? 29. An unknown distribution has a mean of 25 and a standard deviation of six. Let X = one object from this distribution. What is the sample size if the standard deviation of ΣX is 42? 30. An unknown distribution has a mean of 19 and a standard deviation of 20. Let X = the object of interest. What is the sample size if the mean of ΣX is 15,200? Use the following information to answer the next three exercises: A market researcher analyzes how many electronics devices customers buy in a single purchase. The distribution has a mean of three with a standard deviation of 0.7. She samples 400 customers. 31. What is the z-score for Σx = 840? 32. What is the z-score for Σx = 1,186? 33. What is P(Σx < 1186)? Use the following information to answer the next three exercises: An unkwon distribution has a mean of 100, a standard deviation of 100, and a sample size of 100. Let X = one object of interest. 34. What is the mean of ΣX? 35. What is the standard deviation of ΣX? 36. What is P(Σx > 9000)? 7.3 Using the Central Limit Theorem Use the following information to answer the next 10 exercises: A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each
weight is equally likely, so the distribution of weights is uniform. A sample of 100 weights is taken. 37. a. What is the distribution for the weights of one 25-pound lifting weight? What are the mean and standard deivation? b. What is the distribution for the mean weight of 100 25-pound lifting weights? c. Find the probability that the mean actual weight for the 100 weights is less than 24.9. 38. Draw the graph of Exercise 7.37. 39. Find the probability that the mean actual weight for the 100 weights is greater than 25.2. 40. Draw the graph of Exercise 7.39. 41. Find the 90th percentile for the mean weight for the 100 weights. 42. Draw the graph of Exercise 7.41. 43. a. What is the distribution for the sum of the weights of 100 25-pound lifting weights? b. Find P(Σx < 2450). 44. Draw the graph of Exercise 7.43. 45. Find the 90th percentile for the total weight of the 100 weights. 46. Draw the graph of Exercise 7.45. Use the following information to answer the next five exercises: The length of time a particular smartphone's battery lasts 444 Chapter 7 \| The Central Limit Theorem follows an exponential distribution with a mean of ten months. A sample of 64 of these smartphones is taken. 47. a. What is the standard deviation? b. What is the parameter m? 48. What is the distribution for the length of time one battery lasts? 49. What is the distribution for the mean length of time 64 batteries last? 50. What is the distribution for the total length of time 64 batteries last? 51. Find the probability that the sample mean is between 7 and 11. 52. Find the 80th percentile for the total length of time 64 batteries last. 53. Find the interquartile range (IQR) for the mean amount of time 64 batteries last. 54. Find the middle 80 percent for the total amount of time 64 batteries last. Use the following information to answer the next six exercises: A uniform distribution has a minimum of six and a maximum of ten. A sample of 50 is taken. 55. Find P(Σx > 420). 56. Find the 90th percentile for the sums. 57. Find the 15th percentile for the sums. 58. Find the first quartile for the sums. 59. Find the third quartile for the sums. 60. Find the 80th percentile for the sums. HOMEWORK 7.1 The Central Limit Theorem for Sample Means (Averages) 61. Previously, De Anza's statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students. In words, Χ = ____________. a. b. Χ ~ _____(_____, _____) c. ¯ = ____________. In words, X ¯ ~ ______ (______, ______) d. X e. Find the probability that an individual had between $0.80 and $1.00. Graph the situation, and shade in the area to be determined. f. Find the probability that the average amount of change of the 25 students was between $0.80 and $1.00. Graph the situation, and shade in the area to be determined. g. Explain why there is a difference in part (e) and part (f). 62. Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls. ¯ = average distance in feet for 49 fly balls, then X a. b. What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the ¯ ~ _______(_______, _______). If X horizontal axis for X ¯ . Shade the region corresponding to the probability. Find the probability. c. Find the 80th percentile of the distribution of the average of 49 fly balls. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 445 63. According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers. a. b. In words, Χ = _____________. ¯ = _____________. In words, X ¯ ~ _____(_____, _____) c. X d. Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in complete sentences. e. Would you be surprised if one taxpayer finished his or her Form 1040 in more than 12 hours? In a complete sentence, explain why. 64. Suppose that a category of world-class runners are known to run a marathon (26 miles) in an average of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races. Let X ¯ be the average of the 49 races. ¯ ~ _____(_____, _____) a. X b. Find the probability that the runner will average between 142 and 146 minutes in these 49 marathons. c. Find the 80th percentile for the average of these 49 marathons. d. Find the median of the average running times. 65. The length of songs in a collector’s online album collection is uniformly distributed from 2 to 3.5 minutes. Suppose we randomly pick five albums from the collection. There are a total of 43 songs on the five albums. In words, Χ = _________. a. b. Χ ~ _____________ c. ¯ = _____________. In words, X ¯ ~ _____(_____, _____) d. X e. Find the first quartile for the average song length. f. The IQR for the average song length is _______–_______. 66. In 1940, the average size of a U.S. farm was 174 acres. Let’s say that the standard deviation was 55 acres. Suppose we randomly survey 38 farmers from 1940. In words, Χ = _____________. a. ¯ = _____________. b. In words, X ¯ ~ _____(_____, _____) ¯ d. The IQR for X c. X is from _______ acres to _______ acres. 67. Determine which of the following are true and which are false. Then, in complete sentences, justify your answers. ¯ a. When the sample size is large, the mean of X is approximately equal to the mean of Χ. ¯ b. When the sample size is large, X is approximately normally distributed. ¯ c. When the sample size is large, the standard deviation of X is approximately the same as the standard deviation of Χ. 68. The percentage of fat calories that a person in America consumes each day is normally distributed with a mean of about ¯ = average percentage of 36 and a standard deviation of about ten. Suppose that 16 individuals are randomly chosen. Let X fat calories. ¯ ~ ______(______, ______) a. X b. For the group of 16, find the probability that the average percentage of fat calories consumed is more than five. Graph the situation and shade in the area to be determined. c. Find the first quartile for the average percentage of fat calories. 446 Chapter 7 \| The Central Limit Theorem 69. The distribution of income in some economically developing countries is considered wedge shaped (many very poor people, very few middle income people, and even fewer wealthy people). Suppose we pick a country with a wedge-shaped distribution. Let the average salary be $2,000 per year with a standard deviation of $8,000. We randomly survey 1,000 residents of that country. a. b. In words, Χ = _____________. ¯ = _____________. In words, X ¯ ~ _____(_____, _____) c. X d. How is it possible for the standard deviation to be greater than the average? e. Why is it more likely that the average salary of the 1,000 residents will be from $2,000 to $2,100 than from $2,100 to $2,200? 70. Which of the following is NOT true about the distribution for averages? a. The mean, median, and mode are equal. b. The area under the curve is 1. c. The curve never touches the x-axis. d. The curve is skewed to the right. 71. The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. The distribution to use for the average cost of gasoline for the 16 gas stations is: a. X b. X c. X d. X ¯ ~ N(4.59, 0.10) ¯ ~ N ⎛ ⎝4.59, 0.10 16 ⎝4.59, 16 0.10 ⎝4.59, 16 0.10 ¯ ~ .2 The Central Limit Theorem for Sums (Optional) 72. Which of the following is NOT true about the theoretical distribution of sums? a. The mean, median, and mode are equal. b. The area under the curve is one. c. The curve never touches the x-axis. d. The curve is skewed to the right. 73. Suppose that the duration of a particular type of criminal trial is known to have a mean of 21 days and a standard deviation of seven days. We randomly sample nine trials. In words, ΣX = ______________. a. b. ΣX ~ _____(_____, _____) c. Find the probability that the total length of the nine trials is at least 225 days. d. Ninety percent of the total of nine of these types of trials will last at least how long? 74. Suppose that the weight of open boxes of cereal in a home with children is uniformly distributed from two to six pounds with a mean of four pounds and standard deviation of 1.1547. We randomly survey 64 homes with children. In words, X = _____________. In words, ΣX = _______________. a. b. The distribution is _______. c. d. ΣX ~ _____(_____, _____) e. Find the probability that the total weight of the open boxes is less than 250 pounds. f. Find the 35th percentile for the total weight of open boxes of cereal. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 447 75. Salaries for entry-level managers at a restaurant chain are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey 10 managers from these restaurants. In words, X = ______________. In words, ΣX = _____________. a. b. X ~ _____(_____, _____) c. d. ΣX ~ _____(_____, _____) e. Find the probability that the managers earn a total of over $400,000. f. Find the 90th percentile for an individual manager's salary. g. Find the 90th percentile for the sum of ten managers' salary. h. i. If we surveyed 70 managers instead of ten, graphically, how would that change the distribution in part (d)? If
each of the 70 managers received a $3,000 raise, graphically, how would that change the distribution in part (b)? 7.3 Using the Central Limit Theorem 76. The attention span of a two-year-old is exponentially distributed with a mean of about eight minutes. Suppose we randomly survey 60 two-year-olds. a. In words, Χ = _______. b. Χ ~ _____(_____, _____) c. ¯ = ____________. In words, X ¯ ~ _____(_____, _____) d. X e. Before doing any calculations, which do you think will be higher? Explain why. i. The probability that an individual attention span is less than 10 minutes. ii. The probability that the average attention span for the 60 children is less than 10 minutes. f. Calculate the probabilities in part (e). ¯ g. Explain why the distribution for X is not exponential. 448 Chapter 7 \| The Central Limit Theorem 77. The closing stock prices of 35 U.S. semiconductor manufacturers are given as follows: Company Closing Stock Prices 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 Table 7.7 8.625 30.25 27.625 46.75 32.875 18.25 5 0.125 2.9375 6.875 28.25 24.25 21 1.5 30.25 71 43.5 49.25 2.5625 31 16.5 9.5 18.5 18 9 10.5 16.625 1.25 18 12.87 7 12.875 2.875 60.25 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 449 Company Closing Stock Prices 35 Table 7.7 29.25 a. b. In words, Χ = ______________. x¯ = _____ i. ii. sx = _____ iii. n = _____ c. Construct a histogram of the distribution of the averages. Start at x = –0.0005. Use bar widths of 10. d. e. Randomly average five stock prices together. (Use a random number generator.) Continue averaging five prices In words, describe the distribution of the stock prices. together until you have 10 averages. List those 10 averages. f. Use the 10 averages from part (e) to calculate the following: i. ii. x¯ = _____ sx = _____ g. Construct a histogram of the distribution of the averages. Start at x = –0.0005. Use bar widths of 10. h. Does this histogram look like the graph in Part (c)? i. In one or two complete sentences, explain why the graphs either look the same or look different. j. Based on the theory of the central limit theorem, X ¯ ~ _____(_____, ____). Use the following information to answer the next three exercises: Richard’s Furniture Company delivers furniture from 10 a.m. to 2 p.m. continuously and uniformly. We are interested in how long (in hours) past the 10 a.m. start time that individuals wait for their delivery. 78. Χ ~ _____(_____, _____) a. U(0, 4) b. U(10, 2) c. Eχp(2) d. N(2, 1) 79. The average wait time is: a. one hour two hours b. two and a half hours c. four hours d. 80. Suppose that it is now past noon on a delivery day. The probability that a person must wait at least one and a half more hours is a. b. c. d. 1 4 1 2 3 4 3 8 Use the following information to answer the next two exercises: The time to wait for a particular rural bus is distributed uniformly from zero to 75 minutes. One hundred riders are randomly sampled to learn how long they waited. 81. The 90th percentile sample average wait time (in minutes) for a sample of 100 riders is: a. 315.0 b. 40.3 c. 38.5 d. 65.2 450 Chapter 7 \| The Central Limit Theorem 82. Would you be surprised, based on numerical calculations, if the sample average wait time (in minutes) for 100 riders was less than 30 minutes? a. yes b. no c. There is not enough information. Use the following to answer the next two exercises: The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. 83. What's the approximate probability that the average price for 16 gas stations is more than $4.69? a. almost zero b. 0.1587 c. 0.0943 d. unknown 84. Find the probability that the average price for 30 gas stations is less than $4.55. a. 0.6554 b. 0.3446 c. 0.0142 d. 0.9858 e. 0 85. Suppose in a local kindergarten through 12th grade (K–12) school district, 53 percent of the population favor a charter school for grades K through five. A simple random sample of 300 is surveyed. Calculate the following using the normal approximation to the binomial distribtion. a. Find the probability that less than 100 favor a charter school for grades K through 5. b. Find the probability that 170 or more favor a charter school for grades K through 5. c. Find the probability that no more than 140 favor a charter school for grades K through 5. d. Find the probability that there are fewer than 130 that favor a charter school for grades K through 5. e. Find the probability that exactly 150 favor a charter school for grades K through 5. If you have access to an appropriate calculator or computer software, try calculating these probabilities using the technology. 86. Four friends, Janice, Barbara, Kathy, and Roberta, decided to carpool together to get to school. Each day the driver would be chosen by randomly selecting one of the four names. They carpool to school for 96 days. Use the normal approximation to the binomial to calculate the following probabilities. Round the standard deviation to four decimal places. a. Find the probability that Janice is the driver at most 20 days. b. Find the probability that Roberta is the driver more than 16 days. c. Find the probability that Barbara drives exactly 24 of those 96 days. ¯ be the random variable of 87. X ~ N(60, 9). Suppose that you form random samples of 25 from this distribution. Let X averages. Let ΣX be the random variable of sums. For parts (c) through (f), sketch the graph, shade the region, label and scale the horizontal axis for X ¯ , and find the probability. ¯ on the same graph. a. Sketch the distributions of X and X ¯ ~ _____(_____, _____) b. X c. P( x¯ < 60) = _____ d. Find the 30th percentile for the mean. e. P(56 < x¯ < 62) = _____ f. P(18 < x¯ < 58) = _____ g. Σx ~ _____(_____, _____) h. Find the minimum value for the upper quartile for the sum. i. P(1400 < Σx < 1550) = _____ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 451 88. Suppose that the length of research papers is uniformly distributed from 10 to 25 pages. We survey a class in which 55 research papers were turned in to a professor. The 55 research papers are considered a random collection of all papers. We are interested in the average length of the research papers. In words, X = _____________. a. b. X ~ _____(_____, _____) c. μx = _____ d. σx = _____ e. ¯ = ______________. In words, X ¯ ~ _____(_____, _____) In words, ΣX = _____________. f. X g. h. ΣX ~ _____(_____, _____) i. Without doing any calculations, do you think that it’s likely the professor will need to read a total of more than 1,050 pages? Why? j. Calculate the probability that the professor will need to read a total of more than 1,050 pages. k. Why is it so unlikely that the average length of the papers will be less than 12 pages? 89. Salaries for managers in a restaurant chain are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey 10 managers from that district. a. Find the 90th percentile for an individual manager's salary. b. Find the 90th percentile for the average manager's salary. 90. The average length of a maternity stay in a U.S. hospital is said to be 2.4 days with a standard deviation of 0.9 days. We randomly survey 80 women who recently bore children in a U.S. hospital. a. b. In words, X = _____________. ¯ = ___________________. In words, X ¯ ~ _____(_____, _____) In words, ΣX = _______________. c. X d. e. ΣX ~ _____(_____, _____) f. g. h. Which is more likely: Is it likely that an individual stayed more than five days in the hospital? Why or why not? Is it likely that the average stay for the 80 women was more than five days? Why or why not? i. An individual stayed more than five days. ii. The average stay of 80 women was more than five days. i. If we were to sum up the women’s stays, is it likely that collectively, they spent more than a year in the hospital? Why or why not? For each problem, wherever possible, provide graphs and use a calculator. 91. NeverReady batteries has engineered a newer, longer-lasting AAA battery. The company claims this battery has an average life span of 17 hours with a standard deviation of 0.8 hours. Your statistics class questions this claim. As a class, you randomly select 30 batteries and find that the sample mean life span is 16.7 hours. If the process is working properly, what is the probability of getting a random sample of 30 batteries in which the sample mean life span is 16.7 hours or less? Is the company’s claim reasonable? 92. Men have an average weight of 172 pounds with a standard deviation of 29 pounds. a. Find the probability that 20 randomly selected men will have a sum weight greater than 3,600 pounds. b. If 20 men have a sum weight greater than 3,500 pounds, then their total weight exceeds the safety limits for water taxis. Based on (a), is this a safety concern? Explain. 452 Chapter 7 \| The Central Limit Theorem 93. Large bags of a brand of multicolored candies have a claimed net weight of 396.9 g. The standard deviation for the weight of the individual candies is 0.017 g. The following table is from a stats experiment conducted by a statistics class. Red (g) Orange (g) Yellow (g) Brown (g) Blue (g) Green (g) 0.883 0.769 0.859 0.784 0.824 0.858 0.848 0.851 0.696 0.876 0.855 0.806 0.840 0.868 0.859 0.982 0.751 0.841 0.856 0.799 0.966 0.859 0.857 0.942 0.873 0.809 0.890 0.878 0.905 0.735 0.895 0.865 0.864 0.852 0.866 0.859 0.838 0.863 0.888 0.925 0.793 0.977 0.850 0.830 0.856 0.842 0.778 0.786 0.853 0.864 0.873 0.880 0.882 0.931 0.925 0.914 0.881 0.865 0.865 1.015 0.876 0.809 0.865 0.848 0.940 0.833 0.845 0.852 0.778 0.814 0.791 0.810 0.881 0.881 0.863 0.775 0.854 0.810 0.858 0.8
18 0.868 0.803 0.932 0.842 0.832 0.807 0.841 0.932 0.833 0.881 0.818 0.864 0.825 0.855 0.942 0.825 0.869 0.912 0.887 Table 7.8 The bag contained 465 candies and the listed weights in the table came from randomly selected candies. Count the weights. a. Find the mean sample weight and the standard deviation of the sample weights of candies in the table. b. Find the sum of the sample weights in the table and the standard deviation of the sum of the weights. If 465 candies are randomly selected, find the probability that their weights sum to at least 396.9 g. c. Is the candy company's labeling accurate? d. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 453 94. The Screw Right Company claims their 3 4 inch screws are within ±0.23 of the claimed mean diameter of 0.750 inches with a standard deviation of 0.115 inches. The following data were recorded. 0.757 0.723 0.754 0.737 0.757 0.741 0.722 0.741 0.743 0.742 0.740 0.758 0.724 0.739 0.736 0.735 0.760 0.750 0.759 0.754 0.744 0.758 0.765 0.756 0.738 0.742 0.758 0.757 0.724 0.757 0.744 0.738 0.763 0.756 0.760 0.768 0.761 0.742 0.734 0.754 0.758 0.735 0.740 0.743 0.737 0.737 0.725 0.761 0.758 0.756 Table 7.9 The screws were randomly selected from the local home repair store. a. Find the mean diameter and standard deviation for the sample. b. Find the probability that 50 randomly selected screws will be within the stated tolerance levels. Is the company’s diameter claim plausible? 95. Your company has a contract to perform preventive maintenance on thousands of air conditioners in a large city. Based on service records from previous years, the time that a technician spends servicing a unit averages one hour with a standard deviation of one hour. In the coming week, your company will service a simple random sample of 70 units in the city. You plan to budget an average of 1.1 hours per technician to complete the work. Will this be enough time? 96. A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ test, what is the probability that the sample mean scores will be between 85 and 125 points? 97. Certain coins have an average weight of 5.201 g with a standard deviation of 0.065 g. If a vending machine is designed to accept coins whose weights range from 5.111 g to 5.291 g, what is the expected number of rejected coins when 280 randomly selected coins are inserted into the machine? REFERENCES 7.1 The Central Limit Theorem for Sample Means (Averages) Baran, D. (2010). 20 percent of Americans have never used email. WebGuild. Retrieved from http://www.webguild.org/ 20080519/20-percent-of-americans-have-never-used-email The Flurry Blog. (2013). Retrieved from http://blog.flurry.com U.S. Department of Agriculture. (n.d.). Retrieved from https://www.usda.gov/ 7.2 The Central Limit Theorem for Sums (Optional) Farago, P. (2012, Oct. 29). The truth about cats and dogs: Smartphone vs tablet usage differences. Flurry Analytics Blog. Retrieved from http://flurrymobile.tumblr.com/post/113379683050/the-truth-about-cats-and-dogs-smartphone-vs 7.3 Using the Central Limit Theorem The Wall Street Journal. (n.d.). Retrieved from https://www.wsj.com/ Centers for Disease Control and Prevention. (2017, April 16). National health and nutrition examination survey. National Center for Health Statistics. Retrieved from http://www.cdc.gov/nchs/nhanes.htm SOLUTIONS 1 mean = 4 hours, standard deviation = 1.2 hours, sample size = 16 454 Chapter 7 \| The Central Limit Theorem 3 a. Check student's solution. b. 3.5, 4.25, 0.2441 5 The fact that the two distributions are different accounts for the different probabilities. 7 0.3345 9 7833.92 11 0.0089 13 7326.49 15 77.45% 17 0.4207 19 3,888.5 21 0.8186 23 5 25 0.9772 27 The sample size, n, gets larger. 29 49 31 26.00 33 0.1587 35 1000 37 a. U(24, 26), 25, 0.5774 b. N(25, 0.0577) c. 0.0416 39 0.0003 41 25.07 43 a. N(2500, 5.7735) b. 0 45 2507.40 47 a. 10 b. 1 10 49 N ⎛ ⎝10, 10 8 ⎞ ⎠ 51 0.7799 53 1.69 55 0.0072 57 391.54 59 405.51 61 a. Χ = amount of change students carry This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 455 b. Χ ~ E(0.88, 0.88) ¯ = average amount of change carried by a sample of 25 students. c. X ¯ ~ N(0.88, 0.176) d. X e. 0.0819 f. 0.1882 g. The distributions are different. Part (a) is exponential and part (b) is normal. 63 a. length of time for an individual to complete IRS form 1040, in hours b. mean length of time for a sample of 36 taxpayers to complete IRS form 1040, in hours c. N ⎛ ⎝10.53, 1 3 ⎞ ⎠ d. Yes, I would be surprised, because the probability is almost 0. e. No, I would not be totally surprised because the probability is 0.2312. 65 a. the length of a song, in minutes, in the collection b. U(2, 3.5) c. the average length, in minutes, of the songs from a sample of five albums from the collection d. N(2.75, 0.0220) e. 2.74 minutes f. 0.03 minutes 67 a. True. The mean of a sampling distribution of the means is approximately the mean of the data distribution. b. True. According to the central limit theorem, the larger the sample, the closer the sampling distribution of the means becomes normal. c. The standard deviation of the sampling distribution of the means will decrease, making it approximately the same as the standard deviation of X as the sample size increases. 69 a. X = the yearly income of someone in a Third World country b. the average salary from samples of 1,000 residents of a Third World country ¯ ∼ N ⎛ c. X ⎝2,000, 8,000 1,000 ⎞ ⎠ d. Very wide differences in data values can have averages smaller than standard deviations. e. The distribution of the sample mean will have higher probabilities closer to the population mean. P(2,000 < X P(2,100 < X ¯ < 2,100) = 0.1537 ¯ < 2,200) = 0.1317 71 b 73 a. the total length of time for nine criminal trials b. N(189, 21) c. 0.0432 456 Chapter 7 \| The Central Limit Theorem d. 162.09; 90 percent of the total nine trials of this type will last 162 days or more. 75 a. X = the salary of one elementary school teacher in the district b. X ~ N(44000, 6500) c. ΣX ~ sum of the salaries of 10 elementary school teachers in the sample d. ΣX ~ N(44,000, 20,554.80) e. 0.9742 f. $52,330.09 g. 466,342.04 h. Sampling 70 teachers instead of 10 would cause the distribution to be more spread out. It would be a more symmetrical normal curve. i. If every teacher received a $3,000 raise, the distribution of X would shift to the right by $3,000. In other words, it would have a mean of $47,000. 77 a. X = the closing stock prices for U.S. semiconductor manufacturers c. b. i. $20.71, ii. $17.31, iii. 35 d. exponential distribution, Χ ~ Exp ⎛ ⎝ 1 20.71 ⎞ ⎠ e. Answers will vary. f. i. $20.71, ii. $11.14 g. Answers will vary. h. Answers will vary. i. Answers will vary. j. N ⎛ ⎝20.71, 17.31 5 ⎞ ⎠ 79 b 81 b 83 a 85 a. 0 b. 0.1123 c. 0.0162 d. 0.0003 e. 0.0268 87 a. Check student’s solution. ¯ ~ N ⎛ b. X ⎝60, 9 25 ⎞ ⎠ c. 0.5000 d. 59.06 e. 0.8536 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 457 f. 0.1333 g. N(1500, 45) h. 1530.35 i. 0.6877 89 a. $52,330 b. $46,634 91 • We have μ = 17, σ = 0.8, x¯ = 16.7, and n = 30. To calculate the probability, we use normalcdf(lower, upper, μ, σ n ) = normalcdf ⎛ ⎝E – 99,16.7,17, 0.8 30 ⎞ ⎠ = 0.0200. • If the process is working properly, then the probability that a sample of 30 batteries would have at most 16.7 life span hours is only 2%. Therefore, the class was justified to question the claim. 93 a. For the sample, we have n = 100, x¯ = 0.862, and s = 0.05. b. Σ x¯ = 85.65, Σs = 5.18 c. normalcdf(396.9,E99,(465)(0.8565),(0.05)( 465 )) ≈ 1 d. Because the probability of a sample of size of 465 having at least a mean sum of 396.9 is appproximately 1, we can conclude that the company is correctly labeling their candy packages. 95 Use normalcdf ⎛ ⎝E – 99,1.1,1, 1 70 ⎞ ⎠ = 0.7986. This means that there is an 80 percent chance that the service time will be less than 1.1 hours. It may be wise to schedule more time because there is an associated 20 percent chance that the maintenance time will be greater than 1.1 hours. 97 Because we have normalcdf ⎛ ⎝5.111,5.291,5.201,0.065 280 within the limits; therefore, there should be no rejected coins out of a well-selected sample size of 280. ⎞ ⎠ ≈ 1, we can conclude that practically all the coins are 458 Chapter 7 \| The Central Limit Theorem This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 459 8 \| CONFIDENCE INTERVALS Figure 8.1 Have you ever wondered what the average number of chocolate candies in a bag at the grocery store is? You can use confidence intervals to answer this question. (credit: comedy_nose/flickr) Introduction By the end of this chapter, the student should be able to do the following: Chapter Objectives Interpret the Student's t probability distribution as the sample size changes • Calculate and interpret confidence intervals for estimating a population mean and a population proportion • • Discriminate between problems applying the normal and the Student's t-distributions • Calculate the sample size required to estimate a population mean and a population proportion, given a desired confidence level and margin of error Suppose you were trying to determine the mean rent of a two-bedroom apartment in your town. You might look in the classified section of the newspaper, write down several rents listed, and average them together. You would have obtained a point estimate of the true mean. If you are trying to determine the percentage of times you make a basket when shooting a basketball, you might count the number of shots you make and divide that by the number of shots you attempt. In this case, you would have obtained a point estimate for the true proportion. 460 Chapter 8 \| Confidence Interv
als We use sample data to make generalizations about an unknown population. This part of statistics is called inferential statistics. The sample data help us to make an estimate of a population parameter. We realize that the point estimate is most likely not the exact value of the population parameter, but close to it. After calculating point estimates, we construct interval estimates, called confidence intervals. In this chapter, you will learn to construct and interpret confidence intervals. You will also learn a new distribution, the Student's-t, and how it is used with those intervals. Throughout the chapter, it is important to keep in mind that the confidence interval is a random variable. It is the population parameter that is fixed. If you worked in the marketing department of an entertainment company, you might be interested in the mean number of songs a consumer downloads a month from an internet music store. If so, you could conduct a survey and calculate the sample mean, x¯ , and the sample standard deviation, s. You would use x¯ to estimate the population mean and s to estimate the population standard deviation. The sample mean, x¯ , is the point estimate for the population mean, μ. The sample standard deviation, s, is the point estimate for the population standard deviation, σ. Each instance of x¯ and s is called a statistic. A confidence interval is another type of estimate but, instead of being just one number, it is an interval of numbers. The interval of numbers is a range of values calculated from a given set of sample data. The confidence interval is likely to include an unknown population parameter. Suppose, for the internet music example, we do not know the population mean, μ, but we do know that the population standard deviation is σ = 1 and our sample size is 100. Then, by the central limit theorem, the standard deviation for the sample mean is σ n = 1 100 = 0.1. The empirical rule, which applies to bell-shaped distributions, says that in approximately 95 percent of the samples, the sample mean, x¯ , will be within two standard deviations of the population mean, μ. For our internet music example, two standard deviations would be calculated as (2)(0.1) = 0.2. The sample mean, x¯ , is likely to be within 0.2 units of μ. In this example, we do not know the true population mean μ (because we do not have information from all the internet music users!), but we can compute the sample mean x¯ based on our sample of 100 individuals. Because the sample mean is likely to be within 0.2 units of the true population mean 95 percent of the times that we take a sample of 100 users, we can say with 95 percent confidence that μ is within 0.2 units of x¯ . In other words, μ is somewhere between x¯ − 0.2 and x¯ + 0.2 . Suppose that from the sample of 100 internet music customers, we compute a sample mean download of x¯ = 2 songs per month. Since we know that the population standard deviation is σ − 1 , according to the central limit theorem, the standard deviation for the sample means is σ = 1 = 0.1 . 100 We know that there is a 95 percent chance that the true population mean value μ is between two standard deviations from the sample mean. That is, with 95 percent confidence we can say that μ is between x¯ − 2× σ n and x¯ − 2× σ n . Replacing the symbols for their values in this example, we say that we are 95 percent confident that the true average number between x¯ − 2× σ n downloaded = 2 − .02 = 1.8 , and of songs = 2 − 2× σ internet month music from store per an is 100 x¯ + 2× σ n = 2 + 2× σ 100 = 2 + .02 = 2.2. The 95 percent confidence interval for μ is (1.8, 2.2). The 95 percent confidence interval implies two possibilities. Either the interval (1.8, 2.2) contains the true mean, μ, or our sample produced an x¯ all the samples (95–100 percent). that is not within 0.2 units of the true mean μ. The second possibility happens for only 5 percent of Remember that a confidence interval is created for an unknown population parameter like the population mean, μ. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 461 Confidence intervals for some parameters have the form (point estimate – margin of error, point estimate + margin of error). The margin of error depends on the confidence level or percentage of confidence and the standard error of the mean. When you read newspapers and journals, you might notice that some reports use the phrase margin of error. Other reports will not use that phrase, but include a confidence interval as the point estimate plus or minus the margin of error. Those are two ways of expressing the same concept. NOTE Although the text covers only symmetrical confidence intervals, there are non-symmetrical confidence intervals (for example, a confidence interval for the standard deviation). Have your instructor record the number of meals each student in your class eats out in a week. Assume that the standard deviation is known to be three meals. Construct an approximate 95 percent confidence interval for the true mean number of meals students eat out each week. 1. Calculate the sample mean. 2. Let σ = 3 and n = the number of students surveyed. 3. Construct the interval ⎞ ⎞ ⎠, ⎠ ⎛ − ⎝ We say we are approximately 95 percent confident that the true mean number of meals that students eat out in a week is between __________ and ___________. 8.1 \| A Single Population Mean Using the Normal Distribution A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. Suppose that our sample has a mean of x¯ = 10 and we have constructed the 90 percent confidence interval (5, 15), where the margin of error = 5. Calculating the Confidence Interval To construct a confidence interval for a single unknown population mean, μ, where the population standard deviation is known, we need x¯ as an estimate for μ, and we need the margin of error. Here, the margin of error is called the error bound for a population mean (EBM) is called the margin of error for a population mean (EBM). The sample mean, x¯ , is the point estimate of the unknown population mean, μ. The confidence interval (CI) estimate will have the form: (point estimate – error bound, point estimate + error bound) or, in symbols, ( x¯ – EBM, x¯ +EBM ). The margin of error (EBM) depends on the confidence level (CL). The confidence level is often considered the probability that the calculated confidence interval estimate will contain the true population parameter. However, it is more accurate to state that the confidence level is the percentage of confidence intervals that contain the true population parameter when repeated samples are taken. Most often, the person constructing the confidence interval will choose a confidence level of 90 percent or higher, because that person wants to be reasonably certain of his or her conclusions. Another probability, which is called alpha (α) is related to the confidence level, CL. Alpha is the probability that the confidence interval does not contain the unknown population parameter. Mathematically, alpha can be computed as 462 α = 1 − CL . Example 8.1 Chapter 8 \| Confidence Intervals Suppose we have collected data from a sample. We know the sample mean, but we do not know the mean for the entire population. The sample mean is seven, and the error bound for the mean is 2.5. x¯ and EBM = 2.5. The confidence interval is (7 – 2.5, 7 + 2.5), and calculating the values gives (4.5, 9.5). If the confidence level is 95 percent, then we say, "We estimate with 95 percent confidence that the true value of the population mean is between 4.5 and 9.5." 8.1 Suppose we have data from a sample. The sample mean is 15, and the error bound for the mean is 3.2. What is the confidence interval estimate for the population mean? A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. Suppose that our sample has a mean of x¯ = 10, and we have constructed the 90 percent confidence interval (5, 15) where EBM = 5. To get a 90 percent confidence interval, we must include the central 90 percent of the probability of the normal distribution. If we include the central 90 percent, we leave out a total of α = 10 percent in both tails, or 5 percent in each tail, of the normal distribution. Figure 8.2 The critical value 1.645 is the z-score in a standard normal probability distribution that puts an area of 0.90 in the center, an area of 0.05 in the far left tail, and an area of 0.05 in the far right tail. To capture the central 90 percent, we must go out 1.645 standard deviations on either side of the calculated sample mean. The critical value will change depending on the confidence level of the interval. It is important that the standard deviation used be appropriate for the parameter we are estimating, so in this section, we need to use the standard deviation that applies to sample means, which is σ is commonly called the n . The fraction σ n standard error of the mean in order to distinguish clearly the standard deviation for a mean from the population standard deviation, σ. In summary, as a result of the central limit theorem, the following statements apply: ¯ • X is normally distributed, that is, X ¯ ~ N ⎛ ⎝μ X, σ n ⎞ ⎠ . This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 463 • When the population standard deviation σ is known, we use a normal distribution to calculate the error bound. Calculating the Confidence Interval To construct a confidence interval estimate for an unknown population mean, we need data from a random sample. The steps to construct and interpret the confidence interval are as follows: • Calculate the sample mean, x¯ , from the sample data. Remember, in this section, we already know the population
standard deviation, σ. • Find the z-score that corresponds to the confidence level. • Calculate the error bound EBM. • Construct the confidence interval. • If we denote the critical z-score by z a 2 , and the sample size by n, then the formula for the confidence interval with confidence level Cl = 1 − α , is given by ( x¯ + z a 2 × σ n ). • Write a sentence that interprets the estimate in the context of the situation in the problem. (Explain what the confidence interval means, in the words of the problem.) We will first examine each step in more detail and then illustrate the process with some examples. Finding the z-Score for the Stated Confidence Level When we know the population standard deviation, σ, we use a standard normal distribution to calculate the error bound EBM and construct the confidence interval. We need to find the value of z that puts an area equal to the confidence level (in decimal form) in the middle of the standard normal distribution Z ~ N(0, 1). The confidence level, CL, is the area in the middle of the standard normal distribution. CL = 1 – α, so α is the area that is split equally between the two tails. Each of the tails contains an area equal to α 2 . The z-score that has an area to the right of α 2 is denoted by z α 2 . For example, when CL = 0.95, α = 0.05, and α 2 = 0.025, we write z α 2 = z 0.025. The area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 – 0.025 = 0.975. = z0.025 = 1.96 , using a calculator, computer, or standard normal probability table. z α 2 Normal table (see appendices) shows that the probability for 0 to 1.96 is 0.47500, and so the probability to the right tail of the critical value 1.96 is 0.5 – 0.475 = 0.025 invNorm(0.975, 0, 1) = 1.96. In this command, the value 0.975 is the total area to the left of the critical value that we are looking to calculate. The parameters 0 and 1 are the mean value and the standard deviation of the standard normal distribution Z. NOTE Remember to use the area to the LEFT of z α 2 . In this chapter, the last two inputs in the invNorm command are 0, 1, because you are using a standard normal distribution Z with mean 0 and standard deviation 1. Calculating the Margin of Error EBM The error bound formula for an unknown population mean, μ, when the population standard deviation, σ, is known is 464 Chapter 8 \| Confidence Intervals Margin of error = ⎛ ⎝z α 2 ⎞ ⎛ ⎝ ⎠ ⎞ ⎠. σ n Constructing the Confidence Interval The confidence interval estimate has the format sample mean plus or minus the margin of error. The graph gives a picture of the entire situation CL + α 2 + α 2 = CL + α = 1. Figure 8.3 Writing the Interpretation The interpretation should clearly state the confidence level (CL), explain which population parameter is being estimated (here, a population mean), and state the confidence interval (both endpoints): "We estimate with ___percent confidence that the true population mean (include the context of the problem) is between ___ and ___ (include appropriate units)." Example 8.2 Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of three points. A random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68. Find a confidence interval estimate for the population mean exam score (the mean score on all exams). Find a 90 percent confidence interval for the true (population) mean of statistics exam scores. Solution 8.2 • You can use technology to calculate the confidence interval directly. • The first solution is shown step-by-step (Solution A). • The second solution uses the TI-83, 83+, and 84+ calculators (Solution B). Solution A To find the confidence interval, you need the sample mean, x¯ , and the EBM. • x¯ = 68 EBM=(z α 2 )( σ n ) • The confidence level is 90 percent (CL = 0.90). σ= 3; n= 36; CL = 0.90, so α = 1 – CL = 1 – 0.90 = 0.10. α 2 = 0.05, z α 2 = z0.05 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 465 The area to the right of z0.05 is 0.05 and the area to the left of z0.05 is 1 – 0.05 = 0.95. = z0.05 = 1.645 z α 2 using invNorm(0.95, 0, 1) on the TI-83,83+, and 84+ calculators. This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution. EBM = (1.645) ⎛ ⎝ ⎞ ⎠ 3 36 = 0.8225 x¯ – EBM = 68 – 0.8225 = 67.1775 x¯ + EBM = 68 + 0.8225 = 68.8225 The 90 percent confidence interval is (67.1775, 68.8225). Solution 8.2 Solution B Press STAT and arrow over to TESTS. Arrow down to 7:ZInterval. Press ENTER. Arrow to Stats and press ENTER. Arrow down and enter 3 for σ, 68 for x¯ , 36 for n, and .90 for C-level. Arrow down to Calculate and press ENTER. The confidence interval is (to three decimal places)(67.178, 68.822). Interpretation We estimate with 90 percent confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82. Explanation of 90 percent Confidence Level Ninety percent of all confidence intervals constructed in this way contain the true mean statistics exam score. For example, if we constructed 100 of these confidence intervals, we would expect 90 of them to contain the true population mean exam score. 8.2 Suppose average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of 6 minutes. A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 min. Find a 90 percent confidence interval estimate for the population mean delivery time. 466 Chapter 8 \| Confidence Intervals Example 8.3 The specific absorption rate (SAR) for a cell phone measures the amount of radio frequency (RF) energy absorbed by the user’s body when using the handset. Every cell phone emits RF energy. Different phone models have different SAR measures. For certification from the Federal Communications Commission for sale in the United States, the SAR level for a cell phone must be no more than 1.6 watts per kilogram. Table 8.1 shows the highest SAR level for a random selection of cell phone models of a random cell phone company. Phone Model # SAR Phone Model # SAR Phone Model # SAR 800 900 1000 1100 1200 1300 1400 1500 1600 1700 Table 8.1 1.11 1.48 1.43 1.3 1.09 1800 1900 2000 2100 2200 0.455 2300 1.41 0.82 0.78 1.25 2400 2500 2600 2700 1.36 1.34 1.18 1.3 1.26 1.29 0.36 0.52 1.6 1.39 2800 2900 3000 3100 3200 3300 3400 3500 3600 3700 0.74 0.5 0.4 0.867 0.68 0.51 1.13 0.3 1.48 1.38 Find a 98 percent confidence interval for the true (population) mean of the SARs for cell phones. Assume that the population standard deviation is σ = 0.337. Solution 8.3 Solution A To find the confidence interval, start by finding the point estimate: the sample mean, x¯ = 1.024. This is calculated by adding the specific absorption rate for the 30 cell phones in the sample, and dividing the result by 30. Next, find the EBM. Because you are creating a 98 percent confidence interval, CL = 0.98. Figure 8.4 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 467 You need to find z0.01, having the property that the area under the normal density curve to the right of z0.01 is 0.01 and the area to the left is 0.99. Use your calculator, a computer, or a probability table for the standard normal distribution to find z0.01 = 2.326. EBM = (z0.01) σ n = (2.326)0.337 30 = 0.1431 To find the 98 percent confidence interval, find x¯ ± EBM . x¯ – EBM = 1.024 – 0.1431 = 0.8809 x¯ + EBM = 1.024 + 0.1431 = 1.1671 We estimate with 98 percent confidence that the true SAR mean for the population of cell phones in the United States is between 0.8809 and 1.1671 watts per kilogram. Solution 8.3 Solution B Press STAT and arrow over to TESTS. Arrow down to 7:ZInterval. Press ENTER. Arrow to Stats and press ENTER. Arrow down and enter the following values: σ: 0.337 x¯ : 1.024 n: 30 C-level: 0.98 Arrow down to Calculate and press ENTER. The confidence interval is (to three decimal places) (0.881, 1.167). 468 Chapter 8 \| Confidence Intervals 8.3 Table 8.2 shows a different random sampling of 20 cell phone models. Use these data to calculate a 93 percent confidence interval for the true mean SAR for cell phones certified for use in the United States. As previously, assume that the population standard deviation is σ = 0.337. Phone Model SAR Phone Model SAR 450 550 650 750 850 950 1050 1150 1250 1350 Table 8.2 1.48 1450 0.8 1.15 1.36 0.77 1550 1650 1750 1850 0.462 1950 1.36 1.39 1.3 0.7 2050 2150 2250 2350 1.53 0.68 1.4 1.24 0.57 0.2 0.51 0.3 0.73 0.869 Notice the difference in the confidence intervals calculated in Example 8.3 and the following Try It exercise. These intervals are different for several reasons: they are calculated from different samples, the samples are different sizes, and the intervals are calculated for different levels of confidence. Even though the intervals are different, they do not yield conflicting information. The effects of these kinds of changes are the subject of the next section in this chapter. Changing the Confidence Level or Sample Size Example 8.4 Suppose we change the original problem in Example 8.2 by using a 95 percent confidence level. Find a 95 percent confidence interval for the true (population) mean statistics exam score. Solution 8.4 To find the confidence interval, you need the sample mean, x¯ , and the EBM. • x¯ = 68 EBM=(z α 2 )( σ n ) • The confidence level is 95 percent (CL = 0.95). σ= 3; n= 36 CL = 0.95, so α = 1 – CL = 1 – 0.95 = 0.05. α 2 = 0.025 z α 2 = z0.025 The area to the right of z 0.025 is 0.025, and the area to the left of z 0.025 is 1 – 0.025 = 0.975. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 469 = z0.025 = 1.96, z α 2 when using invnorm(0.975,0,1) on the TI-83, 83+, or 84+ calculators. (Thi
s can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution.) 3 36 ⎞ ⎛ ⎠ = 0.98 EBM = (1.96) ⎝ x¯ – EBM = 68 – 0.98 = 67.02 x¯ + EBM = 68 + 0.98 = 68.98 Notice that the EBM is larger for a 95 percent confidence level in the original problem. Interpretation We estimate with 95 percent confidence that the true population mean for all statistics exam scores is between 67.02 and 68.98. Explanation of 95 percent Confidence Level 95 percent of all confidence intervals constructed in this way contain the true value of the population mean statistics exam score. Comparing the Results The 90 percent confidence interval is (67.18, 68.82). The 95 percent confidence interval is (67.02, 68.98). The 95 percent confidence interval is wider. If you look at the graphs, because the area 0.95 is larger than the area 0.90, it makes sense that the 95 percent confidence interval is wider. For more certainty that the confidence interval actually does contain the true value of the population mean for all statistics exam scores, the confidence interval necessarily needs to be wider. Figure 8.5 Summary: Effect of Changing the Confidence Level • Increasing the confidence level increases the error bound, making the confidence interval wider. • Decreasing the confidence level decreases the error bound, making the confidence interval narrower. 8.4 Refer back to the pizza-delivery Try It exercise. The population standard deviation is six minutes and the sample mean deliver time is 36 minutes. Use a sample size of 20. Find a 95 percent confidence interval estimate for the true mean pizza-delivery time. 470 Chapter 8 \| Confidence Intervals Example 8.5 Suppose we change the original problem in Example 8.2 to see what happens to the error bound if the sample size is changed. Leave everything the same except the sample size. Use the original 90 percent confidence level. What happens to the error bound and the confidence interval if we increase the sample size and use n = 100 instead of n = 36? What happens if we decrease the sample size to n = 25 instead of n = 36? • x¯ = 68 • EBM = ⎛ ⎝, the confidence level is 90 percent (CL = 0.90), z α 2 = z0.05 = 1.645. Solution 8.5 Solution A If we increase the sample size n to 100, we decrease the margin of error. When n = 100, EBM = ⎛ ⎝1.645) ⎛ ⎝ ⎞ ⎠ 3 100 = 0.4935. Solution 8.5 Solution B If we decrease the sample size n to 25, we increase the error bound. When n = 25, EBM = ⎛ ⎝1.645) ⎛ ⎝ ⎞ ⎠ 3 25 = 0.987. Summary: Effect of Changing the Sample Size • Increasing the sample size causes the error bound to decrease, making the confidence interval narrower. • Decreasing the sample size causes the error bound to increase, making the confidence interval wider. 8.5 Refer back to the pizza-delivery Try It exercise. The mean delivery time is 36 minutes and the population standard deviation is six minutes. Assume the sample size is changed to 50 restaurants with the same sample mean. Find a 90 percent confidence interval estimate for the population mean delivery time. Working Backward to Find the Error Bound or Sample Mean When we calculate a confidence interval, we find the sample mean, calculate the error bound, and use them to calculate the confidence interval. However, sometimes when we read statistical studies, the study may state the confidence interval only. If we know the confidence interval, we can work backward to find both the error bound and the sample mean. Finding the Error Bound • From the upper value for the interval, subtract the sample mean, • Or, from the upper value for the interval, subtract the lower value. Then divide the difference by 2. Finding the Sample Mean • Subtract the error bound from the upper value of the confidence interval, This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 471 • Or, average the upper and lower endpoints of the confidence interval. Notice that there are two methods to perform each calculation. You can choose the method that is easier to use with the information you know. Example 8.6 Suppose we know that a confidence interval is (67.18, 68.82) and we want to find the error bound. We may know that the sample mean is 68, or perhaps our source only gives the confidence interval and does not tell us the value of the sample mean. Calculate the error bound: • • If we know that the sample mean is 68, EBM = 68.82 – 68 = 0.82. If we do not know the sample mean, EBM = (68.82 − 67.18) = 0.82. The margin of error is the quantity 2 that we add and subtract from the sample mean to obtain the confidence interval. Therefore, the margin of error is half of the length of the interval. Calculate the sample mean: • • If we know the error bound, x¯ = 68.82 – 0.82 = 68. If we do not know the error bound, x¯ = (67.18 + 68.82) 2 = 68. 8.6 Suppose we know that a confidence interval is (42.12, 47.88). Find the error bound and the sample mean. Calculating the Sample Size n If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. In this situation, we are given the desired margin of error, EBM, and we need to compute the sample size n. The formula for sample size is n = z2 σ 2 EBM 2 n to the closest integer. , found by solving the error bound formula for n. Always round up the value of In this formula, z is the critical value z α 2 , corresponding to the desired confidence level. A researcher planning a study who wants a specified confidence level and error bound can use this formula to calculate the size of the sample needed for the study. Example 8.7 The population standard deviation for the age of Foothill College students is 15 years. If we want to be 95 percent confident that the sample mean age is within two years of the true population mean age of Foothill College students, how many randomly selected Foothill College students must be surveyed? From the problem, we know that σ = 15 and EBM = 2. z = z0.025 = 1.96, because the confidence level is 95 percent. n = z2 σ 2 EBM 2 = (1.96)2 (15)2 22 = 216.09 using the sample size equation. Use n = 217. Always round the answer up to the next higher integer to ensure that the sample size is large enough. 472 Chapter 8 \| Confidence Intervals Therefore, 217 Foothill College students should be surveyed in order to be 95 percent confident that we are within two years of the true population mean age of Foothill College students. 8.7 The population standard deviation for the height of high school basketball players is three inches. If we want to be 95 percent confident that the sample mean height is within one inch of the true population mean height, how many randomly selected students must be surveyed? 8.2 \| A Single Population Mean Using the Student's tDistribution In practice, we rarely know the population standard deviation. In the past, when the sample size was large, this unknown number did not present a problem to statisticians. They used the sample standard deviation s as an estimate for σ and proceeded as before to calculate a confidence interval with close-enough results. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval. William S. Gosset (1876–1937) of the Guinness brewery in Dublin, Ireland, ran into this problem. His experiments with hops and barley produced very few samples. Just replacing σ with s did not produce accurate results when he tried to calculate a confidence interval. He realized that he could not use a normal distribution for the calculation; he found that the actual distribution depends on the sample size. This problem led him to discover what is called the Student's t-distribution. The name comes from the fact that Gosset wrote under the pen name Student. Up until the mid-1970s, some statisticians used the normal distribution approximation for large sample sizes and used the Student's t-distribution only for sample sizes of at most 30. With graphing calculators and computers, the practice now is to use the Student's t-distribution whenever s is used as an estimate for σ. If you draw a simple random sample of size n from a population that has an approximately normal distribution with mean μ and unknown population standard deviation σ and calculate the t-score t = x¯ – μ , then the t-scores follow a Student's ⎛ ⎝ ⎞ ⎠ s n t-distribution with n – 1 degrees of freedom. The t-score has the same interpretation as the z-score: It measures how far x¯ is from its mean μ. For each sample size n, there is a different Student's t-distribution. The degrees of freedom (df), n -– 1, are the sample size minus 1. Properties of the Student's t-distribution • The graph for the Student's t-distribution is similar to the standard normal curve. • The mean for the Student's t-distribution is zero, and the distribution is symmetric about zero. • The Student's t-distribution has more probability in its tails than the standard normal distribution. Figure 8.6 shows the graphs of the student t-distribution for 1, 2 and 5 degrees of freedom: (v), compare to the standard normal distribution (in black). This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 473 Figure 8.6 • The exact shape of the Student's t-distribution depends on the degrees of freedom. As the degrees of freedom increase, the graph of the Student's t-distribution becomes more like the graph of the standard normal distribution. • The underlying population of individual observations is assumed to be normally distributed with unknown population mean μ and unknown population standard deviation σ. The size of the underlying population is generally not relevant unless it is very small. If it is bell-shaped (normal), then the assumption is met and does not need discussion. Random sampling is assumed, but that is a comple
tely separate assumption from normality. Calculators and computers can easily calculate any Student's t-probabilities. The TI-83, 83+, and 84+ have a tcdf function to find the probability for given values of t. The grammar for the tcdf command is tcdf(lower bound, upper bound, degrees of freedom). However, for confidence intervals, we need to use inverse probability to find the value of t when we know the probability. For the TI-84+, you can use the invT command on the DISTRibution menu. The invT command works similarly to the invnorm. The invT command requires two inputs: invT(area to the left, degrees of freedom). The output is the t-score that corresponds to the area we specified. The TI-83 and 83+ do not have the invT command. (The TI-89 has an inverse T command.) A probability table for the Student's t-distribution can also be used. The table gives critical t-values that correspond to the confidence level (column) and degrees of freedom (row). (The TI-86 does not have an invT program or command, so if you are using that calculator, you need to use a probability table for the Student's t-distribution.) When using a t-table, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only corresponding area in one or both tails. 474 Chapter 8 \| Confidence Intervals A Student's t-table (see Appendix H) gives t-scores given the degrees of freedom and the right-tailed probability. The table is very limited. Calculators and computers can easily calculate any Student's t-probabilities. If the population standard deviation is not known, the error bound for a population mean is • EBM = ⎛ ⎝ is the t-score with area to the right equal to α 2 , • use df = n – 1 degrees of freedom, and • s = sample standard deviation. The format for the confidence interval is ( x¯ − EBM, x¯ + EBM). To calculate the confidence interval directly, do the following: Press STAT. Arrow over to TESTS. Arrow down to 8:TInterval and press ENTER (or just press 8). Example 8.8 Suppose you do a study of acupuncture to determine how effective it is in relieving pain. You measure sensory rates for 15 subjects with the results given. Use the sample data to construct a 95 percent confidence interval for the mean sensory rate for the population (assumed normal) from which you took the data. The solution is shown step-by-step and by using the TI-83, 83+, or 84+ calculators. 8.6; 9.4; 7.9; 6.8; 8.3; 7.3; 9.2; 9.6; 8.7; 11.4; 10.3; 5.4; 8.1; 5.5; 6.9 Solution 8.8 • The first solution is step-by-step (Solution A). • The second solution uses the TI-83+ and TI-84 calculators (Solution B). To find the confidence interval, you need the sample mean, x¯ , and the EBM. x¯ = 8.6 + 9.4 + 7.9 + 6.8 + 8.3 + 7.3 + 9.2 + 9.6 + 8.7 + 11.4 + 10.3 + 5.4 + 8.1 + 5.5 + 6.9 = 8.2267; 2 + (9.4 − x¯ ) 2 + ⋯ + (5.5 − x¯ ) 14 2 15 + (6.9 − x¯ ) 2 = 1.6722; (8.6 − x¯ ) s = n = 15 df = 15 – 1 = 14 CL, so α = 1 – CL = 1 – 0.95 = 0.05 α 2 = 0.025; t α 2 = t0.025 The area to the right of t0.025 is 0.025, and the area to the left of t0.025 is 1 – 0.025 = 0.975. = t0.025 = 2.14 using invT(.975,14) on the TI-84+ calculator. t α 2 EBM = ⎛ ⎝ ⎛ EBM = (2.14) ⎝ x¯ – EBM = 8.2267 – 0.9240 = 7.3 ⎞ ⎠ = 0.924 1.6722 15 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 475 x¯ + EBM= 8.2267 + 0.9240 = 9.15 The 95 percent confidence interval is (7.30, 9.15). We estimate with 95 percent confidence that the true population mean sensory rate is between 7.30 and 9.15. Solution 8.8 Press STAT and arrow over to TESTS. Arrow down to 8:TInterval and press ENTER (or you can just press 8). Arrow to Data and press ENTER. Arrow down to List and enter the list name where you put the data. There should be a 1 after Freq. Arrow down to C-level and enter 0.95. Arrow down to Calculate and press ENTER. The 95 percent confidence interval is (7.3006, 9.1527). NOTE When calculating the error bound, you can also use a probability table for the Student's t-distribution to find the value of t. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row); the t-score is found where the row and column intersect in the table. 8.8 You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results. Construct a 95 percent confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data. 8.2, 9.1, 7.7, 8.6, 6.9, 11.2, 10.1, 9.9, 8.9, 9.2, 7.5, 10.5 Example 8.9 A group of researchers is working to understand the scope of industrial pollution in the human body. Industrial chemicals may enter the body through pollution or as ingredients in consumer products. In October 2008, the scientists tested cord-blood samples for 20 newborn infants in the United States. The cord blood of the in utero/ newborn group was tested for 430 industrial compounds, pollutants, and other chemicals, including chemicals linked to brain and nervous-system toxicity, immune-system toxicity, reproductive toxicity, and fertility problems. There are health concerns about the effects of some chemicals on the brain and nervous system. Table 8.2 shows how many of the targeted chemicals were found in each infant’s cord blood. 79 145 147 160 116 100 159 151 156 126 137 83 156 94 121 144 123 114 139 99 Table 8.3 476 Chapter 8 \| Confidence Intervals Use this sample data to construct a 90 percent confidence interval for the mean number of targeted industrial chemicals to be found in an infant’s blood. Solution 8.9 Solution A From the sample data, you can calculate x¯ = 79 + 145 + ⋯ + 139 + 99 20 = 127.45 2 (79 − x¯ ) + (145 − x¯ ) 2 + ⋯ + (139 − x¯ ) 2 + (99 − x¯ ) 2 19 s = infants in the sample, so n = 20, and df = 20 – 1 = 19. There are 20 = 25.965. You are asked to calculate a 90 percent confidence interval: CL = 0.90, so α = 1 – CL = 1 – 0.90 = 0.10. α 2 = t0.05 = 0.05,t α 2 By definition, the area to the right of t0.05 is 0.05, and so the area to the left of t0.05 is 1 – 0.05 = 0.95. Use a table, calculator, or computer to find that t0.05 = 1.729. EBM = t α 2 ( s n ) = 1.729 ⎛ ⎝ 25.965 20 ⎞ ⎠ ≈ 10.038 x¯ – EBM = 127.45 – 10.038 = 117.412 x¯ + EBM = 127.45 + 10.038 = 137.488 We estimate with 90 percent confidence that the mean number of all targeted industrial chemicals found in cord blood in the United States is between 117.412 and 137.488. Solution 8.9 Solution B Enter the data as a list. Press STAT and arrow over to TESTS. Arrow down to 8:TInterval and press ENTER (or you can just press 8). Arrow to Data and press ENTER. Arrow down to List and enter the list name where you put the data. Arrow down to Freq and enter 1. Arrow down to C-level and enter 0.90. Arrow down to Calculate and press ENTER. The 90 percent confidence interval is (117.41, 137.49). 8.9 A random sample of statistics students was asked to estimate the total number of hours they spend watching television in an average week. The responses are recorded in Table 8.4. Use the following sample data to construct a 98 percent confidence interval for the mean number of hours statistics students will spend watching television in one This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 477 week. 0 5 3 1 20 9 10 1 10 4 14 2 4 4 5 Table 8.4 8.3 \| A Population Proportion During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40 percent of the vote within 3 percentage points (if the sample is large enough). Often, election polls are calculated with 95 percent confidence, so the pollsters would be 95 percent confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43 (0.40 – 0.03, 0.40 + 0.03). Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers. The procedure to find the confidence interval, the sample size, the error bound for a population (EBP), and the confidence level for a proportion is similar to that for the population mean, but the formulas are different. How do you know you are dealing with a proportion problem? First, the data that you are collecting is categorical, consisting of two categories: Success or Failure, Yes or No. Examples of situations where you are the following trying to estimate the true population proportion are the following: What proportion of the population smoke? What proportion of the population will vote for candidate A? What proportion of the population has a college-level education? The distribution of the sample proportions (based on samples of size n) is denoted by P′ (read “P prime”). The central limit theorem for proportions asserts that the sample proportion distribution P′ follows a normal distribution with mean value p, and standard deviation p • q n , where p is the population proportion and q = 1 -– p. The confidence interval has the form (p′ – EBP, p′ + EBP). EBP is error bound for the proportion. p′ = the estimated proportion of successes (p′ is a point estimate for p, the true proportion.) p′ = x n x = the number of successes n = the size of the sample The error bound for a proportion is EBP = ⎛ ⎝z α 2 ⎛ ⎞ ⎠ ⎝ p′ q′ n ⎞ ⎠, where q′ = 1 – p′. This formula is similar to the error bound formula for a mean, except that the "appropriate standard
deviation" is different. For a mean, when the population standard deviation is known, the appropriate standard deviation that we use is σ . For a n proportion, the appropriate standard deviation is pq n . However, in the error bound formula, we use p′ q′ n as the standard deviation, instead of pq n . In the error bound formula, the sample proportions p′ and q′, are estimates of the unknown population proportions p and q. The estimated proportions p′ and q′ are used because p and q are not known. The sample proportions p′ and q′ are calculated 478 Chapter 8 \| Confidence Intervals from the data: p′ is the estimated proportion of successes, and q′ is the estimated proportion of failures. The confidence interval can be used only if the number of successes np′ and the number of failures nq′ are both greater than five. That is, in order to use the formula for confidence intervals for proportions, you need to verify that both np' ≥ 5 and nq' ≥ 5 . Example 8.10 Suppose that a market research firm is hired to estimate the percentage of adults living in a large city who have cell phones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes, they own cell phones. Using a 95 percent confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones. Solution 8.10 • The first solution is step-by-step (Solution A). • The second solution uses a function of the TI-83, 83+, or 84 calculators (Solution B). ⎛ ⎝500, 421 Let X = the number of people in the sample who have cell phones. X is binomial. X ~ B 500 ⎞ ⎠ . To calculate the confidence interval, you must find p′, q′, and EBP. n = 500 x = the number of successes = 421 p′ = 0.842 is the sample proportion; this is the point estimate of the population proportion. p′ = x n = 421 500 = 0.842 Because CL = 0.95, then α = 1 – CL = 1 – 0.95 = 0.05 ⎛ ⎝ = 0.025. ⎞ ⎠ α 2 q′ = 1 – p′ = 1 – 0.842 = 0.158 Then, z α 2 = z0.025 = 1.96. Use the TI-83, 83+, or 84+ calculator command invNorm(0.975,0,1) to find z0.025. Remember that the area to the right of z0.025is 0.025, and the area to the left of z0.025is 0.975. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table. EBP = ⎞ ⎠ p′ q′ n = (1.96) ⎛ ⎝z α 2 p′ – EBP = 0.842 – 0.032 = 0.81 (0.842)(0.158) 500 = 0.032 The confidence interval for the true binomial population proportion is (p′ – EBP, p′ + EBP) = (0.810, 0.874). p′ + EBP = 0.842 + 0.032 = 0.874 Interpretation We estimate with 95 percent confidence that between 81 percent and 87.4 percent of all adult residents of this city have cell phones. Explanation of 95 percent Confidence Level Ninety-five percent of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have cell phones. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 479 Solution 8.10 Press STAT and arrow over to TESTS. Arrow down to A:1-PropZint. Press ENTER. Arrow down to x and enter 421. Arrow down to n and enter 500. Arrow down to C-Level and enter .95. Arrow down to Calculate and press ENTER. The confidence interval is (0.81003, 0.87397). 8.10 Suppose 250 randomly selected people are surveyed to determine whether they own tablets. Of the 250 surveyed, 98 reported owning tablets. Using a 95 percent confidence level, compute a confidence interval estimate for the true proportion of people who own tablets. Example 8.11 For a class project, a political science student at a large university wants to estimate the percentage of students who are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90 percent confidence interval for the true percentage of students who are registered voters, and interpret the confidence interval. Solution 8.11 • The first solution is step-by-step (Solution A). • The second solution uses a function of the TI-83, 83+, or 84 calculators (Solution B). Solution A x = 300 and n = 500 n = 300 p′ = x 500 q′ = 1 – p′ = 1 - 0.600 = 0.400 = 0.600 Because CL = 0.90, then α = 1 – CL = 1 – 0.90 = 0.10 ⎛ ⎝ = 0.05. ⎞ ⎠ α 2 = z0.05 = 1.645 z α 2 Use the TI-83, 83+, or 84+ calculator command invNorm(0.95,0,1) to find z0.05. Remember that the area to the right of z0.05 is 0.05, and the area to the left of z0.05 is 0.95. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table. EBP = ⎛ ⎝z α 2 ⎞ ⎠ p′ q′ n = (1.645) (0.60)(0.40) 500 = 0.036 p′ – EBP = 0.60 − 0.036 = 0.564 p′ + EBP = 0.60 + 0.036 = 0.636 480 Chapter 8 \| Confidence Intervals The confidence interval for the true binomial population proportion is (p′ – EBP , p′ + EBP) = (0.564, 0.636). Interpretation • We estimate with 90 percent confidence that the true percentage of all students who are registered voters is between 56.4 percent and 63.6 percent. • Alternate wording: We estimate with 90 percent confidence that between 56.4 percent and 63.6 percent of all students are registered voters. Explanation of 90 percent Confidence Level Ninety percent of all confidence intervals constructed in this way contain the true value for the population percentage of students who are registered voters. Solution 8.11 Solution B Press STAT and arrow over to TESTS. Arrow down to A:1-PropZint. Press ENTER. Arrow down to x and enter 300. Arrow down to n and enter 500. Arrow down to C-Level and enter 0.90. Arrow down to Calculate and press ENTER. The confidence interval is (0.564, 0.636). 8.11 A student polls her school to determine whether students in the school district are for or against the new legislation regarding school uniforms. She surveys 600 students and finds that 480 are against the new legislation. a. Compute a 90 percent confidence interval for the true percentage of students who are against the new legislation, and interpret the confidence interval. b. In a sample of 300 students, 68 percent said they own an iPod and a smartphone. Compute a 97 percent confidence interval for the true percentage of students who own an iPod and a smartphone. Plus-Four Confidence Interval for p There is a certain amount of error introduced into the process of calculating a confidence interval for a proportion. Because we do not know the true proportion for the population, we are forced to use point estimates to calculate the appropriate standard deviation of the sampling distribution. Studies have shown that the resulting estimation of the standard deviation can be flawed. Fortunately, there is a simple adjustment that allows us to produce more accurate confidence intervals: We simply pretend that we have four additional observations. Two of these observations are successes, and two are failures. The new sample size, then, is n + 4, and the new count of successes is x + 2. Computer studies have demonstrated the effectiveness of the plus-four confidence interval for p method. It should be used when the confidence level desired is at least 90 percent and the sample size is at least ten. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 481 Example 8.12 A random sample of 25 statistics students was asked: “Have you used a product in the past week?” Six students reported using the product within the past week. Use the plus-four method to find a 95 percent confidence interval for the true proportion of statistics students who use the product weekly. Solution 8.12 Six students out of 25 reported using a product within the past week, so x = 6 and n = 25. Because we are using the plus-four method, we will use x = 6 + 2 = 8, and n = 25 + 4 = 29. p′ = x n = 8 29 q′ = 1 – p′ = 1 – 0.276 = 0.724 ≈ 0.276 Because CL = 0.95, we know α = 1 – 0.95 = 0.05, and α 2 = 0.025. z0.025 = 1.96 EPB = ⎞ ⎠ p′ q′ n = (1.96) ⎛ ⎝z α 2 p′ – EPB = 0.276 – 0.163 = 0.113 p′ + EPB = 0.276 + 0.163 = 0.439 0.276(0.724) 29 ≈ 0.163 We are 95 percent confident that the true proportion of all statistics students who use the product is between 0.113 and 0.439. Solution 8.12 Press STAT and arrow over to TESTS. Arrow down to A:1-PropZint. Press ENTER. Arrow down to x and enter 8. Arrow down to n and enter 29. Arrow down to C-Level and enter 0.95. Arrow down to Calculate and press ENTER. The confidence interval is (0.113, 0.439). REMINDER Remember that the plus-four method assumes an additional four trials: two successes and two failures. You do not need to change the process for calculating the confidence interval; simply update the values of x and n to reflect these additional trials. 8.12 Out of a random sample of 65 freshmen at State University, 31 students have declared their majors. Use the plus-four method to find a 96 percent confidence interval for the true proportion of freshmen at State University who have declared their majors. 482 Chapter 8 \| Confidence Intervals Example 8.13 A group of researchers recently conducted a study analyzing the privacy management habits of teen internet users. In a group of 50 teens, 13 reported having more than 500 friends on a social media site. Use the plus four method to find a 90 percent confidence interval for the true proportion of teens who would report having more than 500 online friends. Solution 8.13 Using plus-four, we have x = 13 + 2 = 15, and n = 50 + 4 = 54. p' = 15 54 ≈ 0.278 q' = 1 – p' = 1 - 0.278 = 0.722 Because CL = 0.90, we know α = 1 – 0.90 = 0.10, and α 2 = 0.05. z0.05 = 1.645 EPB = (z α 2 p′ q′ n ⎞ ⎛ ⎠ = (1.645) ⎝ ⎛ ) ⎝ p′ – EPB = 0.278 – 0.100 = 0.178 p′ + EPB = 0.278 + 0.160 = 0.378 (0.278)(0.722) 54 ⎞ ⎠ ≈ 0.100 We are 90 percent confident that between 17.8 percent and 37.8 percent of all teens would report having more than 500 friends on a social media site. Solution 8.13 Press
STAT and arrow over to TESTS. Arrow down to A:1-PropZint. Press ENTER. Arrow down to x and enter 15. Arrow down to n and enter 54. Arrow down to C-Level and enter 0.90. Arrow down to Calculate and press ENTER. The confidence interval is (0.178, 0.378). 8.13 The research group referenced in Example 8.13 talked to teens in smaller focus groups but also interviewed additional teens over the phone. When the study was complete, 588 teens had answered the question about their social media site friends, with 159 saying that they have more than 500 friends. Use the plus-four method to find a 90 percent confidence interval for the true proportion of teens who would report having more than 500 online friends based on this larger sample. Compare the results to those in Example 8.13. Calculating the Sample Size n If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 483 The margin of error formula for a population proportion is • EBP = z α 2 × p′ • q′ n , where p′ is the sample proportion, q′ = 1 – p′, and n is the sample size. • Solving for n gives you an equation for the sample size. (p′ q′) 2 ⎛ ⎞ ⎝z α ⎠ 2 EBP2 • n = . This formula tells us that we can compute the sample size n required for a confidence level of Cl = 1 − α by taking the square of the critical value z a 2 , multiplying by the point estimate p′, and by q′ = 1 – p′ and finally dividing the result by the square of the margin of error. Always remember to round up the value of n. Example 8.14 Suppose a mobile phone company wants to determine the current percentage of customers ages 50+ who use text messaging on their cell phones. How many customers ages 50+ should the company survey in order to be 90 percent confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of customers ages 50+ who use text messaging on their cell phones? Assume that p′ = 0.5. Solution 8.14 From the problem, we know that EBP = 0.03 (3 percent=0.03), and z α 2 z0.05 = 1.645 because the confidence level is 90 percent. To calculate the sample size n, use the formula and make the substitutions. n = z2 p′ q′ EBP2 gives n = 1.6452(0.5)(0.5) 0.032 = 751.7 Round the answer to the next higher value. The sample size should be 752 cell phone customers ages 50+ in order to be 90 percent confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of all customers ages 50+ who use text messaging on their cell phones. 8.14 An internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. How many customers should the company survey in order to be 90 percent confident that the estimated proportion is within 5 percentage points of the true population proportion of customers who click on ads on their smartphones? Assume that the sample proportion p′ is 0.50. 8.4 \| Confidence Interval (Home Costs) 484 Chapter 8 \| Confidence Intervals 8.1 Confidence Interval (Home Costs) Student Learning Outcomes • The student will calculate the 90 percent confidence interval for the mean cost of a home in the area in which this school is located. • The student will interpret confidence intervals. • The student will determine the effects of changing conditions on the confidence interval. Collect the Data Check the Real Estate section in your local newspaper. Record the sale prices for 35 randomly selected homes recently listed in the county. NOTE Many newspapers list them only one day per week. Also, we will assume that homes come up for sale randomly. 1. Complete the following table: __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Table 8.5 Describe the Data 1. Compute the following: a. b. x¯ = _____ s x = _____ c. n = _____ 2. ¯ . In words, define the random variable X 3. State the estimated distribution to use. Use both words and symbols. Find the Confidence interval 1. Calculate the confidence interval and the error bound. a. Confidence interval: _____ b. Error Bound: _____ 2. How much area is in both tails (combined)? α = _____ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 485 3. How much area is in each tail? α 2 = _____ 4. Fill in the blanks on the graph with the area in each section. Then, fill in the number line with the upper and lower limits of the confidence interval and the sample mean. Figure 8.7 5. Some students think that a 90 percent confidence interval contains 90 percent of the data. Use the list of data on the first page and count how many of the data values lie within the confidence interval. What percentage is this? Is this percentage close to 90 percent? Explain why this percentage should or should not be close to 90 percent. Describe the Confidence Interval 1. In two to three complete sentences, explain what a confidence interval means (in general), as if you were talking to someone who has not taken statistics. 2. In one to two complete sentences, explain what this confidence interval means for this particular study. Use the Data to Construct Confidence Intervals 1. Using the given information, construct a confidence interval for each confidence level given. Confidence Level EBM/Error Bound Confidence Interval 50% 80% 95% 99% Table 8.6 2. What happens to the EBM as the confidence level increases? Does the width of the confidence interval increase or decrease? Explain why this happens. 8.5 \| Confidence Interval (Place of Birth) 486 Chapter 8 \| Confidence Intervals 8.2 Confidence Interval (Place of Birth) Student Learning Outcomes • The student will calculate the 90 percent confidence interval of the proportion of students in this school who were born in this state. • The student will interpret confidence intervals. • The student will determine the effects of changing conditions on the confidence interval. Collect the Data 1. Survey the students in your class, asking them whether they were born in this state. Let X = the number who were born in this state. a. n = ____________ b. x = ____________ 2. In words, define the random variable P′. 3. State the estimated distribution to use. Find the Confidence interval and Error bound 1. Calculate the confidence interval and the error bound. a. Confidence interval: _____ b. Error Bound: _____ 2. How much area is in both tails (combined)? α = _____ 3. How much area is in each tail? α 2 = _____ 4. Fill in the blanks on the graph with the area in each section. Then, fill in the number line with the upper and lower limits of the confidence interval and the sample proportion. Figure 8.8 Describe the Confidence Interval 1. In two to three complete sentences, explain what a confidence interval means (in general), as though you were talking to someone who has not taken statistics. 2. In one to two complete sentences, explain what this confidence interval means for this particular study. 3. Construct a confidence interval for each confidence level given. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 487 Confidence Level EBP/Error Bound Confidence Interval 50% 80% 95% 99% Table 8.7 4. What happens to the EBP as the confidence level increases? Does the width of the confidence interval increase or decrease? Explain why this happens. 8.6 \| Confidence Interval (Women's Heights) 488 Chapter 8 \| Confidence Intervals 8.3 Confidence Interval (Women's Heights) Student Learning Outcomes • The student will calculate a 90 percent confidence interval using the given data. • The student will determine the relationship between the confidence level and the percentage of constructed intervals that contain the population mean. Given: 59.4 71.6 69.3 65.0 62.9 66.5 61.7 55.2 67.5 67.2 63.8 62.9 63.0 63.9 68.7 65.5 61.9 69.6 58.7 63.4 61.8 60.6 69.8 60.0 64.9 66.1 66.8 60.6 65.6 63.8 61.3 59.2 64.1 59.3 64.9 62.4 63.5 60.9 63.3 66.3 61.5 64.3 62.9 60.6 63.8 58.8 64.9 65.7 62.5 70.9 62.9 63.1 62.2 58.7 64.7 66.0 60.5 64.7 65.4 60.2 65.0 64.1 61.1 65.3 64.6 59.2 61.4 62.0 63.5 61.4 65.5 62.3 65.5 64.7 58.8 66.1 64.9 66.9 57.9 69.8 58.5 63.4 69.2 65.9 62.2 60.0 58.1 62.5 62.4 59.1 66.4 61.2 60.4 58.7 66.7 67.5 63.2 56.6 67.7 62.5 Table 8.8 Heights of 100 Women (in Inches) 1. Table 8.8 lists the heights of 100 women. Use a random number generator to select 10 data values randomly. 2. Calculate the sample mean and the sample standard deviation. Assume that the population standard deviation is known to be 3.3 in. With these values, construct a 90 percent confidence interval for your sample of 10 values. Write the confidence interval you obtained in the first space of Table 8.9. 3. Now write your confidence interval on the board. As others in the class write their confidence intervals on the board, copy them into Table 8.9. __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 489 __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Table 8.9 90 percent Confidence Intervals Discussion Questions 1. The actual population mean for the 100 heights
given in Table 8.8 is μ = 63.4. Using the class listing of confidence intervals, count how many of them contain the population mean μ; i.e., for how many intervals does the value of μ lie between the endpoints of the confidence interval? 2. Divide this number by the total number of confidence intervals generated by the class to determine the percentage of confidence intervals that contain the mean μ. Write that percentage here: _____________. 3. Is the percentage of confidence intervals that contain the population mean μ close to 90 percent? 4. Suppose we had generated 100 confidence intervals. What do you think would happen to the percentage of confidence intervals that contained the population mean? 5. When we construct a 90 percent confidence interval, we say that we are 90 percent confident that the true population mean lies within the confidence interval. Using complete sentences, explain what we mean by this phrase. 6. Some students think that a 90 percent confidence interval contains 90 percent of the data. Use the list of data given (the heights of women) and count how many of the data values lie within the confidence interval that you generated based on that data. How many of the 100 data values lie within your confidence interval? What percentage is this? Is this percentage close to 90 percent? 7. Explain why it does not make sense to count data values that lie in a confidence interval. Think about the random variable that is being used in the problem. 8. Suppose you obtained the heights of 10 women and calculated a confidence interval from this information. Without knowing the population mean μ, would you have any way of knowing for certain whether your interval actually contained the value of μ? Explain. 490 Chapter 8 \| Confidence Intervals KEY TERMS binomial distribution independent trials Independent means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances, the binomial RV X is defined as the number of successes in n trials. The notation is X~B(n,p). The mean is μ = np, and the standard deviation is σ = a discrete random variable (RV) that arises from Bernoulli trials; there are a fixed number, n, of npq . The probability of exactly x successes in n trials is P⎛ ⎝X = x⎞ ⎝n ⎠ = ⎛ x ⎞ ⎠p x qn − x . confidence interval (CI) an interval estimate for an unknown population parameter. This depends on the following: • • • the desired confidence level, information that is known about the distribution (for example, known standard deviation), and the sample and its size. confidence level (CL) the percentage expression for the probability that the confidence interval contains the true population parameter; for example, if the CL = 90 percent, then in 90 out of 100 samples, the interval estimate will enclose the true population parameter degrees of freedom (df) the number of objects in a sample that are free to vary error bound for a population mean (EBM) the margin of error; depends on the confidence level, sample size, and known or estimated population standard deviation error bound for a population proportion (EBP) the margin of error; depends on the confidence level, the sample size, and the estimated (from the sample) proportion of successes inferential statistics also called statistical inference or inductive statistics; this facet of statistics deals with estimating a population parameter based on a sample statistic For example, if four out of the 100 calculators sampled are defective, we might infer that 4 percent of the production is defective. normal distribution a bell-shaped continuous random variable X, with center at the mean value (μ) and distance from the center to the inflection points of the bell curve given by the standard deviation (σ). We write X ~ N(μ, σ) . If the mean value is 0 and the standard deviation is 1, the random variable is called the standard normal distribution, and it is denoted with the letter Z parameter a numerical characteristic of a population plus-four confidence interval plus-four confidence interval when you add two imaginary successes and two imaginary failures (four overall) to your sample point estimate a single number computed from a sample and used to estimate a population parameter standard deviation a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation Student's t-distribution investigated and reported by William S. Gossett in 1908 and published under the pseudonym Student the major characteristics of the random variable (RV) are as follows: • It is continuous and assumes any real values. • The pdf is symmetrical about its mean of zero. However, it is more spread out and flatter at the apex than the normal distribution. • It approaches the standard normal distribution as n get larger. • There is a family of t-distributions: Each representative of the family is completely defined by the number of degrees of freedom, which is one less than the number of data. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 491 CHAPTER REVIEW 8.1 A Single Population Mean Using the Normal Distribution In this module, we learned how to calculate the confidence interval for a single population mean where the population standard deviation is known. When estimating a population mean, the margin of error is called the error bound for a population mean (EBM). A confidence interval has the general form (lower bound, upper bound) = (point estimate – EBM, point estimate + EBM). The calculation of EBM depends on the size of the sample and the level of confidence desired. The confidence level is the percentage of all possible samples that can be expected to include the true population parameter. As the confidence level increases, the corresponding EBM increases as well. As the sample size increases, the EBM decreases. By the central limit theorem, EBM = z σ n. Given a confidence interval, you can work backward to find the error bound (EBM) or the sample mean. To find the error bound, find the difference of the upper bound of the interval and the mean. If you do not know the sample mean, you can find the error bound by calculating half of the difference of the upper and lower bounds. To find the sample mean given a confidence interval, find the difference of the upper bound and the error bound. If the error bound is unknown, then average the upper and lower bounds of the confidence interval to find the sample mean. Sometimes researchers know in advance that they want to estimate a population mean within a specific margin of error for a given level of confidence. In that case, solve the EBM formula for n to discover the size of the sample that is needed to achieve this goal: n = z2 σ 2 EBM 2 8.2 A Single Population Mean Using the Student's t-Distribution In many cases, the researcher does not know the population standard deviation, σ, of the measure being studied. In these cases, it is common to use the sample standard deviation, s, as an estimate of σ. The normal distribution creates accurate confidence intervals when σ is known, but it is not as accurate when s is used as an estimate. In this case, the Student’s t-distribution is much better. Define a t-score using the following formula: t = x¯ − μ s n The t-score follows the Student’s t-distribution with n – 1 degrees of freedom. The confidence interval under this distribution is calculated with EBM = ⎛ s n, standard deviation, and n is the sample size. Use a table, calculator, or computer to find t α 2 is the t-score with area to the right equal to α 2 for a given α. where t α 2 , s is the sample ⎝t α 2 ⎞ ⎠ 8.3 A Population Proportion Some statistical measures, like many survey questions, measure qualitative rather than quantitative data. In this case, the population parameter being estimated is a proportion. It is possible to create a confidence interval for the true population proportion by following procedures similar to those used in creating confidence intervals for population means. The formulas are slightly different, but they follow the same reasoning. Let p′ represent the sample proportion, x/n, where x represents the number of successes, and n represents the sample size. Let q′ = 1 – p′. Then the confidence interval for a population proportion is given by the following formula: (lower bound, upper bound) = (p′ – EBP, p′ + EBP) = (p′ – z p′ q′ n , p′ + z p′ q′ n ). The plus–four method for calculating confidence intervals is an attempt to balance the error introduced by using estimates of the population proportion when calculating the standard deviation of the sampling distribution. Simply imagine four 492 Chapter 8 \| Confidence Intervals additional trials in the study; two are successes and two are failures. Calculate p′ = x + 2 n + 4 , and proceed to find the confidence interval. When sample sizes are small, this method has been demonstrated to provide more accurate confidence intervals than the standard formula used for larger samples. FORMULA REVIEW 8.1 A Single Population Mean Using the Normal Distribution ¯ X ~ N ⎛ ⎝μ X, σ n ⎞ ⎠ The distribution of sample means is normally distributed with mean equal to the population mean and standard deviation given by the population standard deviation divided by the square root of the sample size. The general form for a confidence interval for a single population mean, known standard deviation, normal distribution is given by (lower bound, upper bound) = (point estimate – EBM, point estimate + EBM) =( x¯ − EBM, x¯ + EBM) ⎞ ⎠. ⎛ ⎝ x¯ − z σ = n , x¯ + z σ n = the error bound for the mean, or the margin EBM = z σ n of error for a single population mean; this formula is used when the population standard deviation is known. CL = confidence level, or the pr
oportion of confidence intervals created that is expected to contain the true population parameter Single population mean, known standard deviation, normal distribution Use the normal distribution for means; population standard deviation is known: EBM = z α 2 ⋅ σ n The confidence interval has the format ( x¯ − EBM, x¯ + EBM). 8.2 A Single Population Mean Using the Student's t-Distribution s = the standard deviation of sample values t = x¯ − μ s n is the formula for the t-score, which measures how far away a measure is from the population mean in the Student’s t-distribution. df = n – 1; t-distribution, where n represents the size of the sample the degrees of freedom for a Student’s T~tdf the random variable, T, has a Student’s t-distribution with df degrees of freedom EBM = t α 2 s n = the error bound for the population mean when the population standard deviation is unknown α = 1 – CL = the proportion of confidence intervals that will not contain the population parameter t α 2 is the t-score in the Student’s t-distribution with area to = the z-score with the property that the area to the z α 2 right of the z-score is ∝ 2 ; this is the z-score, used in the calculation of EBM, where α = 1 – CL. n = z2 σ 2 EBM 2 = the formula used to determine the sample size (n) needed to achieve a desired margin of error at a given level of confidence the right equal to α 2 . The general form for a confidence interval for a single mean, population standard deviation unknown, Student's t is given by (lower bound, upper bound) = (point estimate – EBM, point estimate + EBM) ⎛ ⎝ x¯ – ts = n , x¯ + ts n ⎞ ⎠. General form of a confidence interval 8.3 A Population Proportion (lower value, upper value) = (point estimate error bound, point estimate + error bound) To find the error bound when you know the confidence interval, error bound = upper value point estimate or error bound = upper value − lower value 2 . p′ = x/n, where x represents the number of successes and n represents the sample size. The variable p′ is the sample proportion and serves as the point estimate for the true population proportion. q′ = 1 – p′ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 493 p′ ~ N ⎛ ⎝p, pq n ⎞ ⎠ The variable p′ has a binomial distribution that can be approximated with the normal distribution shown here, EBP = the error bound for a proportion = z α 2 p′ q′ n . Confidence interval for a proportion: (lower bound, upper bound) = (p′ – EBP, p′ + EBP) = ⎛ ⎝p′ – z p′ q′ n , p′ + z p′ q′ n ⎞ ⎠ n = 2 p′ q′ z α 2 EBP2 provides the number of participants needed to estimate the population proportion with confidence 1 – α and margin of error EBP. PRACTICE Use the normal distribution for a single population proportion p′ = x n. EBP = ⎛ ⎝z α 2 ⎞ ⎠ p′q′ n p′ + q′ = 1 The confidence interval has the format (p′ – EBP, p′ + EBP). is a point estimate for μ. x¯ p′ is a point estimate for ρ. s is a point estimate for σ. 8.1 A Single Population Mean Using the Normal Distribution Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 lb. We wish to construct a 95 percent confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 lb. The sample standard deviation is 11 lb. 1. Identify the following: x¯ = _____ a. b. σ = _____ c. n = _____ ¯ . 2. In words, define the random variables X and X 3. Which distribution should you use for this problem? 4. Construct a 95 percent confidence interval for the population mean weight of newborn elephants. State the confidence interval, sketch the graph, and calculate the error bound. 5. What will happen to the confidence interval obtained, if 500 newborn elephants are weighed instead of 50? Why? Use the following information to answer the next seven exercises: The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal. 6. Identify the following: x¯ = _____ a. b. σ = _____ c. n = _____ ¯ . 7. In words, define the random variables X and X 8. Which distribution should you use for this problem? 9. Construct a 90 percent confidence interval for the population mean time to complete the forms. State the confidence interval, sketch the graph, and calculate the error bound. 10. If the Census wants to increase its level of confidence and keep the error bound the same by taking another survey, what changes should it make? 11. If the Census did another survey, kept the error bound the same, and surveyed only 50 people instead of 200, what would happen to the level of confidence? Why? 494 Chapter 8 \| Confidence Intervals 12. Suppose the Census needed to be 98 percent confident of the population mean length of time. Would the Census have to survey more people? Why or why not? Use the following information to answer the next 10 exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 lb, with a standard deviation of 0.1 lb. The population standard deviation is known to be 0.2 lb. 13. Identify the following: x¯ = ______ a. b. σ = ______ c. n = ______ 14. In words, define the random variable X. ¯ . 15. In words, define the random variable X 16. Which distribution should you use for this problem? 17. Construct a 90 percent confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound. 18. Construct a 95 percent confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound. 19. In complete sentences, explain why the confidence interval in Exercise 8.17 is larger than in Exercise 8.18. 20. In complete sentences, give an interpretation of what the interval in Exercise 8.18 means. 21. What would happen if 40 heads of lettuce were sampled instead of 20 and the error bound remained the same? 22. What would happen if 40 heads of lettuce were sampled instead of 20 and the confidence level remained the same? Use the following information to answer the next 14 exercises: The mean age for all Foothill College students for a recent fall term was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that 25 winter students were randomly selected. The mean age for the sample was 30.4. We are interested in the true mean age for winter Foothill College students. Let X = the age of a winter Foothill College student. 23. x¯ = _____ 24. n = _____ 25. ________ = 15 ¯ . 26. In words, define the random variable X 27. What is x¯ estimating? 28. Is σ x known? 29. As a result of your answer to Exercise 8.26, state the exact distribution to use when calculating the confidence interval. Construct a 95 percent confidence interval for the true mean age of winter Foothill College students by working out and then answering the next eight exercises. 30. How much area is in both tails (combined)? α =________ 31. How much area is in each tail? α 2 =________ 32. Identify the following specifications: a. lower limit b. upper limit c. error bound 33. The 95 percent confidence interval is __________________. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 495 34. Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample mean. Figure 8.9 35. In one complete sentence, explain what the interval means. 36. Using the same mean, standard deviation, and level of confidence, suppose that n were 69 instead of 25. Would the error bound become larger or smaller? How do you know? 37. Using the same mean, standard deviation, and sample size, how would the error bound change if the confidence level were reduced to 90 percent? Why? 8.2 A Single Population Mean Using the Student's t-Distribution Use the following information to answer the next five exercises: A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hr, with a sample standard deviation of 0.5 hr. 38. Identify the following: a. b. x¯ =_______ s x =_______ c. n =_______ d. n – 1 =_______ ¯ 39. Define the random variables X and X in words. 40. Which distribution should you use for this problem? 41. Construct a 95 percent confidence interval for the population mean time spent waiting. State the confidence interval, sketch the graph, and calculate the error bound. 42. Explain in complete sentences what the confidence interval means. Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watch an average of 151 hours each month, with a standard deviation of 32 hours. Assume that the underlying population distribution is normal. 43. Identify the following: a. b. x¯ =_______ s x =_______ c. n =_______ d. n – 1 =_______ 44. Define the random variable X in words. ¯ 45. Define the random variable X in words. 46. Which distribution should you use for this problem? 496 Chapter 8 \| Confidence Intervals 47. Construct a 99 percent confidence interval for the population mean hours spent watching television per month. State the confidence interval, sketch the graph, and calculate the error bound. 48. Why would the error bound change if the confidence level were lowered to 95 percent? Use the following information to answer the
next 13 exercises: The data in Table 8.10 are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let X = the number of colors on a national flag. X Freq. 1 2 3 4 5 1 7 18 7 6 Table 8.10 49. Calculate the following: a. b. x¯ =______ s x =______ c. n =______ ¯ 50. Define the random variable X in words. 51. What is x¯ estimating? 52. Is σ x known? 53. As a result of your answer to Exercise 8.52, state the exact distribution to use when calculating the confidence interval. Construct a 95 percent confidence interval for the true mean number of colors on national flags. 54. How much area is in both tails (combined)? 55. How much area is in each tail? 56. Calculate the following: lower limit a. b. upper limit c. error bound 57. The 95 percent confidence interval is_____. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 497 58. Fill in the blanks on the graph with the areas, the upper and lower limits of the confidence interval, and the sample mean. Figure 8.10 59. In one complete sentence, explain what the interval means. 60. Using the same x¯ , s x , and level of confidence, suppose that n were 69 instead of 39. Would the error bound become larger or smaller? How do you know? 61. Using the same x¯ , s x , and n = 39, how would the error bound change if the confidence level were reduced to 90 percent? Why? 8.3 A Population Proportion Use the following information to answer the next two exercises: Marketing companies are interested in knowing the population percentage of women who make the majority of household purchasing decisions. 62. When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 90 percent confident that the population proportion is estimated to within 0.05? 63. If it were later determined that it was important to be more than 90 percent confident and a new survey were commissioned, how would it affect the minimum number you need to survey? Why? Use the following information to answer the next five exercises: Suppose a marketing company conducted a survey. It randomly surveyed 200 households and found that in 120 of them, the women made the majority of the purchasing decisions. We are interested in the population proportion of households where women make the majority of the purchasing decisions. 64. Identify the following: a. x = ______ b. n = ______ c. p′ = ______ 65. Define the random variables X and P′ in words. 66. Which distribution should you use for this problem? 67. Construct a 95 percent confidence interval for the population proportion of households where the women make the majority of the purchasing decisions. State the confidence interval, sketch the graph, and calculate the error bound. 68. List two difficulties the company might have in obtaining random results if this survey were done by email. Use the following information to answer the next five exercises: Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage earners, 250 identified themselves as midlevel managers, and 160 identified themselves as executives. In the survey, 82 percent of manual laborers preferred trucks, 62 percent of non-manual wage earners preferred trucks, 54 percent of mid-level managers preferred trucks, and 26 percent of executives preferred trucks. 498 Chapter 8 \| Confidence Intervals 69. We are interested in finding the 95 percent confidence interval for the percentage of executives who prefer trucks. Define random variables X and P′ in words. 70. Which distribution should you use for this problem? 71. Construct a 95 percent confidence interval. State the confidence interval, sketch the graph, and calculate the error bound. 72. Suppose we want to lower the sampling error. What is one way to accomplish that? 73. The sampling error given in the survey is ±2 percent. Explain what the ±2 percent means. Use the following information to answer the next five exercises: A poll of 1,200 voters asked what the most significant issue was in the upcoming election. Sixty-five percent answered "the economy." We are interested in the population proportion of voters who believe the economy is the most important. 74. Define the random variable X in words. 75. Define the random variable P′ in words. 76. Which distribution should you use for this problem? 77. Construct a 90 percent confidence interval, and state the confidence interval and the error bound. 78. What would happen to the confidence interval if the level of confidence were 95 percent? Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning iceskating classes. All of the class names are put into a bucket. The 5 p.m., Monday night, ages 8 to 12, beginning ice-skating class is picked. In that class are 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population. 79. What is being counted? 80. In words, define the random variable X. 81. Calculate the following: a. x = _______ b. n = _______ c. p′ = _______ 82. State the estimated distribution of X. X~ ________ 83. Define a new random variable P′. What is p′ estimating? 84. In words, define the random variable P′. 85. State the estimated distribution of P′. Construct a 92 percent confidence interval for the true proportion of girls in the ages 8 to 12 beginning ice-skating classes at the Ice Chalet. 86. How much area is in both tails (combined)? 87. How much area is in each tail? 88. Calculate the following: lower limit a. b. upper limit c. error bound 89. The 92 percent confidence interval is _______. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 499 90. Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample proportion. Figure 8.11 91. In one complete sentence, explain what the interval means. 92. Using the same p′ and level of confidence, suppose that n were increased to 100. Would the error bound become larger or smaller? How do you know? 93. Using the same p′ and n = 80, how would the error bound change if the confidence level were increased to 98 percent? Why? 94. If you decreased the allowable error bound, why would the minimum sample size increase (keeping the same level of confidence)? 500 Chapter 8 \| Confidence Intervals HOMEWORK 8.1 A Single Population Mean Using the Normal Distribution 95. Among various ethnic groups, the standard deviation of heights is known to be approximately three inches. We wish to construct a 95 percent confidence interval for the mean height of male Swedes. 48 male Swedes are surveyed. The sample mean is 71 inches. The sample standard deviation is 2.8 in. x¯ =________ a. i. ii. σ =________ iii. n =________ ¯ . In words, define the random variables X and X b. c. Which distribution should you use for this problem? Explain your choice. d. Construct a 95 percent confidence interval for the population mean height of male Swedes. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. What will happen to the level of confidence obtained if 1,000 male Swedes are surveyed instead of 48? Why? 96. Announcements for 84 upcoming engineering conferences were randomly picked from a stack of IEEE Spectrum magazines. The mean length of the conferences was 3.94 days, with a standard deviation of 1.28 days. Assume the underlying population is normal. ¯ . In words, define the random variables X and X a. b. Which distribution should you use for this problem? Explain your choice. c. Construct a 95 percent confidence interval for the population mean length of engineering conferences. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. 97. Suppose that an accounting firm does a study to determine the time needed to complete one person’s tax forms. It randomly surveys 100 people. The sample mean is 23.6 hours. There is a known standard deviation of 7.0 hours. The population distribution is assumed to be normal. a. x¯ =________ i. ii. σ =________ iii. n =________ ¯ . In words, define the random variables X and X b. c. Which distribution should you use for this problem? Explain your choice. d. Construct a 90 percent confidence interval for the population mean time to complete the tax forms. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. f. If the firm wished to increase its level of confidence and keep the error bound the same by taking another survey, which changes should it make? If the firm did another survey, kept the error bound the same, and only surveyed 49 people, what would happen to the level of confidence? Why? g. Suppose that the firm decided that it needed to be at least 96 percent confident of the population mean length of time to within one hour. How would the number of people the firm surveys change? Why? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 501 98. A sample of 16 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is normal. The weight of each bag was then recorded. The mean weight was two ounces with a standard deviation of 0.12 ounces. The population standard deviation is known to be 0.1 ounce. a. x¯ =________ i. ii. σ =________ sx =________ iii. b. In words, define the random variable X. ¯ . In words, define the random variable X c. d. Which distribution should you use for this problem? Explain your choice. e. Construct a 90 percent c
onfidence interval for the population mean weight of the candies. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. f. Construct a 98 percent confidence interval for the population mean weight of the candies. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. g. h. In complete sentences, explain why the confidence interval in Part f is larger than the confidence interval in Part e. In complete sentences, give an interpretation of what the interval in Part f means. 99. A camp director is interested in the mean number of letters each child sends during his or her camp session. The population standard deviation is known to be 2.5. A survey of 20 campers is taken. The mean from the sample is 7.9, with a sample standard deviation of 2.8. x¯ =________ a. i. ii. σ =________ iii. n =________ ¯ b. Define the random variables X and X c. Which distribution should you use for this problem? Explain your choice. d. Construct a 90 percent confidence interval for the population mean number of letters campers send home. in words. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. What will happen to the error bound and confidence interval if 500 campers are surveyed? Why? 100. What is meant by the term 90 percent confident when constructing a confidence interval for a mean? a. b. c. d. If we took repeated samples, approximately 90 percent of the samples would produce the same confidence interval. If we took repeated samples, approximately 90 percent of the confidence intervals calculated from those samples would contain the sample mean. If we took repeated samples, approximately 90 percent of the confidence intervals calculated from those samples would contain the true value of the population mean. If we took repeated samples, the sample mean would equal the population mean in approximately 90 percent of the samples. 502 Chapter 8 \| Confidence Intervals 101. The Federal Election Commission collects information about campaign contributions and disbursements for candidates and political committees during each election cycle. During the 2012 campaign season, there were 1,619 candidates for the House of Representatives across the United States who received contributions from individuals. Table 8.11 shows the total receipts from individuals for a random selection of 40 House candidates rounded to the nearest $100. The standard deviation for this data to the nearest hundred is σ = $909,200. $3,600 $1,243,900 $10,900 $385,200 $581,500 $7,400 $2,900 $400 $3,714,500 $632,500 $391,000 $467,400 $56,800 $5,800 $405,200 $733,200 $8,000 $468,700 $75,200 $41,000 $13,300 $9,500 $953,800 $1,113,500 $1,109,300 $353,900 $986,100 $88,600 $378,200 $13,200 $3,800 $745,100 $5,800 $3,072,100 $1,626,700 $512,900 $2,309,200 $6,600 $202,400 $15,800 Table 8.11 a. Find the point estimate for the population mean. b. Using 95 percent confidence, calculate the error bound. c. Create a 95 percent confidence interval for the mean total individual contributions. d. Interpret the confidence interval in the context of the problem. 102. The American Community Survey (ACS), part of the U.S. Census Bureau, conducts a yearly census similar to the one taken every 10 years, but with a smaller percentage of participants. The most recent survey estimates with 90 percent confidence that the mean household income in the United States falls between $69,720 and $69,922. Find the point estimate for mean U.S. household income and the error bound for mean U.S. household income. 103. The average height of young adult males has a normal distribution with standard deviation of 2.5 in. You want to estimate the mean height of students at your college or university to within 1 in. with 93 percent confidence. How many male students must you measure? 8.2 A Single Population Mean Using the Student's t-Distribution 104. In six packages of multicolored fruit snacks, there were five red snack pieces. The total number of snack pieces in the six bags was 68. We wish to calculate a 96 percent confidence interval for the population proportion of red snack pieces. a. Define the random variables X and P′ in words. b. Which distribution should you use for this problem? Explain your choice. c. Calculate p′. d. Construct a 96 percent confidence interval for the population proportion of red snack pieces per bag. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. Do you think that six packages of fruit snacks yield enough data to give accurate results? Why or why not? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 503 105. A random survey of enrollment at 35 community colleges across the United States yielded the following figures: 6,414, 1,550, 2,109, 9,350, 21,828, 4,300, 5,944, 5,722, 2,825, 2,044, 5,481, 5,200, 5,853, 2,750, 10,012, 6,357, 27,000, 9,414, 7,681, 3,200, 17,500, 9,200, 7,380, 18,314, 6,557, 13,713, 17,768, 7,493, 2,771, 2,861, 1,263, 7,285, 28,165, 5,080, 11,622. Assume the underlying population is normal. a. i. ii. x¯ = __________ s x = __________ iii. n = __________ iv. n – 1 = __________ ¯ b. Define the random variables X and X c. Which distribution should you use for this problem? Explain your choice. d. Construct a 95 percent confidence interval for the population mean enrollment at community colleges in the in words. United States. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. What will happen to the error bound and confidence interval if 500 community colleges are surveyed? Why? 106. Suppose that a committee is studying whether there is wasted time in our judicial system. It is interested in the mean amount of time individuals waste at the courthouse waiting to be called for jury duty. The committee randomly surveyed 81 people who recently served as jurors. The sample mean wait time was 8 hr, with a sample standard deviation of 4 hr. a. i. ii. x¯ = __________ s x = __________ iii. n = __________ iv. n – 1 = __________ ¯ b. Define the random variables X and X c. Which distribution should you use for this problem? Explain your choice. d. Construct a 95 percent confidence interval for the population mean time wasted. in words. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. Explain in a complete sentence what the confidence interval means. 107. A pharmaceutical company makes a drug used during surgery. It is assumed that the distribution for the length of time the drug lasts is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of the antibiotic drug for each patient (in hours) was as follows: 2.7, 2.8, 3.0, 2.3, 2.3, 2.2, 2.8, 2.1, and 2.4. a. i. ii. x¯ = __________ s x = __________ iii. n = __________ iv. n – 1 = __________ b. Define the random variable X in words. ¯ c. Define the random variable X d. Which distribution should you use for this problem? Explain your choice. e. Construct a 95 percent confidence interval for the population mean length of time. in words. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. f. What does it mean to be 95 percent confident in this problem? 504 Chapter 8 \| Confidence Intervals 108. Suppose that 14 children who were learning to ride two-wheel bikes were surveyed to determine how long they had to use training wheels. It was revealed that they used them an average of six months, with a sample standard deviation of three months. Assume that the underlying population distribution is normal. a. i. ii. x¯ = __________ s x = __________ iii. n = __________ iv. n – 1 = __________ b. Define the random variable X in words. ¯ c. Define the random variable X d. Which distribution should you use for this problem? Explain your choice. e. Construct a 99 percent confidence interval for the population mean length of time using training wheels. in words. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. f. Why would the error bound change if the confidence level were lowered to 90 percent? 109. The Federal Election Commission (FEC) collects information about campaign contributions and disbursements for candidates and political committees during each election cycle. A political action committee (PAC) is a committee formed to raise money for candidates and campaigns. A Leadership PAC is a PAC formed by a federal politician (senator or representative) to raise money to help other candidates’ campaigns. The FEC has reported financial information for 556 Leadership PACs that operated during the 2011–2012 election cycle. The following table shows the total receipts during this cycle for a random selection of 20 Leadership PACs. $46,500.00 $0 $40,966.50 $105,887.20 $5,175.00 $29,050.00 $19,500.00 $181,557.20 $31,500.00 $149,970.80 $2,555,363.20 $12,025.00 $409,000.00 $60,521.70 $18,000.00 $61,810.20 $76,530.80 $119,459.20 $0 $63,520.00 $6,500.00 $502,578.00 $705,061.10 $708,258.90 $135,810.00 $2,000.00 $2,000.00 $0 $1,287,933.80 $219,148.30 Table 8.12 x¯ = $251, 854.23 s = $521, 130.41 Use the sample data to construct a 96 percent confidence interval for the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle. Use the Student's t-distribution. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 505 110. A major business magazine published data on the best small firms in 2012. These were firms that have been publicly traded for at least a year, have a stock price of at least $5 per share, and have reported annual revenue between $5 million and $1 billion. Table 8.13 shows the ages of the corporate CEOs for a random sample of these firms. 48 58 51 61 56 59 74 63 53 50 59 60 60 57 46 55 63 57 47 55 57 43 61 62 49 67 67 55 55 4
9 Table 8.13 Use the sample data to construct a 90 percent confidence interval for the mean age of CEOs for these top small firms. Use the Student's t-distribution. 111. Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected, and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats, and the sample standard deviation is 4.1 seats. a. i. ii. x¯ = __________ s x = __________ iii. n = __________ iv. n – 1 = __________ ¯ b. Define the random variables X and X c. Which distribution should you use for this problem? Explain your choice. d. Construct a 92 percent confidence interval for the population mean number of unoccupied seats per flight. in words. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. 112. In a recent sample of 84 used car sales costs, the sample mean was $6,425, with a standard deviation of $3,156. Assume the underlying distribution is approximately normal. a. Which distribution should you use for this problem? Explain your choice. ¯ b. Define the random variable X c. Construct a 95 percent confidence interval for the population mean cost of a used car. in words. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. d. Explain what a 95 percent confidence interval means for this study. 113. Six different national brands of chocolate chip cookies were randomly selected at the supermarket. The grams of fat per serving are as follows: 8, 8, 10, 7, 9, 9. Assume the underlying distribution is approximately normal. a. Construct a 90 percent confidence interval for the population mean grams of fat per serving of chocolate chip cookies sold in supermarkets. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. b. If you wanted a smaller error bound while keeping the same level of confidence, what should have been changed in the study before it was done? c. Go to the store and record the grams of fat per serving of six brands of chocolate chip cookies. d. Calculate the mean. e. Is the mean within the interval you calculated in Part a? Did you expect it to be? Why or why not? 506 Chapter 8 \| Confidence Intervals 114. A survey of the mean number of cents off given by coupons was conducted by randomly surveying one coupon per page from the coupons section of a local newspaper. The following data were collected: 20¢, 75¢, 50¢, 65¢, 30¢, 55¢, 40¢, 40¢, 30¢, 55¢, $1.50, 40¢, 65¢, 40¢. Assume the underlying distribution is approximately normal. a. i. ii. x¯ = __________ s x = __________ iii. n = __________ iv. n – 1 = __________ ¯ b. Define the random variables X and X c. Which distribution should you use for this problem? Explain your choice. d. Construct a 95 percent confidence interval for the population mean worth of coupons. in words. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. If many random samples were collected with 14 samples as the size, which percentage of the confidence intervals constructed should contain the population mean worth of coupons? Explain why. Use the following information to answer the next two exercises: A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of soda served in the 16-oz serving size. The sample mean is 13.30, with a sample standard deviation of 1.55. Assume the underlying population is normally distributed. 115. Find the 95 percent confidence interval for the true population mean for the amount of soda served. a. b. c. d. (12.42, 14.18) (12.32, 14.29) (12.50, 14.10) Impossible to determine 116. Which of the following is the error bound? a. 0.87 b. 1.98 c. 0.99 d. 1.74 8.3 A Population Proportion 117. Insurance companies are interested in knowing the population percentage of drivers who always buckle up before riding in a car. a. When designing a study to determine this population proportion, what is the minimum number you would need b. to survey to be 95 percent confident that the population proportion is estimated to within 0.03? If it were later determined that it was important to be more than 95 percent confident and a new survey was commissioned, how would that affect the minimum number you would need to survey? Why? 118. Suppose that the insurance companies did conduct a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up. a. i. x = __________ ii. n = __________ iii. p′ = __________ b. Define the random variables X and P′ in words. c. Which distribution should you use for this problem? Explain your choice. d. Construct a 95 percent confidence interval for the population proportion who claim they always buckle up. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. If this survey were done by telephone, list three difficulties the companies might have in obtaining random results. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 507 119. According to a recent survey of 1,200 people, 61 percent believe that the president is doing an acceptable job. We are interested in the population proportion of people who believe the president is doing an acceptable job. a. Define the random variables X and P′ in words. b. Which distribution should you use for this problem? Explain your choice. c. Construct a 90 percent confidence interval for the population proportion of people who believe the president is doing an acceptable job. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. 120. An article regarding dating and marriage recently appeared in a major newspaper. Of the 1,709 randomly selected adults, 315 identified themselves as ethnicity A, 323 identified themselves as ethnicity B, 254 identified themselves as ethnicity C, and 779 identified themselves as ethnicity D. In this survey, 86 percent of ethnicity B said that they would welcome a person of ethnicity A into their families. Among ethnicity C, 77 percent would welcome a person of ethnicity D into their families, 71 percent would welcome a person of ethnicity A, and 66 percent would welcome a person of ethnicity B. a. We are interested in finding the 95 percent confidence interval for the percent of all ethnicity B adults who would welcome a person of ethnicity D into their families. Define the random variables X and P′ in words. b. Which distribution should you use for this problem? Explain your choice. c. Construct a 95 percent confidence interval. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. 121. Refer to the information in Exercise 8.120. a. Construct three 95 percent confidence intervals: i. percentage of all ethnicity C who would welcome a person of ethnicity D into their families ii. percentage of all ethnicity C who would welcome a person of ethnicity A into their families iii. percentage of all ethnicity C who would welcome a person of ethnicity B into their families b. Even though the three point estimates are different, do any of the confidence intervals overlap? Which? c. For any intervals that do overlap, in words, what does this imply about the significance of the differences in the true proportions? d. For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions? 122. Stanford University conducted a study of whether running is healthy for men and women over age 50. During the first eight years of the study, 1.5 percent of the 451 members of the 50-Plus Fitness Association died. We are interested in the proportion of people over 50 who ran and died in the same eight year period. a. Define the random variables X and P′ in words. b. Which distribution should you use for this problem? Explain your choice. c. Construct a 97 percent confidence interval for the population proportion of people over 50 who ran and died in the same 8-year period. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. d. Explain what a 97 percent confidence interval means for this study. 508 Chapter 8 \| Confidence Intervals 123. A telephone poll of 1,000 adult Americans was reported in an issue of a national magazine. One of the questions asked, “What is the main problem facing the country?” Twenty percent responded "crime". We are interested in the population proportion of adult Americans who believe that crime is the main problem. a. Define the random variables X and P′ in words. b. Which distribution should you use for this problem? Explain your choice. c. Construct a 95 percent confidence interval for the population proportion of adult Americans who believe that crime is the main problem. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. d. Suppose we want to lower the sampling error. What is one way to accomplish that? e. The sampling error given by the group of researchers who conducted the poll is ±3 percent. In one to three complete sentences, explain what the ±3 percent represents. 124. Refer to Exercise 8.123. Another question in the poll asked, “[How much are] you worried about the quality of education in our schools?” Sixty-three percent responded “a lot”. We are interested in the population proportion of adult Americans who are worried a lot about the quality of education in our schools. a. Define the random variables X and P′ in words. b. Which distribution should you use for this problem? Explain your choice. c. Construct a 95 percent confidence interval for the population proportion of adult Americans who are worried a lot about the quality
of education in our schools. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. d. The sampling error given by the group of researchers who conducted the poll is ±3 percent. In one to three complete sentences, explain what the ±3 percent represents. Use the following information to answer the next three exercises: According to a Field Poll, 79 percent of California adults (actual results are 400 out of 506 surveyed) believe that education and our schools is one of the top issues facing California. We wish to construct a 90 percent confidence interval for the true proportion of California adults who believe that education and the schools is one of the top issues facing California. 125. A point estimate for the true population proportion is _______. a. 0.90 b. 1.27 c. 0.79 d. 400 126. A 90 percent confidence interval for the population proportion is _______. a. b. c. d. (0.761, 0.820) (0.125, 0.188) (0.755, 0.826) (0.130, 0.183) 127. The error bound is approximately _____. a. 1.581 b. 0.791 c. 0.059 d. 0.030 Use the following information to answer the next two exercises: Five hundred eleven (511) homes in a certain southern California community are randomly surveyed to determine whether they meet minimal earthquake preparedness recommendations. One hundred seventy-three (173) of the homes surveyed meet the minimum recommendations for earthquake preparedness, and 338 do not. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 509 128. Find the confidence interval at the 90 percent confidence level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness. a. b. c. d. (0.2975, 0.3796) (0.6270, 0.6959) (0.3041, 0.3730) (0.6204, 0.7025) 129. The point estimate for the population proportion of homes that do not meet the minimum recommendations for earthquake preparedness is ______. a. 0.6614 b. 0.3386 c. 173 d. 338 130. On May 23, 2013, a polling group reported that of the 1,005 people surveyed, 76 percent of U.S. workers believe that they will continue working past retirement age. The confidence level for this study was reported at 95 percent with a ±3 percent margin of error. a. Determine the estimated proportion from the sample. b. Determine the sample size. c. d. Calculate the error bound based on the information provided. e. Compare the error bound in Part d to the margin of error reported by the polling group. Explain any differences Identify CL and α. between the values. f. Create a confidence interval for the results of this study. g. A reporter is covering the release of this study for a local news station. How should she explain the confidence interval to her audience? 131. A national survey of 1,000 adults was conducted on May 13, 2013, by a group of researchers. It concluded with 95 percent confidence that 49 percent to 55 percent of Americans believe that big-time college sports programs corrupt the process of higher education. a. Find the point estimate and the error bound for this confidence interval. b. Can we (with 95 percent confidence) conclude that more than half of all American adults believe this? c. Use the point estimate from Part a and n = 1,000 to calculate a 75 percent confidence interval for the proportion of American adults who believe that major college sports programs corrupt higher education. d. Can we (with 75 percent confidence) conclude that at least half of all American adults believe this? 132. A polling group recently conducted a survey asking adults across the United States about music preferences. When asked, 80 of the 571 participants download music weekly. a. Create a 99 percent confidence interval for the true proportion of American adults who download music weekly. b. This survey was conducted through automated telephone interviews on May 6 and 7, 2013. The error bound of the survey compensates for sampling error, or natural variability among samples. List some factors that could affect the survey’s outcome that are not covered by the margin of error. c. Without performing any calculations, describe how the confidence interval would change if the confidence level decreased from 99 percent to 90 percent. 133. You plan to conduct a survey on your college campus to learn about the political awareness of students. You want to estimate the true proportion of college students on your campus who voted in the 2012 presidential election with 95 percent confidence and a margin of error no greater than 5 percent. How many students must you interview? 134. In a recent poll, 9 of 48 respondents rated the likelihood of a certain event occurring in their community as likely or very likely. Use the plus-four method to create a 97 percent confidence interval for the proportion of American adults who believe that the event is likely or very likely. Explain what this confidence interval means in the context of the problem. A local poll in a New England town found that nine of 48 households think winter-proofing their cars is very important. Use the plus-four method to create a 97 percent confidence interval for the proportion of town residents who think winterproofing their cars is very important. Explain what this confidence interval means in the context of this scenario. REFERENCES 510 Chapter 8 \| Confidence Intervals 8.1 A Single Population Mean Using the Normal Distribution Centers for Disease Control and Prevention. (n.d.). National health and nutrition examination survey. Retrieved from http://www.cdc.gov/nchs/nhanes.htm Foothill De Anza Community College District. (n.d.). Headcount enrollment trends by student demographics ten-year trends to most recently completed fall. Retrieved from http://research.fhda.edu/factbook/FH_Demo_Trends/ fall FoothillDemographicTrends.htm Kuczmarski, R. J., et al. (2002). 2000 CDC growth charts for the United States: Methods and development. Retrieved from http://www.cdc.gov/growthcharts/2000growthchart-us.pdf La, L., and; German, K. (n.d.). Cell phones with the highest radiation levels. CNET. Retrieved from http://reviews.cnet.com/ cell-phone-radiation-levels/ U.S. Census Bureau. (n.d.). American FactFinder. Retrieved from http://factfinder2.census.gov/faces/nav/jsf/pages/ searchresults.xhtml?refresh=t U.S. Census Bureau. (2011). Mean income in the past 12 months (in 2011 inflation-adjusted dollars): 2011 American Community from http://factfinder2.census.gov/faces/tableservices/jsf/pages/ productview.xhtml?pid=ACS_11_1YR_S1902&prodType=table estimates. Retrieved Survey 1-year U.S. Federal Election Commission. (n.d.). Disclosure data catalog: Candidate summary report 2012. Retrieved from http://www.fec.gov/data/CandidateSummary.do?format=html&election_yr=2012 U.S. Federal Election Commission. http://www.fec.gov/finance/disclosure/metadata/metadataforcandidatesummary.shtml (n.d.). Metadata description of candidate summary file. Retrieved from 8.2 A Single Population Mean Using the Student's t-Distribution Bloomberg Businessweek. (n.d.). Retrieved from http://www.businessweek.com/ Environmental Working Group. (n.d.). Human toxome project: Mapping the pollution in people. Retrieved from http://www.ewg.org/sites/humantoxome/participants/participant-group.php?group=in+utero%2Fnewborn Federal Election Commission. (n.d.). Disclosure data catalog: 2012 leadership PACs and sponsors. Retrieved from http://www.fec.gov/data/index.jsp Federal Election Commission. (n.d.). Metadata description of leadership PAC list. Retrieved from http://www.fec.gov/ finance/disclosure/metadata/metadataLeadershipPacList.shtml Forbes. (2013). America’s best small companies. Retrieved from http://www.forbes.com/best-small-companies/list/ Forbes. (n.d.). Retrieved from http://www.forbes.com/ Microsoft Bookshelf. (n.d.). 8.3 A Population Proportion Jensen, http://www.publicpolicypolling.com/Day2MusicPoll.pdf (2013). Democrats, Republicans T. divided on opinion of music icons. Retrieved from Madden, M., et al. (2013). Teens, social media, and privacy. Retrieved from http://www.pewinternet.org/Reports/2013/ Teens-Social-Media-And-Privacy.aspx Princeton Survey Research Associates International. (2012). 2012 teens and privacy management survey. Retrieved from http://www.pewinternet.org/~/media//Files/Questionnaire/2013/ Methods%20and%20Questions_Teens%20and%20Social%20Media.pdf Rasmussen Reports. http://www.rasmussenreports.com/public_content/lifestyle/sports/may_2013/ 52_say_big_time_college_athletics_corrupt_education_process (2013). 52% say big-time college athletics corrupt education process. Retrieved from Saad, L. (2013). Three in four U.S. workers plan to work past retirement age. Retrieved from http://www.gallup.com/poll/ 162758/three-fourworkers-plan-work-past-retirement-age.aspx The Field Poll. (n.d.). Retrieved from http://field.com/fieldpollonline/subscribers/ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 511 Zogby Analytics. (2013). New SUNYIT/Zogby analytics poll: Few Americans worry about emergency situations occurring in their community; Only one in three have an emergency plan; 70% support infrastructure “investment” for national security. Retrieved from http://www.zogbyanalytics.com/news/299-americans-neither-worried-norprepared-in-case-of-adisaster-sunyit-zogby-analytics-poll SOLUTIONS 1 a. 244 b. 15 c. 50 3 N ⎛ ⎝244, 15 50 ⎞ ⎠ 5 As the sample size increases, there will be less variability in the mean, so the interval size decreases. ¯ 7 X is the time in minutes it takes to complete the U.S. Census short form. X people to complete the U.S. Census short form. is the mean time it took a sample of 200 9 CI: (7.9441, 8.4559) Figure 8.12 EBM = 0.26 11 The level of confidence would decrease, because decreasing n makes the confidence interval wider, so at the same error bound, the confidence level decreases. 13 a. x¯ = 2.2 b. σ =
0.2 c. n = 20 ¯ 15 X is the mean weight of a sample of 20 heads of lettuce. 17 EBM = 0.07 CI: (2.1264, 2.2736) 512 Chapter 8 \| Confidence Intervals Figure 8.13 19 The interval is greater, because the level of confidence increased. If the only change made in the analysis is a change in confidence level, then all we are doing is changing how much area is being calculated for the normal distribution. Therefore, a larger confidence level results in larger areas and larger intervals. 21 The confidence level would increase. 23 30.4 25 σ 27 μ 29 normal 31 0.025 33 (24.52,36.28) 35 We are 95 percent confident that the true mean age for winter Foothill College students is between 24.52 and 36.28. 37 The error bound for the mean would decrease, because as the CL decreases, you need less area under the normal curve (which translates into a smaller interval) to capture the true population mean. ¯ 39 X is the number of hours a patient waits in the emergency room before being called back to be examined. X mean wait time of 70 patients in the emergency room. is the 41 CI: (1.3808, 1.6192) Figure 8.14 EBM = 0.12 43 a. b. x¯ = 151 s x = 32 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 513 c. n = 108 d. n – 1 = 107 ¯ 45 X is the mean number of hours spent watching television per month from a sample of 108 Americans. 47 CI: (142.92, 159.08) Figure 8.15 EBM = 8.08 49 a. 3.26 b. 1.02 c. 39 51 μ 53 t38 55 0.025 57 (2.93, 3.59) 59 We are 95 percent confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors. 60 The error bound would become EBM = 0.245. This error bound decreases, because as sample sizes increase, variability decreases, and we need less interval length to capture the true mean. 63 It would decrease, because the z-score would decrease, which would reduce the numerator and lower the number. 65 X is the number of successes where the woman makes the majority of the purchasing decisions for the household. P′ is the percentage of households sampled where the woman makes the majority of the purchasing decisions for the household. 67 CI: (0.5321, 0.6679) 514 Chapter 8 \| Confidence Intervals Figure 8.16 EBM: 0.0679 69 X is the number of successes where an executive prefers a truck. P′ is the percentage of executives sampled who prefer a truck. 71 CI: (0.19432, 0.33068) Figure 8.17 EBM: 0.0707 73 The sampling error means that the true mean can be 2 percent above or below the sample mean. 75 P′ is the proportion of voters sampled who said the economy is the most important issue in the upcoming election. 77 CI: (0.62735, 0.67265); EBM: 0.02265 79 the number of girls, ages 8 to 12, in the 5 p.m. Monday night beginning ice-skating class 81 a. x = 64 b. n = 80 c. p′ = 0.8 83 p 85 P′ ~ N ⎛ ⎝0.8, (0.8)(0.2) 80 ⎞ ⎠ CI = (0.72171, 0.87829). 87 0.04 89 (0.72; 0.88) 91 With 92 percent confidence, we estimate the proportion of girls, ages 8 to 12, in a beginning ice-skating class at the Ice Chalet to be between 72 percent and 88 percent. 93 The error bound would increase. Assuming all other variables are kept constant, as the confidence level increases, the area under the curve corresponding to the confidence level becomes larger, which creates a wider interval and thus a larger This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 515 error. 95 a. i. 71 ii. 3 iii. 48 b. X is the height of a Swedish male, and is the mean height from a sample of 48 Swedish males. c. Normal. We know the standard deviation for the population, and the sample size is greater than 30. d. i. CI: (70.151, 71.49) ii. Figure 8.18 iii. EBM = 0.849 e. The confidence interval will decrease in size, because the sample size increased. Recall, when all factors remain unchanged, an increase in sample size decreases variability. Thus, we do not need as large an interval to capture the true population mean. 97 a. i. ii. x¯ = 23.6 σ = 7 iii. n = 100 ¯ b. X is the time needed to complete an individual tax form. X is the mean time to complete tax forms from a sample of 100 customers. c. N ⎛ ⎝23.6, ⎞ ⎠ 7 100 because we know sigma. d. ii. i. (22.228, 24.972) 516 Chapter 8 \| Confidence Intervals Figure 8.19 iii. EBM = 1.372 e. It will need to change the sample size. The firm needs to determine what the confidence level should be and then apply the error bound formula to determine the necessary sample size. f. The confidence level would increase as a result of a larger interval. Smaller sample sizes result in more variability. To capture the true population mean, we need to have a larger interval. g. According to the error bound formula, the firm needs to survey 206 people. Because we increase the confidence level, we need to increase either our error bound or the sample size. 99 a. i. 7.9 ii. 2.5 iii. 20 ¯ b. X is the number of letters a single camper will send home. X is the mean number of letters sent home from a sample of 20 campers. c. N 7.9 ⎛ ⎝ ⎞ ⎠ 2.5 20 d. i. CI: (6.98, 8.82) ii. Figure 8.20 iii. EBM: 0.92 e. The error bound and confidence interval will decrease. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 517 101 a. x¯ = $568,873 b. CL = 0.95, α = 1 – 0.95 = 0.05, z α 2 = 1.96 909200 EBM = z0.025 = 1.96 = $281,764 σ n 40 c. x¯ − EBM = 568,873 − 281,764 = 287,109 x¯ + EBM = 568,873 + 281,764 = 850,637 Alternate solution: 1. Press STAT and arrow over to TESTS. 2. Arrow down to 7:ZInterval. 3. Press ENTER. 4. Arrow to Stats and press ENTER. 5. Arrow down and enter the following values: σ : 909,200 x¯ : 568,873 n: 40 CL: 0.95 6. Arrow down to Calculate and press ENTER. 7. The confidence interval is ($287,114, $850,632). 8. Notice the small difference between the two solutions—these differences are simply due to rounding error in the hand calculations. d. We estimate with 95 percent confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and $850,637. 103 Use the formula for EBM, solved for n: n = z2 σ 2 EBM 2 (This is the value of z for which the area under From the statement of the problem, you know that σ = 2.5, and you need EBM = 1. z = z0.035 = ≈ 20.52. You need to measure at least 21 male students to achieve your goal. the density curve to the right of z is 0.035.) 1.812. n = z2 σ 2 EBM 2 = 1.8122 2.52 12 105 a. i. 8,629 ii. 6,944 iii. 35 iv. 34 t34 i. CI: (6244, 11,014) b. c. 518 Chapter 8 \| Confidence Intervals ii. Figure 8.21 iii. EB = 2385 d. It will become smaller. 107 a. i. ii. x¯ = 2.51 s x = 0.318 iii. n = 9 iv. n - 1 = 8 b. The effective length of time for a tranquilizer c. The mean effective length of time of tranquilizers from a sample of nine patients d. We need to use a Student’s t-distribution, because we do not know the population standard deviation. e. i. CI: (2.27, 2.76) ii. Check student's solution. iii. EBM: 0.25 f. If we were to sample many groups of nine patients, 95 percent of the samples would contain the true population mean length of time. 109 x¯ = $251, 854.23; s = $521, 130.41. Note that we are not given the population standard deviation, only the standard deviation of the sample. There are 30 measures in the sample, so n = 30, and df = 30 - 1 = 29. CL = 0.96, so α = x¯ ⎞ ⎠ = 2.150 ⎞ ⎠ ~ $204, 561.66. 1 - CL = 1 - 0.96 = 0.04. α 2 = 2.150. EBM = t α 2 521, 130.41 30 = 0.02t α 2 = t0.02 s n ⎛ ⎝ ⎛ ⎝ - EBM = $251,854.23 - $204,561.66 = $47,292.57. x¯ + EBM = $251,854.23 + $204,561.66 = $456,415.89. We estimate with 96 percent confidence that the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle lies between $47,292.57 and $456,415.89. Alternate Solution STATTESTS8:TIntervalENTERENTERFreqC-LevelCalculateEnter The difference between solutions arises from rounding differences. 111a. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 519 i. ii. x¯ = s x = iii. n = iv. n - 1 = ¯ b. X is the number of unoccupied seats on a single flight. X is the mean number of unoccupied seats from a sample of 225 flights. c. We will use a Student’s t-distribution, because we do not know the population standard deviation. d. i. CI: (11.12 , 12.08) ii. Check student's solution. iii. EBM: 0.48 113 a. i. CI: (7.64, 9.36) ii. Figure 8.22 iii. EBM: 0.86 b. The sample should have been increased. c. Answers will vary. d. Answers will vary. e. Answers will vary. 115 b 117 a. 1,068 b. The sample size would need to be increased, because the critical value increases as the confidence level increases. 119 a. X = the number of people who believe that the president is doing an acceptable job; P′ = the proportion of people in a sample who believe that the president is doing an acceptable job. b. N ⎛ ⎝0.61, (0.61)(0.39) 1200 ⎞ ⎠ c. i. CI: (0.59, 0.63) ii. Check student’s solution. iii. EBM: 0.02 520 121 a. i. ii. (0.72, 0.82) (0.65, 0.76) iii. (0.60, 0.72) Chapter 8 \| Confidence Intervals b. Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap. c. We can say that there does not appear to be a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a Latino person into their families. d. We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a black person into their families. 123 a. X = the number of adult Americans who believe that crime is the main problem; P′ = the proportion of adult Americans who believe that crime is the main problem. b. Because we are estimating a proportion, ⎞ ⎠ (0
.2)(0.8) 1000 ⎛ ⎝0.2, N . that P′ = 0.2 and n = 1,000, the distribution we should use is c. i. CI: (0.18, 0.22) ii. Check student’s solution. iii. EBM: 0.02 d. One way to lower the sampling error is to increase the sample size. e. The stated ± 3 percent represents the maximum error bound. This means that those doing the study are reporting a maximum error of 3 percent. Thus, they estimate the percentage of adult Americans who the percentage of adult Americans who that crime is the main problem to be between 18 percent and 22 percent. 125 c 127 d 129 a 131 a. p′ = (0.55 + 0.49) 2 = 0.52; EBP = 0.55 – 0.52 = 0.03 b. No, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this. c. CL = 0.75, so α = 1 – 0.75 = 0.25 and α 2 = 0.125. z α 2 = 1.150 . (The area to the right of this z is 0.125, so the area to the left is 1 – 0.125 = 0.875.) EBP = (1.150) 0.52(0.48) 1, 000 ≈ 0.018 (p′ - EBP, p′ + EBP) = (0.52 – 0.018, 0.52 + 0.018) = (0.502, 0.538) Alternate Solution STAT TESTS A: 1-PropZinterval with x = (0.52)(1,000), n = 1,000, CL = 0.75. Answer is (0.502, 0.538). This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 521 d. Yes, this interval does not fall below 0.50, so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75 percent confidence. 133 CL = 0.95; α = 1 – 0.95 = 0.05; α 2 = 0.025; z α 2 = 1.96. Use p′ = q′ = 0.5. n = 2 p′ q′ z α 2 EBP2 = 1.962(0.5)(0.5) 0.052 = 384.16. You need to interview at least 385 students to estimate the proportion to within 5 percent at 95 percent confidence. 522 Chapter 8 \| Confidence Intervals This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 523 9 \| HYPOTHESIS TESTING WITH ONE SAMPLE Figure 9.1 You can use a hypothesis test to decide if a dog breeder’s claim that every Dalmatian has 35 spots is statistically sound. (credit: Robert Neff) Introduction By the end of this chapter, the student should be able to do the following: Chapter Objectives • Differentiate between Type I and Type II errors • Describe hypothesis testing in general and in practice • Conduct and interpret hypothesis tests for a single population mean, population standard deviation known • Conduct and interpret hypothesis tests for a single population mean, population standard deviation unknown • Conduct and interpret hypothesis tests for a single population proportion One job of a statistician is to make statistical inferences about populations based on samples taken from the population. Confidence intervals are one way to estimate a population parameter. Another way to make a statistical inference is to 524 Chapter 9 \| Hypothesis Testing with One Sample make a decision about a parameter. For instance, a car dealer advertises that its new small truck gets 35 miles per gallon, on average. A tutoring service claims that its method of tutoring helps 90 percent of its students get an A or a B. A company says that women managers in their company earn an average of $60,000 per year. A statistician will make a decision about these claims. This process is called hypothesis testing. A hypothesis test involves collecting data from a sample and evaluating the data. Then, the statistician makes a decision as to whether or not there is sufficient evidence, based upon analyses of the data, to reject the null hypothesis. In this chapter, you will conduct hypothesis tests on single means and single proportions. You will also learn about the errors associated with these tests. Hypothesis testing consists of two contradictory hypotheses or statements, a decision based on the data, and a conclusion. To perform a hypothesis test, a statistician will do the following: 1. Set up two contradictory hypotheses. 2. Collect sample data. In homework problems, the data or summary statistics will be given to you. 3. Determine the correct distribution to perform the hypothesis test. 4. Analyze sample data by performing the calculations that ultimately will allow you to reject or decline to reject the null hypothesis. 5. Make a decision and write a meaningful conclusion. NOTE To do the hypothesis test homework problems for this chapter and later chapters, make copies of the appropriate special solution sheets. See Appendix E. 9.1 \| Null and Alternative Hypotheses The actual test begins by considering two hypotheses. They are called the null hypothesis and the alternative hypothesis. These hypotheses contain opposing viewpoints. H0, the —null hypothesis: a statement of no difference between sample means or proportions or no difference between a sample mean or proportion and a population mean or proportion. In other words, the difference equals 0. Ha—, the alternative hypothesis: a claim about the population that is contradictory to H0 and what we conclude when we reject H0. Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data. After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are reject H0 if the sample information favors the alternative hypothesis or do not reject H0 or decline to reject H0 if the sample information is insufficient to reject the null hypothesis. Mathematical Symbols Used in H0 and Ha: H0 equal (=) Ha not equal (≠) or greater than (>) or less than (<) greater than or equal to (≥) less than (<) less than or equal to (≤) more than (>) Table 9.1 NOTE H0 always has a symbol with an equal in it. Ha never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers use = in the null hypothesis, even with This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 525 > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis. Example 9.1 H0: No more than 30 percent of the registered voters in Santa Clara County voted in the primary election. p ≤ 30 Ha: More than 30 percent of the registered voters in Santa Clara County voted in the primary election. p > 30 9.1 A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25 percent. State the null and alternative hypotheses. Example 9.2 We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are the following: H0: μ = 2.0 Ha: μ ≠ 2.0 9.2 We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. a. H0: μ __ 66 b. Ha: μ __ 66 Example 9.3 We want to test if college students take fewer than five years to graduate from college, on the average. The null and alternative hypotheses are the following: H0: μ ≥ 5 Ha: μ < 5 9.3 We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. a. H0: μ __ 45 b. Ha: μ __ 45 526 Chapter 9 \| Hypothesis Testing with One Sample Example 9.4 An article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third of the students pass. The same article stated that 6.6 percent of U.S. students take advanced placement exams and 4.4 percent pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6 percent. State the null and alternative hypotheses. H0: p ≤ 0.066 Ha: p > 0.066 9.4 On a state driver’s test, about 40 percent pass the test on the first try. We want to test if more than 40 percent pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. a. H0: p __ 0.40 b. Ha: p __ 0.40 Bring to class a newspaper, some news magazines, and some internet articles. In groups, find articles from which your group can write null and alternative hypotheses. Discuss your hypotheses with the rest of the class. 9.2 \| Outcomes and the Type I and Type II Errors When you perform a hypothesis test, there are four possible outcomes depending on the actual truth, or falseness, of the null hypothesis H0 and the decision to reject or not. The outcomes are summarized in the following table: ACTION H0 IS ACTUALLY ... True False Do not reject H0 Correct outcome Type II error Reject H0 Table 9.2 Type I error Correct outcome The four possible outcomes in the table are as follows: 1. The decision is not to reject H0 when H0 is true (correct decision). 2. The decision is to reject H0 when, in fact, H0 is true (incorrect decision known as a Type I error). 3. The decision is not to reject H0 when, in fact, H0 is false (incorrect decision known as a Type II error). 4. The decision is to reject H0 when H0 is false (correct decision whose probability is called the Power of the Test). Each of the errors occurs with a particular probability. The Greek letters α and β represent the probabilities. α = probability of a Type I error = P(Type I error) = probability of rejecting the null hypothesis when the null hypothesis is true. β = probability of a Type II error = P(Type II error) = probability of not rejecting the null hypothesis when the null hypothesis is false. α and β should be as small as possible because they are probabilities of errors. They are rarely zero. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sa
mple 527 The Power of the Test is 1 – β. Ideally, we want a high power that is as close to one as possible. Increasing the sample size can increase the Power of the Test. The following are examples of Type I and Type II errors. Example 9.5 Suppose the null hypothesis, H0, is: Frank's rock climbing equipment is safe. Type I error: Frank does not go rock climbing because he considers that the equipment is not safe, when in fact, the equipment is really safe. Frank is making the mistake of rejecting the null hypothesis, when the equipment is actually safe! Type II error: Frank goes climbing, thinking that his equipment is safe, but this is a mistake, and he painfully realizes that his equipment is not as safe as it should have been. Frank assumed that the null hypothesis was true, when it was not. α = probability that Frank thinks his rock climbing equipment may not be safe when, in fact, it really is safe. β = probability that Frank thinks his rock climbing equipment may be safe when, in fact, it is not safe. Notice that, in this case, the error with the greater consequence is the Type II error. (If Frank thinks his rock climbing equipment is safe, he will go ahead and use it.) 9.5 Suppose the null hypothesis, H0, is: the blood cultures contain no traces of pathogen X. State the Type I and Type II errors. Example 9.6 Suppose the null hypothesis, H0, is: a tomato plant is alive when a class visits the school garden. Type I error: The null hypothesis claims that the tomato plant is alive, and it is true, but the students make the mistake of thinking that the plant is already dead. Type II error: The tomato plant is already dead (the null hypothesis is false), but the students do not notice it, and believe that the tomato plant is alive. α = probability that the class thinks the tomato plant is dead when, in fact, it is alive = P(Type I error). β = probability that the class thinks the tomato plant is alive when, in fact, it is dead = P(Type II error). The error with the greater consequence is the Type I error. (If the class thinks the plant is dead, they will not water it.) 9.6 Suppose the null hypothesis, H0, is: a patient is not sick. Which type of error has the greater consequence, Type I or Type II? Example 9.7 It’s a Boy Genetic Labs, a genetics company, claims to be able to increase the likelihood that a pregnancy will result in a boy being born. Statisticians want to test the claim. Suppose that the null hypothesis, H0, is: It’s a Boy 528 Chapter 9 \| Hypothesis Testing with One Sample Genetic Labs has no effect on gender outcome. Type I error: This error results when a true null hypothesis is rejected. In the context of this scenario, we would state that we believe that It’s a Boy Genetic Labs influences the gender outcome, when in fact it has no effect. The probability of this error occurring is denoted by the Greek letter alpha, α. Type II error: This error results when we fail to reject a false null hypothesis. In context, we would state that It’s a Boy Genetic Labs does not influence the gender outcome of a pregnancy when, in fact, it does. The probability of this error occurring is denoted by the Greek letter beta, β. The error with the greater consequence would be the Type I error since couples would use the It’s a Boy Genetic Labs product in hopes of increasing the chances of having a boy. 9.7 Red tide is a bloom of poison-producing algae—a few different species of a class of plankton called dinoflagellates. When the weather and water conditions cause these blooms, shellfish such as clams living in the area develop dangerous levels of a paralysis-inducing toxin. In Massachusetts, the Division of Marine Fisheries montors levels of the toxin in shellfish by regular sampling of shellfish along the coastline. If the mean level of toxin in clams exceeds 800 μg (micrograms) of toxin per kilogram of clam meat in any area, clam harvesting is banned there until the bloom is over and levels of toxin in clams subside. Describe both a Type I and a Type II error in this context, and state which error has the greater consequence. Example 9.8 A certain experimental drug claims a cure rate of at least 75 percent for males with a disease. Describe both the Type I and Type II errors in context. Which error is the more serious? Type I: A patient believes the cure rate for the drug is less than 75 percent when it actually is at least 75 percent. Type II: A patient believes the experimental drug has at least a 75 percent cure rate when it has a cure rate that is less than 75 percent. In this scenario, the Type II error contains the more severe consequence. If a patient believes the drug works at least 75 percent of the time, this most likely will influence the patient’s (and doctor’s) choice about whether to use the drug as a treatment option. 9.8 Determine both Type I and Type II errors for the following scenario: Assume a null hypothesis, H0, that states the percentage of adults with jobs is at least 88 percent. Identify the Type I and Type II errors from these four possible choices. a. Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88 percent when that percentage is actually less than 88 percent b. Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88 percent when the percentage is actually at least 88 percent c. Reject the null hypothesis that the percentage of adults who have jobs is at least 88 percent when the percentage is actually at least 88 percent d. Reject the null hypothesis that the percentage of adults who have jobs is at least 88 percent when that percentage is actually less than 88 percent This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 529 9.3 \| Distribution Needed for Hypothesis Testing Earlier in the course, we discussed sampling distributions. Particular distributions are associated with hypothesis testing. Perform tests of a population mean using a normal distribution or a Student's t-distribution. (Remember, use a Student's t-distribution when the population standard deviation is unknown and the distribution of the sample mean is approximately normal.) We perform tests of a population proportion using a normal distribution (usually n is large). Assumptions When you perform a hypothesis test of a single population mean μ using a Student's t-distribution (often called a t-test), there are fundamental assumptions that need to be met in order for the test to work properly. Your data should be a simple random sample that comes from a population that is approximately normally distributed. You use the sample standard deviation to approximate the population standard deviation. Note that if the sample size is sufficiently large, a t-test will work even if the population is not approximately normally distributed. When you perform a hypothesis test of a single population mean μ using a normal distribution (often called a z-test), you take a simple random sample from the population. The population you are testing is normally distributed or your sample size is sufficiently large. You know the value of the population standard deviation which, in reality, is rarely known. When you perform a hypothesis test of a single population proportion p, you take a simple random sample from the population. You must meet the conditions for a binomial distribution, which are the following: there are a certain number n of independent trials, the outcomes of any trial are success or failure, and each trial has the same probability of a success p. The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five (np > 5 and nq > 5). Then the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = p and σ = pq n . Remember that q = 1 – p. 9.4 \| Rare Events, the Sample, and the Decision and Conclusion Establishing the type of distribution, sample size, and known or unknown standard deviation can help you figure out how to go about a hypothesis test. However, there are several other factors you should consider when working out a hypothesis test. Rare Events The thinking process in hypothesis testing can be summarized as follows: You want to test whether or not a particular property of the population is true. You make an assumption about the true population mean for numerical data or the true population proportion for categorical data. This assumption is the null hypothesis. Then you gather sample data that is representative of the population. From this sample data you compute the sample mean (or the sample proportion). If the value that you observe is very unlikely to occur (a rare event) if the null hypothesis is true, then you wonder why this is happening. A plausible explanation is that the null hypothesis is false. For example, Didi and Ali are at a birthday party of a very wealthy friend. They hurry to be first in line to grab a prize from a tall basket that they cannot see inside because they will be blindfolded. There are 200 plastic bubbles in the basket, and Didi and Ali have been told that there is only one with a $100 bill. Didi is the first person to reach into the basket and pull out a bubble. Her bubble contains a $100 bill. The probability of this happening is 1 200 = 0.005. Because this is so unlikely, Ali is hoping that what the two of them were told is wrong and there are more $100 bills in the basket. A rare event has occurred (Didi getting the $100 bill) so Ali doubts the assumption about only one $100 bill being in the basket. Using the Sample to Test the Null Hypothesis After you collect data and obtain the test statistic (the sample mean, sample proportion, or other test statistic), you can determine the probability of obtaining that test statistic when the null hypothesis is true. This pro
bability is called the p-value. When the p-value is very small, it means that the observed test statistic is very unlikely to happen if the null hypothesis is true. This gives significant evidence to suggest that the null hypothesis is false, and to reject it in favor of the alternative hypothesis. In practice, to reject the null hypothesis we want the p-value to be smaller than 0.05 (5 percent) or sometimes even smaller than 0.01 (1 percent). 530 Chapter 9 \| Hypothesis Testing with One Sample Example 9.9 Suppose a baker claims that his bread height is more than 15 cm, on average. Several of his customers do not believe him. To persuade his customers that he is right, the baker decides to do a hypothesis test. He bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm. The baker knows from baking hundreds of loaves of bread that the standard deviation for the height is 0.5 cm and the distribution of heights is normal. The null hypothesis could be H0: μ ≤ 15. The alternate hypothesis is Ha: μ > 15. The words is more than translates as a ">" so "μ > 15" goes into the alternate hypothesis. The null hypothesis must contradict the alternate hypothesis. Since σ is known (σ = 0.5 cm), the distribution for the population is known to be normal with mean μ = 15 and standard deviation σ n = 0.16 . = 0.5 10 Suppose the null hypothesis is true (which is that the mean height of the loaves is no more than 15 cm). Then is the mean height (17 cm) calculated from the sample unexpectedly large? The hypothesis test works by asking the question how unlikely the sample mean would be if the null hypothesis were true. The graph shows how far out the sample mean is on the normal curve. The p-value is the probability that, if we were to take other samples, any other sample mean would fall at least as far out as 17 cm. The p-value, then, is the probability that a sample mean is the same or greater than 17 cm when the population mean is, in fact, 15 cm. We can calculate this probability using the normal distribution for means. In Figure 9.2, the p-value is the area under the normal curve to the right of 17. Using a normal distribution table or a calculator, we can compute that this probability is practically zero. Figure 9.2 p-value = P( x¯ > 17), which is approximately zero. Because the p-value is almost 0, we conclude that obtaining a sample height of 17 cm or higher from 10 loaves of bread is very unlikely if the true mean height is 15 cm. We reject the null hypothesis and conclude that there is sufficient evidence to claim that the true population mean height of the baker’s loaves of bread is higher than 15 cm. 9.9 A normal distribution has a standard deviation of 1. We want to verify a claim that the mean is greater than 12. A sample of 36 is taken with a sample mean of 12.5. H0: μ ≤ 12 Ha: μ > 12 The p-value is 0.0013. Draw a graph that shows the p-value. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 531 Decision and Conclusion A systematic way to make a decision of whether to reject or not reject the null hypothesis is to compare the p-value and a preset or preconceived α, also called the level of significance of the test. A preset α is the probability of a Type I error (rejecting the null hypothesis when the null hypothesis is true). It may or may not be given to you at the beginning of the problem. When you make a decision to reject or not reject H0, do as follows: • • If p-value < α , reject H0. The results of the sample data are significant. There is sufficient evidence to conclude that H0 is an incorrect belief and that the alternative hypothesis, Ha, may be correct. If p-value ≥ α , do not reject H0. The results of the sample data are not significant.There is not sufficient evidence to conclude that the alternative hypothesis, Ha, may be correct. • When you do not reject H0, it does not mean that you should believe that H0 is true. It simply means that the sample data have failed to provide sufficient evidence to cast serious doubt about the truthfulness of H0. Conclusion: After you make your decision, write a thoughtful conclusion about the hypotheses in terms of the given problem. Example 9.10 When using the p-value to evaluate a hypothesis test, you might find it useful to use the following mnemonic device: If the p-value is low, the null must go. If the p-value is high, the null must fly. This memory aid relates a p-value less than the established alpha (the p is low) as rejecting the null hypothesis and, likewise, relates a p-value higher than the established alpha (the p is high) as not rejecting the null hypothesis. Fill in the blanks. Reject the null hypothesis when ______________________________________. The results of the sample data _____________________________________. Do not reject the null hypothesis when __________________________________________. The results of the sample data ____________________________________________. Solution 9.10 Reject the null hypothesis when the p-value is less than the established alpha value. The results of the sample data support the alternative hypothesis. Do not reject the null hypothesis when the p-value is greater or equal to the established alpha value. The results of the sample data do not support the alternative hypothesis. 9.10 It’s a Boy Genetics Labs, a genetics company, claims their procedures improve the chances of a boy being born. The results for a test of a single population proportion are as follows: H0: p = 0.50, Ha: p > 0.50 α = 0.01 p-value = 0.025 Interpret the results and state a conclusion in simple, nontechnical terms. 532 Chapter 9 \| Hypothesis Testing with One Sample 9.5 \| Additional Information and Full Hypothesis Test Examples • In a hypothesis test problem, you may see words such as "the level of significance is 1 percent". The "1 percent" is the preconceived or preset α. • The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data. • If no level of significance is given, a common standard to use is α = 0.05. • When you calculate the p-value and draw the picture, the p-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed. • The alternative hypothesis, Ha , tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test. • Ha never has a symbol that contains an equal sign. • Thinking about the meaning of the p-value: A data analyst should have more confidence that he made the correct decision to reject the null hypothesis with a smaller p-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p-value such as 0.4, as opposed to a p-value of 0.056 (alpha = 0.05 is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules. The following examples illustrate a left-, right-, and two-tailed test. Example 9.11 H0: μ = 5 Test of a single population mean. Ha tells you the test is left-tailed. The picture of the p-value is as follows: Ha: μ < 5 Figure 9.3 9.11 H0: μ = 10 Assume the p-value is 0.0935. What type of test is this? Draw the picture of the p-value. Ha: μ < 10 Example 9.12 H0: p ≤ 0.2 Ha: p > 0.2 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 533 This is a test of a single population proportion. Ha tells you the test is right-tailed. The picture of the p-value is as follows: Figure 9.4 9.12 H0: μ ≤ 1 Assume the p-value is 0.1243. What type of test is this? Draw the picture of the p-value. Ha: μ > 1 Example 9.13 Ha: p ≠ 50 H0: p = 50 This is a test of a single population mean. Ha tells you the test is two-tailed. The picture of the p-value is as follows. Figure 9.5 9.13 H0: p = 0.5 Assume the p-value is 0.2564. What type of test is this? Draw the picture of the p-value. Ha: p ≠ 0.5 534 Chapter 9 \| Hypothesis Testing with One Sample Full Hypothesis Test Examples Example 9.14 Jeffrey, as an eight-year-old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal. Solution 9.14 Set up the hypothesis test: Since the problem is about a mean, this is a test of a single population mean. H0: μ = 16.43 For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is left-tailed. Ha: μ < 16.43 Determine the distribution needed: ¯ Random variable: X = the mean time to swim the 25-yard freestyle. ¯ Distribution for the test: X is normal (population standard deviation is known: σ = 0.8) with mean μ = 16.43 and standard error of 0.8 15 ; μ = 16.43 comes from H0 and not the data. σ = 0.8, and n = 15. Using a table or a calculator, we can calculate the p-value as the area to the left of 16 under the normal curve: p-value = P( x¯ < 16) = 0.0187 where the sample mean in the problem is given as 16. p-value = 0.0187. The p-value is the area to the left of the sample mean given as 16. Graph: Figure 9.6 μ = 16.43 comes from H0. Our assumption is μ = 16.43. Interpretation of the p-value: If H0 is true, there is a 0.0187 probability (1.87 percent), that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87 percent chance is smal
l, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event. Compare α and the p-value: α = 0.05 p-value = 0.0187 α > p-value This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 535 Make a decision: Since α > p-value, reject H0. An alternative approach is to find the z-test corresponding to the sample mean x¯ = 16. This is z-test = x¯ − μ X σ X n = 16 − 16.43 0.8 15 = - 2.081729 . The critical z-value = –1.645 for this test has probability 0.05 to its left tail, according to the Normal Table (see Appendices). Because the z-test is to the left of the critical z-value, we reject the null hypothesis. This means that you reject μ = 16.43. In other words, you do not think Jeffrey swims the 25-yard freestyle in 16.43 seconds but instead that he swims faster with the new goggles. Conclusion: At the 5 percent significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds. The p-value can easily be calculated. Press STAT and arrow over to TESTS. Press 1:z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 16.43 for μ0 (null hypothesis), .8 for σ, 16 for the sample mean, and 15 for n. Arrow down to μ : (alternate hypothesis) and arrow over to < μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p = 0.0187) but it also calculates the test statistic (z-score) for the sample mean. μ < 16.43 is the alternative hypothesis. Do this set of instructions again except arrow to Draw(instead of Calculate). Press ENTER. A shaded graph appears with z = -2.08 (test statistic) and p = 0.0187 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. When the calculator does a z-Test, the z-Test function finds the p-value by doing a normal probability calculation: P( x¯ < 16) = 2nd DISTR normcdf ⎛ ⎝ − 10 ^ 99, 16, 16.43, 0.8 / 15⎞ ⎠. The Type I and Type II errors for this problem are as follows: The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.) The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard freestyle, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.) HISTORICAL NOTE (EXAMPLE 9.11) The traditional way to compare the two probabilities, α and the p-value, is to compare the critical value (z-score from α) to the test statistic (z-score from data). The calculated test statistic for the p-value is –2.08. (From the central limit theorem, the test statistic formula is z = x¯ − μ X σ X n ) ( . For this problem, x¯ = 16, μX = 16.43 from the null hypothesis, σX = 0.8, and n = 15.) You can find the critical value for α = 0.05 in the normal table (see Appendix H: Tables). The z-score for an area to the left equal to 0.05 is midway between –1.65 and –1.64 (0.05 is midway between 0.0505 and 0.0495). The z-score is –1.645. Since –1.645 > –2.08 (which demonstrates that α > p-value), reject H0. Traditionally, the decision to reject or not reject was done in this way. Today, comparing the two probabilities α and the p-value is 536 Chapter 9 \| Hypothesis Testing with One Sample very common. For this problem, the p-value, 0.0187, is considerably smaller than α, 0.05. You can be confident about your decision to reject. The graph shows α, the p-value, and the test statistic and the critical value. Figure 9.7 9.14 The mean throwing distance of a football by Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal. First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion. Press STAT and arrow over to TESTS. Press 1: z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 40 for μ0 (null hypothesis), 2 for σ, 45 for the sample mean, and 20 for n. Arrow down to μ: (alternative hypothesis) and set it either as <, ≠, or >. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value but it also calculates the test statistic (z-score) for the sample mean. Select <, ≠, or > for the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with test statistic and p-value. Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. Example 9.15 A college football coach records the mean weight that his players can bench press as 275 pounds, with a standard deviation of 55 pounds. Three of his players thought that the mean weight was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) 205(3); 215(3); 225(1); 241(2); 252(2); 265(2); 275(2); 313(2); 316(5); 338(2); 341(1); 345(2); 368(2); 385(1). Conduct a hypothesis test using a 2.5 percent level of significance to determine if the bench press mean is more than 275 pounds. Solution 9.15 Set up the hypothesis test: This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 537 Since the problem is about a mean weight, this is a test of a single population mean. H0: μ = 275 Calculating the distribution needed: Ha: μ > 275 This is a right-tailed test. Random variable: X ¯ = the mean weight, in pounds, lifted by the football players. Distribution for the test: It is normal because σ is known. ¯ X ~ N ⎛ ⎝275, 55 30 ⎞ ⎠ x¯ = 286.2 pounds (from the data). σ = 55 pounds. Always use σ if you know it. We assume μ = 275 pounds unless our data shows us otherwise. First, we compute the sample mean: x− = 205 + 205 + 205 + 215 + ⋯ + 385 30 = 286.2. Next, we compute the z-test: z-test = 286.2 − 275 55 30 = 1.115362 Finally, the p-value is the probability to the right tail of the z-test, which we can compute from the table of z-scores as 0.5 –- 0.36650 = 0.1335. () p-value = P( x¯ > 286.2) = 0.1323 Interpretation of the p-value: If H0 is true, then there is a 0.1331 probability, 13.23 percent, that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23 percent chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event. Figure 9.8 Compare α and the p-value: α = 0.025 p-value = 0.1323 Make a decision: Since α < p-value, do not reject H0. Conclusion: At the 2.5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds. The p-value can easily be calculated. 538 Chapter 9 \| Hypothesis Testing with One Sample Put the data and frequencies into lists. Press STAT and arrow over to TESTS. Press 1:z-Test. Arrow over to Data and press ENTER. Arrow down and enter 275 for μ0, 55 for σ, the name of the list where you put the data, and the name of the list where you put the frequencies. Arrow down to μ: and arrow over to > μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p = 0.1331, a little different from the previous calculation—in it we used the sample mean rounded to one decimal place instead of the data), but also the test statistic (z-score) for the sample mean, the sample mean, and the sample standard deviation. μ > 275 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with z = 1.112 (test statistic) and p = 0.1331 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. Example 9.16 Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples 10 statistics students and obtains the scores 65; 65; 70; 67; 66; 63; 63; 68; 72; 71. He performs a hypothesis test using a 5 percent level of significance. The data are assumed to be from a normal distribution. Solution 9.16 Set up the hypothesis test: A 5 percent level of significance means that α = 0.05. This is a test of a single population mean. H0: μ = 65 Since the instructor thinks the average score is higher, use a ">". The ">" means the test is right-tailed. Ha: μ > 65 Determine the distribution needed: Random variable: X ¯ = average score on the first statistics test. Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given. You are only given n = 10 sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a Student's t-distribution. Use t-distribution. Therefore, the distribution for the test is t with nine degrees of freedom. Calculate the p-value using the Student's t-distribution: First, we compute the sample mean as Next, we compute the t-test as x¯ = 65 + 65 + ⋯ + 71 10 = 67. t-test = x¯ − μ X s X n = 67 − 65 3.12 10 ≈ 1.98. The p
-value is the probability to the right tail of 1.98 in a t-distribution with nine degrees of freedom. p-value = P( x¯ > 67) = 0.0396 where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data. Interpretation of the p-value: If the null hypothesis is true, then there is a 0.0396 probability— (3.96 percent—) This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 539 that the sample mean is 65 or more. Figure 9.9 Compare α and the p-value: Since α = 0.05 and p-value = 0.0396, α > p-value. Make a decision: Since α > p-value, reject H0. Alternatively, according to a Student's t-distribution table (see Appendices), the critical t-value is 1.833. Since the t-test (1.98) is to the right of the critical t-value 1.833, we reject the null hypothesis. This decision means we reject μ = 65. In other words, we believe the average test score is more than 65. Conclusion: At a 5 percent level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks. The p-value can easily be calculated. Put the data into a list. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 65 for μ0, the name of the list where you put the data, and 1 for Freq:. Arrow down to μ: and arrow over to > μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p = 0.0396) but it also calculates the test statistic (t-score) for the sample mean, the sample mean, and the sample standard deviation. μ > 65 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with t = 1.9781 (test statistic) and p = 0.0396 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. 9.16 It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price are recorded for 10 weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5 percent level of significance. State the null and alternative hypotheses, find the p-value, state your conclusion, and identify the Type I and Type II errors. 540 Chapter 9 \| Hypothesis Testing with One Sample Example 9.17 Joon believes that 50 percent of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50 percent. Joon samples 100 firsttime brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1 percent level of significance. Solution 9.17 Set up the hypothesis test: The 1 percent level of significance means that α = 0.01. This is a test of a single population proportion. H0: p = 0.50 The words is the same or different from tell you this is a two-tailed test. Ha: p ≠ 0.50 Calculate the distribution needed: Random variable: P′ = the percentage of first-time brides who are younger than their grooms. Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for P′, the estimated proportion. P′ follows a normal distribution with mean value μ = p, and standard error σ = p ⋅ q n . In our example, p = q = 0.5, and n = 100, where p = 0.50, q = 1 – p = 0.50, and n = 100. Calculate the p-value using the normal distribution for proportions: First, we compute the sample proportion as p^ = 53 100 = 0.53. Next, the z-test is given by z-test = p^ − p p ⋅ q n = 0.53 − 0.50 0.50×0.50 100 = 0.6. is positive, we compute the area to the right Since the z-test tail of 0.6 in a normal distribution, P(Z > 0.6) = 0.2742531. Finally, because this is a two-sided test of significance, we multiply this probability times two to account for the left tail, and obtain () p-value = 2×0.2742531 = 0.5485062 where x = 53, p′ = x n = 53 100 = 0.53. Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability, (54.85 percent) that the sample (estimated) proportion p′ is 0.53 or more OR 0.47 or less (see the graph in Figure 9.9). Figure 9.10 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 541 μ = p = 0.50 comes from H0, the null hypothesis. p′ = 0.53. Since the curve is symmetrical and the test is two-tailed, the p′ for the left tail is equal to 0.50 – 0.03 = 0.47 where μ = p = 0.50. (0.03 is the difference between 0.53 and 0.50.) Compare α and the p-value: Since α = 0.01 and p-value = 0.5485, α < p-value. Make a decision: Since α < p-value, you cannot reject H0. Conclusion: At the 1 percent level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50 percent. The p-value can easily be calculated. Press STAT and arrow over to TESTS. Press 5:1-PropZTest. Enter .5 for p0, 53 for x and 100 for n. Arrow down to Prop and arrow to not equals p0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator calculates the p-value (p = 0.5485) and the test statistic (z-score). Prop not equals .5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with z = 0.6 (test statistic) and p = 0.5485 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. The Type I and Type II errors are as follows: The Type I error is to conclude that the proportion of first-time brides who are younger than their grooms is different from 50 percent when, in fact, the proportion is actually 50 percent. Reject the null hypothesis when the null hypothesis is true. The Type II error is there is not enough evidence to conclude that the proportion of first-time brides who are younger than their grooms differs from 50 percent when, in fact, the proportion does differ from 50 percent. Do not reject the null hypothesis when the null hypothesis is false. 9.17 A teacher believes that 85 percent of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85 percent. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1 percent level of significance. First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion. Example 9.18 Suppose a consumer group suspects that the proportion of households that have three cell phones is 30 percent. A cell phone company has reason to believe that the proportion is not 30 percent. Before the cell phone company starts a big advertising campaign, it conducts a hypothesis test. The company's marketing people survey 150 households with the result that 43 of the households have three cell phones. Solution 9.18 Set up the hypothesis test: 542 Chapter 9 \| Hypothesis Testing with One Sample H0: p = 0.30 Determine the distribution needed: Ha: p ≠ 0.30 The random variable is P′ = proportion of households that have three cell phones. The distribution for the hypothesis test is P ' ~ N ⎛ ⎝0.30, (0.30) ⋅ (0.70) 150 ⎞ ⎠. a. The value that helps determine the p-value is p′. Calculate p′. Solution 9.18 a. p′ = x n where x is the number of successes and n is the total number in the sample. x = 43, n = 150 p′ = 43 150. b. What is a success for this problem? Solution 9.18 b. A success is having three cell phones in a household. c. What is the level of significance? Solution 9.18 c. The level of significance is the preset α. Since α is not given, assume that α = 0.05. d. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately. Calculate the p-value. Solution 9.18 d. First we compute the sample proportion p^ = 43 150 = 0.287. Next, the z-test is given by z-test = p^ − p p ⋅ q n = 0.287 – 0.30 0.30×0.70 150 ≈ –0.36. is negative, we compute the area to the left tail of –0.36 in a normal distribution, Since the z-test P(Z < − 0.36) ≈ 0.3607902. Finally, because this is a two-sided test of significance, we multiply this probability times two to account for the right tail, and obtain p-value = 2×0.3607902 = 0.7215804. e. Make a decision. _____________(Reject/Do not reject) H0 because____________. Solution 9.18 e. Assuming that α = 0.05, α < p-value. The decision is do not reject H0 because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30 percent. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 543 9.18 Marketers believe that 92 percent of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. Two hundred American adults are surveyed, of which 174 report having cell phones. Use a 5 percent level of significance. State the null and alternative hypotheses, find the p-value, state your conclusion, and identify the Type I and Type II errors. The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter p. The distribution for the test is normal. The estimated proportion p′ is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived α = 0.01, for comparison, and a 95 percent c
onfidence interval computation. The poem is clever and humorous, so please enjoy it! Example 9.19 My dog has so many fleas, They do not come off with ease. As for shampoo, I have tried many types Even one called Bubble Hype, Which only killed 25 percent of the fleas, Unfortunately I was not pleased. I've used all kinds of soap, Until I had given up hope Until one day I saw An ad that put me in awe. A shampoo used for dogs Called GOOD ENOUGH to Clean a Hog Guaranteed to kill more fleas. I gave Fido a bath And after doing the math His number of fleas Started dropping by 3's! Before his shampoo I counted 42. At the end of his bath, I redid the math And the new shampoo had killed 17 fleas. So now I was pleased. Now it is time for you to have some fun With the level of significance being .01, You must help me figure out Use the new shampoo or go without? Solution 9.19 Set up the hypothesis test: H0: p ≤ 0.25 Determine the distribution needed: Ha: p > 0.25 544 Chapter 9 \| Hypothesis Testing with One Sample In words, clearly state what your random variable X ¯ or P′ represents. P′ = The proportion of fleas that are killed by the new shampoo State the distribution to use for the test. Normal: ⎛ ⎝0.25, N (0.25)(1 − 0.25) 42 ⎞ ⎠ The z-test is given by z-test = p¯ − p p ⋅ q n = 0.4048 − 0.25 42 ≈ 2.316834. Because this is a hypothesis test one-sided to the right, we compute the p-value as the area to the right tail of the z-test in a standard normal distribution, P(Z > 3.32) ≈ 0.0103. In one to two complete sentences, explain what the p-value means for this problem. If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 ⎛ ⎝ or more. ⎞ ⎠ 17 42 Use the previous information to sketch a picture of this situation. Clearly label and scale the horizontal axis and shade the region(s) corresponding to the p-value. Figure 9.11 Compare α and the p-value: Indicate the correct decision (reject or do not reject the null hypothesis) and the reason for it, and write an appropriate conclusion, using complete sentences. Alpha Decision Reason for Decision 0.01 Do not reject H0 α < p-value Table 9.3 Conclusion: At the 1 percent level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25 percent. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 545 Construct a 95 percent confidence interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. Figure 9.12 Confidence Interval: (0.26, 0.55). We are 95 percent confident that the true population proportion p of fleas that are killed by the new shampoo is between 26 percent and 55 percent. NOTE This test result is not very definitive since the p-value is very close to alpha. In reality, one would probably do more tests by giving the dog another bath after the fleas have had a chance to return. Example 9.20 The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass: 1.11, 1.07, 1.11, 1.07, 1.12, 1.08, 0.98, 0.98, 1.02, 0.95, 0.95 Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal. Solution 9.20 Let’s follow a four-step process to answer this statistical question. 1. State the question: We need to determine if, at a 0.05 significance level, the average conductivity of the selected glass is greater than one. Our hypotheses will be as follows: a. H0: μ ≤ 1 b. Ha: μ > 1 2. Plan: We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's t-distribution. Assume the underlying population is normal. 3. Do the calculations: We will input the sample data into the TI-83 as follows. 546 Chapter 9 \| Hypothesis Testing with One Sample Figure 9.13 Figure 9.14 Figure 9.15 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 547 Figure 9.16 4. State the conclusions: Since the p-value (p = 0.036) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data support the claim that the average conductivity level is greater than one. Example 9.21 In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users. The rate of brain cancer for noncell phone users is 0.0340 percent. Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error. Solution 9.21 We will follow the four-step process. 1. We need to conduct a hypothesis test on the claimed cancer rate. Our hypotheses will be as follows: a. H0: p ≤ 0.00034 b. Ha: p > 0.00034 If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancercausing environments, we want to minimize the chances of incorrectly identifying causes of cancer. 2. We will be testing a sample proportion with x = 172 and n = 420,019. The sample is sufficiently large because we have np = 420,019(0.00034) = 142.8, nq = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success p = 0.00034. Thus we will be able to generalize our results to the population. 3. The associated TI results are shown in the following figures. 548 Chapter 9 \| Hypothesis Testing with One Sample Figure 9.17 Figure 9.18 4. Since the p-value = 0.0073 is greater than our alpha value = 0.005, we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users. 9.6 \| Hypothesis Testing of a Single Mean and Single Proportion This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 549 9.1 Hypothesis Testing of a Single Mean and Single Proportion Student Learning Outcomes • The student will select the appropriate distributions to use in each case. • The student will conduct hypothesis tests and interpret the results. Television Survey In a recent survey, it was stated that Americans watch television on average four hours per day. Assume that σ = 2. Using your class as the sample, conduct a hypothesis test to determine if the average for students at your school is lower. 1. H0: _____________ 2. Ha: _____________ 3. In words, define the random variable. __________ = ______________________ 4. The distribution to use for the test is _______________________. 5. Determine the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: Figure 9.19 b. Determine the p-value. 7. Do you or do you not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. Language Survey About 42.3 percent of Californians and 19.6 percent of all Americans over age five speak a language other than English at home. Using your class as the sample, conduct a hypothesis test to determine if the percentage of the students at your school who speak a language other than English at home is different from 42.3 percent. 1. H0: ___________ 2. Ha: ___________ 550 Chapter 9 \| Hypothesis Testing with One Sample 3. In words, define the random variable. __________ = _______________ 4. The distribution to use for the test is ________________. 5. Determine the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: Figure 9.20 b. Determine the p-value. 7. Do you or do you not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. Jeans Survey You've read in an article that young adults own an average of three pairs of jeans. Survey eight people from your class to determine if the average is higher than three. Assume the population is normal. 1. H0: _____________ 2. Ha: _____________ 3. In words, define the random variable. __________ = ______________________ 4. The distribution to use for the test is _______________________. 5. Determine the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 551 Figure 9.21 b. Determine the p-value. 7. Do you or do you not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. 552 Chapter 9 \| Hypothesis Testing with One Sample KEY TERMS binomial distribution independent trials Independent means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV Χ is defined as the number of successes in n trials. The notation is: X ~ B(n, p) μ = np and the standard deviation is σ = npq . a discrete random variable (RV) that arises from Bernoulli trials; there are a fixed number, n, of The probability of exactly x successes in n trials is P(X = x) = ⎞ ⎠p x qn − x . n ⎛ ⎝ x confidence interval (CI) an interval estimate for an unknown population parameter This depends on the following: • The desired confidence level. • Information that is known about the distribution (for example, known standard deviation). • The sample and its size. hypothesis a statement about the value of a population parameter; in the case of two hypotheses, the statement assumed
to be true is called the null hypothesis (notation H0) and the contradictory statement is called the alternative hypothesis (notation Ha) hypothesis testing based on sample evidence, a procedure for determining whether the hypothesis stated is a reasonable statement and should not be rejected, or is unreasonable and should be rejected level of significance of the test probability of a Type I error (reject the null hypothesis when it is true) Notation: α. In hypothesis testing, the level of significance is called the preconceived α or the preset α. normal distribution a bell-shaped continuous random variable X, with center at the mean value (μ) and distance from the center to the inflection points of the bell curve given by the standard deviation (σ) We write X ~ N(μ, σ) . If the mean value is 0 and the standard deviation is 1, the random variable is called the standard normal distribution, and it is denoted with the letter Z. p-value the probability that an event will happen purely by chance assuming the null hypothesis is true; the smaller the p-value, the stronger the evidence is against the null hypothesis standard deviation a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation Student's t-distribution investigated and reported by William S. Gosset in 1908 and published under the pseudonym Student The major characteristics of the random variable (RV) are as follows • It is continuous and assumes any real values. • The pdf is symmetrical about its mean of zero. However, it is more spread out and flatter at the apex than the normal distribution. • It approaches the standard normal distribution as n gets larger. • There is a family of t-distributions: every representative of the family is completely defined by the number of degrees of freedom, which is one less than the number of data items. Type 1 error the decision is to reject the null hypothesis when, in fact, the null hypothesis is true Type 2 error the decision is not to reject the null hypothesis when, in fact, the null hypothesis is false CHAPTER REVIEW 9.1 Null and Alternative Hypotheses In a hypothesis test, sample data are evaluated in order to arrive at a decision about some type of claim. If certain conditions This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 553 about the sample are satisfied, then the claim can be evaluated for a population. In a hypothesis test, we do the following: 1. Evaluate the null hypothesis, typically denoted with H0. The null is not rejected unless the hypothesis test shows otherwise. The null statement must always contain some form of equality (=, ≤, or ≥). 2. Always write the alternative hypothesis, typically denoted with Ha or H1, using less than, greater than, or not equals symbols, i.e., (≠, >, or <). 3. If we reject the null hypothesis, then we can assume there is enough evidence to support the alternative hypothesis. 4. Never state that a claim is proven true or false. Keep in mind the underlying fact that hypothesis testing is based on probability laws; therefore, we can talk only in terms of non-absolute certainties. 9.2 Outcomes and the Type I and Type II Errors In every hypothesis test, the outcomes are dependent on a correct interpretation of the data. Incorrect calculations or misunderstood summary statistics can yield errors that affect the results. A Type I error occurs when a true null hypothesis is rejected. A Type II error occurs when a false null hypothesis is not rejected. The probabilities of these errors are denoted by the Greek letters α and β, for a Type I and a Type II error respectively. The power of the test, 1 – β, quantifies the likelihood that a test will yield the correct result of a true alternative hypothesis being accepted. A high power is desirable. 9.3 Distribution Needed for Hypothesis Testing In order for a hypothesis test’s results to be generalized to a population, certain requirements must be satisfied. When testing for a single population mean: 1. A Student's t-test should be used if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with an unknown standard deviation. 2. The normal test will work if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with a known standard deviation. When testing a single population proportion use a normal test for a single population proportion if the data come from a simple, random sample, fill the requirements for a binomial distribution, and the mean number of success and the mean number of failures satisfy the conditions: np > 5 and nq > n where n is the sample size, p is the probability of a success, and q is the probability of a failure. 9.4 Rare Events, the Sample, and the Decision and Conclusion When the probability of an event occurring is low, and it happens, it is called a rare event. Rare events are important to consider in hypothesis testing because they can inform your willingness not to reject or to reject a null hypothesis. To test a null hypothesis, find the p-value for the sample data and graph the results. When deciding whether or not to reject the null the hypothesis, keep these two parameters in mind: 1. α > p-value, reject the null hypothesis. 2. α ≤ p-value, do not reject the null hypothesis. 9.5 Additional Information and Full Hypothesis Test Examples The hypothesis test itself has an established process. This can be summarized as follows: 1. Determine H0 and Ha. Remember, they are contradictory. 2. Determine the random variable. 3. Determine the distribution for the test. 4. Draw a graph, calculate the test statistic, and use the test statistic to calculate the p-value. (A z-score and a t-score are examples of test statistics.) 5. Compare the preconceived α with the p-value, make a decision (reject or do not reject H0), and write a clear conclusion using English sentences. Notice that in performing the hypothesis test, you use α and not β. β is needed to help determine the sample size of the data that are used in calculating the p-value. Remember that the quantity 1 – β is called the Power of the Test. A high power is 554 Chapter 9 \| Hypothesis Testing with One Sample desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same. If the power is low, the null hypothesis might not be rejected when it should be. FORMULA REVIEW 9.1 Null and Alternative Hypotheses H0 and Ha are contradictory. If H0 has: equal (=) then Ha has: not equal (≠) or greater than (>) or less than (<) Table 9.4 greater than or equal to (≥) less than or equal to (≤) less than (<) greater than (>) If α ≤ p-value, then do not reject H0. If α > p-value, then reject H0. α is preconceived. Its value is set before the hypothesis test starts. The p-value is calculated from the data. 9.2 Outcomes and the Type I and Type II Errors α = probability of a Type I error = P(Type I error) = probability of rejecting the null hypothesis when the null hypothesis is true. β = probability of a Type II error = P(Type II error) = PRACTICE 9.1 Null and Alternative Hypotheses probability of not rejecting the null hypothesis when the null hypothesis is false. 9.3 Distribution Needed for Hypothesis Testing If there is no given preconceived α, then use α = 0.05. Types of Hypothesis Tests • Single population mean, known population variance (or standard deviation): Normal test. • Single population mean, unknown population variance (or standard deviation): Student's t-test. • Single population proportion: Normal test. • For a single population mean, we may use a normal distribution with the following mean and standard deviation. Means: μ = μ x¯ and σ x¯ = . σ x n • For a single population proportion, we may use a normal distribution with the following mean and standard deviation. Proportions: µ = p and σ = pq n . 1. You are testing that the mean speed of your cable internet connection is more than three megabits per second. What is the random variable? Describe it in words. 2. You are testing that the mean speed of your cable internet connection is more than three megabits per second. State the null and alternative hypotheses. 3. The American family has an average of two children. What is the random variable? Describe in words. 4. The mean entry level salary of an employee at a company is $58,000. You believe it is higher for IT professionals in the company. State the null and alternative hypotheses. 5. A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the proportion is actually less. What is the random variable? Describe in words. 6. A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the claim is correct. State the null and alternative hypotheses. 7. In a population of fish, approximately 42 percent are female. A test is conducted to see if, in fact, the proportion is less. State the null and alternative hypotheses. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 555 8. Suppose that a recent article stated that the mean time students spend doing homework each week is 2.5 hours. A study was then done to see if the mean time has increased in the new century. A random sample of 26 students. The mean length of time the students spent on homework was 3 hours with a standard deviation of 1.8 hours. Suppose that it is somehow known that the population standard deviation is 1.5. If you were conducting a hypothesis test to determine if the mean length of homework has increased, what w
ould the null and alternative hypotheses be? The distribution of the population is normal. a. H0: ________ b. Ha: ________ 9. A random survey of 75 long-term marathon runners revealed that the mean length of time they've been running is 17.4 years with a standard deviation of 6.3 years. If you were conducting a hypothesis test to determine if the population mean time for these runners could likely be 15 years, what would the null and alternative hypotheses be? a. H0: __________ b. Ha: __________ 10. Researchers published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from a particular type of disease. Suppose that in a survey of 100 people in a certain town, seven of them suffered from this disease. If you were conducting a hypothesis test to determine if the true proportion of people in that town suffering from this disease is lower than the percentage in the general adult American population, what would the null and alternative hypotheses be? a. H0: ________ b. Ha: ________ 9.2 Outcomes and the Type I and Type II Errors 11. The mean price of mid-sized cars in a region is $32,000. A test is conducted to see if the claim is true. State the Type I and Type II errors in complete sentences. 12. A sleeping bag is tested to withstand temperatures of –15 °F. You think the bag cannot stand temperatures that low. State the Type I and Type II errors in complete sentences. 13. For Exercise 9.12, what are α and β in words? 14. In words, describe 1 – β for Exercise 9.12. 15. A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, H0, is: the surgical procedure will go well. State the Type I and Type II errors in complete sentences. 16. A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, H0, is: the surgical procedure will go well. Which is the error with the greater consequence? 17. The power of a test is 0.981. What is the probability of a Type II error? 18. A group of divers is exploring an old sunken ship. Suppose the null hypothesis, H0, is the sunken ship does not contain buried treasure. State the Type I and Type II errors in complete sentences. 19. A microbiologist is testing a water sample for E. coli. Suppose the null hypothesis, H0, is the sample does not contain E. coli. The probability that the sample does not contain E. coli, but the microbiologist thinks it does is 0.012. The probability that the sample does contain E. coli, but the microbiologist thinks it does not is 0.002. What is the power of this test? 20. A microbiologist is testing a water sample for E. coli. Suppose the null hypothesis, H0, is the sample contains E-coli. Which is the error with the greater consequence? 9.3 Distribution Needed for Hypothesis Testing 21. Which two distributions can you use for hypothesis testing for this chapter? 22. Which distribution do you use when the standard deviation is not known? Assume sample size is large. 23. Which distribution do you use when the standard deviation is not known and you are testing one population mean? Assume sample size is large. 24. A population mean is 13. The sample mean is 12.8, and the sample standard deviation is two. The sample size is 20. What distribution should you use to perform a hypothesis test? Assume the underlying population is normal. 25. A population has a mean of 25 and a standard deviation of five. The sample mean is 24, and the sample size is 108. What distribution should you use to perform a hypothesis test? 556 Chapter 9 \| Hypothesis Testing with One Sample 26. It is thought that 42 percent of respondents in a taste test would prefer Brand A. In a particular test of 100 people, 39 percent preferred Brand A. What distribution should you use to perform a hypothesis test? 27. You are performing a hypothesis test of a single population mean using a Student’s t-distribution. What must you assume about the distribution of the data? 28. You are performing a hypothesis test of a single population mean using a Student’s t-distribution. The data are not from a simple random sample. Can you accurately perform the hypothesis test? 29. You are performing a hypothesis test of a single population proportion. What must be true about the quantities of np and nq? 30. You are performing a hypothesis test of a single population proportion. You find out that np is less than five. What must you do to be able to perform a valid hypothesis test? 31. You are performing a hypothesis test of a single population proportion. The data come from which distribution? 9.4 Rare Events, the Sample, and the Decision and Conclusion 32. When do you reject the null hypothesis? 33. The probability of winning the grand prize at a particular carnival game is 0.005. Is the outcome of winning very likely or very unlikely? 34. The probability of winning the grand prize at a particular carnival game is 0.005. Michele wins the grand prize. Is this considered a rare or common event? Why? 35. It is believed that the mean height of high school students who play basketball on the school team is 73 inches with a standard deviation of 1.8 inches. A random sample of 40 players is chosen. The sample mean was 71 inches, and the sample standard deviation was 1.5 inches. Do the data support the claim that the mean height is less than 73 inches? The p-value is almost zero. State the null and alternative hypotheses and interpret the p-value. 36. The mean age of graduate students at a university is at most 31 years with a standard deviation of two years. A random sample of 15 graduate students is taken. The sample mean is 32 years and the sample standard deviation is three years. Are the data significant at the 1 percent level? The p-value is 0.0264. State the null and alternative hypotheses and interpret the p-value. 37. Does the shaded region represent a low or a high p-value compared to a level of significance of 1 percent? Figure 9.22 38. What should you do when α > p-value? 39. What should you do if α = p-value? 40. If you do not reject the null hypothesis, then it must be true. Is that statement correct? State why or why not in complete sentences. Use the following information to answer the next seven exercises: Suppose that a recent article stated that the mean time students spend doing homework each week is 2.5 hours. A study was then done to see if the mean time has increased in the new century. A random sample of 26 students was taken. The mean length of time they did homework each week was three hours with a standard deviation of 1.8 hours. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of time doing homework each week has increased. Assume This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 557 the distribution of homework times is approximately normal. 41. Is this a test of means or proportions? 42. What symbol represents the random variable for this test? 43. In words, define the random variable for this test. 44. Is σ known and, if so, what is it? 45. Calculate the following: x¯ _______ a. b. σ _______ sx _______ c. d. n _______ 46. Since both σ and s x are given, which should be used? In one to two complete sentences, explain why. 47. State the distribution to use for the hypothesis test. 48. A random survey of 75 long-term marathon runners revealed that the mean length of time they have been running is 17.4 years with a standard deviation of 6.3 years. Conduct a hypothesis test to determine if the population mean time is likely to be 15 years. Is this a test of one mean or proportion? a. b. State the null and alternative hypotheses. H0: ____________________ Ha : ____________________ Is this a right-tailed, left-tailed, or two-tailed test? c. d. What symbol represents the random variable for this test? e. f. g. Calculate the following: In words, define the random variable for this test. Is the population standard deviation known and, if so, what is it? x¯ = _____________ i. ii. s = ____________ iii. n = ____________ h. Which test should be used? i. State the distribution to use for the hypothesis test. j. Find the p-value. k. At a pre-conceived α = 0.05, give your answer for each of the following: i. Decision: ii. Reason for the decision: iii. Conclusion (write out in a complete sentence): 9.5 Additional Information and Full Hypothesis Test Examples 49. Assume H0: μ = 9 and Ha: μ < 9. Is this a left-tailed, right-tailed, or two-tailed test? 50. Assume H0: μ ≤ 6 and Ha: μ > 6. Is this a left-tailed, right-tailed, or two-tailed test? 51. Assume H0: p = 0.25 and Ha: p ≠ 0.25. Is this a left-tailed, right-tailed, or two-tailed test? 52. Draw the general graph of a left-tailed test. 53. Draw the graph of a two-tailed test. 54. A bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use? 55. Your friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use? 56. A bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use? 57. You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50 percent, but you think it is less for this particular coin. What type of test would you use? 558 Chapter 9 \| Hypothesis Testing with One Sample 58. If the alternative hypothesis has a not equals ( ≠ ) symbol, you know to use which type of test? 59. Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test? 60. Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test? 61. Assume the null hypothesis states that the mean is equal to 88
. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test? HOMEWORK 9.1 Null and Alternative Hypotheses 62. Some of the following statements refer to the null hypothesis, some to the alternate hypothesis. State the null hypothesis, H0, and the alternative hypothesis. Ha, in terms of the appropriate parameter (μ or p). a. The mean number of years Americans work before retiring is 34. b. At most 60 percent of Americans vote in presidential elections. c. The mean starting salary for San Jose State University graduates is at least $100,000 per year. d. Twenty-nine percent of high school students take physical education daily. e. Less than 5 percent of adults ride the bus to work in Los Angeles. f. The mean number of cars a person owns in her lifetime is not more than 10. g. About half of Americans prefer to live away from cities, given the choice. h. Europeans have a mean paid vacation each year of six weeks. i. The chance of developing breast cancer is under 11 percent for women. j. Private universities' mean tuition cost is more than $20,000 per year. 63. A recent survey of 273 randomly selected teens living in Massachusetts asked about social media. Sixty-three said that they routinely use a certain app to share pictures. The researchers want to determine if there is good evidence that more than 30 percent of teens use this app. The alternative hypothesis is as follows: a. p < 0.30 b. p ≤ 0.30 c. p ≥ 0.30 d. p > 0.30 64. A statistics instructor believes that fewer than 20 percent of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is as follows: a. p = 0.20 b. p > 0.20 c. p < 0.20 d. p ≤ 0.20 65. Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are as follows: a. Ho: x¯ = 4.5, Ha : x¯ > 4.5 b. Ho: μ ≥ 4.5, Ha: μ < 4.5 c. Ho: μ = 4.75, Ha: μ > 4.75 d. Ho: μ = 4.5, Ha: μ > 4.5 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 559 9.2 Outcomes and the Type I and Type II Errors 66. State the Type I and Type II errors in complete sentences given the following statements. a. The mean number of years Americans work before retiring is 34. b. At most 60 percent of Americans vote in presidential elections. c. The mean starting salary for San Jose State University graduates is at least $100,000 per year. d. 29 percent of high school students take physical education every day. e. Less than 5 percent of adults ride the bus to work in Los Angeles. f. The mean number of cars a person owns in his or her lifetime is not more than 10. g. About half of Americans prefer to live away from cities, given the choice. h. Europeans have a mean paid vacation each year of six weeks. i. The chance of developing breast cancer is under 11 percent for women. j. Private universitie' mean tuition cost is more than $20,000 per year. 67. For Statements A–J in Exercise 9.66, answer the following in complete sentences. a. State a consequence of committing a Type I error. b. State a consequence of committing a Type II error. 68. When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the U.S. Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is the drug is unsafe. What is the Type II error? a. To conclude the drug is safe when, in fact, it is unsafe. b. Not to conclude the drug is safe when, in fact, it is safe. c. To conclude the drug is safe when, in fact, it is safe. d. Not to conclude the drug is unsafe when, in fact, it is unsafe. 69. A statistics instructor believes that fewer than 20 percent of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________. a. at least 20 percent, when, in fact, it is less than 20 percent. b. 20 percent, when, in fact, it is 20 percent. c. d. less than 20 percent, when, in fact, it is at least 20 percent. less than 20 percent, when, in fact, it is less than 20 percent. 70. It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5 percent, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours a. b. c. d. is more than seven hours. is at most seven hours. is at least seven hours. is less than seven hours. 71. Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The Type I error is a. b. c. d. to conclude that the current mean hours per week is higher than 4.5, when, in fact, it is higher. to conclude that the current mean hours per week is higher than 4.5, when, in fact, it is the same. to conclude that the mean hours per week currently is 4.5, when, in fact, it is higher. to conclude that the mean hours per week currently is no higher than 4.5, when, in fact, it is not higher. 560 Chapter 9 \| Hypothesis Testing with One Sample 9.3 Distribution Needed for Hypothesis Testing 72. It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5 percent, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The distribution to be used for this test is X ¯ ~ ________________ ) a. N(7.24, 1.93 22 b. N(7.24, 1.93) c. d. t22 t21 9.4 Rare Events, the Sample, and the Decision and Conclusion 73. The National Institute of Mental Health published an article stating that in any one-year period approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population. a. Is this a test of one mean or proportion? b. State the null and alternative hypotheses. H0: ____________________ Ha: ____________________ Is this a right-tailed, left-tailed, or two-tailed test? c. d. What symbol represents the random variable for this test? e. f. Calculate the following: In words, define the random variable for this test. i. x = ________________ ii. n = ________________ iii. p′ = _____________ g. Calculate σx = __________. Show the formula setup. h. State the distribution to use for the hypothesis test. i. Find the p-value. j. At a pre-conceived α = 0.05, give your answer for each of the following: i. Decision: ii. Reason for the decision: iii. Conclusion (write out in a complete sentence): 9.5 Additional Information and Full Hypothesis Test Examples For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E, Solution Sheets. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. NOTE If you are using a Student's-t-distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. In general, you must first prove that assumption, however. 74. A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, are the data highly inconsistent with the claim? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 561 75. In 2009, President Barack Obama announced a new national fuel economy and emissions policy for cars and light trucks. It stated that the combined fleet fuel economy for an auto manufacturer of cars and light trucks will have to average 35.5 mpg or better by 2016. From past studies on fuel economy, it is known that the standard deviation of a typical fleet is 7.6 mpg. An auto manufacturer selects a random sample of 55 cars and light trucks and finds the sample mean fuel economy to be 34.6 mpg with a standard deviation of 10.3 mpg. Can the manufacturer claim that their fleet meets the fuel economy standard in the 2016 policy at the 5 percent level? 76. The cost of a daily newspaper varies from city to city. However,
the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1 percent level? 77. An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1 percent level? 78. The mean number of sick days an employee takes per year is believed to be about 10. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let x = the number of sick days they took for the past year. Should the personnel team believe that the mean number is 10? 79. In 1955, Life Magazine reported that the 25-year-old mother of three worked, on average, an 80-hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. Eighty-one women were surveyed with the following results. The sample mean was 83; the sample standard deviation was 10. Does it appear that the mean work week has increased for women at the 5 percent level? 80. Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think? 81. A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5, 4, 7, 3, 6, 4, 5, 3, 6, 3, 8, 5. Conduct a hypothesis test of your belief. 82. Refer to Exercise 9.81. Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four. 83. According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7 percent girls). Suppose you don’t believe the reported figures of the percentage of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percentage of girls born in China is 46.7? 84. A group of researchers research a common contagious disease. A newspaper found that 13 percent of Americans have been diagnosed with the disease in the last year. The researchers doubt that the percentage is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had been diagnosed with the disease. Would you agree with the newspaper's poll? In complete sentences, give three reasons why polls might give different results. 85. The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks 10 engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours? Data (length of mean work week): 70, 45, 55, 60, 65, 55, 55, 60, 50, 55. 86. Use the Lap time data for Lap 4 (see Appendix C: Data Sets) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all 20 races given. 87. Use the Initial Public Offering data (see Appendix C: Data Sets) to test the claim that the mean offer price was $18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices. 562 NOTE Chapter 9 \| Hypothesis Testing with One Sample The following questions were written by past students. They are excellent problems! 88. "Asian Family Reunion," by Chau Nguyen Every two years it comes around. We all get together from different towns. In my honest opinion, It's not a typical family reunion. Not forty, or fifty, or sixty, But how about seventy companions! The kids would play, scream, and shout One minute they're happy, another they'll pout. The teenagers would look, stare, and compare From how they look to what they wear. The men would chat about their business That they make more, but never less. Money is always their subject And there's always talk of more new projects. The women get tired from all of the chats They head to the kitchen to set out the mats. Some would sit and some would stand Eating and talking with plates in their hands. Then come the games and the songs And suddenly, everyone gets along! With all that laughter, it's sad to say That it always ends in the same old way. They hug and kiss and say "good-bye" And then they all begin to cry! I say that 60 percent shed their tears But my mom counted 35 people this year. She said that boys and men will always have their pride, So we won't ever see them cry. I myself don't think she's correct, So could you please try this problem to see if you object? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 563 89. "Blowing Bubbles," by Sondra Prull Studying stats just made me tense, I had to find some sane defense. Some light and lifting simple play To float my math anxiety away. Blowing bubbles lifts me high Takes my troubles to the sky. POIK! They're gone, with all my stress Bubble therapy is the best. The label said each time I blew The average number of bubbles would be at least 22. I blew and blew and this I found From 64 blows, they all are round! But the number of bubbles in 64 blows Varied widely, this I know. 20 per blow became the mean They deviated by 6, and not 16. From counting bubbles, I sure did relax But now I give to you your task. Was 22 a reasonable guess? Find the answer and pass this test! 564 Chapter 9 \| Hypothesis Testing with One Sample 90. "Dalmatian Darnation," by Kathy Sparling A greedy dog breeder named Spreckles Bred puppies with numerous freckles The Dalmatians he sought Possessed spot upon spot The more spots, he thought, the more shekels. His competitors did not agree That freckles would increase the fee. They said, “Spots are quite nice But they don't affect price; One should breed for improved pedigree.” The breeders decided to prove This strategy was a wrong move. Breeding only for spots Would wreak havoc, they thought. His theory they want to disprove. They proposed a contest to Spreckles Comparing dog prices to freckles. In records they looked up One hundred one pups: Dalmatians that fetched the most shekels. They asked Mr. Spreckles to name An average spot count he'd claim To bring in big bucks. Said Spreckles, “Well, shucks, It's for one hundred one that I aim.” Said an amateur statistician Who wanted to help with this mission. “Twenty-one for the sample Standard deviation's ample.” They examined one hundred and one Dalmatians that fetched a good sum. They counted each spot, Mark, freckle, and dot And tallied up every one. Instead of one hundred one spots They averaged ninety-six dots Can they muzzle Spreckles’ Obsession with freckles Based on all the dog data they've got? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 565 91. Macaroni and Cheese, please!! by Nedda Misherghi and Rachelle Hall As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value. One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook: Price per box of Mac and Cheese • 5 stores @ $2.02 • 15 stores @ $0.25 • 3 stores @ $1.29 • 6 stores @ $0.35 • 4 stores @ $2.27 • 7 stores @ $1.50 • 5 stores @ $1.89 • 8 stores @ $0.75 I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. After all, as a poor starving student I can't be expected to feed our class of animals! 566 Chapter 9 \| Hypothesis Testing with One Sample 92. "William Shakespeare: The Tragedy of Hamlet, Prince of Denmark," by Jacqueline Ghodsi THE CHARACTERS (in order of appearance): • HAMLET, Prince of Denmark and student of statistics • POLONIUS, Hamlet’s tutor • HORATIO, friend to Hamlet and fellow student Scene: The great library of the castle, in which Hamlet does his lessons Act I The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprim
anding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage. POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of your imagination! HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable. POLONIUS: If thou dost insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What a piece of work is man, how noble in reason, how infinite in faculties”? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true. HORATIO (to Hamlet): What should we do, my Lord? HAMLET: Go to thy purpose, Horatio. HORATIO: To what end, my Lord? HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no. Horatio exits, followed by Polonius, leaving Hamlet to ponder alone. Act II The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters. POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations? HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes. POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.) HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.) (Curtain falls) 93. "Untitled," by Stephen Chen I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1 percent. So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 567 94. "Japanese Girls’ Names" by Kumi Furuichi It used to be very typical for Japanese girls’ names to end with “ko.” The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation. “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters that have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on. However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names that end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children. I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, coworkers’, and acquaintances’ names that I could remember. Following are the names. Some are repeats. Test to see if the proportion has dropped for this generation. Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko. 568 Chapter 9 \| Hypothesis Testing with One Sample 95. "Phillip’s Wish," by Suzanne Osorio My nephew likes to play Chasing the girls makes his day. He asked his mother If it is okay To get his ear pierced. She said, “No way!” To poke a hole through your ear, Is not what I want for you, dear. He argued his point quite well, Says even my macho pal, Mel, Has gotten this done. It’s all just for fun. C’mon please, mom, please, what the hell. Again Phillip complained to his mother, Saying half his friends (including their brothers) Are piercing their ears And they have no fears He wants to be like the others. She said, “I think it’s much less. We must do a hypothesis test. And if you are right, I won’t put up a fight. But, if not, then my case will rest.” We proceeded to call fifty guys To see whose prediction would fly. Nineteen of the fifty Said piercing was nifty And earrings they’d occasionally buy. Then there’s the other thirty-one, Who said they’d never have this done. So now this poem’s finished. Will his hopes be diminished, Or will my nephew have his fun? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 569 96. "The Craven," by Mark Salangsang Once upon a morning dreary In stats class I was weak and weary. Pondering over last night’s homework Whose answers were now on the board This I did and nothing more. While I nodded nearly napping Suddenly, there came a tapping. As someone gently rapping, Rapping my head as I snore. Quoth the teacher, “Sleep no more.” “In every class you fall asleep,” The teacher said, his voice was deep. “So a tally I’ve begun to keep Of every class you nap and snore. The percentage being forty-four.” “My dear teacher I must confess, While sleeping is what I do best. The percentage, I think, must be less, A percentage less than forty-four.” This I said and nothing more. “We’ll see,” he said and walked away, And fifty classes from that day He counted till the month of May The classes in which I napped and snored. The number he found was twenty-four. At a significance level of 0.05, Please tell me am I still alive? Or did my grade just take a dive Plunging down beneath the floor? Upon thee I hereby implore. 97. Toastmasters International cites a report by Gallup Poll that 40 percent of Americans fear public speaking. A student believes that less than 40 percent of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percentage at her school is less than 40. 98. Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68 percent also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68 percent represents California. Note: For more accurate results, use more California community colleges and this past year's data. 99. According to an article in a local poll, a city found that 14 percent of its residents walk for exercise. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen city residents replied that they walk for exercise. Conduct a hypothesis test to determine if the rate is still 14 percent or if it has decreased. 570 Chapter 9 \| Hypothesis Testing with One Sample 100. The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test. 101. Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test. 102. La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of
4 months. Conduct a hypothesis test to determine if the mean weaning age in the United States is less than four years old. 103. Harley Davidson motorcycles are the largest selling motorcycle in the United States, with 14 percent of all motorcycles sold in 2012. Interestingly, a random sample of 1,945 stolen motorcycles was selected, and it was found that just 8 percent of them were Harleys. Is there good evidence that the proportion of Harleys among stolen motorcycles is significantly less than their share of all motorcycles? After conducting the test, what decision and conclusion would you make? a. Reject H0: There is sufficient evidence to conclude that the proportion of Harleys stolen is significantly less than their share of all motorcycles b. Do not reject H0: There is not sufficient evidence to conclude that the proportion of Harleys stolen is significantly less than their share of all motorcycles c. Do not reject H0: There is sufficient evidence to conclude that the proportion of Harleys stolen is significantly more than their share of all motorcycles d. Reject H0: There is sufficient evidence to conclude that the proportion of Harleys stolen is significantly more than their share of all motorcycles 104. A statistics instructor believes that fewer than 20 percent of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. At a 1 percent level of significance, what is an appropriate conclusion? a. There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20 percent. b. There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is more than 20 percent. c. There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20 percent. d. There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20 percent. 105. Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. At a significance level of a = 0.05, what is the correct conclusion? a. There is enough evidence to conclude that the mean number of hours is more than 4.75. b. There is enough evidence to conclude that the mean number of hours is more than 4.5. c. There is not enough evidence to conclude that the mean number of hours is more than 4.5. d. There is not enough evidence to conclude that the mean number of hours is more than 4.75. Hypothesis testing: For the following 10 exercises, answer each question. a. State the null and alternate hypotheses. b. State the p-value. c. State alpha. d. What is your decision? e. Write a conclusion. f. Answer any other questions asked in the problem. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 571 106. A research group is studying a particular infectious disease. In 2011 at least 18 percent of nursing home residents had the disease. An Introduction to Statistics class in Daviess County, KY, conducted a hypothesis test at the nursing home (approximately 1,200 residents) to determine if the local nursing home's incidence was lower. One hundred fifty residents were chosen at random and surveyed. Of the 150 residents surveyed, 82 have the disease. Use a significance level of 0.05 and, using appropriate statistical evidence, conduct a hypothesis test and state the conclusions. 107. A recent survey in the New York Times Almanac indicated that 48.8 percent of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate? 108. Driver error can be listed as the cause of approximately 54 percent of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using α = 0.05, is the AAA proportion accurate? 109. The U.S. Department of Energy reported that 51.7 percent of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the α = 0.05 level? Are the results applicable across the country? Why? 110. For Americans using library services, the American Library Association claims that at most 67 percent of patrons borrow books. The library director in Owensboro, KY, feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use α = 0.01 level of significance. What is the possible proportion of patrons who do borrow books from the Owensboro Library? 111. The Weather Underground reported that the mean amount of summer rainfall for the northeastern United States is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the α = 0.05 level, can it be concluded that the mean rainfall was below the reported average? What if α = 0.01? Assume the amount of summer rainfall follows a normal distribution. 112. A survey in the New York Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX, chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the α = 0.10 level, is the Austin, TX, commute significantly less than the mean commute time for the 15 largest U.S. cities? 113. A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals: 3; 2; 1; 3; 7; 2; 9; 4; 6; 6; 8; 0; 5; 6; 4; 2; 1; 3; 4; 1 At the α = 0.05 level, can it be concluded that the sample mean is higher than 5.8 visits per year? 114. According to the New York Times Almanac the mean family size in the United States is 3.18. A sample of a college math class resulted in the following family sizes: 5; 4; 5; 4; 4; 3; 6; 4; 3; 3; 5; 5; 6; 3; 3; 2; 7; 4; 5; 2; 2; 2; 3; 2 At α = 0.05, is the class’s mean family size greater than the national average? Does the Almanac result remain valid? Why? 115. The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct? REFERENCES 9.1 Null and Alternative Hypotheses Centers for Disease Control and Prevention. (n.d.). Physical activity facts. Retrieved from http://www.cdc.gov/ healthyschools/physicalactivity/facts.htm National Institute of Mental Health. (n.d.). Publications about depression. Retrieved from http://www.nimh.nih.gov/ publicat/depression.cfm 572 Chapter 9 \| Hypothesis Testing with One Sample 9.5 Additional Information and Full Hypothesis Test Examples American Automobile Association. (n.d.). Retrieved from www.aaa.com American Library Association. (n.d.). Retrieved from www.ala.org Amit Schitai. (n.d.). Data. Bureau of Labor Statistics. (n.d.). Occupational employment statistics. Retrieved from http://www.bls.gov/oes/current/ oes291111.htm Centers for Disease Control and Prevention. (n.d.). Retrieved from www.cdc.gov De Anza College. (2006). Foothill-De Anza Community College District. Retrieved from http://research.fhda.edu/factbook/ DAdemofs/Fact_sheet_da_2006w.pdf Federal Bureau of Investigation. (n.d.). Uniform crime reports and index of crime in Daviess in the state of Kentucky enforced by Daviess County from 1985 to 2005. Retrieved from http://www.disastercenter.com/kentucky/crime/3868.htm Gallup. (n.d.). Retrieved from www.gallup.com Johansen, C., Boice, Jr. J., McLaughlin, J., & Olsen, J. (2001, Feb. 7). Cellular telephones and cancer—a nationwide cohort study in Denmark. Journal of National Cancer Institute, 93(3), 203–7. Retrieved from http://www.ncbi.nlm.nih.gov/ pubmed/11158188 La Leche League International. (n.d.). Retrieved from http://www.lalecheleague.org/Law/BAFeb01.html Online Learning Consortium. (2005 Nov.). Growing by degrees: Online education in the United States, 2005. Newburyport, MA: Allen, I. E., & Seaman, J. Available at http://files.eric.ed.gov/fulltext/ED530062.pdf Toastmasters detail.asp?CategoryID=1&SubCategoryID=10&ArticleID=429&Page=1 International. Retrieved (n.d.). from http://toastmasters.org/artisan/ U.S. Census Bureau. (n.d.). Language use. Retrieved from https://www.census.gov/topics/population/language-use.html U.S. Census Bureau. (n.d.). QuickFacts. Retrieved from https://www.census.gov/quickfacts/table/PST045216/00 U.S. Department of Energy. (n.d.). Retrieved from http://energy.gov Weather Underground. (n.d.). Retrieved from www.wunderground.com SOLUTIONS 1 The random variable is the mean Internet speed in megabits per second. 3 The random variable is the mean number of children an American family has. 5 The random variable is the proportion of peopl
e picked at random in Times Square visiting the city. 7 a. H0: p = 0.42 b. Ha: p < 0.42 9 a. H0: μ = 15 b. Ha: μ ≠ 15 11 Type I: The mean price of mid-sized cars is $32,000, but we conclude that it is not $32,000. Type II: The mean price of mid-sized cars is not $32,000, but we conclude that it is $32,000. 13 α = the probability that you think the bag cannot withstand –15 degrees F, when, in fact, it can. β = the probability that you think the bag can withstand –15 degrees F, when, in fact, it cannot. 15 Type I: The procedure will go well, but the doctors think it will not. Type II: The procedure will not go well, but the doctors think it will. 17 0.019 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 573 19 0.998 21 A normal distribution or a Student’s t-distribution 23 Use a Student’s t-distribution 25 a normal distribution for a single population mean 27 It must be approximately normally distributed. 29 They must both be greater than five. 31 binomial distribution 33 The outcome of winning is very unlikely. 35 H0: μ > = 73 Ha: μ < 73 The p-value is almost zero, which means there is sufficient data to conclude that the mean height of high school students who play basketball on the school team is less than 73 inches at the 5 percent level. The data do support the claim. 37 The shaded region shows a low p-value. 39 Do not reject H0. 41 means 43 the mean time spent on homework for 26 students 45 a. 3 b. 1.5 c. 1.8 d. 26 ¯ 47 X ~ N ⎛ ⎝2.5, 1.5 26 ⎞ ⎠ 49 This is a left-tailed test. 51 This is a two-tailed test. 53 Figure 9.23 55 a right-tailed test 57 a left-tailed test 59 This is a left-tailed test. 61 This is a two-tailed test. Chapter 9 \| Hypothesis Testing with One Sample 574 62 a. H0: μ = 34; Ha: μ ≠ 34 b. H0: p ≤ 0.60; Ha: p > 0.60 c. H0: μ ≥ 100,000; Ha: μ < 100,000 d. H0: p = 0.29; Ha: p ≠ 0.29 e. H0: p = 0.05; Ha: p < 0.05 f. H0: μ ≤ 10; Ha: μ > 10 g. H0: p = 0.50; Ha: p ≠ 0.50 h. H0: μ = 6; Ha: μ ≠ 6 i. H0: p ≥ 0.11; Ha: p < 0.11 j. H0: μ ≤ 20,000; Ha: μ > 20,000 64 c 66 a. Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years. b. Type I error: We conclude that more than 60 percent of Americans vote in presidential elections, when the actual percentage is at most 60 percent.Type II error: We conclude that at most 60 percent of Americans vote in presidential elections when, in fact, more than 60 percent do. c. Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000. d. Type I error: We conclude that the proportion of high school seniors who take physical education daily is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who take physical education daily is 29% when, in fact, it is not 29%. e. Type I error: We conclude that fewer than 5 percent of adults ride the bus to work in Los Angeles, when the percentage that do is really 29%. Type II error: We conclude that 29%. or more adults ride the bus to work in Los Angeles when, in fact, fewer that 29% do. f. Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10. g. Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half. h. Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not. i. Type I error: We conclude that the proportion is less than 11 percent, when it is really at least 11 percent. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11 percent, when in fact it is less than 11 percent. j. Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000. 68 b 70 d 72 d 74 a. H0: μ ≥ 50,000 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 575 b. Ha: μ < 50,000 c. Let X ¯ = the average lifespan of a brand of tires. d. normal distribution e. z = -2.315 f. p-value = 0.0103 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is less than 0.05. iv. Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles. i. (43,537, 49,463) 75 a. H0: μ ≥ 35.5 b. Ha: μ < 35.5 c. Let x¯ = the average mpg for the sample of cars and trucks in the fleet d. normal distribution e. z = -0.648 f. p-value = 0.2578 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. Conclusion: There is sufficient evidence to support the claim that the manufacturer’s fleet meets the fuel economy standards in the 2016 policy. i. (31.88 mpg, 37.32 mpg) 76 a. H0: μ = $1.00 b. Ha: μ ≠ $1.00 c. Let x¯ = the average cost of a daily newspaper. d. normal distribution e. z = –0.866 f. p-value = 0.3865 g. Check student’s solution. h. i. Alpha: 0.01 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.01. iv. Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1. i. ($0.84, $1.06) Chapter 9 \| Hypothesis Testing with One Sample 576 78 a. H0: μ = 10 b. Ha: μ ≠ 10 c. Let X ¯ = the mean number of sick days an employee takes per year. d. Student’s t-distribution e. t = –1.12 f. p-value = 0.300 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. Conclusion: At the 5 percent significance level, there is insufficient evidence to conclude that the mean number of sick days is not 10. i. (4.9443, 11.806) 80 a. H0: p ≥ 0.6 b. Ha: p < 0.6 c. Let P′ = the proportion of students who feel more enriched as a result of taking elementary statistics. d. normal for a single proportion e. 1.12 f. p-value = 0.1308 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched. i. Confidence interval: (0.409, 0.654) The “plus-4s” confidence interval is (0.411, 0.648) 82 a. H0: μ = 4 b. Ha: μ ≠ 4 ¯ c. Let X the average I.Q. of a set of brown trout. d. e. two-tailed Student's t-test t = 1.95 f. p-value = 0.076 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 577 iv. Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four. i. (3.8865, 5.9468) 84 a. H0: p ≥ 0.13 b. Ha: p < 0.13 c. Let P′ = the proportion of Americans who have the disease d. normal for a single proportion e. –2.688 f. p-value = 0.0036 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is less than 0.05. iv. Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have been diagnosed with the disease is less than 13 percent. i. (0, 0.0623). The plus-4s confidence interval is (0.0022, 0.0978) 86 a. H0: μ ≥ 129 b. Ha: μ < 129 c. Let X ¯ = the average time in seconds that Terri finishes Lap 4. d. Student's t-distribution e. t = 1.209 f. 0.8792 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. Conclusion: There is insufficient evidence to conclude that Terri’s mean lap time is less than 129 seconds. i. (128.63, 130.37) 88 a. H0: p = 0.60 b. Ha: p < 0.60 c. Let P′ = the proportion of family members who shed tears at a reunion. d. normal for a single proportion e. –1.71 f. 0.0438 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. 578 Chapter 9 \| Hypothesis Testing with One Sample iii. Reason for decision: p-value < alpha iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the p-value and alpha are quite close, so other tests should be done. i. We are 95 percent confident that between 38.29 percent and 61.71 percent of family members will shed tears at a family reunion. (0.3829, 0.6171). The plus-4s confidence interval (see chapter 8) is (0.3861, 0.6139) Note that here the large-sample 1 – PropZTest provides the approximate p-value of 0.0438. Whenever a p-value based on a normal approximation is close to the level of significance, the exact p-value based on binomial probabilities should be calculated wh
enever possible. This is beyond the scope of this course. 89 a. H0: μ ≥ 22 b. Ha: μ < 22 c. Let X ¯ = the mean number of bubbles per blow. d. Student's t-distribution e. –2.667 f. p-value = 0.00486 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is less than 0.05. iv. Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22. i. (18.501, 21.499) 91 a. H0: μ ≤ 1 b. Ha: μ > 1 c. Let X ¯ = the mean cost in dollars of macaroni and cheese in a certain town. d. Student's t-distribution e. t = 0.340 f. p-value = 0.36756 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05 iv. Conclusion: The mean cost could be $1, or less. At the 5 percent significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1. i. (0.8291, 1.241) 93 a. H0: p = 0.01 b. Ha: p > 0.01 c. Let P′ = the proportion of errors generated d. Normal for a single proportion This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 579 e. 2.13 f. 0.0165 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is less than 0.05. iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01. i. Confidence interval: (0, 0.094). The plus-4s confidence interval is (0.004, 0.144). 95 a. H0: p = 0.50 b. Ha: p < 0.50 c. Let P′ = the proportion of friends that has a pierced ear. d. normal for a single proportion e. –1.70 f. p-value = 0.0448 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is less than 0.05. (However, they are very close.) iv. Conclusion: There is sufficient evidence to support the claim that less than 50 percent of his friends have pierced ears. i. Confidence interval: (0.245, 0.515): The plus-4s confidence interval is (0.259, 0.519). 97 a. H0: p = 0.40 b. Ha: p < 0.40 c. Let P′ = the proportion of schoolmates who fear public speaking. d. normal for a single proportion e. –1.01 f. p-value = 0.1563 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. Conclusion: There is insufficient evidence to support the claim that less than 40 percent of students at the school fear public speaking. i. Confidence interval: (0.3241, 0.4240): The plus-4s confidence interval is (0.3257, 0.4250). 99 a. H0: p = 0.14 b. Ha: p < 0.14 580 Chapter 9 \| Hypothesis Testing with One Sample c. Let P′ = the proportion of nursing home residents that have the disease. d. normal for a single proportion e. –0.2756 f. p-value = 0.3914 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. At the 5 percent significance level, there is insufficient evidence to conclude that the proportion of nursing home residents that have the disease is less than 0.14. i. Confidence interval: (0.0502, 0.2070): The plus-4s confidence interval (see chapter 8) is (0.0676, 0.2297). 101 a. H0: μ = 69,110 b. Ha: μ > 69,110 c. Let X ¯ = the mean salary in dollars for California registered nurses. d. Student's t-distribution e. t = 1.719 f. p-value: 0.0466 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is less than 0.05. iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110. i. ($68,757, $73,485) 103 a. H0: p ≥ 0.14, Ha: p < 0.14 b. p-value < 0.0002 c. Alpha: 0.05 d. Reject the null hypothesis. e. At the 5 percent significance level, there is sufficient evidence to conclude that the proportion of Harleys stolen is significantly less than their share of all motorcycles. (conclusion a) 105 c 107 a. H0: p = 0.488 Ha: p ≠ 0.488 b. p-value = 0.0114 c. alpha = 0.05 d. Reject the null hypothesis. e. At the 5 percent level of significance, there is enough evidence to conclude that 48.8 percent of families own stocks. f. The survey does not appear to be accurate. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 581 109 a. H0: p = 0.517 Ha: p ≠ 0.517 b. p-value = 0.9203. c. alpha = 0.05. d. Do not reject the null hypothesis. e. At the 5 percent significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517. f. However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation. 111 a. H0: µ ≥ 11.52 Ha: µ < 11.52 b. p-value = 0.000002 which is almost 0. c. alpha = 0.05. d. Reject the null hypothesis. e. At the 5 percent significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average. f. We would make the same conclusion if alpha was 1 percent because the p-value is almost 0. 113 a. H0: µ ≤ 5.8 Ha: µ > 5.8 b. p-value = 0.9987 c. alpha = 0.05 d. Do not reject the null hypothesis. e. At the 5 percent level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year. 115 a. H0: µ ≥ 150 Ha: µ < 150 b. p-value = 0.0622 c. alpha = 0.01 d. Do not reject the null hypothesis. e. At the 1 percent significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average. f. The student academic group’s claim appears to be correct. 582 Chapter 9 \| Hypothesis Testing with One Sample This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 583 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES Figure 10.1 If you want to test a claim that involves two groups (the types of breakfasts eaten east and west of the Mississippi River), you can use a slightly different technique when conducting a hypothesis test. (credit: Chloe Lim) Introduction By the end of this chapter, the student should be able to do the following: Chapter Objectives • Classify hypothesis tests by type • Conduct and interpret hypothesis tests for two population means, population standard deviations known • Conduct and interpret hypothesis tests for two population means, population standard deviations unknown • Conduct and interpret hypothesis tests for two population proportions • Conduct and interpret hypothesis tests for matched or paired samples Studies often compare two groups. For example, researchers are interested in the effect aspirin has in preventing heart 584 Chapter 10 \| Hypothesis Testing with Two Samples attacks. Over the last few years, newspapers and magazines have reported various aspirin studies involving two groups. Typically, one group is given aspirin and the other group is given a placebo. Then, the heart attack rate is studied over several years. There are other situations that deal with the comparison of two groups. For example, studies compare various diet and exercise programs. Politicians compare the proportion of individuals from different income brackets who might vote for them. Students are interested in whether the SAT or GRE preparatory courses really help raise their scores. You have learned to conduct hypothesis tests on single means and single proportions. You will expand upon that in this chapter. You will compare two means or two proportions to each other. The general procedure is the same, just expanded. To compare two means or two proportions, you work with two groups. The groups are classified as independent groups or matched pairs. Independent groups consist of two samples that are independent, that is, sample values selected from one population are not related in any way to sample values selected from the other population. Matched pairs consist of two samples that are dependent. The parameter tested using matched pairs is the population mean. The parameters tested using independent groups are either population means or population proportions. NOTE This chapter relies on either a calculator or a computer to calculate the degrees of freedom, the test statistics, and p values. TI-83+ and TI-84 instructions are included, as well as the test statistic formulas. When using a TI-83+ or TI-84 calculator, we do not need to separate two population means, independent groups, or population variances unknown into large and small sample sizes. However, most statistical computer software has the ability to differentiate these tests. This chapter deals with the following hypothesis tests: • Independent groups (samples are independent) ◦ Test of two population means ◦ Test of two population proportions • Matched or paired samples (samples are dependent) ◦ Test of the two population proportions by testing one population mean of differences 10.1 \| Two Population Means with Unknown Standard Deviations 1. The two independent samples are simple random samples from two distinct populations. 2. For the two distinct populations ◦ ◦ if the sample sizes are small, the distributions are important (should be normal), and if the sample sizes are large, the distributions are not important (need not be normal) The test comparing
two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch t-test. The degrees of freedom formula was developed by Aspin-Welch. The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual ¯ samples. To account for the variation, we take the difference of the sample means, X error to standardize the difference. The result is a t-score test statistic. ¯ 1 − X 2 , and divide by the standard Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error, ¯ of the difference in sample means, X ¯ 1 − X 2 . The standard error is calculated as follows: This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 585 (s1 )2 n1 (s2 )2 n2 + The test statistic (t-score) is calculated as follows: ( x¯ 1 – x¯ 2 ) – (μ1 – μ2 ) (s2 )2 n2 + (s1 )2 n1 where • s1 and s2, the sample standard deviations, are estimates of σ1 and σ2, respectively, • σ1 and σ1 are the unknown population standard deviations, • x¯ 1 and x¯ 2 are the sample means, and • μ1 and μ2 are the population means. The number of degrees of freedom (df) requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The df are not always a whole number. The test statistic calculated previously is approximated by the Student’s t-distribution with df as follows: Degrees of freedom d f = 2 ⎛ ⎜ ⎝ (s1)2 n1 (s2)2 n2 ⎞ ⎟ ⎠ + 2 ⎛ ⎝ 1 n1 – 1 ⎛ ⎞ ⎜ ⎠ ⎝ (s1)2 n1 ⎞ ⎟ ⎠ + ⎛ ⎝ 1 n2 – 1 ⎛ ⎞ ⎜ ⎠ ⎝ (s2)2 n2 ⎞ ⎟ ⎠ 2 When both sample sizes n1 and n2 are five or larger, the Student’s t approximation is very good. Notice that the sample variances (s1)2 and (s2)2 are not pooled. (If the question comes up, do not pool the variances.) It is not necessary to compute this by hand. A calculator or computer easily computes it. Example 10.1 Independent groups The average amount of time boys and girls aged 7 to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in Table 10.1. Each populations has a normal distribution. Sample Size Average Number of Hours Playing Sports per Day Sample Standard Deviation Girls 9 Boys 16 Table 10.1 2 3.2 0.866 1.00 Is there a difference in the mean amount of time boys and girls aged 7 to 11 play sports each day? Test at the 5 percent level of significance. Solution 10.1 The population standard deviations are not known. Let g be the subscript for girls and b be the subscript for boys. Then, μg is the population mean for girls and μb is the population mean for boys. This is a test of two independent groups, two population means. 586 Chapter 10 \| Hypothesis Testing with Two Samples b ¯ g − X = difference in the sample mean amount of time girls and boys play sports each ¯ Random variable: X day. H0: μg = μb Ha: μg ≠ μb The words the same tell you H0 has an "=". Since there are no other words to indicate Ha, assume it says is different. This is a two-tailed test. H0: μg – μb = 0 Ha: μg – μb ≠ 0 Distribution for the test: Use tdf where df is calculated using the df formula for independent groups, two population means. Using a calculator, df is approximately 18.8462. Do not pool the variances. Calculate the p-value using a Student’s t-distribution: p-value = 0.0054 Graph: Figure 10.2 sg = 0.866 sb = 1 So, x¯ Half the p-value is below –1.2, and half is above 1.2. = 2 – 3.2 = –1.2 g – x¯ b Make a decision: Since α > p-value, reject H0. This means you reject μg = μb. The means are different. Press STAT. Arrow over to TESTS and press 4:2-SampTTest. Arrow over to Stats and press ENTER. Arrow down and enter 2 for the first sample mean, 0.866 for Sx1, 9 for n1, 3.2 for the second sample mean, 1 for Sx2, and 16 for n2. Arrow down to μ1: and arrow to does not equal μ2. Press ENTER. Arrow down to Pooled: and No. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.0054, the dfs are approximately 18.8462, and the test statistic is –3.14. Do the procedure again, but instead of Calculate do Draw. Conclusion—: At the 5 percent level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that girls and boys aged 7 to 11 play sports per day is different (mean number of hours boys aged 7 to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged 7 to 11 play sports per day is greater than the mean number of hours played by boys). This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 587 10.1 Two samples are shown in Table 10.2. Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5 percent level of significance. Sample Size Sample Mean Sample Standard Deviation Population A 25 Population B 16 Table 10.2 5 4.7 1 1.2 NOTE When the sum of the sample sizes is larger than 30 (n1 + n2 > 30), you can use the normal distribution to approximate the Student’s t. Example 10.2 A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is 4 math classes with a standard deviation of 1.5 math classes. College B samples nine graduates. Their average is 3.5 math classes with a standard deviation of 1 math class. The community group believes that a student who graduates from College A has taken more math classes, on average. Both populations have a normal distribution. Test at a 1 percent significance level. Answer the following questions: a. Is this a test of two means or two proportions? Solution 10.2 a. two means b. Are the populations standard deviations known or unknown? Solution 10.2 b. unknown c. Which distribution do you use to perform the test? Solution 10.2 c. Student’s t d. What is the random variable? Solution 10.2 ¯ d. X ¯ A - X B e. What are the null and alternate hypotheses? Write the null and alternate hypotheses in symbols. 588 Chapter 10 \| Hypothesis Testing with Two Samples Solution 10.2 e. Ho : μ A ≤ μB Ha : μ A > μB f. Is this test right-, left-, or two-tailed? Solution 10.2 f. Figure 10.3 right g. What is the p-value? Solution 10.2 g. 0.1928 h. Do you reject or not reject the null hypothesis? Solution 10.2 h. do not reject i. Conclusion: Solution 10.2 i. At the 1 percent level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from College A has taken more math classes, on average, than a student who graduates from College B. 10.2 A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is 5 years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed. a. Are the population standard deviations known? b. Conduct an appropriate hypothesis test. At the 5 percent significance level, what is your conclusion? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 589 Example 10.3 A professor at a large community college wanted to determine whether there is a difference in the means of final exam scores between students who took his statistics course online and the students who took his face-to-face statistics class. He believed that the mean of the final exam scores for the online class would be lower than that of the face-to-face class. Was the professor correct? The randomly selected 30 final exam scores from each group are listed in Table 10.3 and Table 10.4. 67.6 41.2 85.3 55.9 82.4 91.2 73.5 94.1 64.7 64.7 70.6 38.2 61.8 88.2 70.6 58.8 91.2 73.5 82.4 35.5 94.1 88.2 64.7 55.9 88.2 97.1 85.3 61.8 79.4 79.4 Table 10.3 Online Class 77.9 95.3 81.2 74.1 98.8 88.2 85.9 92.9 87.1 88.2 69.4 57.6 69.4 67.1 97.6 85.9 88.2 91.8 78.8 71.8 98.8 61.2 92.9 90.6 97.6 100 95.3 83.5 92.9 89.4 Table 10.4 Face-to-Face Class Is the mean of the final exam scores of the online class lower than the mean of the final exam scores of the faceto-face class? Test at a 5 percent significance level. Answer the following questions: a. Is this a test of two means or two proportions? b. Are the population standard deviations known or unknown? c. Which distribution do you use to perform the test? d. What is the random variable? e. What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols. f. Is this test right-, left-, or two-tailed? g. What is the p-value? h. Do you reject or not reject the null hypothesis? i. At the _____ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______. (See the conclusion in Example 10.2, and write yours in a similar fashion.) First put the data for each group into two lists (such as L1 and L2). Press STAT. Arrow over to TESTS and press 4:2SampTTest. Make sure Data is highlighted and press ENTER. Arrow down and enter L1 for the first list and L2 for the second list. Arrow down to μ1: and arrow to ≠ μ2 (does not equal). Press ENTER. Arrow down to Pooled: No. Press ENTER. Arrow down to Calculate and press ENTER. NOTE Be careful not to mix up the information for Group 1 and Group 2! 590 Chapter 10 \| Hypothesis Testing with Two Samples Solution 10.3 two means a. b. unknown c. Student’s t ¯ d. X ¯ 1 – X 2 e. 1. H0: μ1 = μ2 Null
hypothesis: The means of the final exam scores are equal for the online and face-to- face statistics classes. 2. Ha: μ1 < μ2 Alternative hypothesis: The mean of the final exam scores of the online class is less than the mean of the final exam scores of the face-to-face class. f. left-tailed g. p-value = 0.0011 Figure 10.4 h. Reject the null hypothesis. i. The professor was correct. The evidence shows that the mean of the final exam scores for the online class is lower than that of the face-to-face class. At the 5 percent level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean of the final exam scores for the online class is less than the mean of final exam scores of the face-to-face class. Cohen’s Standards for Small, Medium, and Large Effect Sizes Cohen’s d is a measure of effect size based on the differences between two means. Cohen’s d, named for U.S. statistician Jacob Cohen, measures the relative strength of the differences between the means of two populations based on sample data. The calculated value of effect size is then compared to Cohen’s standards of small, medium, and large effect sizes. Size of Effect d Small medium Large 0.2 0.5 0.8 Table 10.5 Cohen’s Standard Effect Sizes Cohen’s d is the measure of the difference between two means divided by the pooled standard deviation: d = x¯ 1 – x¯ s pooled 2 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 591 where s pooled = (n1 – 1)s1 2 2 + (n2 – 1)s2 n1 + n2 – 2 . Example 10.4 Calculate Cohen’s d for Example 10.2. Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem. Solution 10.4 μ1 = 4 s1 = 1.5 n1 = 11 μ2 = 3.5 s2 = 1 n2 = 9 d = 0.384 The effect is small because 0.384 is between Cohen’s value of 0.2 for small effect size and 0.5 for medium effect size. The size of the differences of the means for the two colleges is small, indicating that there is not a significant difference between them. Example 10.5 Calculate Cohen’s d for Example 10.3. Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem. Solution 10.5 d = 0.834; large, because 0.834 is greater than Cohen’s 0.8 for a large effect size. The size of the differences between the means of the final exam scores of online students and students in a face-to-face class is large, indicating a significant difference. 10.5 Weighted alpha is a measure of risk-adjusted performance of stocks over a period of a year. A high positive weighted alpha signifies a stock whose price has risen, while a small positive weighted alpha indicates an unchanged stock price during the time period. Weighted alpha is used to identify companies with strong upward or downward trends. The weighted alpha for the top 30 stocks of banks in the Northeast and in the West as identified by Nasdaq on May 24, 2013 are listed in Table 10.6 and Table 10.7, respectively. 94.2 75.2 69.6 52.0 48.0 41.9 36.4 33.4 31.5 27.6 77.3 71.9 67.5 50.6 46.2 38.4 35.2 33.0 28.7 26.5 76.3 71.7 56.3 48.7 43.2 37.6 33.7 31.8 28.5 26.0 Table 10.6 Northeast 126.0 70.6 65.2 51.4 45.5 37.0 33.0 29.6 23.7 22.6 116.1 70.6 58.2 51.2 43.2 36.0 31.4 28.7 23.5 21.6 78.2 68.2 55.6 50.3 39.0 34.1 31.0 25.3 23.4 21.5 Table 10.7 West Is there a difference in the weighted alpha of the top 30 stocks of banks in the Northeast and in the West? Test at a 5 percent significance level. Answer the following questions: 592 Chapter 10 \| Hypothesis Testing with Two Samples a. Is this a test of two means or two proportions? b. Are the population standard deviations known or unknown? c. Which distribution do you use to perform the test? d. What is the random variable? e. What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols. f. Is this test right-, left-, or two-tailed? g. What is the p-value? h. Do you reject or not reject the null hypothesis? i. At the _____ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______. j. Calculate Cohen’s d and interpret it. 10.2 \| Two Population Means with Known Standard Deviations Even though this situation is not likely (knowing the population standard deviations), the following example illustrates hypothesis testing for independent means, known population standard deviations. The sampling distribution for the ¯ difference between the means is normal, and both populations must be normal. The random variable is X normal distribution has the following format: Normal distribution is ¯ 1 – X 2 . The ¯ μ1 – μ2, ⎣ (σ1)2 n1 + (σ2)2 n2 ⎤ ⎥. ⎦ The standard deviation is (σ1)2 n1 (σ2)2 n2 . + The test statistic (z-score) is z = Example 10.6 ( x¯ 1 – x¯ 2) – (μ1 – μ2) (σ2)2 n2 + (σ1)2 n1 . Independent groups, population standard deviations known: The mean lasting time of two competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. Both populations have a normal distribution. The data are recorded in Table 10.8. Wax Sample Mean Number of Months Floor Wax Lasts Population Standard Deviation 1 2 3 2.9 Table 10.8 0.33 0.36 Does the data indicate that Wax 1 is more effective than Wax 2? Test at a 5 percent level of significance. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 593 Solution 10.6 This is a test of two independent groups, two population means, population standard deviations known. ¯ Random Variable: X ¯ 1 – X 2 = difference in the mean number of months the competing floor waxes last. H0: μ1 ≤ μ2 Ha: μ1 > μ2 The words is more effective says that Wax 1 lasts longer than Wax 2, on average. Longer is a > symbol and goes into Ha. Therefore, this is a right-tailed test. Distribution for the test: The population standard deviations are known, so the distribution is normal. Using the formula, the distribution is ¯ 0, 0.332 20 + 0.362 20 ⎞ ⎠. Since μ1 ≤ μ2, then μ1 – μ2 ≤ 0 and the mean for the normal distribution is zero. Calculate the p value using the normal distribution: p value = 0.1799 Graph: Figure 10..9 = 0.1 Compare α and the p value: α = 0.05 and p value = 0.1799. Therefore, α < p value. Make a decision: Since α < p value, do not reject H0. Conclusion: At the 5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the mean time Wax 1 lasts is longer (Wax 1 is more effective) than the mean time Wax 2 lasts. Press STAT. Arrow over to TESTS and press 3:2-SampZTest. Arrow over to Stats and press ENTER. Arrow down and enter .33 for sigma1, .36 for sigma2, 3 for the first sample mean, 20 for n1, 2.9 for the second sample mean, and 20 for n2. Arrow down to μ1: and arrow to > μ2. Press ENTER. Arrow down to Calculate and press ENTER. The p value is p = 0.1799, and the test statistic is 0.9157. Do the procedure again, but instead of Calculate do Draw. 594 Chapter 10 \| Hypothesis Testing with Two Samples 10.6 The means of the number of revolutions per minute of two competing engines are to be compared. Thirty engines are randomly assigned to be tested. Both populations have normal distributions. Table 10.9 shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1? Test at a 5 percent level of significance. Engine Sample Mean Number of RPM Population Standard Deviation 1 2 1,500 1,600 Table 10.9 50 60 Example 10.7 An interested citizen wanted to know if Democratic U.S. senators are older than Republican U.S. senators, on average. On May 26, 2013, the mean age of 30 randomly selected Republican senators was 61 years 247 days (61.675 years) with a standard deviation of 10.17 years. The mean age of 30 randomly selected Democratic senators was 61 years 257 days (61.704 years) with a standard deviation of 9.55 years. Do the data indicate that Democratic senators are older than Republican senators, on average? Test at a 5 percent level of significance. Solution 10.7 This is a test of two independent groups, two population means. The population standard deviations are unknown, but the sum of the sample sizes is 30 + 30 = 60, which is greater than 30, so we can use the normal approximation to the Student’s-t distribution. Subscripts: 1: Democratic senators; 2: Republican senators ¯ Random variable: X ¯ 1 – X 2 = difference in the mean age of Democratic and Republican U.S. senators. H0: µ1 ≤ µ2 H0: µ1 – µ2 ≤ 0 Ha: µ1 > µ2 Ha: µ1 – µ2 > 0 The words older than translates as a > symbol and goes into Ha. Therefore, this is a right-tailed test. Distribution for the test: The distribution is the normal approximation to the Student’s t for means, independent groups. Using the formula, the distribution is ¯ () X ¯ 1 – X 2 ∼ N[0, (9.55)2 30 + (10.17)2 30 ] Since µ1 ≤ µ2, µ1 – µ2 ≤ 0 and the mean for the normal distribution is zero. Calculating the p value using the normal distribution gives p value = 0.4040. Graph: This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 595 Figure 10.6 Compare α and the p value: α = 0.05 and p value = 0.4040. Therefore, α < p value. Make a decision: Since α < p value, do not reject H0. Conclusion: At the 5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of Democratic senators is greater than the mean age of the Republican senators. 10.3 \| Comparing Two Independent Population Proportions When conducting a hypothesis test that compares two independent population proportions, the following characteristics should be present: 1. The two independent samples are simple random samples that are independent. 2. The number of successes is at least five, and the number of failures is at least five, for each of the samples. 3. Growing literature states that the population mus
t be at least 10 or 20 times the size of the sample. This keeps each population from being over-sampled and causing incorrect results. Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populations or it may be due to chance. A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the population proportions. The difference of two proportions follows an approximate normal distribution. Generally, the null hypothesis states that the two proportions are the same. That is, H0: pA = pB. To conduct the test, we use a pooled proportion, pc. The pooled proportion is calculated as follows: The distribution for the differences is pc = x A + xB n A + nB . P′ A − P′B ~ N[0, pc(1 − pc)( 1 n A + 1 nB )]. The test statistic (z-score) is z = Example 10.8 (p′ A − p′B) − (p A − pB) ) pc(1 − pc)( 1 n A + 1 nB . Two types of medication for hives are being tested to determine if there is a difference in the proportions of adult patient reactions. Twenty out of a random sample of 200 adults given Medication A still had hives 30 minutes 596 Chapter 10 \| Hypothesis Testing with Two Samples after taking the medication. Twelve out of another random sample of 200 adults given Medication B still had hives 30 minutes after taking the medication. Test at a 1 percent level of significance. Solution 10.8 The problem asks for a difference in proportions, making it a test of two proportions. Let A and B be the subscripts for Medication A and Medication B, respectively. Then, pA and pB are the desired population proportions. Random Variable: P′A – P′B = difference in the proportions of adult patients who did not react after 30 minutes to Medication A and to Medication B. H0: pA = pB pA – pB = 0 Ha: pA ≠ pB pA – pB ≠ 0 The words is a difference tell you the test is two-tailed. Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal: pc = x A + xB n A + nB = 20 + 12 200 + 200 = 0.08 1 – pc = 0.92 P′ A – P′B ~ N ⎡ ⎣0, (0.08)(0.92)( 1 200 + 1 200 ⎤ ) ⎦ P′A – P′B follows an approximate normal distribution. Calculate the p-value using the normal distribution: p-value = 0.1404. Estimated proportion for group A: p′ A = Estimated proportion for group B: p′B = x A n A xB nB = 20 200 = 12 200 = 0.1 = 0.06 Graph: Figure 10.7 P′A – P′B = 0.1 – 0.06 = 0.04. Half the p-value is below –0.04, and half is above 0.04. Compare α and the p-value: α = 0.01 and the p-value = 0.1404. α < p-value. Make a decision: Since α < p-value, do not reject H0. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 597 Conclusion: At a 1 percent level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the proportions of adult patients who did not react after 30 minutes to Medication A and Medication B. Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 20 for x1, 200 for n1, 12 for x2, and 200 for n2. Arrow down to p1: and arrow to not equal p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1404, and the test statistic is 1.47. Do the procedure again, but instead of Calculate do Draw. 10.8 Two types of valves are being tested to determine if there is a difference in pressure tolerances. Fifteen out of a random sample of 100 of Valve A cracked under 4,500 psi. Six out of a random sample of 100 of Valve B cracked under 4,500 psi. Test at a 5 percent level of significance. Example 10.9 A research study was conducted about gender differences in texting. The researcher believed that the proportion of girls involved in texting is less than the proportion of boys involved. The data collected in spring 2010 among a random sample of middle and high school students in a large school district in the southern United States is summarized in Table 10.9. Is the proportion of girls sending texts less than the proportion of boys texting? Test at a 1 percent level of significance. Males Females Sent texts 183 Total number surveyed 2231 156 2169 Table 10.10 Solution 10.9 This is a test of two population proportions. Let M and F be the subscripts for males and females. Then, pM and pF are the desired population proportions. Random variable: p′F − p′M = difference in the proportions of males and females who sent texts. H0: pF = pM H0: pF – pM = 0 Ha: pF < pM Ha: pF – pM < 0 The words less than tell you the test is left-tailed. Distribution for the test: Since this is a test of two population proportions, the distribution is normal: 598 Chapter 10 \| Hypothesis Testing with Two Samples = 156 + 183 2169 + 2231 = 0.077 pc = xF + x M nF + n M 1 − pc = 0.923 Therefore, ⎛ ⎝0, + 1 1 2169 p′F – p′ M ∼ N ⎛ (0.077)(0.923) ⎝ ⎞ ⎞ ⎠ ⎠ 2231 p′F – p′M follows an approximate normal distribution. Calculate the p-value using the normal distribution: p-value = 0.1045 Estimated proportion for females: 0.0719 Estimated proportion for males: 0.082 Graph: Figure 10.8 Decision: Since α < p-value, do not reject H0. Conclusion: At the 1 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the proportion of girls sending texts is less than the proportion of boys sending texts. Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 156 for x1, 2169 for n1, 183 for x2, and 2231 for n2. Arrow down to p1: and arrow to less than p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1045 and the test statistic is z = –1.256. Example 10.10 Researchers conducted a study of smartphone use (Phone A versus Phone B) among adults. A cell phone company claimed that Phone B smartphones are more popular with whites (non-Hispanic) than with African Americans. The results of the survey indicate that of the 232 African American cell phone owners randomly sampled, 5 percent own Phone B. Of the 1,343 white cell phone owners randomly sampled, 10 percent own Phone B. Test at the 5 percent level of significance. Is the proportion of white Phone B owners greater than the proportion of African American Phone B owners? Solution 10.10 This is a test of two population proportions. Let W and A be the subscripts for the whites and African Americans. Then, pW and pA are the desired population proportions. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 599 Random variable: p′W – p′A = difference in the proportions of Phone A and Phone B users. H0: pW = pA H0: pW – pA = 0 Ha: pW > pA Ha: pW – pA > 0 The words more popular indicate that the test is right-tailed. Distribution for the test: The distribution is approximately normal. pc = xW + x A nW + n A = 134 + 12 1343 + 232 = 0.0927 1 − pc = 0.9073 Therefore, p′W – p′ A ∽ N ⎛ ⎝0, ⎛ (0.0927)(0.9073) ⎝ 1 1343 + 1 232 ⎞ ⎞ ⎠ ⎠ p′W – p′ A follows an approximate normal distribution. Calculate the p-value using the normal distribution: p-value = 0.0077 Estimated proportion for group A: 0.10 Estimated proportion for group B: 0.05 Graph: Figure 10.9 Decision: Since α > p-value, reject the H0. Conclusion: At the 5 percent level of significance, from the sample data, there is sufficient evidence to conclude that a larger proportion of white cell phone owners use Phone B than African Americans. TI-83+ and TI-84: Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 135 for x1, 1343 for n1, 12 for x2, and 232 for n2. Arrow down to p1: and arrow to greater than p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.0092, and the test statistic is z = 2.33. 600 Chapter 10 \| Hypothesis Testing with Two Samples 10.10 A group of citizens wanted to know if the proportion of homeowners in their small city was different in 2011 than in 2010. Their research showed that of the 113,231 available homes in their city in 2010, 7,622 of them were owned by the families who live there. In 2011, 7,439 of the 104,873 of the available homes were owned by city residents. Test at a 5 percent significance level. Answer the following questions: a. Is this a test of two means or two proportions? b. Which distribution do you use to perform the test? c. What is the random variable? d. What are the null and alternative hypotheses? Write the null and alternative hypotheses in symbols. e. Is this test right-, left-, or two-tailed? f. What is the p-value? g. Do you reject or not reject the null hypothesis? h. At the ______ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______. 10.4 \| Matched or Paired Samples (Optional) When using a hypothesis test for matched or paired samples, the following characteristics should be present: 1. Simple random sampling is used. 2. Sample sizes are often small. 3. Two measurements (samples) are drawn from the same pair of individuals or objects. 4. Differences are calculated from the matched or paired samples. 5. The differences form the sample that is used for the hypothesis test. 6. Either the matched pairs have differences that come from a population that is normal or the number of differences is sufficiently large so that distribution of the sample mean of differences is approximately normal. In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated. The differences are the data. The population mean for the differences, μd, is then tested using a Student’s-t test for a single population mean with n – 1 degrees of freedom, where n is the number of differences. The test statistic (t-score) is x¯ t = d − μd sd ⎞ ⎛ ⎠ ⎝ n . Example 10.11 A study was conducted to investigate the effectiveness of pain-reducing medication. Results for randomly selected subjects are shown in Table 10.10. A lower score
indicates less pain. The before value is matched to an after value, and the differences are calculated. The differences have a normal distribution. Are the sensory measurements, on average, lower after the medication? Test at a 5 percent significance level. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 601 Subject: A B C D E F G H Before 6.6 6.5 9.0 10.3 11.3 8.1 6.3 11.6 After 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2.0 Table 10.11 Solution 10.11 Corresponding before and after values form matched pairs. (Calculate after – before.) After Data Before Data Difference 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2 Table 10.12 6.6 6.5 9 10.3 11.3 8.1 6.3 11.6 0.2 –4.1 –1.6 –1.8 –3.2 –2 –2.9 –9.6 The data for the test are the differences: {0.2, –4.1, –1.6, –1.8, –3.2, –2, –2.9, –9.6} The sample mean and sample standard deviation of the differences are: xd = –3.13 and sd = 2.91 Verify these values. Let μd be the population mean for the differences. We use the subscript d to denote differences. ¯ Random variable: X d = the mean difference of the sensory measurements. H0: μd ≥ 0 The null hypothesis is zero or positive, meaning that there is the same or more pain felt after taking the medication. That means the subject shows no improvement. μd is the population mean of the differences. Ha: μd < 0 The alternative hypothesis is negative, meaning there is less pain felt after taking the medication. That means the subject shows improvement. The score should be lower after taking the medication, so the difference ought to be negative to indicate improvement. Distribution for the test: The distribution is a Student’s t with df = n – 1 = 8 – 1 = 7. Use t7. Note —that the test is for a single population mean. Calculate the p-value using the Student’s-t distribution: p-value = 0.0095 Graph: 602 Chapter 10 \| Hypothesis Testing with Two Samples Figure 10.10 ¯ X d is the random variable for the differences. The sample mean and sample standard deviation of the differences are as follows: x¯ s¯ d d = –3.13 = 2.91 Compare α and the p-value: α = 0.05 and p-value = 0.0095. α > p-value. Make a decision: Since α > p-value, reject H0. This means that μd < 0 and there is improvement. Conclusion: At a 5 percent level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after taking the medication. The medication appears to be effective in reducing pain. NOTE For the TI-83+ and TI-84 calculators, you can either calculate the differences ahead of time (after before) and put the differences into a list or you can put the after data into a first list and the before data into a second list. Then, go to a third list and arrow up to the name. Enter 1st list name – 2nd list name. The calculator will do the subtraction, and you will have the differences in the third list. Use your list of differences as the data. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 0 for μ0 , the name of the list where you put the data, and 1 for Freq:. Arrow down to μ: and arrow over to < μ0. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is 0.0094, and the test statistic is –3.04. Do these instructions again except, arrow to Draw instead of Calculate. Press ENTER. 10.11 A study was conducted to investigate how effective a new diet was in lowering cholesterol. Results for the randomly selected subjects are shown in the table. The differences have a normal distribution. Are the subjects’ cholesterol levels lower on average after the diet? Test at the 5 percent level. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 603 Subject A B C D E F G H I Before 209 210 205 198 216 217 238 240 222 After 199 207 189 209 217 202 211 223 201 Table 10.13 Example 10.12 A college football coach was interested in whether the college’s strength development class increased his players’ maximum lift (in pounds) on the bench press exercise. He asked four of his players to participate in a study. The amount of weight they could each lift was recorded before they took the strength development class. After completing the class, the amount of weight they could each lift was again measured. The data are as follows: Weight (in pounds) Player 1 Player 2 Player 3 Player 4 Amount of weight lifted prior to the class 205 Amount of weight lifted after the class 295 241 252 338 330 368 360 Table 10.14 The coach wants to know if the strength development class makes his players stronger, on average. Record the differences data. Calculate the differences by subtracting the amount of weight lifted prior to the class from the weight lifted after completing the class. The data for the differences are: {90, 11, -8, -8}. Assume the differences have a normal distribution. Using the differences data, calculate the sample mean and the sample standard deviation. x¯ d = 21.3, sd = 46.7 NOTE The data given here would indicate that the distribution is right-skewed. The difference 90 may be an extreme outlier. It is pulling the sample mean to be 21.3 (positive). The means of the other three data values are negative. Using the difference data, this becomes a test of a single __________. ¯ Define the random variable: X d is the mean difference in the maximum lift per player. The distribution for the hypothesis test is t3. H0: μd ≤ 0, Ha: μd > 0 Graph: 604 Chapter 10 \| Hypothesis Testing with Two Samples Figure 10.11 Calculate the p-value: The p-value is 0.2150. Decision: If the level of significance is 5 percent, the decision is not to reject the null hypothesis, because α < p-value. What is the conclusion? At a 5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the strength development class helped make the players stronger, on average. 10.12 A new prep class was designed to improve SAT test scores. Five students were selected at random. Their scores on two practice exams were recorded, one before the class and one after. The data are recorded in Table 10.15. Are the scores, on average, higher after the class? Test at a 5 percent level. SAT Scores Student 1 Student 2 Student 3 Student 4 Score before class 1840 Score after class 1920 1960 2160 1920 2200 2150 2100 Table 10.15 Example 10.13 Seven eighth-graders at Kennedy Middle School measured how far they could push the shot put with their dominant (writing) hand and their weaker (nonwriting) hand. They thought that they could push equal distances with both hands. The data are collected and recorded in Table 10.16. Distance (in feet) using Student 1 Student 2 Student 3 Student 4 Student 5 Student 6 Student 7 Dominant Hand Weaker Hand 30 28 26 14 34 27 17 18 19 17 26 26 20 16 Table 10.16 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 605 Conduct a hypothesis test to determine whether the mean difference in distances between the children’s dominant versus weaker hands is significant. Record the differences data. Calculate the differences by subtracting the distances with the weaker hand from the distances with the dominant hand. The data for the differences are: {2, 12, 7, –1, 2, 0, 4}. The differences have a normal distribution. Using the differences data, calculate the sample mean and the sample standard deviation. x¯ = 3.71, sd d = 4.5. ¯ Random variable: X d = mean difference in the distances between the hands. Distribution for the hypothesis test: t6 H0: μd = 0 Ha: μd ≠ 0 Graph: Figure 10.12 Calculate the p-value: The p-value is 0.0716 (using the data directly). Test statistic = 2.18. p-value = 0.0719 using ⎛ ⎝ x¯ d = 3.71, sd = 4.5 ⎞ ⎠ . Decision: Assume α = 0.05. Since α < p-value, do not reject H0. Conclusion: At the 5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the children’s weaker and dominant hands to push the shot put. 10.13 Five ball players think they can throw the same distance with their dominant hand (throwing) and off-hand (catching hand). The data were collected and recorded in Table 10.17. Conduct a hypothesis test to determine whether the mean 5 difference in distances between the dominant and off-hand is significant. Test at the 5 percent level. Player 1 Player 2 Player 3 Player 4 Player 5 Dominant Hand 120 Off-Hand 105 111 109 135 98 140 111 125 99 Table 10.17 606 Chapter 10 \| Hypothesis Testing with Two Samples 10.5 \| Hypothesis Testing for Two Means and Two Proportions This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 607 10.1 Hypothesis Testing for Two Means and Two Proportions Student Learning Outcomes • The student will select the appropriate distributions to use in each case. • The student will conduct hypothesis tests and interpret the results. Supplies: • The business section from two consecutive days’ newspapers • Three small packages of multicolored chocolates • Five small packages of peanut butter candies Increasing Stocks Survey Look at yesterday’s newspaper business section. Conduct a hypothesis test to determine if the proportion of New York Stock Exchange (NYSE) stocks that increased is greater than the proportion of NASDAQ stocks that increased. As randomly as possible, choose 40 NYSE stocks and 32 NASDAQ stocks and complete the following statements. 1. H0: _________ 2. Ha: _________ 3. In words, define the random variable. 4. The distribution to use for the test is _____________. 5. Calculate the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: Figure 10.13 b. Calculate the p value. 7. Do you reject or not reject the null hypothesis? Why? 8. Write a clear conclusion using a c
omplete sentence. 608 Chapter 10 \| Hypothesis Testing with Two Samples Decreasing Stocks Survey Randomly pick eight stocks from the newspaper. Using two consecutive days’ business sections, test whether the stocks went down, on average, for the second day. 1. H0: ________ 2. Ha: ________ 3. In words, define the random variable. 4. The distribution to use for the test is _____________. 5. Calculate the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph Figure 10.14 b. Calculate the p value: 7. Do you reject or not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. Candy Survey Buy three small packages of multicolored chocolates and five small packages of peanut butter candies (same net weight as the multicolored chocolates). Test whether the mean number of candy pieces per package is the same for the two brands. 1. H0: ________ 2. Ha: ________ 3. In words, define the random variable. 4. What distribution should be used for this test? 5. Calculate the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 609 Figure 10.15 b. Calculate the p value. 7. Do you reject or not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. Shoe Survey Test whether women have, on average, more pairs of shoes than men. Include all forms of sneakers, shoes, sandals, and boots. Use your class as the sample. 1. H0: ________ 2. Ha: ________ 3. In words, define the random variable. 4. The distribution to use for the test is ________________. 5. Calculate the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph Figure 10.16 b. Calculate the p value. 7. Do you reject or not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. 610 Chapter 10 \| Hypothesis Testing with Two Samples KEY TERMS degrees of freedom (df) the number of objects in a sample that are free to vary pooled proportion estimate of the common value of p1 and p2 standard deviation a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation variable (random variable) a characteristic of interest in a population being studied. Common notation for variables are uppercase Latin letters X, Y, Z,... Common notation for a specific value from the domain (set of all possible values of a variable) are lowercase Latin letters x, y, z,.... For example, if X is the number of children in a family, then x represents a specific integer 0, 1, 2, 3, .... Variables in statistics differ from variables in intermediate algebra in two ways: • The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if X = hair color, then the domain is {black, blond, gray, green, orange}. • We can tell what specific value x of the random variable X takes only after performing the experiment. CHAPTER REVIEW 10.1 Two Population Means with Unknown Standard Deviations Two population means from independent samples where the population standard deviations are not known ¯ • Random variable: X ¯ 1 − X 2 = the difference of the sampling means • Distribution: Student’s t-distribution with degrees of freedom (variances not pooled) 10.2 Two Population Means with Known Standard Deviations A hypothesis test of two population means from independent samples where the population standard deviations are known (typically approximated with the sample standard deviations) will have these characteristics: ¯ • Random variable: X ¯ 1 − X 2 = the difference of the means • Distribution: normal distribution 10.3 Comparing Two Independent Population Proportions Test of two population proportions from independent samples • Random variable: p^ A – p^ B = difference between the two estimated proportions • Distribution: normal distribution 10.4 Matched or Paired Samples (Optional) A hypothesis test for matched or paired samples (t-test) has these characteristics: • Test the differences by subtracting one measurement from the other measurement • Random variable: x¯ d = mean of the differences. • Distribution: Student’s t distribution with n – 1 degrees of freedom. • If the number of differences is small (less than 30), the differences must follow a normal distribution. • Two samples are drawn from the same set of objects. • Samples are dependent. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 611 FORMULA REVIEW 10.1 Two Population Means with Unknown Standard Deviations Standard error: SE = (s1)2 n1 (s2)2 n2 + Test statistic (t-score): t = ( x¯ Degrees of freedom: 1 − x¯ 2) − (μ1 − μ2) (s2)2 n2 + (s1)2 n1 ⎛ ⎜ ⎝ (s1)2 n1 + 2 (s2)2 n2 ⎞ ⎟ ⎠ 2 ⎛ ⎝ 1 n1 − 1 ⎛ ⎞ ⎜ ⎠ ⎝ (s1)2 n1 ⎞ ⎟ ⎠ + ⎛ ⎝ 1 n2 − 1 ⎛ ⎞ ⎜ ⎠ ⎝ (s2)2 n2 ⎞ ⎟ ⎠ 2 d f = where: Generally, µ1 - µ2 = 0. where σ1 and σ2 are the known population standard deviations, n1 and n2 are the sample sizes, x¯ are the sample means, and μ1 and μ2 are the population means. and x¯ 1 2 10.3 Comparing Two Independent Population Proportions Pooled proportion: pc = xF + x M nF + n M Distribution for the differences: ⎡ ⎛ ⎣0, pc(1 − pc) ⎝ p′ A − p′B ∼ N 1 n A + 1 nB ⎤ ⎞ ⎦ ⎠ where the null hypothesis is H0: pA = pB = 0 or H0: pA – pB s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes. Test statistic (z-score): z = (p′ A − p′ B) ⎛ pc(1 − pc) ⎝ 1 n A + 1 nB ⎞ ⎠ x¯ 1 and x¯ 2 are the sample means. Cohen’s d is the measure of effect size: d = 1 − x¯ x¯ s pooled 2 where s pooled = (n1 − 1)s1 2 2 + (n2 − 1)s2 n1 + n2 − 2 . 10.2 Two Population Means with Known Standard Deviations Normal distributionμ1 − μ2, ⎣ (σ1)2 n1 + (σ2)2 n2 ⎤ ⎥ . ⎦ Generally, µ1 – µ2 = 0. Test statistic (z-score): where the null hypothesis is H0: pA = pB = 0 or H0: pA − pB and where p′A and p′B are the sample proportions, pA and pB are the population proportions, Pc is the pooled proportion, and nA and nB are the sample sizes. 10.4 Matched or Paired Samples (Optional) Test statistic (t-score): t = x¯ d − μd sd ⎞ ⎛ ⎠ ⎝ n where: d is the mean of the sample differences, μd is the mean x¯ of the population differences, sd is the sample standard deviation of the differences, and n is the sample size. z = ( x¯ 1 − x¯ 2) − (μ1 − μ2) (σ2)2 n2 + (σ1)2 n1 PRACTICE 10.1 Two Population Means with Unknown Standard Deviations Use the following information to answer the next 15 exercises. Indicate if the hypothesis test is for a. independent group means, population standard deviations, and/or variances known, 612 Chapter 10 \| Hypothesis Testing with Two Samples b. independent group means, population standard deviations, and/or variances unknown, c. matched or paired samples, d. e. f. single mean, two proportions, or single proportion. 1. It is believed that 70 percent of males pass their drivers test in the first attempt, while 65 percent of females pass the test in the first attempt. Of interest is whether the proportions are equal. 2. A new laundry detergent is tested on consumers. Of interest is the proportion of consumers who prefer the new brand over the leading competitor. A study is done to test this. 3. A new windshield treatment claims to repel water more effectively. Ten windshields are tested by simulating rain without the new treatment. The same windshields are then treated, and the experiment is run again. A hypothesis test is conducted. 4. The known standard deviation in salary for all mid-level professionals in the financial industry is $11,000. Company A and Company B are in the financial industry. Suppose samples are taken of mid-level professionals from Company A and from Company B. The sample mean salary for mid-level professionals in Company A is $80,000. The sample mean salary for mid-level professionals in Company B is $96,000. Company A and Company B management want to know if their midlevel professionals are paid differently, on average. 5. The average worker in Germany gets eight weeks of paid vacation. 6. According to a television commercial, 80% of dentists agree that a brand of fluoridated toothpaste is the best on the market. 7. It is believed that the average grade on an English essay in a particular school system is higher for females than for males. A random sample of 31 females had a mean score of 82 with a standard deviation of 3, and a random sample of 25 males had a mean score of 76 with a standard deviation of 4. 8. The league mean batting average is 0.280 with a known standard deviation of 0.06. The Rattlers and the Vikings belong to the league. The mean batting average for a sample of eight Rattlers is 0.210, and the mean batting average for a sample of eight Vikings is 0.260. There are 24 players on the Rattlers and 19 players on the Vikings. Are the batting averages of the Rattlers and Vikings statistically different? 9. In a random sample of 100 forests in the United States, 56 were coniferous or contained conifers. In a random sample of 80 forests in Mexico, 40 were coniferous or contained conifers. Is the proportion of conifers in the United States statistically more than the proportion of conifers in Mexico? 10. A new medicine is said to help improve sleep. Eight subjects are picked at random and given the medicine. The mean hours slept for each person were recorded before starting the medication and after. 11. It is thought that teenagers sleep more than adults on average. A study is done to verify this. A sample of 16 teenagers has a mean of 8.9 hours slept and a standard deviation of 1.2. A sample of 12 adults has a mean of 6.9 hours slept and a standard deviation of
0.6. 12. Varsity athletes practice five times a week, on average. 13. A sample of 12 in-state graduate school programs at School A has a mean tuition of $64,000 with a standard deviation of $8,000. At School B, a sample of 16 in-state graduate programs has a mean tuition of $80,000 with a standard deviation of $6,000. On average, are the mean tuitions different? 14. A new WiFi range booster is being offered to consumers. A researcher tests the native range of 12 different routers under the same conditions. The ranges are recorded. Then, the researcher uses the new WiFi range booster and records the new ranges. Does the new WiFi range booster do a better job? 15. A high school principal claims that 30 percent of student athletes drive themselves to school, while 4 percent of nonathletes drive themselves to school. In a sample of 20 student athletes, 45 percent drive themselves to school. In a sample of 35 nonathlete students, 6 percent drive themselves to school. Is the percent of student athletes who drive themselves to school more than the percent of nonathletes? Use the following information to answer the next three exercises: A study is done to determine which of two soft drinks has more sugar. There are 13 cans of Beverage A in a sample and six cans of Beverage B. The mean amount of sugar in Beverage A is 36 grams with a standard deviation of 0.6 grams. The mean amount of sugar in Beverage B is 38 grams with This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 613 a standard deviation of 0.8 grams. The researchers believe that Beverage B has more sugar than Beverage A, on average. Both populations have normal distributions. 16. Are standard deviations known or unknown? 17. What is the random variable? 18. Is this a one-tailed or two-tailed test? Use the following information to answer the next 12 exercises. The U.S. Centers for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites. 19. Is this a test of means or proportions? 20. State the null and alternative hypotheses. a. H0: __________ b. Ha: __________ 21. Is this a right-tailed, left-tailed, or two-tailed test? 22. In symbols, what is the random variable of interest for this test? 23. In words, define the random variable of interest for this test. 24. Which distribution (normal or Student’s t) would you use for this hypothesis test? 25. Explain why you chose the distribution you did for Exercise 10.24. 26. Calculate the test statistic and p-value. 27. Sketch a graph of the situation. Label the horizontal axis. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value. 28. Find the p-value. 29. At a preconceived α = 0.05, write the following: a. Your decision: b. The reason for your decision: c. Your conclusion (write out in a complete sentence): 30. Does it appear that the means are the same? Why or why not? 10.2 Two Population Means with Known Standard Deviations Use the following information to answer the next five exercises. The mean speeds of fastball pitches from two different baseball pitchers are to be compared. A sample of 14 fastball pitches is measured from each pitcher. The populations have normal distributions. Table 10.18 shows the result. Scouters believe that Rodriguez pitches a speedier fastball. Pitcher Sample Mean Speed of Pitches (mph) Population Standard Deviation 3 7 Wesley 86 Rodriguez 91 Table 10.18 31. What is the random variable? 32. State the null and alternative hypotheses. 33. What is the test statistic? 34. What is the p value? 35. At the 1 percent significance level, what is your conclusion? Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of 614 Chapter 10 \| Hypothesis Testing with Two Samples the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller. Plant Group Sample Mean Height of Plants (inches) Population Standard Deviation Food No food Table 10.19 16 14 2.5 1.5 36. Is the population standard deviation known or unknown? 37. State the null and alternative hypotheses. 38. What is the p value? 39. Draw the graph of the p value. 40. At the 1 percent significance level, what is your conclusion? Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. Fifteen pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point. Sample Mean Melting Temperatures (°F) Population Standard Deviation Alloy Gamma 800 Alloy Zeta 900 Table 10.20 95 105 41. State the null and alternative hypotheses. 42. Is this a right-, left-, or two-tailed test? 43. What is the p value? 44. Draw the graph of the p value. 45. At the 1 percent significance level, what is your conclusion? 10.3 Comparing Two Independent Population Proportions Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS2 had system failures within the first eight hours of operation. OS2 is believed to be more stable (have fewer crashes) than OS1. 46. Is this a test of means or proportions? 47. What is the random variable? 48. State the null and alternative hypotheses. 49. What is the p-value? 50. What can you conclude about the two operating systems? Use the following information to answer the next 12 exercises. In the recent U.S. Census, 3 percent of the U.S. population reported being of two or more races. However, the percent varies tremendously from state to state. Suppose that two random surveys are conducted. In the first random survey, out of 1,000 North Dakotans, only 9 people reported being of two or more races. In the second random survey, out of 500 Nevadans, 17 people reported being of two or more races. Conduct This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 615 a hypothesis test to determine if the population percents are the same for the two states or if the percent for Nevada is statistically higher than for North Dakota. 51. Is this a test of means or proportions? 52. State the null and alternative hypotheses. a. H0: _________ b. Ha: _________ 53. Is this a right-tailed, left-tailed, or two-tailed test? How do you know? 54. What is the random variable of interest for this test? 55. In words, define the random variable for this test. 56. Which distribution (normal or Student’s t) would you use for this hypothesis test? 57. Explain why you chose the distribution you did for the Exercise 10.56. 58. Calculate the test statistic. 59. Sketch a graph of the situation. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value. Figure 10.17 60. Find the p-value. 61. At a preconceived α = 0.05, write the following: a. Your decision: b. The reason for your decision: c. Your conclusion (write out in a complete sentence): 62. Does it appear that the proportion of Nevadans who are two or more races is higher than the proportion of North Dakotans? Why or why not? 10.4 Matched or Paired Samples (Optional) Use the following information to answer the next five exercises. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in Table 10.21. The before value is matched to an after value, and the differences are calculated. The differences have a normal distribution. Test at the 1 percent significance level. Installation A B C D E F G H Before After Table 10.21 63. What is the random variable? 64. State the null and alternative hypotheses. 65. What is the p-value? 66. Draw the graph of the p-value. 67. What conclusion can you draw about the software patch? Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal 616 Chapter 10 \| Hypothesis Testing with Two Samples distribution. Test at the 1 percent significance level. Subject A B C D E F Before After Table 10.22 68. State the null and alternative hypotheses. 69. What is the p-value? 70. What is the sample mean difference? 71. Draw the graph of the p-value. 72. What conclusion can you draw about the juggling class? Use the following information to answer the next five exercises. A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded. After twelve weeks on the medication, the same six subjects have their blood pressure recorded again. For this test, only systolic pressure is
of concern. Test at the 1 percent significance level. Patient A B C D E F Before 161 162 165 162 166 171 After 158 159 166 160 167 169 Table 10.23 73. State the null and alternative hypotheses. 74. What is the test statistic? 75. What is the p-value? 76. What is the sample mean difference? 77. What is the conclusion? HOMEWORK 10.1 Two Population Means with Unknown Standard Deviations DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. NOTE If you are using a Student’s t-distribution for a homework problem in what follows, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption.) This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 617 78. The mean number of English courses taken in a two-year period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of 3 English courses with a standard deviation of 0.8. The females took an average of 4 English courses with a standard deviation of 1.0. Are the means statistically the same? 79. A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the United States. Two surveys are conducted. Of the 35 two-year colleges surveyed, the mean enrollment was 5,068 with a standard deviation of 4,777. Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard deviation of 8,191. 80. At Rachel’s eleventh birthday party, eight girls were timed to see how long (in seconds) they could sit perfectly still in a relaxed position. After a two-minute rest, they timed themselves while jumping. The girls thought that the mean difference between their jumping and relaxed times would be zero. Test their hypothesis. Relaxed time (seconds) Jumping time (seconds) 26 47 30 22 23 45 37 29 Table 10.24 21 40 28 21 25 43 35 32 81. Mean entry-level salaries for college graduates with mechanical engineering degrees and electrical engineering degrees are believed to be approximately the same. A recruiting office thinks that the mean mechanical engineering salary is lower than the mean electrical engineering salary. The recruiting office randomly surveys 50 entry-level mechanical engineers and 60 entry-level electrical engineers. Their mean salaries were $46,100 and $46,700, respectively. Their standard deviations were $3,450 and $4,210, respectively. Conduct a hypothesis test to determine if you agree that the mean entry-level mechanical engineering salary is lower than the mean entry-level electrical engineering salary. 82. Marketing companies have collected data implying that teenage girls use more ringtones on their smartphones than teenage boys do. In one study of 40 randomly chosen teenage girls and boys (20 of each) with smartphones, the mean number of ringtones for the girls was 3.2 with a standard deviation of 1.5. The mean for the boys was 1.7 with a standard deviation of 0.8. Conduct a hypothesis test to determine if the means are approximately the same or if the girls’ mean is higher than the boys’ mean. Use the information from Appendix C to answer the next four exercises. 83. Using the data from Lap 1 only, conduct a hypothesis test to determine if the mean time for completing a lap in races is the same as it is in practices. 84. Repeat the test in Exercise 10.83, but use Lap 5 data this time. 85. Repeat the test in Exercise 10.83, but this time combine the data from Laps 1 and 5. 86. In two to three complete sentences, explain in detail how you might use Terri Vogel’s data to answer the following question: Does Terri Vogel drive faster in races than she does in practices? Use the following information to answer the next two exercises. The Eastern and Western Major League Soccer conferences have a new Reserve Division that allows new players to develop their skills. Data for a randomly picked date showed the following annual goals. 618 Chapter 10 \| Hypothesis Testing with Two Samples Western Eastern Los Angeles 9 D United 9 FC Dallas 3 Chicago 8 Chivas USA 4 Columbus 7 Real Salt Lake 3 New England 6 Colorado 4 MetroStars 5 San Jose 4 Kansas City 3 Table 10.25 Conduct a hypothesis test to answer the next two exercises. 87. The exact distribution for the hypothesis test is a. b. c. d. the normal distribution the Student’s t-distribution the uniform distribution the exponential distribution 88. If the level of significance is 0.05, the conclusion is: a. There is sufficient evidence to conclude that the W Division teams score fewer goals, on average, than the E teams. b. There is insufficient evidence to conclude that the W Division teams score more goals, on average, than the E teams. c. There is insufficient evidence to conclude that the W teams score fewer goals, on average, than the E teams. d. There is not sufficient evidence to determine a conclusion. 89. Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The day subscript refers to the statistics day students. The night subscript refers to the statistics night students. Which of the following is a concluding statement: a. There is sufficient evidence to conclude that statistics night students’ mean on Exam 2 is better than the statistics day students’ mean on Exam 2. b. There is insufficient evidence to conclude that the statistics day students’ mean on Exam 2 is better than the statistics night students’ mean on Exam 2. c. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2. d. There is sufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2. 90. Researchers interviewed people in a certain industry in Canada and the United States. The mean age of the 100 Canadians upon entering this industry was 18 with a standard deviation of 6. The mean age of the 130 Americans upon entering this industry was 20 with a standard deviation of 8. Is the mean age of entering this industry in Canada lower than the mean age in the United States? Test at a 1 percent significance level. 91. A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. The powder diet group had a mean weight loss of 42 pounds with a standard deviation of 12 pounds. The liquid diet group had a mean weight loss of 45 pounds with a standard deviation of 14 pounds. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 619 92. Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91, respectively. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The day subscript refers to the statistics day students. The night subscript refers to the statistics night students. An appropriate alternative hypothesis for the hypothesis test is a. μday > μnight b. μday < μnight c. μday = μnight d. μday ≠ μnight 10.2 Two Population Means with Known Standard Deviations DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. NOTE If you are using a Student’s t-distribution for one of the following homework problems, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption.) 93. A study is done to determine if students in the California state university system take longer to graduate, on average, than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. Suppose that from years of research, it is known that the population standard deviations are 1.5811 years and 1 year, respectively. The following data are collected. The California state university system students took on average 4.5 years with a standard deviation of 0.8. The private university students took on average 4.1 years with a standard deviation of 0.3. 94. Parents of teenage boys often complain that auto insurance costs more, on average, for teenage boys than for teenage girls. A group of concerned parents examines a random sample of insurance bills. The mean annual cost for 36 teenage boys was $679. For 23 teenage girls, it was $559. From past years, it is known that the population standard deviation for each group is $180. Determine whether you believe that the mean cost for auto insurance for teenage boys is greater than that for teenage girls. 95. A group of transfer-bound students wondered if they will spend the sa
me mean amount on texts and supplies each year at their four-year university as they have at their community college. They conducted a random survey of 54 students at their community college and 66 students at their local four-year university. The sample means were $947 and $1,011, respectively. The population standard deviations are known to be $254 and $87, respectively. Conduct a hypothesis test to determine if the means are statistically the same. 96. Some manufacturers claim that nonhybrid sedan cars have a lower mean miles per gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 31 mpg with a standard deviation of 7 mpg. Thirty-one nonhybrid sedans get a mean of 22 mpg with a standard deviation of 4 mpg. Suppose that the population standard deviations are known to be 6 and 3, respectively. Conduct a hypothesis test to evaluate the manufacturers’ claim. 97. A baseball fan wanted to know if there is a difference between the number of games played in a World Series when the American League won the series versus when the National League won the series. From 1922 to 2012, the population standard deviation of games won by the American League was 1.14, and the population standard deviation of games won by the National League was 1.11. Of 19 randomly selected World Series games won by the American League, the mean number of games won was 5.76. The mean number of 17 randomly selected games won by the National League was 5.42. Conduct a hypothesis test. 620 Chapter 10 \| Hypothesis Testing with Two Samples 98. One of the questions in a study of marital satisfaction of dual-career couples was to rate the statement “I’m pleased with the way we divide the responsibilities for childcare.” The ratings went from 1 (strongly agree) to 5 (strongly disagree). Table 10.26 contains 10 of the paired responses for husbands and wives. Conduct a hypothesis test to see if the mean difference in the husband’s versus the wife’s satisfaction level is negative (meaning that, within the partnership, the husband is happier than the wife). Wife’s Score Husband’s Score Table 10.26 10.3 Comparing Two Independent Population Proportions DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. NOTE If you are using a Student’s t-distribution for one of the following homework problems, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption.) 99. A recent drug survey showed an increase in the use of prescription medication among local senior citizens as compared to the national percent. Suppose that a survey of 100 local seniors and 100 national seniors is conducted to see if the proportion of prescription medication use is higher locally or nationally. Locally, 65 senior citizens reported taking prescription medication within the past month, while 60 national seniors reported using them. larger + smaller dimension larger dimension 100. Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, ⎛ ⎝ to 1942. Thirty-seven early works were sampled, averaging 1.74 with a standard deviation of 0.11. Sixty-five of the later works were sampled, averaging 1.746 with a standard deviation of 0.1064. Do you think that there is a significant difference in the Golden Ratio calculation? , was the same in the Whitney Exhibit for works from 1900 to 1919 as for works from 1920 ⎞ ⎠ 101. A year was randomly picked from 1985 to the present. In that year, there were 2,051 Hispanic students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe College, there were 321 Hispanic students out of a total of 2,441 students. In general, do you think that the percent of Hispanic students at the two colleges is basically the same or different? Use the following information to answer the next three exercises. Neuroinvasive West Nile virus is a severe disease that affects a person’s nervous system. It is spread by the Culex species of mosquito. In the United States in 2010, there were 629 reported cases of neuroinvasive West Nile virus out of a total of 1,021 reported cases, and there were 486 neuroinvasive reported cases out of a total of 712 cases reported in 2011. Is the 2011 proportion of neuroinvasive West Nile virus cases more than the 2010 proportion of neuroinvasive West Nile virus cases? Using a 1 percent level of significance, conduct an appropriate hypothesis test. • 2011 subscript: 2011 group. • 2010 subscript: 2010 group 102. This is a. a test of two proportions b. a test of two independent means c. a test of a single mean d. a test of matched pairs. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 621 103. An appropriate null hypothesis is a. p2011 ≤ p2010 b. p2011 ≥ p2010 c. μ2011 ≤ μ2010 d. p2011 > p2010 104. The p-value is 0.0022. At a 1 percent level of significance, what is the appropriate conclusion? a. There is sufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile virus is less than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile virus. b. There is insufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile virus is more than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile virus. c. There is insufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile virus is less than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile virus. d. There is sufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile virus is more than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile virus. 105. Researchers conducted a study to find out if there is a difference in the use of e-readers by different age groups. Randomly selected participants were divided into two age groups. In the 16- to 29-year-old group, 7 percent of the 628 surveyed use e-readers, while 11 percent of the 2,309 participants 30 years old and older use e-readers. 106. Adults aged 18 years and older were randomly selected for a survey about a specific disease. The researchers wanted to determine if the proportion of women who have the disease is less than the proportion of southern men who do. The results are shown in Table 10.27. Test at the 1 percent level of significance. Number diagnosed with disease Sample size Men 42,769 Women 67,169 Table 10.27 155,525 248,775 107. Two computer users were discussing tablet computers. A higher proportion of people ages 16 to 29 use tablets than of people age 30 and older. Table 10.28 details the number of tablet owners for each age group. Test at the 1 percent level of significance. 16–29 year olds 30 years and older Own a Tablet 69 Sample Size 628 Table 10.28 231 2,309 108. A group of friends debated whether more men use smartphones than women. They consulted a research study of smartphone use among adults. The results of the survey indicate that of the 973 men randomly sampled, 379 use smartphones. For women, 404 of the 1,304 who were randomly sampled use smartphones. Test at the 5 percent level of significance. 109. While her husband spent 2.5 hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who do. The population was Saturday afternoon shoppers. Out of 67 men, 24 said they enjoyed the activity. Eight of the 24 women surveyed claimed to enjoy the activity. Interpret the results of the survey. 622 Chapter 10 \| Hypothesis Testing with Two Samples 110. We are interested in whether children’s educational computer software costs less, on average, than children’s entertainment software. Thirty-six educational software titles were randomly picked from a catalog. The mean cost was $31.14 with a standard deviation of $4.69. Thirty-five entertainment software titles were randomly picked from the same catalog. The mean cost was $33.86 with a standard deviation of $10.87. Decide whether children’s educational software costs less, on average, than children’s entertainment software. 111. A researcher recently claimed that the proportion of college-age males who wear at least one piece of jewelery is as high as the proportion of college-age females. She conducted a survey in her classes. Out of 107 males, 20 wear at least one piece of jewelery. Out of 92 females, 47 wear at least one piece of jewelery. Do you believe that the proportion of males has reached the proportion of females? 112. Use the data sets found in Appendix C to answer this exercise. Is the proportion of race laps Terri completes slower than 130 seconds less than the proportion of practice laps she completes slower than 135 seconds? 113. To Breakfast or Not to Breakfast? by Richard Ayore In the American society, birthdays are one of those days that everyone looks forward to. People of different ages and peer groups gather to mark the 18th, 20th, …, birthdays. During this time, one looks back to see what he or she has achieved for the past year and also focuses ahead for more to come. If, by any chance, I am invited to one of these parties, my experience is always different. Instead of dancing around with my friends while the music is booming, I get carried away by memories of my family back home in Kenya. I remember the good times I had w
ith my brothers and sister while we did our daily routine. Every morning, I remember we went to the shamba (garden) to weed our crops. I remember one day arguing with my brother as to why he always remained behind just to join us an hour later. In his defense, he said that he preferred waiting for breakfast before he came to weed. He said, “This is why I always work more hours than you guys!” And so, to prove him wrong or right, we decided to give it a try. One day we went to work as usual without breakfast, and recorded the time we could work before getting tired and stopping. On the next day, we all ate breakfast before going to work. We recorded how long we worked again before getting tired and stopping. Of interest was our mean increase in work time. Though not sure, my brother insisted that it was more than two hours. Using the data in Table 10.29, solve our problem. Work hours with breakfast Work hours without breakfast 8 7 9 5 9 8 10 Table 10.29 10.4 Matched or Paired Samples (Optional) DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 623 NOTE If you are using a Student’s t-distribution for the homework problems, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption.) 114. Ten individuals went on a low-fat diet for 12 weeks to lower their cholesterol. The data are recorded in Table 10.30. Do you think that their cholesterol levels were significantly lowered? Starting cholesterol level Ending cholesterol level 140 220 110 240 200 180 190 360 280 260 Table 10.30 140 230 120 220 190 150 200 300 300 240 Use the following information to answer the next two exercises. A new preventative medication was tried on a group of 224 patients who had the same risk factors for a disease. 45 patients developed the disease after four years. In a control group of 224 patients, 68 developed the disease after four years. We want to test whether the method of treatment reduces the proportion of patients who develop the disease after four years. Let the subscript t = treated patient and ut = untreated patient. 115. The appropriate hypotheses are a. H0: pt < put and Ha: pt ≥ put b. H0: pt ≤ put and Ha: pt > put c. H0: pt = put and Ha: pt ≠ put d. H0: pt = put and Ha: pt < put 116. If the p-value is 0.0062, what is the conclusion? Use α = 0.05. a. The method has no effect. b. There is sufficient evidence to conclude that the method reduces the proportion of patients who develop the disease after four years. c. There is sufficient evidence to conclude that the method increases the proportion of patients who develop the disease after four years. d. There is insufficient evidence to conclude that the method reduces the proportion of patients who develop the disease after four years. Use the following information to answer the next two exercises. An experiment is conducted to show that blood pressure can be consciously reduced in people trained in a biofeedback exercise program. Six subjects were randomly selected, and blood pressure measurements were recorded before and after the training. The difference between blood pressures was calculated (after – before), producing the following results: x¯ pressure has decreased after the training. = −10.2 sd = 8.4. Using the data, test the hypothesis that the blood d 624 Chapter 10 \| Hypothesis Testing with Two Samples 117. The distribution for the test is t5 t6 a. b. c. N(−10.2, 8.4) d. N(−10.2, 8.4 6 ) 118. If α = 0.05, the p-value and the conclusion are a. 0.0014; There is sufficient evidence to conclude that the blood pressure decreased after the training. b. 0.0014; There is sufficient evidence to conclude that the blood pressure increased after the training. c. 0.0155; There is sufficient evidence to conclude that the blood pressure decreased after the training. d. 0.0155; There is sufficient evidence to conclude that the blood pressure increased after the training. 119. A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four new students. She records their 18-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as follows. Player 1 Player 2 Player 3 Player 4 Mean score before class 83 Mean score after class 80 78 80 93 86 87 86 Table 10.31 The correct decision is reject H0. a. b. do not reject H0. 120. A local research group is studying a chronic disease. They believe the number of cases of the disease is higher in 2013 than in 2012 in the southern United States. The group compared the estimates of new cases by southern state in 2012 and 2013. The results are in Table 10.32. Southern States 2012 2013 Alabama Arkansas Florida Georgia Kentucky Louisiana 3,450 3,720 2,150 2,280 15,540 15,710 6,970 7,310 3,160 3,300 3,320 3,630 Mississippi 1,990 2,080 North Carolina 7,090 7,430 Oklahoma 2,630 2,690 South Carolina 3,570 3,580 Tennessee 4,680 5,070 Texas Virginia Table 10.32 15,050 14,980 6,190 6,280 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 625 121. A traveler wanted to know if the prices of hotels are different in the 10 cities that he visits the most often. The list of the cities with the corresponding prices for his two favorite hotel chains is in Table 10.33. Test at the 1 percent level of significance. Hyatt Regency prices in dollars Hilton prices in dollars Cities Atlanta Boston Chicago Dallas Denver Indianapolis Los Angeles New York City Philadelphia 107 358 209 209 167 179 179 625 179 Washington, DC 245 Table 10.33 169 289 299 198 169 214 169 459 159 239 122. A politician asked his staff to determine whether the underemployment rate in the Northeast decreased from 2011 to 2012. The results are in Table 10.34. Northeastern States 2011 2012 Connecticut Delaware Maine Maryland Massachusetts New Hampshire New Jersey New York Ohio Pennsylvania Rhode Island Vermont West Virginia Table 10.34 17.3 17.4 19.3 16.0 17.6 15.4 19.2 18.5 18.2 16.5 20.7 14.7 15.5 16.4 13.7 16.1 15.5 18.2 13.5 18.7 18.7 18.8 16.9 22.4 12.3 17.3 BRINGING IT TOGETHER: HOMEWORK Use the following information to answer the next 10 exercises. Indicate which of the following choices best identifies the hypothesis test. 626 Chapter 10 \| Hypothesis Testing with Two Samples A. Independent group means, population standard deviations and/or variances known B. Independent group means, population standard deviations and/or variances unknown C. Matched or paired samples D. Single mean E. Two proportions F. Single proportion 123. A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. The population standard deviations are two pounds and three pounds, respectively. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. 124. A new chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children who like the new chocolate bar is greater than the proportion of adults who like it. 125. The mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 9 males and 16 females. 126. A football league reported that the mean number of touchdowns per game was five. A study is done to determine if the mean number of touchdowns has decreased. 127. A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. From years of research, it is known that the population standard deviations are 1.5811 years and 1 year, respectively. 128. According to a doctor’s magazine, 75 percent of senior citizens think that yearly checkups are very important. A study is done to verify this. 129. According to a recent study, U.S. companies have a mean maternity leave of six weeks. 130. A recent survey showed an increase in use of prescription medication among local senior citizens as compared to the national percent. Suppose that a survey of 100 local senior citizens and 100 national senior citizens is conducted to see if the proportion of prescription medication use is higher locally than nationally. 131. A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are recorded. Of interest is the mean increase in SAT scores. The following data are collected: Pre-course score Post-course score 1 960 1010 840 1100 1250 860 1330 790 990 1110 740 Table 10.35 300 920 1100 880 1070 1320 860 1370 770 1040 1200 850 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 627 132. According to a statistics college professor, 68 percent of his students pass the final exam. A graduate researcher designs a study to determine if this claim is true. 133. Lesley E. Tan investigated the relationship between left-handedness versus right-handedness and motor competence in preschool children. Random samples of 41 left-handed preschool children and 41 right-handed preschool children were given several tests of motor skills to determine if there is evidence of a difference between the children based on this experiment. The experiment produced the means and standard deviations shown in Table 10.36. Determine the appropriate test and best distribution to
use for that test. Left-handed Right-handed Sample size Sample mean 41 97.5 Sample standard deviation 17.5 41 98.1 19.2 Table 10.36 a. Two independent means, normal distribution b. Two independent means, Student’s t-distribution c. Matched or paired samples, Student’s t-distribution d. Two population proportions, normal distribution 134. A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four new students. She records their 18-hole scores before learning the technique and after having taken her class. She conducts a hypothesis test. The data are shown in Table 10.37. Player 1 Player 2 Player 3 Player 4 Mean score before class 83 Mean score after class 80 78 80 93 86 87 86 Table 10.37 This is a. a test of two independent means. b. a test of two proportions. c. a test of a single mean. d. a test of a single proportion. REFERENCES 10.1 Two Population Means with Unknown Standard Deviations Baseball-Almanac. (2013). World series history. Retrieved from http://www.baseball-almanac.com/ws/wsmenu.shtml Graduating Engineer + Computer Careers. (n.d.). Retrieved from http://www.graduatingengineer.com Microsoft Bookshelf. (n.d.). Nasdaq. (n.d.). Sectoring by industry groups. Retrieved from http://www.nasdaq.com/markets/barchart-sectors.aspx Prostitution Research and Education. (2013). Strip clubs: Where prostitution and trafficking happen. Retrieved from www.prostitutionresearch.com/ProsViolPosttrauStress.html U.S. Senate. (n.d.). Retrieved from www.senate.gov Wikipedia. List_of_current_United_States_Senators_by_age (n.d.). List of current United States Senators by age. Retrieved from http://en.wikipedia.org/wiki/ 628 Chapter 10 \| Hypothesis Testing with Two Samples 10.2 Two Population Means with Known Standard Deviations Centers for Disease Control and Prevention. (2008, July 18). State-specific prevalence of obesity among adults—United States, 2007. MMWR, 57(28), 765–768. Retrieved from http://www.cdc.gov/mmwr/preview/mmwrhtml/mm5728a1.htm Federal Bureau of Investigation. (n.d.). Texas Crime Rates 1960–1012. Available at http://www.disastercenter.com/crime/ txcrime.htm Hinduja, S. http://cyberbullying.us/blog/sexting-research-and-gender-differences/ (2013). Sexting Research and Gender Differences. Cyberbulling Research Center. Retrieved from Humes, K. R., Jones, N. A., & Ramirez, R. R. (2011 March). Overview of race and Hispanic origin: 2010 (2010 Census Briefs). Washington, DC: U.S. Census Bureau. Available online at http://www.census.gov/prod/cen2010/briefs/ c2010br-02.pdf Smith, A. http://www.pewinternet.org/~/media/Files/Reports/2011/PIP_Smartphones.pdf (2011, July 11). 35% of American adults own a smartphone. Pew Internet. Available online at Visually. (2013). Smart phone users, by the numbers. Retrieved from http://visual.ly/smart-phone-users-numbers 10.3 Comparing Two Independent Population Proportions American Cancer Society. (n.d.). Retrieved from http://www.cancer.org/index Centers for Disease Control and Prevention. (n.d.). West Nile virus. Retrieved from http://www.cdc.gov/ncidod/dvbid/ westnile/index.htm Chancellor’s Office, California Community Colleges. (1994, Nov.). Educational Resources. (n.d.). Gallup. (2013). State of the states. Retrieved from http://www.gallup.com/poll/125066/State-States.aspx?ref=interactive Hilton Hotels. (n.d.). Retrieved from http://www.hilton.com Hyatt Hotels. Retrieved from http://hyatt.com San Jose Museum of Art. (n.d.). Whitney exhibit (on loan). U.S. Department of Health and Human Services. (n.d). Statistics. Retrieved from https://www.hhs.gov/ SOLUTIONS 1 two proportions 3 matched or paired samples 5 single mean 7 independent group means, population standard deviations and/or variances unknown 9 two proportions 11 independent group means, population standard deviations and/or variances unknown 13 independent group means, population standard deviations and/or variances unknown 15 two proportions 17 The random variable is the difference between the mean amounts of sugar in the two soft drinks. 19 means 21 two-tailed 23 the difference between the mean life spans of whites and nonwhites 25 This is a comparison of two population means with unknown population standard deviations. 27 Check student’s solution. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 629 29 a. Reject the null hypothesis. b. p-value < 0.05 c. There is not enough evidence at the 5 percent level of significance to support the claim that life expectancy in the 1900s is different between whites and nonwhites. 31 the difference in mean speeds of the fastball pitches of the two pitchers 33 –2.46 35 At the 1 percent significance level, we can reject the null hypothesis. There is sufficient data to conclude that the mean speed of Rodriguez’s fastball is faster than Wesley’s. 37 Subscripts: 1 = Food, 2 = No Food H0: μ1 ≤ μ2 Ha: μ1 > μ2 39 Figure 10.18 41 Subscripts: 1 = Gamma, 2 = Zeta H0: μ1 = μ2 Ha: μ1 ≠ μ2 43 0.0062 45 There is sufficient evidence to reject the null hypothesis. The data support that the melting point for Alloy Zeta is different from the melting point of Alloy Gamma. 47 P′OS1 – P′OS2 = difference in the proportions of phones that had system failures within the first eight hours of operation with OS1 and OS2. 49 0.1018 51 proportions 53 right-tailed 55 The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota. 57 Our sample sizes are much greater than five each, so we use the normal for two proportions distribution for this hypothesis test. 59 Check student’s solution. 61 a. Reject the null hypothesis. b. p-value < alpha c. At the 5 percent significance level, there is sufficient evidence to conclude that the proportion (percent) of the population that is of two or more races in Nevada is statistically higher than that in North Dakota. 630 Chapter 10 \| Hypothesis Testing with Two Samples 63 the mean difference of the system failures 65 0.0067 67 With a p-value 0.0067, we can reject the null hypothesis. There is enough evidence to support that the software patch is effective in reducing the number of system failures. 69 0.0021 71 Figure 10.19 73 H0: μd ≥ 0 Ha: μd < 0 75 0.0699 77 We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective. 79 Subscripts: 1: two-year colleges, 2: four-year colleges a. H0: μ1 ≥ μ2 b. Ha: μ1 < μ2 ¯ c. X ¯ 1 – X 2 d. Student’s t is the difference between the mean enrollments of the two-year colleges and the four-year colleges. e. test statistic: -0.2480 f. p-value: 0.4019 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject. iii. Reason for Decision: p-value > alpha iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean enrollment at four-year colleges is higher than at two-year colleges. 81 Subscripts: 1: mechanical engineering, 2: electrical engineering a. H0: µ1 ≥ µ2 b. Ha: µ1 < µ2 ¯ c. X ¯ 1 − X 2 is the difference between the mean entry-level salaries of mechanical engineers and electrical engineers. d. e. t108 test statistic: t = –0.82 f. p-value: 0.2061 g. Check student’s solution. h. i. Alpha: 0.05 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 631 ii. Decision: Do not reject the null hypothesis. iii. Reason for Decision: p-value > alpha iv. Conclusion: At the 5 percent significance level, there is insufficient evidence to conclude that the mean entry- level salaries of mechanical engineers is lower than that of electrical engineers. 83 a. H0: µ1 = µ2 b. Ha: µ1 ≠ µ2 ¯ c. X ¯ 1 − X 2 is the difference between the mean times for completing a lap in races and in practices. d. e. t20.32 test statistic: –4.70 f. p-value: 0.0001 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for Decision: p-value < alpha iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices. 85 a. H0: µ1 = µ2 b. Ha: µ1 ≠ µ2 c. is the difference between the mean times for completing a lap in races and in practices. d. e. t40.94 test statistic: –5.08 f. p-value: zero g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for Decision: p-value < alpha iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices. 88 c ¯ 90 Test: two independent sample means, population standard deviations unknown. Random variable: X Distribution: H0: μ1 = μ2, Ha: μ1 < μ2 The mean age of entering the industry in Canada is lower than the mean age in the United States. ¯ 1 − X 2 632 Chapter 10 \| Hypothesis Testing with Two Samples Figure 10.20 Graph: left-tailed p-value : 0.0151 Decision: Do not reject H0. Conclusion: At the 1 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of entering the industry in Canada is lower than the mean age in the United States. 92 d 94 Subscripts: 1 = boys, 2 = girls a. H0: µ1 ≤ µ2 b. Ha: µ1 > µ2 c. The random variable is the difference in the mean auto insurance costs for boys and girls. d. normal e. test statistic: z = 2.50 f. p value: 0.0062 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for Decision: p value < alpha iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean cost of auto insurance for teenage boys is greater than that for girls. 96 Subscripts: 1 = non-hybrid sedans, 2 = hybrid sedans a. H0: µ1 ≥ µ2 b. Ha: µ1 < µ2 c. The rando
m variable is the difference in the mean miles per gallon of nonhybrid sedans and hybrid sedans. d. normal e. test statistic: 6.36 f. p-value: 0 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p value < alpha iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean miles per gallon of non-hybrid sedans is less than that of hybrid sedans. 98 a. H0: µd = 0 b. Ha: µd < 0 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 633 c. The random variable Xd is the average difference between husband’s and wife’s satisfaction level. d. t9 test statistic: t = –1.86 e. f. p value: 0.0479 g. Check student’s solution h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis, but run another test. iii. Reason for Decision: p value < alpha iv. Conclusion: This is a weak test because alpha and the p value are close. However, there is insufficient evidence to conclude that the mean difference is negative. 101 Subscripts: 1 = Cabrillo College, 2 = Lake Tahoe College a. H0: p1 = p2 b. Ha: p1 ≠ p2 c. The random variable is the difference between the proportions of Hispanic students at Cabrillo College and Lake Tahoe College. d. normal for two proportions e. test statistic: 4.29 f. p-value: 0.00002 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: There is sufficient evidence to conclude that the proportions of Hispanic students at Cabrillo College and Lake Tahoe College are different. 103 a 105 Test: two independent sample proportions. Random variable: p′1 - p′2 Distribution: H0: p1 = p2 Ha: p1 ≠ p2 The proportion of e-reader users is different for the 16- to 29-year-old users from that of the 30 and older users. Graph: two-tailed Figure 10.21 p-value : 0.0033 Decision: Reject the null hypothesis. Conclusion: At the 5 percent level of significance, from the sample data, there is sufficient evidence to conclude that the proportion of e-reader users 16 to 29 years old is different from the proportion of e-reader users 30 and older. 107 Test: two independent sample proportions Random variable: p′1 − p′2 Distribution: H0: p1 = p2 Ha: p1 > p2 A higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older. Graph: righttailed 634 Chapter 10 \| Hypothesis Testing with Two Samples Figure 10.22 p-value: 0.2354 Decision: Do not reject the H0. Conclusion: At the 1 percent level of significance, from the sample data, there is not sufficient evidence to conclude that a higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older. 109 Subscripts: 1: men; 2: women a. H0: p1 ≤ p2 b. Ha: p1 > p2 c. P′1 − P′2 is the difference between the proportions of men and women who enjoy shopping for electronic equipment. d. normal for two proportions e. test statistic: 0.22 f. p-value: 0.4133 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for Decision: p-value > alpha iv. Conclusion: At the 5 percent significance level, there is insufficient evidence to conclude that the proportion of men who enjoy shopping for electronic equipment is more than the proportion of women. 111 a. H0: p1 = p2 b. Ha: p1 ≠ p2 c. P′1 − P′2 is the difference between the proportions of men and women that have at least one pierced ear. d. normal for two proportions e. test statistic: –4.82 f. p-value: zero g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for Decision: p-value < alpha iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the proportions of males and females with at least one pierced ear is different. 113 a. H0: µd = 0 b. Ha: µd > 0 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 \| Hypothesis Testing with Two Samples 635 c. The random variable Xd is the mean difference in work times on days when eating breakfast and on days when not eating breakfast. d. e. t9 test statistic: 4.8963 f. p-value: 0.0004 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for Decision: p-value < alpha iv. Conclusion: At the 5 percent level of significance, there is sufficient evidence to conclude that the mean difference in work times on days when eating breakfast and on days when not eating breakfast has increased. 114 p-value = 0.1494 At the 5 percent significance level, there is insufficient evidence to conclude that the medication lowered cholesterol levels after 12 weeks. 116 b 118 c ¯ 120 Test: two matched pairs or paired samples (t-test) Random variable: X mean of the differences of new female breast cancer cases in the south between 2013 and 2012 is greater than zero. The estimate for new female breast cancer cases in the south is higher in 2013 than in 2012. Graph: right-tailed p-value: 0.0004 Distribution: t12 H0: μd = 0 Ha: μd > 0 The d Figure 10.23 Decision: Reject H0. Conclusion: At the 5 percent level of significance, from the sample data, there is sufficient evidence to conclude that there was a higher estimate of new female breast cancer cases in 2013 than in 2012. 122 Test: matched or paired samples (t-test) Difference data: {–0.9, –3.7, –3.2, –0.5, 0.6, –1.9, –0.5, 0.2, 0.6, 0.4, 1.7, –2.4, ¯ 1.8} Random Variable: X in the northeastern states between 2012 and 2011 is less than zero. The underemployment rate went down from 2011 to 2012. Graph: left-tailed. Distribution: H0: μd = 0 Ha: μd < 0 The mean of the differences of the rate of underemployment d Figure 10.24 636 Chapter 10 \| Hypothesis Testing with Two Samples p-value: 0.1207 Decision: Do not reject H0. Conclusion: At the 5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that there was a decrease in the underemployment rates of the northeastern states from 2011 to 2012. 124 e 126 d 128 f 130 e 132 f The graduate researcher will be comparing a sample proportion to a population proportion or claim. Thus, the study includes the hypothesis test of a single proportion. A two proportion hypothesis test compares two sample proportions. 134 a This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 637 11 \| THE CHI-SQUARE DISTRIBUTION Figure 11.1 The chi-square distribution can be used to find relationships between two things, like grocery prices at different stores. (credit: Pete/flickr) Introduction Chapter Objectives By the end of this chapter, the student should be able to do the following: Interpret the chi-square probability distribution as the sample size changes • • Conduct and interpret chi-square goodness-of-fit hypothesis tests • Conduct and interpret chi-square test of independence hypothesis tests • Conduct and interpret chi-square homogeneity hypothesis tests • Conduct and interpret chi-square single variance hypothesis tests Have you ever wondered if lottery numbers were evenly distributed or if some numbers occurred with a greater frequency? How about if the types of movies people preferred were different across different age groups? What about if a coffee machine was dispensing approximately the same amount of coffee each time? You could answer these questions by conducting a hypothesis test. 638 Chapter 11 \| The Chi-Square Distribution You will now study a new distribution, one that is used to determine the answers to such questions. This distribution is called the chi-square distribution. In this chapter, you will learn the three major applications of the chi-square distribution: • The goodness-of-fit test, which determines if data fit a particular distribution, such as in the lottery example • The test of independence, which determines if events are independent, such as in the movie example • The test of a single variance, which tests variability, such as in the coffee example NOTE Though the chi-square distribution depends on calculators or computers for most of the calculations, there is a table available (see Appendix G). TI-83+ and TI-84 calculator instructions are included in the text. Look in the sports section of a newspaper or on the internet for some sports data: baseball averages, basketball scores, golf tournament scores, football odds, swimming times, and the like. Plot a histogram and a boxplot using your data. See if you can determine a probability distribution that your data fits. Have a discussion with the class about your choice. 11.1 \| Facts About the Chi-Square Distribution The notation for the chi-square distribution is 2 χ ∼ χd f where df = degrees of freedom, which depends on how chi-square is being used. If you want to practice calculating chisquare probabilities then use df = n – –1. The degrees of freedom for the three major uses are calculated differently. For the χ2 distribution, the population mean is μ = df, and the population standard deviation is σ = 2(d f ) . The random variable is shown as χ2, but it may be any uppercase letter. The random variable for a chi-square distribution with k degrees of freedom is the sum of k independent, squared standard normal variables is χ2 = (Z1)2 + (Z2)2 + ... + (Zk)2, where the following are true: • The curve is nonsymmetrical and skewed to the right. • There is a different chi-square curve for each df. Figure 11.2 • The test statistic for any test is always greater than or equal to zero. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 639 • When df > 90, the chi-square curve approximates the normal distribution. For X ~ χ1,000 2 , the mean, μ = df = 1,000 and the standard deviation, σ = 2(1,000) = 44.7. Therefore, X ~ N(1,000, 44.7), approximately. • The mean,
μ, is located just to the right of the peak. Figure 11.3 11.2 \| Goodness-of-Fit Test In this type of hypothesis test, you determine whether the data fit a particular distribution. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test, meaning the distribution for the hypothesis test is chi-square, to determine if there is a fit. The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities. The test statistic for a goodness-of-fit test is: (O − E)2 E Σ k where • O = observed values (data), • E = expected values (from theory), and • k = the number of different data cells or categories. The observed values are the data values, and the expected values are the values you would expect to get if the null hypothesis were true. There are n terms of the form (O − E)2 . E The number of degrees of freedom is df = (number of categories – 1). The goodness-of-fit test is almost always right-tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve. NOTE The expected value for each cell needs to be at least five for you to use this test. Example 11.1 Absenteeism of college students from math classes is a major concern to math instructors because missing class appears to increase the drop rate. Suppose that a study was done to determine if the actual student absenteeism rate follows faculty perception. The faculty expected that a group of 100 students would miss class according to Table 11.1. 640 Chapter 11 \| The Chi-Square Distribution Number of Absences per Term Expected Number of Students 0–2 3–5 6–8 9–11 12+ Table 11.1 50 30 12 6 2 A random survey across all mathematics courses was then done to determine the number of observed absences in a course. Table 11.2 displays the results of that survey. Number of Absences per Term Actual Number of Students 0–2 3–5 6–8 9–11 12+ Table 11.2 35 40 20 1 4 Determine the null and alternative hypotheses needed to conduct a goodness-of-fit test. H0: Student absenteeism fits faculty perception. The alternative hypothesis is the opposite of the null hypothesis. Ha: Student absenteeism does not fit faculty perception. a. Can you use the information as it appears in the charts to conduct the goodness-of-fit test? Solution 11.1 a. No. Notice that the expected number of absences for the 12+ entry is less than five; it is two. Combine that group with the 9–11 group to create new tables where the number of students for each entry is at least five. The new results are in Table 11.2 and Table 11.3. Number of Absences per Term Expected Number of Students 0–2 3–5 6–8 9+ Table 11.3 50 30 12 8 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 641 Number of Absences per Term Actual Number of Students 0–2 3–5 6–8 9+ Table 11.4 35 40 20 5 b. What is the number of degrees of freedom (df)? Solution 11.1 b. There are four cells or categories in each of the new tables. df = number of cells – 1 = 4 – 1 = 3. 11.1 A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in Table 11.5. Number Produced Number Defective 0–100 101–200 201–300 301–400 401–500 Table 11.5 5 6 7 8 10 A random sample was taken to determine the actual number of defects. Table 11.6 shows the results of the survey. Number Produced Number Defective 0–100 101–200 201–300 301–400 401–500 Table 11.6 5 7 8 9 11 State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom. 642 Chapter 11 \| The Chi-Square Distribution Example 11.2 Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 60 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as in Table 11.6. For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five-day work week? Test at a 5 percent significance level. Monday Tuesday Wednesday Thursday Friday Number of Absences 15 12 9 9 15 Table 11.7 Day of the Week Employees Were Most Absent Solution 11.2 The null and alternative hypotheses are as follows: • H0: The absent days occur with equal frequencies; that is, they fit a uniform distribution. • Ha: The absent days occur with unequal frequencies; that is, they do not fit a uniform distribution. If the absent days occur with equal frequencies, then, out of 60 absent days (the total in the sample: 15 + 12 + 9 + 9 + 15 = 60) there would be 12 absences on Monday, 12 on Tuesday, 12 on Wednesday, 12 on Thursday, and 12 on Friday. These numbers are the expected (E) values. The values in the table are the observed (O) values or data. This time, calculate the χ2 test statistic by hand. Make a chart with the following headings and fill in the columns: • Expected (E) values (12, 12, 12, 12, 12) • Observed (O) values (15, 12, 9, 9, 15) • • • (O – E) (O – E)2 (O – E)2 E Now add (sum) the last column. The sum is three. This is the χ2 test statistic. To find the p-value, calculate P(χ2 > 3). This test is right-tailed. Use a computer or calculator to find the p-value. You should get p-value = 0.5578. The dfs are the number of cells – 1 = 5 – 1 = 4. Press 2nd DISTR. Arrow down to χ2cdf. Press ENTER. Enter (3,10^99,4). Rounded to four decimal places, you should see .5578, which is the p-value. Next, complete a graph like the following one with the proper labeling and shading. You should shade the right tail. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 643 Figure 11.4 The decision is not to reject the null hypothesis. Conclusion: At a 5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies. TI-83+ and some TI-84 calculators do not have a special program for the test statistic for the goodness-of-fit test. The next example, Example 11.3, has the calculator instructions. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values—the data—into a first list and the expected values—the values you expect if the null hypothesis is true—into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press Calculate or Draw. Make sure you clear any lists before you start. To Clear Lists in the calculators: Go into STAT EDIT and arrow up to the list name area of the particular list. Press CLEAR and then arrow down. The list will be cleared. Alternatively, you can press STAT and press 4 for ClrList. Enter the list name and press ENTER. 11.2 Teachers want to know which night each week their students are doing most of their homework. Most teachers think that students do homework equally throughout the week. Suppose a random sample of 56 students were asked on which night of the week they did the most homework. The results were distributed as in Table 11.8. Sunday Monday Tuesday Wednesday Thursday Friday Saturday 11 8 10 7 10 5 5 Number of Students Table 11.8 From the population of students, do the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week? What type of hypothesis test should you use? 644 Chapter 11 \| The Chi-Square Distribution Example 11.3 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 645 One study indicates that the number of televisions that American families have is distributed (this is the given distribution for the American population) as in Table 11.9. Number of Televisions Percent 0 1 2 3 4+ Table 11.9 10 16 55 11 8 The table contains expected (E) percents. A random sample of 600 families in the far western U.S. resulted in the data in Table 11.10. Number of Televisions Frequency 0 1 2 3 4+ Table 11.10 66 119 340 60 15 Total = 600 The table contains observed (O) frequency values. At the 1 percent significance level, does it appear that the distribution number of televisions of far western U.S. families is different from the distribution for the American population as a whole? Solution 11.3 This problem asks you to test whether the far western U.S. families distribution fits the distribution of the American families. This test is always right-tailed. The first table contains expected percentages. To get expected (E) frequencies, multiply the percentage by 600. The expected frequencies are shown in Table 11.10. Number of Televisions Percent Expected Frequency 0 1 2 3 more than 3 10 16 55 11 8 (0.10)(600) = 60 (0.16)(600) = 96 (0.55)(600) = 330 (0.11)(600) = 66 (0.08)(600) = 48 646 Chapter 11 \| The Chi-Square Distribution Therefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI calculators, you can let the calculator do the math. For example, instead of 60, enter 0.10 * 600. H0: The number of televisions distribution of far western U.S. families is the same as the number of televisions distribution of the American population. Ha: The number of televisions distribution of far western U.S. families is different from the number of televisions distribution of the American population. Distribution for the test: χ4 2 where df = (the number of cells) – 1 = 5 – 1 = 4. NOTE df ≠ 600 – 1 Calculate the test statistic: χ2 = 29.65 Graph Figure 11.5 Probability statement: p-value = P(χ2 > 29.65) = .000006 Compare α and the p-value: • α = .01 • p-value = 0.000006 So, α > p-value. Make a decision: Since α > p-val
ue, reject Ho. This means you reject the hypothesis that the distribution for the far western states is the same as that of the American population as a whole. Conclusion: At the 1 percent significance level, from the data, there is sufficient evidence to conclude that the number of televisions distribution for the far western United States is different from the number of televisions distribution for the American population as a whole. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 647 Press STAT and ENTER. Make sure to clear lists L1, L2, and L3 if they have data in them—see the note at the end of Example 11.2. Into L1, put the observed frequencies 66, 119, 349, 60, 15. Into L2, put the expected frequencies .10600, .16600, .55600, .11600, .08*600. Arrow over to list L3 and up to the name area L3. Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see sum (Enter L3). Rounded to two decimal places, you should see 29.65. Press 2nd DISTR. Press 7 or Arrow down to 7:χ2cdf and press ENTER. Enter (29.65,1E99,4). Rounded to four places, you should see 5.77E-6 = .000006 (rounded to six decimal places), which is the p-value. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values—the values you expect if the null hypothesis is true—into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press Calculate or Draw. Make sure you clear any lists before you start. 11.3 The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in Table 11.12. Number of Pets Percent 0 1 2 3 4+ Table 11.12 18 25 30 18 9 A random sample of 1,000 students from the eastern United States resulted in the data in Table 11.13. Number of Pets Frequency 0 1 2 3 4+ Table 11.13 210 240 320 140 90 At the 1 percent significance level, does it appear that the distribution number of pets of students in the eastern United 648 Chapter 11 \| The Chi-Square Distribution States is different from the distribution for the United States student population as a whole? What is the p-value? Example 11.4 Suppose you flip two coins 100 times. The results are 20 HH, 27 HT, 30 TH, and 23 TT. Are the coins fair? Test at a 5 percent significance level. Solution 11.4 This problem can be set up as a goodness-of-fit problem. The sample space for flipping two fair coins is {HH, HT, TH, TT}. Out of 100 flips, you would expect 25 HH, 25 HT, 25 TH, and 25 TT. This is the expected distribution. The question, “Are the coins fair?” is the same as saying, “Does the distribution of the coins (20 HH, 27 HT, 30 TH, 23 TT) fit the expected distribution?” Random variable: Let X = the number of heads in one flip of the two coins. X takes on the values 0, 1, 2. There are 0, 1, or 2 heads in the flip of two coins. Therefore, the number of cells is three. Since X = the number of heads, the observed frequencies are 20 for two heads, 57 for one head, and 23 for zero heads or both tails. The expected frequencies are 25 for two heads, 50 for one head, and 25 for zero heads or both tails. This test is right-tailed. H0: The coins are fair. Ha: The coins are not fair. Distribution for the test: χ2 Calculate the test statistic: χ2 = 2.14. 2 where df = 3 – 1 = 2. Graph Figure 11.6 Probability statement: p-value = P(χ2 > 2.14) = 0.3430. Compare α and the p-value: • α = .05 • p-value = 0.3430 α < p-value. Make a decision: Since α < p-value, do not reject H0. Conclusion: There is insufficient evidence to conclude that the coins are not fair. Press STAT and ENTER. Make sure you clear lists L1, L2, and L3 if they have data in them. Into L1, put the observed frequencies 20, 57, 23. Into L2, put the expected frequencies 25, 50, 25. Arrow over to list L3 and up to the name area L3. Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 649 arrow over to MATH. Press 5. You should see sum. Enter L3. Rounded to two decimal places, you should see 2.14. Press 2nd DISTR. Arrow down to 7:χ2cdf—or press 7. Press ENTER. Enter 2.14,1E99,2). Rounded to four places, you should see .3430, which is the p-value. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values—the data—into a first list and the expected values—the values you expect if the null hypothesis is true—into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press Calculate or Draw. Make sure you clear any lists before you start. 11.4 Students in a social studies class hypothesize that the literacy rates around the world for every region are 82 percent. Table 11.14 shows the actual literacy rates around the world broken down by region. What are the test statistic and the degrees of freedom? MDG Region Developed regions Adult Literacy Rate (%) 99 Commonwealth of Independent States 99.5 Northern Africa Sub-Saharan Africa Latin America and the Caribbean Eastern Asia Southern Asia Southeastern Asia Western Asia Oceania Table 11.14 67.3 62.5 91 93.8 61.9 91.9 84.5 66.4 11.3 \| Test of Independence Tests of independence involve using a contingency table of observed (data) values. The test statistic for a test of independence is similar to that of a goodness-of-fit test (O – E)2 E Σ (i ⋅ j) where • O = observed values, • E = expected values, • i = the number of rows in the table, and 650 Chapter 11 \| The Chi-Square Distribution • j = the number of columns in the table. There are i ⋅ j terms of the form (O – E)2 E . A test of independence determines whether two factors are independent. You first encountered the term independence in Probability Topics. As a review, consider the following example. NOTE The expected value for each cell needs to be at least five for you to use this test. Example 11.5 Suppose A = a speeding violation in the last year and B = a cell phone user while driving. If A and B are independent, then P(A AND B) = P(A)P(B). A AND B is the event that a driver received a speeding violation last year and also used a cell phone while driving. Suppose, in a study of drivers who received speeding violations in the last year, and who used cell phones while driving, that 755 people were surveyed. Out of the 755, 70 had a speeding violation and 685 did not; 305 used cell phones while driving and 450 did not. Let y = expected number of drivers who used a cell phone while driving and received speeding violations. If A and B are independent, then P(A AND B) = P(A)P(B). By substitution, y 755 = ⎛ ⎝ 70 755 ⎞ ⎛ ⎝ ⎠ 305 755 ⎞ ⎠. Solve for y: y = (70)(305) 755 = 28.3. About 28 people from the sample are expected to use cell phones while driving and to receive speeding violations. In a test of independence, we state the null and alternative hypotheses in words. Since the contingency table consists of two factors, the null hypothesis states that the factors are independent and the alternative hypothesis states that they are not independent (dependent). If we do a test of independence using the example, then the null hypothesis is the following: H0: Being a cell phone user while driving and receiving a speeding violation are independent events. If the null hypothesis were true, we would expect about 28 people to use cell phones while driving and to receive a speeding violation. The test of independence is always right-tailed because of the calculation of the test statistic. If the expected and observed values are not close together, then the test statistic is very large and way out in the right tail of the chisquare curve, as it is in a goodness-of-fit. The number of degrees of freedom for the test of independence is df = (number of columns – 1)(number of rows – 1). The following formula calculates the expected number (E): E = (row total)(column total) total number surveyed 11.5 A sample of 300 students is taken. Of the students surveyed, 50 were music students, while 250 were not. 97 were on the honor roll, while 203 were not. If we assume being a music student and being on the honor roll are independent events, what is the expected number of music students who are also on the honor roll? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 651 Example 11.6 In a volunteer group, adults 21 and older volunteer from one to nine hours each week to spend time with a disabled senior citizen. The program recruits among community college students, four-year college students, and non-students. In Table 11.15 is a sample of the adult volunteers and the number of hours they volunteer per week. Type of Volunteer 1–3 Hours 4–6 Hours 7–9 Hours Row Total Community College Students 111 Four-year College Students Non-students Column Total 96 91 298 96 133 150 379 48 61 53 162 255 290 294 839 Table 11.15 Number of Hours Worked per Week by Volunteer Type (Observed) The table contains observed (O) values (data). Is the number of hours volunteered independent of the type of volunteer? Solution 11.6 The observed values and the question at the end of the problem, “Is the number of hours volunteered independent of the type of volunteer?” tell you this is a test of independence. The two factors are number of hours volunteered and type of volunteer. This test is always right-tailed. H0: The number of hours volunteered is independent of the type of volunteer. Ha: The number of hours volunteered is dependent on the type of volunteer. The expected result are in Table 11.15. Type of Volunteer 1-3 Hours 4–6 Hours 7-9 Hours Comm
unity College Students 90.57 115.19 49.24 Four-Year College Students 103 131 56 Nonstudents 104.42 132.81 56.77 Table 11.16 Number of Hours Worked per Week by Volunteer Type (Expected) The table contains expected (E) values (data). For example, the calculation for the expected frequency for the top-left cell is E = (row total)(column total) total number surveyed = (255)(298) 839 = 90.57. Calculate the test statistic: χ2 = 12.99 (calculator or computer) 2 Distribution for the test: χ4 df = (3 columns – 1)(3 rows – 1) = (2)(2) = 4 Graph 652 Chapter 11 \| The Chi-Square Distribution Figure 11.7 Probability statement: p-value = P(χ2 > 12.99) = 0.0113 Compare α and the p-value: Since no α is given, assume α = 0.05. p-value = 0.0113. α > p-value. Make a decision: Since α > p-value, reject H0. This means that the factors are not independent. Conclusion: At a 5 percent level of significance, from the data, there is sufficient evidence to conclude that the number of hours volunteered and the type of volunteer are dependent on each other. For the example in Table 11.15, if there had been another type of volunteer, teenagers, what would the degrees of freedom be? Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 3 ENTER 3 ENTER. Enter the table values by row from Table 11.15. Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is 12.9909 and the p-value = .0113. Do the procedure a second time, but arrow down to Draw instead of Calculate. 11.6 The Bureau of Labor Statistics gathers data about employment in the United States. A sample is taken to calculate the number of U.S. citizens working in one of several industry sectors over time. Table 11.17 shows the results: Industry Sector Non-agriculture Wage and Salary Goods-producing, Excluding Agriculture Services-providing Agriculture, Forestry, Fishing, and Hunting Non-agriculture Self-employed and Unpaid Family Worker Secondary Wage and Salary Jobs in Agriculture and Private Household Industries 2000 2010 2020 Total 13,243 13,044 15,018 41,305 2,457 1,771 1,950 6,178 10,786 11,273 13,068 35,127 240 931 214 894 201 972 655 2,797 14 11 11 36 Secondary Jobs as a Self-employed or Unpaid Family Worker 196 144 152 492 Table 11.17 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 653 Industry Sector Total Table 11.17 2000 2010 2020 Total 27,867 27,351 31,372 86,590 We want to know if the change in the number of jobs is independent of the change in years. State the null and alternative hypotheses and the degrees of freedom. Example 11.7 De Anza College is interested in the relationship between anxiety level and the need to succeed in school. A random sample of 400 students took a test that measured anxiety level and need to succeed in school. Table 11.18 shows the results. De Anza College wants to know if anxiety level and need to succeed in school are independent events. Need to Succeed in School High Anxiety MedHigh Anxiety Medium Anxiety MedLow Anxiety Low Anxiety Row Total High Need Medium Need Low Need Column Total 35 18 4 57 42 48 5 95 53 63 11 127 15 33 15 63 10 31 17 58 155 193 52 400 Table 11.18 Need to Succeed in School vs. Anxiety Level a. How many high anxiety level students are expected to have a high need to succeed in school? Solution 11.7 a. The column total for a high anxiety level is 57. The row total for high need to succeed in school is 155. The sample size or total surveyed is 400. E = (row total)(column total) total surveyed = 155 ⋅ 57 400 = 22.09 The expected number of students who have a high anxiety level and a high need to succeed in school is about 22. b. If the two variables are independent, how many students do you expect to have a low need to succeed in school and a med-low level of anxiety? Solution 11.7 b. The column total for a med-low anxiety level is 63. The row total for a low need to succeed in school is 52. The sample size or total surveyed is 400. c. E = (row total)(column total) total surveyed = ________ 654 Chapter 11 \| The Chi-Square Distribution Solution 11.7 c. E = (row total)(column total) total surveyed = 8.19 d. The expected number of students who have a med-low anxiety level and a low need to succeed in school is about ________. Solution 11.7 d. 8 11.7 Refer back to the information in Try It. How many services-providing jobs are there expected to be in 2020? How many nonagriculture wage and salary jobs are there expected to be in 2020? 11.4 \| Test for Homogeneity The goodness-of-fit test can be used to decide whether a population fits a given distribution, but it will not suffice to decide whether two populations follow the same unknown distribution. A different test, called the test for homogeneity, can be used to draw a conclusion about whether two populations have the same distribution. To calculate the test statistic for a test for homogeneity, follow the same procedure as with the test of independence. NOTE The expected value for each cell needs to be at least five for you to use this test. Hypotheses H0: The distributions of the two populations are the same. Ha: The distributions of the two populations are not the same. Test Statistic Use a χ 2 test statistic. It is computed in the same way as the test for independence. Degrees of freedom (df) df = number of columns – 1 Requirements All values in the table must be greater than or equal to five. Common Uses Comparing two populations. For example: men vs. women, before vs. after, east vs. west. The variable is categorical with more than two possible response values. Example 11.8 Do male and female college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected male college students and 300 randomly selected female college students were asked about their living arrangements: dormitory, apartment, with parents, other. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 655 The results are shown in Table 11.18. Do male and female college students have the same distribution of living arrangements? Dormitory Apartment With Parents Other Males 72 Females 91 84 86 49 88 45 35 Table 11.19 Distribution of Living Arragements for College Males and College Females Solution 11.8 H0: The distribution of living arrangements for male college students is the same as the distribution of living arrangements for female college students. Ha: The distribution of living arrangements for male college students is not the same as the distribution of living arrangements for female college students. Degrees of freedom (df): df = number of columns – 1 = 4 – 1 = 3 2 Distribution for the test: χ3 Calculate the test statistic: χ2 = 10.1287 (calculator or computer) Probability statement: p-value = P(χ2 >10.1287) = 0.0175 Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 2 ENTER 4 ENTER. Enter the table values by row. Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is 10.1287 and the p-value = 0.0175. Do the procedure a second time but arrow down to Draw instead of Calculate. Compare α and the p-value: Since no α is given, assume α = 0.05. p-value = 0.0175. α > p-value. Make a decision: Since α > p-value, reject H0. This means that the distributions are not the same. Conclusion: At a 5 percent level of significance, from the data, there is sufficient evidence to conclude that the distributions of living arrangements for male and female college students are not the same. Notice that the conclusion is only that the distributions are not the same. We cannot use the test for homogeneity to draw any conclusions about how they differ. 11.8 Do families and singles have the same distribution of cars? Suppose that 100 randomly selected families and 200 656 Chapter 11 \| The Chi-Square Distribution randomly selected singles were asked what type of car they drove: sport, sedan, hatchback, truck, van/SUV. The results are shown in Table 11.20. Do families and singles have the same distribution of cars? Test at a level of significance of 0.05. Sport Sedan Hatchback Truck Van/SUV Family 5 Single 45 Table 11.20 15 65 35 37 17 46 28 7 Example 11.9 Both before and after a recent earthquake, surveys were conducted asking voters which of the three candidates they planned on voting for in the upcoming city council election. Has there been a change since the earthquake? Use a level of significance of 0.05. Table 11.20 shows the results of the survey. Has there been a change in the distribution of voter preferences since the earthquake? Perez Chung Stevens Before 167 After 214 128 197 135 225 Table 11.21 Solution 11.9 H0: The distribution of voter preferences was the same before and after the earthquake. Ha: The distribution of voter preferences was not the same before and after the earthquake. Degrees of freedom (df): df = number of columns – 1 = 3 – 1 = 2 2 Distribution for the test: χ2 Calculate the test statistic: χ2 = 3.2603 (calculator or computer) Probability statement: p-value=P(χ2 > 3.2603) = 0.1959 Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 2 ENTER 3 ENTER. Enter the table values by row. Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is 3.2603 and the p-value = 0.1959. Do the procedure a second time but arrow down to Draw instead of Calculate. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 657 Compare α
and the p-value: α = 0.05 and the p-value = 0.1959. α < p-value. Make a decision: Since α < p-value, do not reject Ho. Conclusion: At a 5 percent level of significance, from the data, there is insufficient evidence to conclude that the distribution of voter preferences was not the same before and after the earthquake. 11.9 Ivy League schools receive many applications, but only some can be accepted. At the schools listed in Table 11.22, two types of applications are accepted: regular and early decision. Application Type Accepted Brown Columbia Cornell Dartmouth Penn Yale Regular Early Decision Table 11.22 2,115 1,792 577 627 5,306 1,228 1,734 444 2,685 1,245 1,195 761 We want to know if the number of regular applications accepted follows the same distribution as the number of early applications accepted. State the null and alternative hypotheses, the degrees of freedom and the test statistic, sketch the graph of the p-value, and draw a conclusion about the test of homogeneity. 11.5 \| Comparison of the Chi-Square Tests You have seen the χ2 test statistic used in three different circumstances. The following bulleted list is a summary that will help you decide which χ2 test is the appropriate one to use. • Goodness-of-Fit: Use the goodness-of-fit test to decide whether a population with an unknown distribution fits a known distribution. In this case there will be a single qualitative survey question or a single outcome of an experiment from a single population. Goodness-of-fit is typically used to see if the population is uniform (all outcomes occur with equal frequency), the population is normal, or the population is the same as another population with a known distribution. The null and alternative hypotheses are as follows: H0: The population fits the given distribution. Ha: The population does not fit the given distribution. Independence: Use the test for independence to decide whether two variables (factors) are independent or dependent. In this case there will be two qualitative survey questions or experiments and a contingency table will be constructed. The goal is to see if the two variables are unrelated/independent or related/dependent. The null and alternative hypotheses are as follows: H0: The two variables (factors) are independent. Ha: The two variables (factors) are dependent. • • Homogeneity: Use the test for homogeneity to decide if two populations with unknown distributions have the same distribution. In this case there will be a single qualitative survey question or experiment given to two different populations. The null and alternative hypotheses are as follows: H0: The two populations follow the same distribution. Ha: The two populations have different distributions. 11.6 \| Test of a Single Variance A test of a single variance assumes that the underlying distribution is normal. The null and alternative hypotheses are stated in terms of the population variance or population standard deviation. The test statistic is 658 where • n = the total number of data, s2 = sample variance, and • • σ2 = population variance. Chapter 11 \| The Chi-Square Distribution ⎞ ⎠s2 ⎛ ⎝n - 1 σ 2 You may think of s as the random variable in this test. The number of degrees of freedom is df = n – 1. A test of a single variance may be right-tailed, left-tailed, or two-tailed. Example 11.10 will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance. Example 11.10 Math instructors are not only interested in how their students do on exams, on average, but how the exam scores vary. To many instructors, the variance, or standard deviation, may be more important than the average. Suppose a math instructor believes that the standard deviation for his final exam is five points. One of his best students thinks otherwise. The student claims that the standard deviation is more than five points. If the student were to conduct a hypothesis test, what would the null and alternative hypotheses be? Solution 11.10 Even though we are given the population standard deviation, we can set up the test using the population variance as follows: • H0: σ2 = 52 • Ha: σ2 > 52 11.10 A scuba instructor wants to record the collective depths each of his students dives during their checkout. He is interested in how the depths vary, even though everyone should have been at the same depth. He believes the standard deviation is three feet. His assistant thinks the standard deviation is less than three feet. If the instructor were to conduct a test, what would the null and alternative hypotheses be? Example 11.11 With individual lines at its various windows, a post office finds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes. With a significance level of 5 percent, test the claim that a single line causes lower variation among waiting times (shorter waiting times) for customers. Solution 11.11 Since the claim is that a single line causes less variation, this is a test of a single variance. The parameter is the population variance, σ2, or the population standard deviation, σ. Random variable: The sample standard deviation, s, is the random variable. Let s = standard deviation for the waiting times. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 659 • H0: σ2 = 7.22 • Ha: σ2 < 7.22 The word less tells you this is a left-tailed test. Distribution for the test: χ24 2 , where • n = the number of customers sampled, and • df = n – 1 = 25 – 1 = 24. Calculate the test statistic: χ 2 = (n − 1)s2 σ 2 = (25 − 1)(3.5)2 7.22 = 5.67 where n = 25, s = 3.5, and σ = 7.2. Graph Figure 11.8 Probability statement: p-value = P ( χ2 < 5.67) = 0.000042 Compare α and the p-value: α = 0.05 p-value = 0.000042 α > p-value Make a decision: Since α > p-value, reject H0. This means that you reject σ2 = 7.22. In other words, you do not think the variation in waiting times is 7.2 minutes; you think the variation in waiting times is less. Conclusion: At a 5 percent level of significance, from the data, there is sufficient evidence to conclude that a single line causes a lower variation among the waiting times or with a single line, the customer waiting times vary less than 7.2 minutes. In 2nd DISTR, use 7:χ2cdf. The syntax is (lower, upper, df) for the parameter list. For Example 11.11, χ2cdf(-1E99,5.67,24). The p-value = 0.000042. 11.11 The FCC conducts broadband speed tests to measure how much data per second passes between a consumer’s computer and the internet. As of August 2012, the standard deviation of internet speeds across internet service providers (ISPs) was 12.2 percent. Suppose a sample of 15 ISPs is taken, and the standard deviation is 13.2. An analyst 660 Chapter 11 \| The Chi-Square Distribution claims that the standard deviation of speeds is more than what was reported. State the null and alternative hypotheses, compute the degrees of freedom, calculate the test statistic, sketch the graph of the p-value, and draw a conclusion. Test at the 1 percent significance level. 11.7 \| Lab 1: Chi-Square Goodness-of-Fit This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 661 11.1 Lab 1: Chi-Square Goodness-of-Fit Student Learning Outcome • The student will evaluate data collected to determine if they fit either the uniform or exponential distributions. Collect the Data Go to your local supermarket. Ask 30 people as they leave for the total amount on their grocery receipts. Or, ask 3 cashiers for the last 10 amounts. Be sure to include the express lane, if it is open. NOTE You may need to combine two categories so that each cell has an expected value of at least five. 1. Record the values. __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Table 11.23 2. Construct a histogram of the data. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes. Figure 11.9 3. Calculate the following: a. b. c. x¯ = ________ s = ________ s2 = ________ 662 Chapter 11 \| The Chi-Square Distribution Uniform Distribution Test to see if grocery receipts follow the uniform distribution. 1. Using your lowest and highest values, X ~ U (_______, _______). 2. Divide the distribution into fifths. 3. Calculate the following: a. lowest value = _________ b. 20th percentile = _________ c. 40th percentile = _________ d. 60th percentile = _________ e. 80th percentile = _________ f. highest value = _________ 4. For each fifth, count the observed number of receipts and record it. Then determine the expected number of receipts and record that. Fifth Observed Expected 1st 2nd 3rd 4th 5th Table 11.24 5. H0: ________ 6. Ha: ________ 7. What distribution should you use for a hypothesis test? 8. Why did you choose this distribution? 9. Calculate the test statistic. 10. Find the p-value. 11. Sketch a graph of the situation. Label and scale the x-axis. Shade the area corresponding to the p-value. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 663 Figure 11.10 12. State your decision. 13. State your conclusion in a complete sentence. Exponential Distribution Test to see if grocery receipts follow the exponential distribution with decay parameter 1 ¯ x . as the decay parameter, X ~ Exp(_________). 1. Using 1 x¯ 2. Calculate the following: a. lowest value = ________ b. first quartile
= ________ c. 37th percentile = ________ d. median = ________ e. 63rd percentile = ________ f. 3rd quartile = ________ g. highest value = ________ 3. For each cell, count the observed number of receipts and record it. Then determine the expected number of receipts and record that. Cell Observed Expected 1st 2nd 3rd 4th 5th 6th Table 11.25 4. H0: ________ 664 Chapter 11 \| The Chi-Square Distribution 5. Ha: ________ 6. What distribution should you use for a hypothesis test? 7. Why did you choose this distribution? 8. Calculate the test statistic. 9. Find the p-value. 10. Sketch a graph of the situation. Label and scale the x-axis. Shade the area corresponding to the p-value. Figure 11.11 11. State your decision. 12. State your conclusion in a complete sentence. Discussion Questions 1. Did your data fit either distribution? If so, which? 2. In general, do you think it’s likely that data could fit more than one distribution? In complete sentences, explain why or why not. 11.8 \| Lab 2: Chi-Square Test of Independence This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 665 11.2 Lab 2: Chi-Square Test of Independence Student Learning Outcome • The student will evaluate if there is a significant relationship between favorite type of snack and gender. Collect the Data 1. Using your class as a sample, complete the following chart. Ask one another what your favorite snack is, then total the results. NOTE You may need to combine two food categories so that each cell has an expected value of at least five. Sweets (candy & baked goods) Ice Cream Chips & Pretzels Fruits & Vegetables Total Male Female Total Table 11.26 Favorite Type of Snack 2. Looking at Table 11.26, does it appear to you that there is a dependence between gender and favorite type of snack food? Why or why not? Hypothesis Test Conduct a hypothesis test to determine if the factors are independent: 1. H0: ________ 2. Ha: ________ 3. What distribution should you use for a hypothesis test? 4. Why did you choose this distribution? 5. Calculate the test statistic. 6. Find the p value. 7. Sketch a graph of the situation. Label and scale the x axis. Shade the area corresponding to the p value. 666 Chapter 11 \| The Chi-Square Distribution Figure 11.12 8. State your decision. 9. State your conclusion in a complete sentence. Discussion Questions 1. Is the conclusion of your study the same as or different from your answer to answer to Question 2 under Collect the Data? 2. Why do you think that occurred? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 667 KEY TERMS contingency table a table that displays sample values for two different factors that may be dependent or contingent on each other; facilitates determining conditional probabilities CHAPTER REVIEW 11.1 Facts About the Chi-Square Distribution The chi-square distribution is a useful tool for assessment in a series of problem categories. These problem categories include primarily (i) whether a data set fits a particular distribution, (ii) whether the distributions of two populations are the same, (iii) whether two events might be independent, and (iv) whether there is a different variability than expected within a population. An important parameter in a chi-square distribution is the degrees of freedom df in a given problem. The random variable in the chi-square distribution is the sum of squares of df standard normal variables, which must be independent. The key characteristics of the chi-square distribution also depend directly on the degrees of freedom. The chi-square distribution curve is skewed to the right, and its shape depends on the degrees of freedom df. For df > 90, the curve approximates the normal distribution. Test statistics based on the chi-square distribution are always greater than or equal to zero. Such application tests are almost always right-tailed tests. 11.2 Goodness-of-Fit Test To assess whether a data set fits a specific distribution, you can apply the goodness-of-fit hypothesis test that uses the chi-square distribution. The null hypothesis for this test states that the data come from the assumed distribution. The test compares observed values against the values you would expect to have if your data followed the assumed distribution. The test is almost always right-tailed. Each observation or cell category must have an expected value of at least five. 11.3 Test of Independence To assess whether two factors are independent, you can apply the test of independence that uses the chi-square distribution. The null hypothesis for this test states that the two factors are independent. The test compares observed values to expected values. The test is right-tailed. Each observation or cell category must have an expected value of at least five. 11.4 Test for Homogeneity To assess whether two data sets are derived from the same distribution, which need not be known, you can apply the test for homogeneity that uses the chi-square distribution. The null hypothesis for this test states that the populations of the two data sets come from the same distribution. The test compares the observed values against the expected values if the two populations followed the same distribution. The test is right-tailed. Each observation or cell category must have an expected value of at least five. 11.5 Comparison of the Chi-Square Tests The goodness-of-fit test is typically used to determine if data fits a particular distribution. The test of independence makes use of a contingency table to determine the independence of two factors. The test for homogeneity determines whether two populations come from the same distribution, even if this distribution is unknown. 11.6 Test of a Single Variance To test variability, use the chi-square test of a single variance. The test may be left-, right-, or two-tailed, and its hypotheses are always expressed in terms of the variance or standard deviation. FORMULA REVIEW 11.1 Facts About the Chi-Square Distribution χ2 = (Z1)2 + (Z2)2 + . . . (Zdf)2 chi-square distribution random variable μχ2 = df chi-square distribution population mean 668 Chapter 11 \| The Chi-Square Distribution ⎠ chi-square distribution population standard ⎝d f ⎞ σ χ 2 = 2⎛ deviation 11.2 Goodness-of-Fit Test (O − E)2 E ∑ k goodness-of-fit test statistic where O: observed values E: expected values k: number of different data cells or categories df = k − 1 degrees of freedom 11.4 Test for Homogeneity Homogeneity test statistic where O = (O − E)2 E ∑ i ⋅ j observed values E = expected values i = number of rows in data contingency table j = number of columns in data contingency table df = (i −1)(j −1) degrees of freedom 11.6 Test of a Single Variance χ 2 = (n − 1) ⋅ s2 σ 2 Test of a single variance statistic 11.3 Test of Independence Test of Independence • The number of degrees of freedom is equal to (number of columns–1)(number of rows–1). where n: sample size s: sample standard deviation σ: population standard deviation df = n – 1 degrees of freedom (O – E)2 E • The test statistic is Σ (i ⋅ j) observed values, E = expected values, i = the number of rows in the table, and j = the number of columns in the table. where O = • If the null hypothesis is true, the expected number E = (row total)(column total) total surveyed . PRACTICE 11.1 Facts About the Chi-Square Distribution Test of a Single Variance • Use the test to determine variation. • The degrees of freedom is the number of samples – 1. • The test statistic is (n – 1) ⋅ s2 , where n = the total σ 2 number of data, s2 = sample variance, and σ2 = population variance. • The test may be left-, right-, or two-tailed. 1. If the number of degrees of freedom for a chi-square distribution is 25, what is the population mean and standard deviation? 2. If df > 90, the distribution is _____________. If df = 15, the distribution is ________________. 3. When does the chi-square curve approximate a normal distribution? 4. Where is μ located on a chi-square curve? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 669 5. Is it more likely the df is 90, 20, or 2 in the graph? Figure 11.13 11.2 Goodness-of-Fit Test Determine the appropriate test to be used in the next three exercises. 6. An archeologist is calculating the distribution of the frequency of the number of artifacts she finds in a dig site. Based on previous digs, the archeologist creates an expected distribution broken down by grid sections in the dig site. Once the site has been fully excavated, she compares the actual number of artifacts found in each grid section to see if her expectation was accurate. 7. An economist is deriving a model to predict outcomes on the stock market. He creates a list of expected points on the stock market index for the next two weeks. At the close of each day’s trading, he records the actual points on the index. He wants to see how well his model matched what actually happened. 8. A personal trainer is putting together a weight-lifting program for her clients. For a 90-day program, she expects each client to lift a specific maximum weight each week. As she goes along, she records the actual maximum weights her clients lifted. She wants to know how well her expectations met with what was observed. Use the following information to answer the next five exercises. A teacher predicts the distribution of grades on the final exam. The predictions are shown in Table 11.27. Grade Proportion A B C D 0.25 0.30 0.35 0.10 Table 11.27 The actual distribution for a class of 20 is in Table 11.28. Grade Frequency A 7 Table 11.28 670 Chapter 11 \| The Chi-Square Distribution Grade Frequency B C D 7 5 1 Table 11.28 9. d f = ______ 10. State the null and alternative hypotheses. 11. χ2 test statistic = ______ 12. p-value = ______ 13. At the 5 percent significance level, what can you conclude? Use the
following information to answer the next nine exercises. The cumulative number of cases of a chronic disease reported for Santa Clara County is broken down by ethnicity as in Table 11.29. Ethnicity White Hispanic Number of Cases 2,229 1,157 Black/African American 457 Asian, Pacific Islander 232 Total = 4,075 Table 11.29 The percentage of each ethnic group in Santa Clara County is as in Table 11.30. Ethnicity White Hispanic Black/African American Asian, Pacific Islander Table 11.30 % of Total County Population Number Expected (round to two decimal places) 1,748.18 42.9% 26.7% 2.6% 27.8% Total = 100% 14. If the ethnicities of patients followed the ethnicities of the total county population, fill in the expected number of cases per ethnic group. Perform a goodness-of-fit test to determine whether the occurrence of disease cases follows the ethnicities of the general population of Santa Clara County. 15. H0: _______ 16. Ha: _______ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 671 17. Is this a right-tailed, left-tailed, or two-tailed test? 18. degrees of freedom = _______ 19. χ2 test statistic = _______ 20. p-value = _______ 21. Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the p-value. Figure 11.14 Let α = 0.05. Decision: ________________ Reason for the decision: ________________ Conclusion (write out in complete sentences): ________________ 22. Does it appear that the pattern of disease cases in Santa Clara County corresponds to the distribution of ethnic groups in this county? Why or why not? 11.3 Test of Independence Determine the appropriate test to be used in the next three exercises. 23. A pharmaceutical company is interested in the relationship between age and presentation of symptoms for a common viral infection. A random sample is taken of 500 people with the infection across different age groups. 24. The owner of a baseball team is interested in the relationship between player salaries and team winning percentage. He takes a random sample of 100 players from different organizations. 25. A marathon runner is interested in the relationship between the brand of shoes runners wear and their run times. She takes a random sample of 50 runners and records their run times and the brand of shoes they were wearing. Use the following information to answer the next seven exercises: Transit Railroads is interested in the relationship between travel distance and the ticket class purchased. A random sample of 200 passengers is taken. Table 11.31 shows the results. The railroad wants to know if a passenger’s choice in ticket class is independent of the distance the passenger must travel. Traveling Distance Third Class Second Class First Class Total 1–100 miles 101–200 miles 201–300 miles Table 11.31 21 18 16 14 16 17 6 8 15 41 42 48 672 Chapter 11 \| The Chi-Square Distribution Traveling Distance Third Class Second Class First Class Total 301–400 miles 401–500 miles Total Table 11.31 12 6 73 14 6 67 21 10 60 47 22 200 26. State the hypotheses. H0: _______ Ha: _______ 27. df = _______ 28. How many passengers are expected to travel between 201 and 300 miles and purchase second-class tickets? 29. How many passengers are expected to travel between 401 and 500 miles and purchase first-class tickets? 30. What is the test statistic? 31. What is the p-value? 32. What can you conclude at the 5 percent level of significance? Use the following information to answer the next ten exercises. An article in the New England Journal of Medicine discussed a study on people who used a certain product in California and Hawaii. In one part of the report, the self-reported ethnicity and product-use levels per day were given. Of the people using the product at most 10 times per day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans, and 7,650 whites. Of the people using the product 11 to 20 times per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 whites. Of the people using the product 21 to 30 times per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 whites. Of the people using the product at least 31 times per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 whites. 33. Complete the table. Product use Per Day African American Native Hawaiian Latino Japanese American White TOTALS 1–10 11–20 21–30 31+ TOTALS Table 11.32 34. State the hypotheses. H0: _______ Ha: _______ 35. Enter expected values in Table 11.32. Round to two decimal places. Calculate the following values. 36. df = _______ 37. χ 2 test statistic = ______ 38. p-value = ______ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 673 39. Is this a right-tailed, left-tailed, or two-tailed test? Explain why. 40. Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the p-value. Figure 11.15 State the decision and conclusion (in a complete sentence) for the following levels of α. 41. α = 0.05 a. Decision: ___________________ b. Reason for the decision: ___________________ c. Conclusion (write out in a complete sentence): ___________________ 42. α = 0.01 a. Decision: ___________________ b. Reason for the decision: ___________________ c. Conclusion (write out in a complete sentence): ___________________ 11.4 Test for Homogeneity 43. A math teacher wants to see if two of her classes have the same distribution of test scores. What test should she use? 44. What are the null and alternative hypotheses for Exercise 11.43? 45. A market researcher wants to see if two different stores have the same distribution of sales throughout the year. What type of test should he use? 46. A meteorologist wants to know if East and West Australia have the same distribution of storms. What type of test should she use? 47. What condition must be met to use the test for homogeneity? Use the following information to answer the next five exercises. Do private practice doctors and hospital doctors have the same distribution of working hours? Suppose that a sample of 100 private practice doctors and 150 hospital doctors are selected at random and asked about the number of hours a week they work. The results are shown in Table 11.33. 20–30 30–40 40–50 50–60 Private Practice 16 Hospital 8 40 44 38 59 6 39 Table 11.33 48. State the null and alternative hypotheses. 49. df = _______ 50. What is the test statistic? 674 Chapter 11 \| The Chi-Square Distribution 51. What is the p-value? 52. What can you conclude at the 5 percent significance level? 11.5 Comparison of the Chi-Square Tests 53. Which test do you use to decide whether an observed distribution is the same as an expected distribution? 54. What is the null hypothesis for the type of test from Exercise 11.53? 55. Which test would you use to decide whether two factors have a relationship? 56. Which test would you use to decide if two populations have the same distribution? 57. How are tests of independence similar to tests for homogeneity? 58. How are tests of independence different from tests for homogeneity? 11.6 Test of a Single Variance Use the following information to answer the next three exercises. An archer’s standard deviation for his hits is six, where the data are measured in distance from the center of the target. An observer claims the standard deviation is less than six. 59. What type of test should be used? 60. State the null and alternative hypotheses. 61. Is this a right-tailed, left-tailed, or two-tailed test? Use the following information to answer the next three exercises. The standard deviation of heights for students in a school is 0.81. A random sample of 50 students is taken, and the standard deviation of heights of the sample is 0.96. A researcher in charge of the study believes the standard deviation of heights for the school is greater than 0.81. 62. What type of test should be used? 63. State the null and alternative hypotheses. 64. df = ________ Use the following information to answer the next four exercises: The average waiting time in a doctor’s office varies. The standard deviation of waiting times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the variance of waiting times is greater than originally thought. 65. What type of test should be used? 66. What is the test statistic? 67. What is the p-value? 68. What can you conclude at the 5 percent significance level? HOMEWORK 11.1 Facts About the Chi-Square Distribution Decide whether the following statements are true or false. 69. As the number of degrees of freedom increases, the graph of the chi-square distribution looks more and more symmetrical. 70. The standard deviation of the chi-square distribution is twice the mean. 71. The mean and the median of the chi-square distribution are the same if df = 24. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 675 11.2 Goodness-of-Fit Test For each problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places. 72. A six-sided die is rolled 120 times. Fill in the expected frequency column. Then, conduct a hypothesis test to determine if the die is fair. The data in Table 11.34 are the result of the 120 rolls. Face Value Frequency Expected Frequency 1 2 3 4 5 6 Table 11.34 15 29 16 15 30 15 73. The marital status distribution of the U.S. male population, ages 15 and older, is as shown in Table 11.35. Marital
Status % Expected Frequency Never Married Married Widowed 31.3% 56.1% 2.5% Divorced/Separated 10.1% Table 11.35 Suppose that a random sample of 400 U.S. males, 18 to 24 years old, yielded the following frequency distribution. We are interested in whether this age group of males fits the distribution of the U.S. adult population. Calculate the frequency one would expect when surveying 400 people. Fill in Table 11.35, rounding to two decimal places. Marital Status Frequency Never Married Married Widowed 140 238 2 Divorced/Separated 20 Table 11.36 Use the following information to answer the next two exercises. The columns in Table 11.37 contain the Race/Ethnicity of U.S. Public Schools for a recent year, the percentages for the Advanced Placement Examinee Population for that class, and the Overall Student Population. Suppose the right column contains the results of a survey of 1,000 local students from that year who took an AP exam. 676 Chapter 11 \| The Chi-Square Distribution Race/Ethnicity AP Examinee Population Overall Student Population Survey Frequency Asian, Asian American, or Pacific Islander Black or African American Hispanic or Latino 10.2% 8.2% 15.5% American Indian or Alaska Native 0.6% White Not Reported/Other Table 11.37 59.4% 6.1% 5.4% 14.5% 15.9% 1.2% 61.6% 1.4% 113 94 136 10 604 43 74. Perform a goodness-of-fit test to determine whether the local results follow the distribution of the U.S. overall student population based on ethnicity. 75. Perform a goodness-of-fit test to determine whether the local results follow the distribution of U.S. AP examinee population, based on ethnicity. 76. The city of South Lake Tahoe, California, has an Asian population of 1,419 out of a total population of 23,609. Suppose that a survey of 1,419 self-reported Asians in the borough of Manhattan in the New York City area yielded the data in Table 11.38. Conduct a goodness-of-fit test to determine if the self-reported subgroups of Asians in Manhattan fit that of the South Lake Tahoe area. Race South Lake Tahoe Frequency Manhattan Frequency Asian Indian 131 Chinese 118 Filipino 1,045 Japanese Korean 80 12 Vietnamese 9 Other 24 Table 11.38 174 557 518 54 29 21 66 Use the following information to answer the next two exercises. UCLA conducted a survey of more than 263,000 college freshmen from 385 colleges in fall 2005. The results of students’ expected majors by gender were reported in The Chronicle of Higher Education (2/2/2006). Suppose a survey of 5,000 graduating females and 5,000 graduating males was done as a follow-up last year to determine what their actual majors were. The results are shown in the tables for Exercise 11.77 and Exercise 11.78. The second column in each table does not add to 100 percent because of rounding. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 677 77. Conduct a goodness-of-fit test to determine if the actual college majors of graduating females fit the distribution of their expected majors. Major Females—Expected Major Females—Actual Major Arts & Humanities 14% Biological Sciences 8.4% Business Education Engineering 13.1% 13% 2.6% Physical Sciences 2.6% Professional 18.9% Social Sciences Technical Other Undecided Table 11.39 13% 0.4% 5.8% 8% 670 410 685 650 145 125 975 605 15 300 420 78. Conduct a goodness-of-fit test to determine if the actual college majors of graduating males fit the distribution of their expected majors. Major Males—Expected Major Males—Actual Major Arts & Humanities 11% Biological Sciences 6.7% Business Education Engineering 22.7% 5.8% 15.6% Physical Sciences 3.6% Professional Social Sciences Technical Other Undecided Table 11.40 9.3% 7.6% 1.8% 8.2% 6.6% 600 330 1,130 305 800 175 460 370 90 400 340 Read the statement and decide whether it is true or false. 79. In a goodness-of-fit test, the expected values are the values we would expect if the null hypothesis were true. 80. In general, if the observed values and expected values of a goodness-of-fit test are not close together, then the test statistic can get very large and on a graph will be way out in the right tail. 81. Use a goodness-of-fit test to determine if high school principals believe that students are absent equally during the week. 82. The test to use to determine if a six-sided die is fair is a goodness-of-fit test. 83. In a goodness-of-fit test, if the p-value is 0.0113, in general, do not reject the null hypothesis. 678 Chapter 11 \| The Chi-Square Distribution 84. A sample of 212 commercial businesses was surveyed for recycling one commodity; a commodity here means any one type of recyclable material such as plastic or aluminum. Table 11.41 shows the business categories in the survey, the sample size of each category, and the number of businesses in each category that recycle one commodity. Based on the study, on average half of the businesses were expected to be recycling one commodity. As a result, the last column shows the expected number of businesses in each category that recycle one commodity. At the 5 percent significance level, perform a hypothesis test to determine if the observed number of businesses that recycle one commodity follows the uniform distribution of the expected values. Business Type Office Retail/ Wholesale Food/ Restaurants Manufacturing/ Medical Hotel/Mixed Table 11.41 Number in Class Observed Number that Recycle One Commodity Expected Number that Recycle One Commodity 35 48 53 52 24 19 27 35 21 9 17.5 24 26.5 26 12 85. Table 11.42 contains information from a survey of 499 participants classified according to their age groups. The researchers making the survey wanted to find out how many people were diagnosed with a particular disease within the last year. The second column shows the percentage of people with the disease per age class among the study participants. The last column comes from a different study at the national level that shows the corresponding percentages of people with the disease in the same age classes in the United States. Perform a hypothesis test at the 5 percent significance level to determine whether the survey participants are a representative sample of the people with the disease nationwide. Age Class (years) % of People Diagnosed % of Expected U.S. Average 20–30 31–40 41–50 51–60 61–70 Table 11.42 75% 26.5% 13.6% 21.9% 21% 32.6% 32.6% 36.6% 36.6% 39.7% 11.3 Test of Independence For each problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 679 86. A recent debate about where in the U.S. skiers believe the skiing is best prompted the following survey. Test to see if the best ski area is independent of the level of the skier. U.S. Ski Area Beginner Intermediate Advanced Tahoe Utah Colorado Table 11.43 20 10 10 30 30 40 40 60 50 87. Car manufacturers are interested in whether there is a relationship between the size of car an individual drives and the number of people in the driver’s family—that is, whether car size and family size are independent. To test this, suppose that 800 car owners were randomly surveyed with the results in Table 11.44. Conduct a test of independence. Family Size Sub & Compact Mid-Size Full-Size Van & Truck 1 2 3–4 5+ Table 11.44 20 20 20 20 35 50 50 30 40 70 100 70 35 80 90 70 88. College students may be interested in whether their majors have any effect on starting salaries after graduation. Suppose that 300 recent graduates were surveyed as to their majors in college and their starting salaries after graduation. Table 11.45 shows the data. Conduct a test of independence. Major < $50,000 $50,000–$68,999 $69,000 + English 5 Engineering 10 Nursing Business 10 10 Psychology 20 Table 11.45 20 30 15 20 30 5 60 15 30 20 89. Some travel agents claim that honeymoon hotspots vary according to age of the bride. Suppose that 280 recent brides were interviewed as to where they spent their honeymoons. The information is given in Table 11.46. Conduct a test of independence. Location 20–29 30–39 40–49 50+ Niagara Falls 15 Poconos Europe 15 10 Virgin Islands 20 Table 11.46 25 25 25 25 25 25 15 15 20 10 5 5 680 Chapter 11 \| The Chi-Square Distribution 90. A manager of a sports club keeps information concerning the main sport in which members participate and their ages. To test whether there is a relationship between the age of a member and his or her choice of sport, 643 members of the sports club are randomly selected. Conduct a test of independence. Sport 18–25 26–30 31–40 41+ Racquetball 42 Tennis Swimming 58 72 Table 11.47 58 76 60 30 38 65 46 65 33 91. A major food manufacturer is concerned that the sales for its skinny french fries have been decreasing. As a part of a feasibility study, the company conducts research into the types of fries sold across the country to determine if the type of fries sold is independent of the area of the country. The results of the study are shown in Table 11.48. Conduct a test of independence. Type of Fries Northeast South Central West Skinny Fries Curly Fries Steak Fries 70 100 20 Table 11.48 50 60 40 20 15 10 25 30 10 92. According to Dan Leonard, an independent insurance agent in the Buffalo, New York area, the following is a breakdown of the amount of life insurance purchased by males in the following age groups. He is interested in whether the age of the male and the amount of life insurance purchased are independent events. Conduct a test for independence. Age of Males None < $200,000 $200,000–$400,000 $401,001–$1,000,000 $1,000,001+ 20–29 30–39 40–49 50+ Table 11.49 40 35 20 40 15 5 0 30 40 20 30 15 0 20 0 15 5 10 30 10 93. Suppose that 600 thirty-year-olds were surveyed to determine whether there is a relationship between the level of education an individual has and salary. Conduct a test of ind
ependence. Annual Salary Not a High School Graduate High School Graduate College Graduate Masters or Doctorate < $30,000 15 $30,000–$40,000 20 $40,000–$50,000 10 $50,000–$60,000 5 $60,000+ 0 Table 11.50 25 40 20 10 5 10 70 40 20 10 5 30 55 60 150 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 681 Read the statement and decide whether it is true or false. 94. The number of degrees of freedom for a test of independence is equal to the sample size minus one. 95. The test for independence uses tables of observed and expected data values. 96. The test to use when determining if the college or university a student chooses to attend is related to his or her socioeconomic status is a test for independence. 97. In a test of independence, the expected number is equal to the row total multiplied by the column total divided by the total surveyed. 98. An ice cream maker performs a nationwide survey about favorite flavors of ice cream in different geographic areas of the United States. Based on Table 11.51, do the numbers suggest that geographic location is independent of favorite ice cream flavors? Test at the 5 percent significance level. U.S. Region/ Flavor West Midwest East South Column Total Table 11.51 Strawberry Chocolate Vanilla Rocky Road Mint Chocolate Chip Pistachio Row Total 12 10 8 15 45 21 32 31 28 22 22 27 30 112 101 19 11 8 8 46 15 15 15 15 60 8 6 7 6 27 97 96 96 102 391 99. Table 11.52 provides results of a recent survey of the youngest online entrepreneurs whose net worth is estimated at one million dollars or more. Their ages range from 17 to 30. Each cell in the table illustrates the number of entrepreneurs who correspond to the specific age group and their net worth. Are the ages and net worth independent? Perform a test of independence at the 5 percent significance level. Age Group/Net Worth Value (in millions of U.S. dollars) 1–5 6–24 ≥25 Row Total 17–25 26–30 Column Total Table 11.52 8 6 7 5 14 12 5 9 14 20 20 40 100. A 2013 poll in California surveyed people about a new tax. The results are presented in Table 11.53 and are classified by ethnic group and response type. Are the poll responses independent of the participants’ ethnic group? Conduct a test of independence at the 5 percent significance level. Opinion/ Ethnicity Asian American White/NonHispanic African American Against Tax In Favor of Tax No Opinion 48 54 16 Column Total 118 Table 11.53 433 234 43 710 41 24 16 81 Latino 160 147 19 326 Row Total 682 459 94 1,235 682 Chapter 11 \| The Chi-Square Distribution 11.4 Test for Homogeneity For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places. 101. A psychologist is interested in testing whether there is a difference in the distribution of personality types for business majors and social science majors. The results of the study are shown in Table 11.54. Conduct a test of homogeneity. Test at a 5 percent level of significance. Open Conscientious Extrovert Agreeable Neurotic Business 41 Social Science 72 52 75 46 63 61 80 58 65 Table 11.54 102. Do men and women select different breakfasts? The breakfasts ordered by randomly selected men and women at a popular breakfast place are shown in Table 11.55. Conduct a test for homogeneity at a 5 percent level of significance. French Toast Pancakes Waffles Omelettes Men 47 Women 65 Table 11.55 35 59 28 55 53 60 103. A fisherman is interested in whether the distribution of fish caught in Green Valley Lake is the same as the distribution of fish caught in Echo Lake. Of the 191 randomly selected fish caught in Green Valley Lake, 105 were rainbow trout, 27 were other trout, 35 were bass, and 24 were catfish. Of the 293 randomly selected fish caught in Echo Lake, 115 were rainbow trout, 58 were other trout, 67 were bass, and 53 were catfish. Perform a test for homogeneity at a 5 percent level of significance. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 683 104. In 2007, the United States had 1.5 million homeschooled students, according to the U.S. National Center for Education Statistics. In Table 11.56, you can see that parents decide to homeschool their children for different reasons, and some reasons are ranked by parents as more important than others. According to the survey results shown in the table, is the distribution of applicable reasons the same as the distribution of the most important reason? Provide your assessment at the 5 percent significance level. Did you expect the result you obtained? Applicable Reason (in thousands of respondents) Most Important Reason (in thousands of respondents) Reasons for Homeschooling Concern About the Environment of Other Schools Dissatisfaction with Academic Instruction at Other Schools To Provide Religious or Moral Instruction Child Has Special Needs, Other Than Physical or Mental Nontraditional Approach to Child’s Education Other Reasons (e.g., finances, travel, family time, etc.) 1,321 1,096 1,257 315 984 485 Column Total 5,458 Table 11.56 Row Total 1,630 1,354 1,797 370 1,083 701 6,935 309 258 540 55 99 216 1,477 105. When looking at energy consumption, we are often interested in detecting trends over time and how they correlate among different countries. The information in Table 11.57 shows the average energy use in units of kg of oil equivalent per capita in the United States and the joint European Union countries (EU) for the six-year period 2005 to 2010. Do the energy use values in these two areas come from the same distribution? Perform the analysis at the 5 percent significance level. European Union United States Row Total Year 2010 2009 2008 2007 2006 2005 3,413 3,302 3,505 3,537 3,595 3,613 7,164 7,057 7,488 7,758 7,697 7,847 Column Total 20,965 45,011 Table 11.57 10,557 10,359 10,993 11,295 11,292 11,460 65,976 684 Chapter 11 \| The Chi-Square Distribution 106. The Insurance Institute for Highway Safety collects safety information about all types of cars every year and publishes a report of top safety picks among all cars, makes, and models. Table 11.58 presents the number of top safety picks in six car categories for the two years 2009 and 2013. Analyze the table data to conclude whether the distribution of cars that earned the top safety picks safety award has remained the same between 2009 and 2013. Derive your results at the 5 percent significance level. Year/Car Type 2009 2013 Column Total Table 11.58 Small 12 31 43 MidSize 22 30 52 Large Small SUV Mid-Size SUV Large SUV 10 19 29 10 11 21 27 29 56 6 4 10 Row Total 87 124 211 11.5 Comparison of the Chi-Square Tests For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places. 107. Is there a difference between the distribution of community college statistics students and the distribution of university statistics students in what technology they use on their homework? Of some randomly selected community college students, 43 used a computer, 102 used a calculator with built-in statistics functions, and 65 used a table from the textbook. Of some randomly selected university students, 28 used a computer, 33 used a calculator with built-in statistics functions, and 40 used a table from the textbook. Conduct an appropriate hypothesis test using a 0.05 level of significance. Read the statement and decide whether it is true or false. 108. If df = 2, the chi-square distribution has a shape that reminds us of the exponential. 11.6 Test of a Single Variance Use the following information to answer the next 12 exercises. Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes. 109. Is the traveler disputing the claim about the average or about the variance? 110. A sample standard deviation of 15 minutes is the same as a sample variance of __________ minutes. 111. Is this a right-tailed, left-tailed, or two-tailed test? 112. H0: __________ 113. df = ________ 114. chi-square test statistic = ________ 115. p-value = ________ 116. Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade the p-value. 117. Let α = 0.05 Decision: ________ Conclusion (write out in a complete sentence): ________ 118. How did you know to test the variance instead of the mean? 119. If an additional test were done on the claim of the average delay, which distribution would you use? 120. If an additional test were done on the claim of the average delay, but 45 flights were surveyed, which distribution would you use? For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 685 121. A plant manager is concerned her equipment may need recalibrating. It seems that the actual weight of the 15-ounce cereal boxes it fills has been fluctuating. The standard deviation should be at most 0.5 ounces. To determine if the machine needs to be recalibrated, 84 randomly selected boxes of cereal from the next day’s production were weighed. The standard deviation of the 84 boxes was 0.54. Does the machine need to be recalibrated? 122. Consumers may be interested in whether the cost of a particular calculator varies from store to store. Based on surveying 43 stores, which yielded a sample
mean of $84 and a sample standard deviation of $12, test the claim that the standard deviation is greater than $15. 123. Isabella, an accomplished Bay-to-Breakers runner, claims that the standard deviation for her time to run the 7.5 mile race is at most 3 minutes. To test her claim, Isabella looks up five of her race times. They are 55 minutes, 61 minutes, 58 minutes, 63 minutes, and 57 minutes. 124. Airline companies are interested in the consistency of the number of babies on each flight so that they have adequate safety equipment. They are also interested in the variation of the number of babies. Suppose that an airline executive believes the average number of babies on flights is six with a variance of nine at most. The airline conducts a survey. The results of the 18 flights surveyed give a sample average of 6.4 with a sample standard deviation of 3.9. Conduct a hypothesis test of the airline executive’s belief. 125. The number of births per woman in China is 1.6, down from 5.91 in 1966. This fertility rate has been attributed to the law passed in 1979 restricting births to one per woman. Suppose that a group of students studied whether the standard deviation of births per woman was greater than 0.75. They asked 50 women across China the number of births they had. The results are shown in Table 11.59. Does the students’ survey indicate that the standard deviation is greater than 0.75? # of Births Frequency 0 1 2 3 Table 11.59 5 30 10 5 126. According to an avid aquarist, the average number of fish in a 20-gallon tank is 10, with a standard deviation of two. His friend, also an aquarist, does not believe that the standard deviation is two. She counts the number of fish in 15 other 20-gallon tanks. Based on the results that follow, do you think that the standard deviation is different from two? Data: 11; 10; 9; 10; 10; 11; 11; 10; 12; 9; 7; 9; 11; 10; and 11. 127. The manager of Frenchies is concerned that patrons are not consistently receiving the same amount of French fries with each order. The chef claims that the standard deviation for a 10-ounce order of fries is at most 1.5 ounces, but the manager thinks that it may be higher. He randomly weighs 49 orders of fries, which yields a mean of 11 ounces and a standard deviation of 2 ounces. 128. You want to buy a specific computer. A sales representative of the manufacturer claims that retail stores sell this computer at an average price of $1,249 with a very narrow standard deviation of $25. You find a website that has a price comparison for the same computer at a series of stores as follows: $1,299; $1,229.99; $1,193.08; $1,279; $1,224.95; $1,229.99; $1,269.95; and $1,249. Can you argue that pricing has a larger standard deviation than claimed by the manufacturer? Use the 5 percent significance level. As a potential buyer, what would be the practical conclusion from your analysis? 129. A company packages apples by weight. One of the weight grades is Class A apples. Class A apples have a mean weight of 150 grams, and there is a maximum allowed weight tolerance of 5 percent above or below the mean for apples in the same consumer package. A batch of apples is selected to be included in a Class A apple package. Given the following apple weights of the batch, does the fruit comply with the Class A grade weight tolerance requirements? Conduct an appropriate hypothesis test. (a) At the 5 percent significance level (b) At the 1 percent significance level Weights in selected apple batch (in grams): 158; 167; 149; 169; 164; 139; 154; 150; 157; 171; 152; 161; 141; 166; and 172. 686 Chapter 11 \| The Chi-Square Distribution BRINGING IT TOGETHER: HOMEWORK 130. a. Explain why a goodness-of-fit test and a test of independence are generally right-tailed tests. b. If you did a left-tailed test, what would you be testing? REFERENCES 11.1 Facts About the Chi-Square Distribution Parade Magazine. (n.d.). Retrieved from https://parade.com/ Santa Clara County Public Health Department. (2011, May). HIV/AIDS epidemiology Santa Clara County. Retrieved from http://sccgov.iqm2.com/Citizens/FileOpen.aspx?Type=4&ID=32762 11.2 Goodness-of-Fit Test College Board. (n.d.). Retrieved from http://www.collegeboard.com Ma, Y., et al. (2003). Association between eating patterns and obesity in a free-living US adult population. American Journal of Epidemiology 158(1), 85–92. Ogden, C. L., et al. (2012, January). Prevalence of obesity in the United States, 2009–2010 (NCHS Data Brief No. 82). Hyattsville, MD: National Center for Health Statistics. Retrieved from http://www.cdc.gov/nchs/data/databriefs/db82.pdf Stevens, B. J. (n.d.). Multi-family and commercial solid waste and recycling survey. Arlington County, VA. Retrieved from http://www.arlingtonva.us/departments/EnvironmentalServices/SW/file84429.pdf U.S. Census Bureau. (n.d.). Current population reports. Retrieved from https://www.census.gov/main/www/cprs.html U.S. Census Bureau. (n.d). Retrieved from https://www.census.gov/ 11.3 Test of Independence Harris Interactive. (n.d.). Retrieved from http://www.statisticbrain.com/favorite-flavor-of-ice-cream/ Statistics Brain. (2016, June 29). Youngest online entrepreneurs list. Retrieved from http://www.statisticbrain.com/ youngest-online-entrepreneur-list 11.4 Test for Homogeneity Bielick, S. (2008, December). 1.5 million homeschooled students in the United States in 2007 (NCES 2009030). Washington, DC: National Center pubsinfo.asp?pubid=2009030 for Education Statistics. Retrieved from http://nces.ed.gov/pubsearch/ Bielick, S. (2008, December). 1.5 million homeschooled students in the United States in 2007—supplemental tables (NCES 2009030). Washington, DC: National Center for Education Statistics. Retrieved from http://nces.ed.gov/pubs2009/ 2009030_sup.pdf Insurance Institute for Highway Safety. (n.d.). Ratings. Retrieved from www.iihs.org/iihs/ratings World Bank Group. (2014). Energy use (kg of oil equivalent per capita). Retrieved from http://data.worldbank.org/indicator/ EG.USE.PCAP.KG.OE/countries 11.6 Test of a Single Variance Apple Insider. (n.d.). Retrieved from http://appleinsider.com/mac_price_guide World Bank. (n.d.). Retrieved from http://www.worldbank.org/ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 687 SOLUTIONS 1 mean = 25 and standard deviation = 7.0711 3 when the number of degrees of freedom is greater than 90 5 df = 2 7 a goodness-of-fit test 9 3 11 2.04 13 We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores. 15 H0: the distribution of disease cases follows the ethnicities of the general population of Santa Clara County. 17 right-tailed 19 2016.136 21 Graph: Check student’s solution. Decision: Reject the null hypothesis. Reason for decision: p-value < alpha Conclusion: The make-up of cases does not fit the ethnicities of the general population of Santa Clara County. 23 a test of independence 25 a test of independence 27 8 29 6.6 31 0.0435 33 Product-use Per Day African American Native Hawaiian Latino Japanese Americans 9,886 6,514 1,671 759 18,830 2,745 3,062 1,419 788 8,014 12,831 8,378 4,932 1,406 800 10,680 4,715 2,305 19,969 26,078 27,559 10,0450 White Totals 7,650 9,877 6,062 3,970 41,490 35,065 15,273 8,622 African American 7,777.57 6,573.16 Native Hawaiian 3,310.11 2797.52 Latino Japanese Americans 8,248.02 10,771.29 6970.76 9,103.29 White 11,383.01 9,620.27 1–10 11–20 21–30 31+ Totals Table 11.60 35 Product Use Per Day 1-10 11-20 Table 11.61 688 Chapter 11 \| The Chi-Square Distribution African American 2,863.02 1,616.25 Native Hawaiian 1,218.49 687.87 Latino Japanese Americans 3,036.20 3,965.05 1,714.01 2,238.37 White 4,190.23 2,365.49 Product Use Per Day 21-30 31+ Table 11.61 37 10,301.8 39 right-tailed 41 a. Reject the null hypothesis. b. p-value < alpha c. There is sufficient evidence to conclude that product use is dependent on ethnic group. 43 test for homogeneity 45 test for homogeneity 47 All values in the table must be greater than or equal to five. 49 3 51 0.00005 53 a goodness-of-fit test 55 a test for independence 57 Answers will vary. Sample answer: Tests of independence and tests for homogeneity both calculate the test statistic the same way ∑ (i j) (O - E)2 E . In addition, all values must be greater than or equal to five. 59 a test of a single variance 61 a left-tailed test 63 H0: σ2 = 0.812; Ha: σ2 > 0.812 65 a test of a single variance 67 0.0542 69 true 71 false 73 Marital Status % Expected Frequency Never Married 31.3% 125.2 Married Widowed 56.1% 224.4 2.5% 10 Divorced/Separated 10.1% 40.4 Table 11.62 a. The data fit the distribution. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 689 b. The data do not fit the distribution. c. 3 d. chi-square distribution with df = 3 e. 19.27 f. 0.0002 g. Check student’s solution. h. i. Alpha = 0.05 ii. Decision: Reject null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: Data do not fit the distribution. 75 a. H0: The local results follow the distribution of the U.S. AP examinee population. b. Ha: The local results do not follow the distribution of the U.S. AP examinee population. c. df = 5 d. chi-square distribution with df = 5 e. chi-square test statistic = 13.4 f. p-value = 0.0199 g. Check student’s solution. h. i. Alpha = 0.05 ii. Decision: Reject null when a = 0.05. iii. Reason for decision: p-value < alpha iv. Conclusion: Local data do not fit the AP examinee distribution. v. Decision: Do not reject null when a = 0.01 vi. Conclusion: There is insufficient evidence to conclude that local data do not follow the distribution of the U.S. AP examinee distribution. 77 a. H0: The actual college majors of graduating females fit the distribution of their expected majors. b. Ha: The actual college majors of graduating females do not f
it the distribution of their expected majors. c. df = 10 d. chi-square distribution with df = 10 e. test statistic = 11.48 f. p-value = 0.3211 g. Check student’s solution. h. i. Alpha = 0.05 ii. Decision: Do not reject null hypothesis when a = 0.05 and a = 0.01. iii. Reason for decision: p-value > alpha iv. Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating females do not fit the distribution of their expected majors. 79 true 81 true 83 false Chapter 11 \| The Chi-Square Distribution 690 85 a. H0: Surveyed individuals fit the distribution of expected patients. b. Ha: The surveyed individuals do not fit the distribution of patients. c. df = 4 d. chi-square distribution with df = 4 e. test statistic = 54.01 f. p-value = 0 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: At the 5 percent level of significance from the data, there is sufficient evidence to conclude that the surveyed patients with the disease do not fit the distribution of expected patients. 87 a. H0: Car size is independent of family size. b. Ha: Car size is dependent on family size. c. df = 9 d. chi-square distribution with df = 9 e. test statistic = 15.8284 f. p-value = 0.0706 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: At the 5 percent significance level, there is insufficient evidence to conclude that car size and family size are dependent. 89 a. H0: Honeymoon locations are independent of bride’s age. b. Ha: Honeymoon locations are dependent on bride’s age. c. df = 9 d. chi-square distribution with df = 9 e. test statistic = 15.7027 f. p-value = 0.0734 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: At the 5 percent significance level, there is insufficient evidence to conclude that honeymoon location and bride age are dependent. 91 a. H0: The types of fries sold are independent of the location. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 691 b. Ha: The types of fries sold are dependent on the location. c. df = 6 d. chi-square distribution with df = 6 e. test statistic =18.8369 f. p-value = 0.0044 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: At the 5 percent significance level, there is sufficient evidence that types of fries and location are dependent. 93 a. H0: Salary is independent of level of education. b. Ha: Salary is dependent on level of education. c. df = 12 d. chi-square distribution with df = 12 e. test statistic = 255.7704 f. p-value = 0 g. Check student’s solution. h. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: p-value < alpha Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that salary and level of education are dependent. 95 true 97 true 99 a. H0: Age is independent of the youngest online entrepreneurs’ net worth. b. Ha: Age is dependent on the net worth of the youngest online entrepreneurs. c. df = 2 d. chi-square distribution with df = 2 e. test statistic = 1.76 f. p-value = 0.4144 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: At the 5 percent significance level, there is insufficient evidence to conclude that age and net worth for the youngest online entrepreneurs are dependent. 101 a. H0: The distribution for personality types is the same for both majors. 692 Chapter 11 \| The Chi-Square Distribution b. Ha: The distribution for personality types is not the same for both majors. c. df = 4 d. chi-square with df = 4 e. test statistic = 3.01 f. p-value = 0.5568 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: There is insufficient evidence to conclude that the distribution of personality types is different for business and social science majors. 103 a. H0: The distribution for fish caught is the same in Green Valley Lake and in Echo Lake. b. Ha: The distribution for fish caught is not the same in Green Valley Lake and in Echo Lake. c. 3 d. chi-square with df = 3 e. 11.75 f. p-value = 0.0083 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: There is evidence to conclude that the distribution of fish caught is different in Green Valley Lake and in Echo Lake. 105 a. H0: The distribution of average energy use in the United States is the same as in Europe between 2005 and 2010. b. Ha: The distribution of average energy use in the United States is not the same as in Europe between 2005 and 2010. c. df = 4 d. chi-square with df = 4 e. test statistic = 2.7434 f. p-value = 0.7395 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: At the 5 percent significance level, there is insufficient evidence to conclude that the average energy use values in the United States and EU are not derived from different distributions for the period from 2005 to 2010. 107 a. H0: The distribution for technology use is the same for community college students and university students. b. Ha: The distribution for technology use is not the same for community college students and university students. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 11 \| The Chi-Square Distribution 693 c. 2 d. chi-square with df = 2 e. 7.05 f. p value = 0.0294 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p value < alpha iv. Conclusion: There is sufficient evidence to conclude that the distribution of technology use for statistics homework is not the same for statistics students at community colleges and at universities. 110 225 112 H0: σ2 ≤ 150 114 36 116 Check student’s solution. 118 The claim is that the variance is no more than 150 minutes. 120 a student's t or normal distribution 122 a. H0: σ = 15 b. Ha: σ > 15 c. df = 42 d. chi-square with df = 42 e. test statistic = 26.88 f. p-value = 0.9663 g. Check student’s solution. h. i. Alpha = 0.05 ii. Decision: Do not reject null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15. 124 a. H0: σ ≤ 3 b. Ha: σ > 3 c. df = 17 d. chi-square distribution with df = 17 e. test statistic = 28.73 f. p-value = 0.0371 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: There is sufficient evidence to conclude that the standard deviation is greater than three. Chapter 11 \| The Chi-Square Distribution 694 126 a. H0: σ = 2 b. Ha: σ ≠ 2 c. df = 14 d. chi-square distiribution with df = 14 e. chi-square test statistic = 5.2094 f. p-value = 0.0346 g. Check student’s solution. h. i. Alpha = 0.05 ii. Decision: Reject the null hypothesis iii. Reason for decision: p-value < alpha iv. Conclusion: There is sufficient evidence to conclude that the standard deviation is different than two. 128 The sample standard deviation is $34.29. H0 : σ2 = 252 Ha : σ2 > 252 df = n – 1 = 7 Test statistic: x2 = x7 2 = (n – 1)s2 252 ⎛ ⎞ ⎝x7 ⎠ = 1 – P = (8 – 1)(34.29)2 252 ⎞ ⎠ = .0681 = 13.169 ; 2 > 13.169 2 ≤ 13.169 ⎛ p-value: P ⎝x7 Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: p-value > alpha Conclusion: At the 5 percent level, there is insufficient evidence to conclude that the variance is more than 625. 130 a. The test statistic is always positive and if the expected and observed values are not close together, the test statistic is large and the null hypothesis will be rejected. b. Testing to see if the data fits the distribution too well or is too perfect. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 695 12 \| LINEAR REGRESSION AND CORRELATION Figure 12.1 Linear regression and correlation can help you determine whether an auto mechanic’s salary is related to his work experience. (credit: Joshua Rothhaas) Introduction Chapter Objectives By the end of this chapter, the student should be able to do the following: • Discuss basic ideas of linear regression and correlation • Create and interpret a line of best fit • Calculate and interpret the correlation coefficient • Calculate and interpret outliers Professionals often want to know how two or more numeric variables are related. For example, is there a relationship between the grade on the second math exam a student takes and the grade on the final exam? If there is a relationship, what is the relationship, and how strong is it? In another example, your income may be determined by your education, your profession, your years of experience, and your ability. The amount you pay a repair person for labor is often determined by an initial amount plus an hourly fee. 696 Chapter 12 \| Linear Regression and Correlation The type of data described in the examples is bivariate data—bi—for two variables. In reality, statisticians use multivariate data, meaning many variables. In this chapter, you will study the simplest form of regression—linear regression—with one independent variable (x). This involves data that fit a line in two dimensions. You will also study correlation, which measures the strength of a relationship. 12.1 \| Linear Equations Linear regression for two variables is based on a linear equa
tion with one independent variable. The equation has the form y = a + bx where a and b are constant numbers. The variable x is the independent variable; y is the dependent variable. Typically, you choose a value to substitute for the independent variable and then solve for the dependent variable. Example 12.1 The following examples are linear equations. y = 3 + 2x y = –0.01 + 1.2x 12.1 Is the following an example of a linear equation? y = –0.125 – 3.5x The graph of a linear equation of the form y = a + bx is a straight line. Any line that is not vertical can be described by this equation. Example 12.2 Graph the equation y = –1 + 2x. Figure 12.2 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 697 12.2 Is the following an example of a linear equation? Why or why not? Figure 12.3 Example 12.3 Aaron’s Word Processing Service does word processing. The rate for services is $32 per hour plus a $31.50 onetime charge. The total cost to a customer depends on the number of hours it takes to complete the job. Find the equation that expresses the total cost in terms of the number of hours required to complete the job. Solution 12.3 Let x = the number of hours it takes to get the job done. Let y = the total cost to the customer. The $31.50 is a fixed cost. If it takes x hours to complete the job, then (32)(x) is the cost of the word processing only. The total cost is y = 31.50 + 32x. 12.3 Emma’s Extreme Sports hires hang-gliding instructors and pays them a fee of $50 per class, as well as $20 per student in the class. The total cost Emma pays depends on the number of students in a class. Find the equation that expresses the total cost in terms of the number of students in a class. Slope and y-interceptof a Linear Equation For the linear equation y = a + bx, b = slope and a = y-inttercept. From algebra, recall that the slope is a number that describes the steepness of a line; the y-intercept is the y-coordinate of the point (0, a), where the line crosses the y-axis. Please note that in previous courses you learned y = mx + b was the slope-intercept form of the equation, where m represented the slope and b represented the y-intercept. In this text, the form y = a + bx is used, where a is the y-intercept and b is the slope. The key is remembering the coefficient of x is the slope, and the constant number is the y-intercept. 698 Chapter 12 \| Linear Regression and Correlation Figure 12.4 Three possible graphs of y = a + bx. (a) If b > 0, the line slopes upward to the right. (b) If b = 0, the line is horizontal. (c) If b < 0, the line slopes downward to the right. Example 12.4 Svetlana tutors to make extra money for college. For each tutoring session, she charges a one-time fee of $25 plus $15 per hour of tutoring. A linear equation that expresses the total amount of money Svetlana earns for each session she tutors is y = 25 + 15x. What are the independent and dependent variables? What is the y-intercept, and what is the slope? Interpret them using complete sentences. Solution 12.4 The independent variable (x) is the number of hours Svetlana tutors each session. The dependent variable (y) is the amount, in dollars, Svetlana earns for each session. The y-intercept is 25 (a = 25). At the start of the tutoring session, Svetlana charges a one-time fee of $25 (this is when x = 0). The slope is 15 (b = 15). For each session, Svetlana earns $15 for each hour she tutors. 12.4 Ethan repairs household appliances such as dishwashers and refrigerators. For each visit, he charges $25 plus $20 per hour of work. A linear equation that expresses the total amount of money Ethan earns per visit is y = 25 + 20x. What are the independent and dependent variables? What is the y-intercept, and what is the slope? Interpret them using complete sentences. 12.2 \| The Regression Equation Data rarely fit a straight line exactly. Usually, you must be satisfied with rough predictions. Typically, you have a set of data with a scatter plot that appear to fit a straight line. This is called a line of best fit or least-squares regression line. If you know a person’s pinky (smallest) finger length, do you think you could predict that person’s height? Collect data from your class (pinky finger length, in inches). The independent variable, x, is pinky finger length and the dependent variable, y, is height. For each set of data, plot the points on graph paper. Make your graph big enough and use a ruler. Then, by eye, draw a line that appears to fit the data. For your line, pick two convenient points and use them to find the slope of the line. Find the y-intercept of the line by extending your line so it crosses the y-axis. Using the slopes and the y-intercepts, write your equation of best fit. Do you think everyone will have the same equation? Why or why not? According to your equation, what is the predicted height for a pinky length of 2.5 inches? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 699 Example 12.5 A random sample of 11 statistics students produced the data in Table 12.1, where x is the third exam score out of 80 and y is the final exam score out of 200. Can you predict the final exam score of a random student if you know the third exam score? x (third exam score) y (final exam score) 65 67 71 71 66 75 67 70 71 69 69 Table 12.1 175 133 185 163 126 198 153 163 159 151 159 Figure 12.5 Using the x- and y-coordinates in the table, we plot the points on a graph to create the scatter plot showing the scores on the final exam based on scores from the third exam. 12.5 SCUBA divers have maximum dive times they cannot exceed when going to different depths. The data in Table 12.2 show different depths in feet, with the maximum dive times in minutes. Use your calculator to find the least squares regression line and predict the maximum dive time for 110 feet. 700 Chapter 12 \| Linear Regression and Correlation x (depth) y (maximum dive time) 50 60 70 80 90 100 Table 12.2 80 55 45 35 25 22 The third exam score, x, is the independent variable, and the final exam score, y, is the dependent variable. We will plot a regression line that best fits the data. If each of you were to fit a line by eye, you would draw different lines. We can obtain a line of best fit using either the median-–median line approach or by calculating the least-squares regression line. Let'’s first find the line of best fit for the relationship between the third exam score and the final exam score using the median-median line approach. Remember that this is the data from Example 12.5 after the ordered pairs have been listed by ordering x values. If multiple data points have the same y values, then they are listed in order from least to greatest y (see data values where x = 71). We first divide our scores into three groups of approximately equal numbers of x values per group. The first and third groups have the same number of x values. We must remember first to put the x values in ascending order. The corresponding y values are then recorded. However, to find the median, we first must rearrange the y values in each group from the least value to the greatest value. Table 12.3 shows the correct ordering of the x values but does not show a reordering of the y values. x (third exam score) y (final exam score) 65 66 67 67 69 69 70 71 71 71 75 Table 12.3 175 126 133 153 151 159 163 159 163 185 198 With this set of data, the first and last groups each have four x values and four corresponding y values. The second group has three x values and three corresponding y values. We need to organize the x and y values per group and find the median x and y values for each group. Let’s now write out our y values for each group in ascending order. For group 1, the y values in order are 126, 133, 153, and 175. For group 2, the y values are already in order. For group 3, the y values are also already in order. We can represent these data as shown in Table 12.4, but notice that we have broken the ordered pairs; (65, 126) is not a data point in our original set: This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 701 Group x (third exam score) y (final exam score) Median x value Median y value 65 66 67 67 69 69 70 71 71 71 75 1 2 3 Table 12.4 126 133 153 175 151 159 163 159 163 185 198 66.5 143 69 71 159 174 When this is completed, we can write the ordered pairs for the median values. This allows us to find the slope and y-intercept of the –median-median line. The ordered pairs are (66.5, 143), (69, 159), and (71, 174). The slope can be calculated using the formula m − y2 − y1 x2 − x1 . Substituting the median x and y values from the first and third groups gives m = 174 − 143 71 − 66.5 , which simplifies to m ≈ 6.9. The y-intercept may be found using the formula b = Σy − mΣx 3 , which means the quantity of the sum of the median y values minus the slope times the sum of the median x values divided by three. The sum of the median x values is 206.5, and the sum of the median y values is 476. Substituting these sums and the slope into the formula gives b = 476 − 6.9(206.5) 3 , which simplifies to b ≈ − 316.3. The line of best fit is represented as y = mx + b. Thus, the equation can be written as y = 6.9x − 316.3. The median–median line may also be found using your graphing calculator. You can enter the x and y values into two separate lists; choose Stat, Calc, Med-Med, and press Enter. The slope, a, and y-intercept, b, will be provided. The calculator shows a slight deviation from the previous manual calculation as a result of rounding. Rounding to the nearest tenth, the calculator gives the –median-median line of y = 6.9x − 315.5. Each point of data is of the the form (x, y), and each point of the line of best fit using least-squares linear regression has the form (x, ŷ). The ŷ is read y hat and i
s the estimated value of y. It is the value of y obtained using the regression line. It is not generally equal to y from data, but it is still important because it can help make predictions for other values. Figure 12.6 702 Chapter 12 \| Linear Regression and Correlation The term y0 – ŷ0 = ε0 is called the error or residual. It is not an error in the sense of a mistake. The absolute value of a residual measures the vertical distance between the actual value of y and the estimated value of y. In other words, it measures the vertical distance between the actual data point and the predicted point on the line, or it measures how far the estimate is from the actual data value. If the observed data point lies above the line, the residual is positive and the line underestimates the actual data value for y. If the observed data point lies below the line, the residual is negative and the line overestimates that actual data value for y. In Figure 12.6, y0 – ŷ0 = ε0 is the residual for the point shown. Here the point lies above the line and the residual is positive. ε = the Greek letter epsilon For each data point, you can calculate the residuals or errors, yi – ŷi = εi for i = 1, 2, 3, . . . , 11. Each \|ε\| is a vertical distance. For the example about the third exam scores and the final exam scores for the 11 statistics students, there are 11 data points. Therefore, there are 11 ε values. If you square each ε and add them, you get the sum of ε squared from i = 1 to i = 11, as shown below. 11 (ε1)2 + (ε2)2 + ... + (ε11)2 = Σ i = 1 ε2 . This is called the sum of squared errors (SSE). Using calculus, you can determine the values of a and b that make the SSE a minimum. When you make the SSE a minimum, you have determined the points that are on the line of best fit. It turns out that the line of best fit has the equation ŷ = a + bx where a = y¯ − b x¯ and b = ∑ ⎝y − y¯ ⎞ ⎛ ⎝x − x¯ ⎞ ⎛ ⎠ ⎠ 2 ⎝x − x¯ ⎞ ⎛ ⎠ ∑ . The sample means of the x values and the y values are x¯ and y¯ , respectively. The best-fit line always passes through the point ( x¯ , y¯ ) . The slope (b) can be written as b = r ⎛ ⎝ s y s x ⎞ ⎠ where sy = the standard deviation of the y values and sx = the standard deviation of the x values. r is the correlation coefficient, which shows the relationship between the x and y values. This will be discussed in more detail in the next section. Least-Squares Criteria for Best Fit The process of fitting the best-fit line is called linear regression. We assume that the data are scattered about a straight line. To find that line, we minimize the sum of the squared errors (SSE), or make it as small as possible. Any other line you might choose would have a higher SSE than the best-fit line. This best-fit line is called the least-squares regression line. NOTE Computer spreadsheets, statistical software, and many calculators can quickly calculate the best-fit line and create the graphs. The calculations tend to be tedious if done by hand. Instructions to use the TI-83, TI-83+, and TI-84+ calculators to find the best-fit line and create a scatter plot are shown at the end of this section. Third Exam vs. Final Exam Example The graph of the line of best fit for the third exam/final exam example is as follows: This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 703 Figure 12.7 The least-squares regression line (best-fit line) for the third exam/final exam example has the equation Understanding and Interpreting the y-intercept ŷ = −173.51 + 4.83x. The y-intercept, a, of the line describes where the plot line crosses the y-axis. The y-intercept of the best-fit line tells us the best value of the relationship when x is zero. In some cases, it does not make sense to figure out what y is when x = 0. For example, in the third exam vs. final exam example, the y-intercept occurs when the third exam score, or x, is zero. Since all the scores are grouped around a passing grade, there is no need to figure out what the final exam score, or y, would be when the third exam was zero. However, the y-intercept is very useful in many cases. For many examples in science, the y-intercept gives the baseline reading when the experimental conditions aren’'t applied to an experimental system. This baseline indicates how much the experimental condition affects the system. It could also be used to ensure that equipment and measurements are calibrated properly before starting the experiment. In biology, the concentration of proteins in a sample can be measured using a chemical assay that changes color depending on how much protein is present. The more protein present, the darker the color. The amount of color can be measured by the absorbance reading. Table 12.5 shows the expected absorbance readings at different protein concentrations. This is called a standard curve for the assay. Concentration (mM) Absorbance (mAU) 125 250 500 750 1,000 1,500 2,000 Table 12.5 0.021 0.023 0.068 0.086 0.105 0.124 0.146 The scatter plot Figure 12.8 includes the line of best fit. 704 Chapter 12 \| Linear Regression and Correlation Figure 12.8 The y-intercept of this line occurs at 0.0226 mAU. This means the assay gives a reading of 0.0226 mAU when there is no protein present. That is, it is the baseline reading that can be attributed to something else, which, in this case, is some other non-protein chemicals that are absorbing light. We can tell that this line of best fit is reasonable because the y-intercept is small, close to zero. When there is no protein present in the sample, we expect the absorbance to be very small, or close to zero, as well. Understanding Slope The slope of the line, b, describes how changes in the variables are related. It is important to interpret the slope of the line in the context of the situation represented by the data. You should be able to write a sentence interpreting the slope in plain English. Interpretation of the Slope: The slope of the best-fit line tells us how the dependent variable (y) changes for every one unit increase in the independent (x) variable, on average. Third Exam vs. Final Exam Example Slope: The slope of the line is b = 4.83. Interpretation: For a 1-point increase in the score on the third exam, the final exam score increases by 4.83 points, on average. Using the Linear Regression T Test: LinRegTTest 1. In the STAT list editor, enter the x data in list L1 and the y data in list L2, paired so that the corresponding (x, y) values are next to each other in the lists. (If a particular pair of values is repeated, enter it as many times as it appears in the data.) 2. On the STAT TESTS menu, scroll down and select LinRegTTest. (Be careful to select LinRegTTest. Some calculators may also have a different item called LinRegTInt.) 3. On the LinRegTTest input screen, enter Xlist: L1, Ylist: L2, and Freq: 1. 4. On the next line, at the prompt β or ρ, highlight ≠ 0 and press ENTER. 5. Leave the line for RegEQ: blank. 6. Highlight Calculate and press ENTER. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 705 Figure 12.9 The output screen contains a lot of information. For now, let’s focus on a few items from the output and return to the other items later. The second line says y = a + bx. Scroll down to find the values a = –173.513 and b = 4.8273. The equation of the best-fit line is ŷ = –173.51 + 4.83x. The two items at the bottom are r2 = .43969 and r = .663. For now, just note where to find these values; we examine them in the next two sections. Graphing the Scatter Plot and Regression Line 1. We are assuming the x data are already entered in list L1 and the y data are in list L2. 2. Press 2nd STATPLOT ENTER to use Plot 1. 3. On the input screen for PLOT 1, highlight On, and press ENTER. 4. For TYPE, highlight the first icon, which is the scatter plot, and press ENTER. 5. Indicate Xlist: L1 and Ylist: L2. 6. For Mark, it does not matter which symbol you highlight. 7. Press the ZOOM key and then the number 9 (for menu item ZoomStat); the calculator fits the window to the data. 8. To graph the best-fit line, press the Y= key and type the equation –173.5 + 4.83X into equation Y1. (The X key is immediately left of the STAT key.) Press ZOOM 9 again to graph it. 9. Optional: If you want to change the viewing window, press the WINDOW key. Enter your desired window using Xmin, Xmax, Ymin, and Ymax. NOTE Another way to graph the line after you create a scatter plot is to use LinRegTTest. 1. Make sure you have done the scatter plot. Check it on your screen. 2. Go to LinRegTTest and enter the lists. 3. At RegEq, press VARS and arrow over to Y-VARS. Press 1 for 1:Function. Press 1 for 1:Y1. Then, arrow down to Calculate and do the calculation for the line of best fit. 4. Press Y= (you will see the regression equation). 5. Press GRAPH, and the line will be drawn. 706 Chapter 12 \| Linear Regression and Correlation The Correlation Coefficient r Besides looking at the scatter plot and seeing that a line seems reasonable, how can you determine whether the line is a good predictor? Use the correlation coefficient as another indicator (besides the scatter plot) of the strength of the relationship between x and y. The correlation coefficient, r, developed by Karl Pearson during the early 1900s, is numeric and provides a measure of the strength and direction of the linear association between the independent variable x and the dependent variable y. If you suspect a linear relationship between x and y, then r can measure the strength of the linear relationship. What the Value of r Tells Us • The value of r is always between –1 and +1. In other words, –1 ≤ r ≤ 1. • The size of the correlation r indicates the strength of the linear relationship between x and y. Values of r close to –1 or to +1 indicate a stronger linear relationship between x and y. • • If r = 0, there is absolutely no linear relationship be
tween x and y (no linear correlation). If r = 1, there is perfect positive correlation. If r = –1, there is perfect negative correlation. In both these cases, all the original data points lie on a straight line. Of course, in the real world, this does not generally happen. What the Sign of r Tells Us • A positive value of r means that when x increases, y tends to increase and when x decreases, y tends to decrease (positive correlation). • A negative value of r means that when x increases, y tends to decrease and when x decreases, y tends to increase (negative correlation). • The sign of r is the same as the sign of the slope, b, of the best-fit line. NOTE A strong correlation does not suggest that x causes y or y causes x. We say correlation does not imply causation. The correlation coefficient is calculated as the quantity of data points times the sum of the quantity of the x-coordinates times the y-coordinates, minus the quantity of the sum of the x-coordinates times the sum of the y-coordinates, all divided by the square root of the quantity of data points times the sum of the x-coordinates squared minus the square of the sum of the x-coordinates, times the number of data points times the sum of the y-coordinates squared minus the square of the sum of the y-coordinates. It can be summarized by the following equation: r = nΣ(xy) − (Σx)(Σy) ⎣nΣy2 − (Σy)2⎤ ⎡ ⎣nΣx2 − (Σx)2⎤ ⎡ ⎦ ⎦ where n is the number of data points. Figure 12.10 (a) A scatter plot showing data with a positive correlation: 0 < r < 1. (b) A scatter plot showing data with a negative correlation: –1 < r < 0. (c) A scatter plot showing data with zero correlation: r = 0. The formula for r looks formidable. However, computer spreadsheets, statistical software, and many calculators can calculate r quickly. The correlation coefficient, r, is the bottom item in the output screens for the LinRegTTest on the TI-83, TI-83+, or TI-84+ calculator (see previous section for instructions). The Coefficient of Determination The variable r2 is called the coefficient of determination and it is the square of the correlation coefficient, but it is usually stated as a percentage, rather than in decimal form. It has an interpretation in the context of the data: This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 707 • r 2 , when expressed as a percent, represents the percentage of variation in the dependent (predicted) variable y that can be explained by variation in the independent (explanatory) variable x using the regression (best-fit) line. • 1 – r 2 , when expressed as a percentage, represents the percentage of variation in y that is not explained by variation in x using the regression line. This can be seen as the scattering of the observed data points about the regression line. Consider the third exam/final exam example introduced in the previous section. • The line of best fit is: ŷ = –173.51 + 4.83x. • The correlation coefficient is r = .6631. • The coefficient of determination is r2 = .66312 = .4397. Interpret r2 in the context of this example. • Approximately 44 percent of the variation (0.4397 is approximately 0.44) in the final exam grades can be explained by the variation in the grades on the third exam, using the best-fit regression line. • Therefore, the rest of the variation (1 – 0.44 = 0.56 or 56 percent) in the final exam grades cannot be explained by the variation of the grades on the third exam with the best-fit regression line. These are the variation of the points that are not as close to the regression line as others. 12.3 \| Testing the Significance of the Correlation Coefficient (Optional) The correlation coefficient, r, tells us about the strength and direction of the linear relationship between x and y. However, the reliability of the linear model also depends on how many observed data points are in the sample. We need to look at both the correlation coefficient r and the sample size n, together. We perform a hypothesis test of the significance of the correlation coefficient to decide whether the linear relationship in the sample data is strong enough to use to model the relationship in the population. The sample data are used to compute r, the correlation coefficient for the sample. If we had data for the entire population, we could find the population correlation coefficient. But, because we have only sample data, we cannot calculate the population correlation coefficient. The sample correlation coefficient, r, is our estimate of the unknown population correlation coefficient. The symbol for the population correlation coefficient is ρ, the Greek letter rho. ρ = population correlation coefficient (unknown). r = sample correlation coefficient (known; calculated from sample data). The hypothesis test lets us decide whether the value of the population correlation coefficient ρ is close to zero or significantly different from zero. We decide this based on the sample correlation coefficient r and the sample size n. If the test concludes the correlation coefficient is significantly different from zero, we say the correlation coefficient is significant. • Conclusion: There is sufficient evidence to conclude there is a significant linear relationship between x and y because the correlation coefficient is significantly different from zero. • What the conclusion means: There is a significant linear relationship between x and y. We can use the regression line to model the linear relationship between x and y in the population. If the test concludes the correlation coefficient is not significantly different from zero (it is close to zero), we say the correlation coefficient is not significant. • Conclusion: There is insufficient evidence to conclude there is a significant linear relationship between x and y because the correlation coefficient is not significantly different from zero. • What the conclusion means: There is not a significant linear relationship between x and y. Therefore, we cannot use the regression line to model a linear relationship between x and y in the population. NOTE • If r is significant and the scatter plot shows a linear trend, the line can be used to predict the value of y for values of x that are within the domain of observed x values. 708 • • Chapter 12 \| Linear Regression and Correlation If r is not significant or if the scatter plot does not show a linear trend, the line should not be used for prediction. If r is significant and the scatter plot shows a linear trend, the line may not be appropriate or reliable for prediction outside the domain of observed x values in the data. Performing the Hypothesis Test • Null hypothesis: H0: ρ = 0. • Alternate hypothesis: Ha: ρ ≠ 0. What the Hypothesis Means in Words: • Null hypothesis H0: The population correlation coefficient is not significantly different from zero. There is not a significant linear relationship (correlation) between x and y in the population. • Alternate hypothesis Ha: The population correlation coefficient is significantly different from zero. There is a significant linear relationship (correlation) between x and y in the population. Drawing a Conclusion: There are two methods to make a conclusion. The two methods are equivalent and give the same result. • Method 1: Use the p-value. • Method 2: Use a table of critical values. In this chapter, we will always use a significance level of 5 percent, α = 0.05. NOTE Using the p-value method, you could choose any appropriate significance level you want; you are not limited to using α = 0.05. But, the table of critical values provided in this textbook assumes we are using a significance level of 5 percent, α = 0.05. If we wanted to use a significance level different from 5 percent with the critical value method, we would need different tables of critical values that are not provided in this textbook. METHOD 1: Using a p-value to Make a Decision To calculate the p-value using LinRegTTEST: 1. Complete the same steps as the LinRegTTest performed previously in this chapter, making sure on the line prompt for β or σ, ≠ 0 is highlighted. 2. When looking at the output screen, the p-value is on the line that reads p =. If the p-value is less than the significance level (α = 0.05): • Decision: Reject the null hypothesis. • Conclusion: There is sufficient evidence to conclude there is a significant linear relationship between x and y because the correlation coefficient is significantly different from zero. If the p-value is not less than the significance level (α = 0.05): • Decision: Do not reject the null hypothesis. • Conclusion: There is insufficient evidence to conclude there is a significant linear relationship between x and y because the correlation coefficient is not significantly different from zero. You will use technology to calculate the p-value, but it is useful to know that the p-value is calculated using a t distribution with n – 2 degrees of freedom and that the p-value is the combined area in both tails. An alternative way to calculate the p-value (p) given by LinRegTTest is the command 2*tcdf(abs(t),10^99, n–2) in 2nd DISTR. Third Exam vs. Final Exam Example: p-value Method • Consider the third exam/final exam example. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 709 • The line of best fit is ŷ = –173.51 + 4.83x, with r = 0.6631, and there are n = 11 data points. • Can the regression line be used for prediction? Given a third exam score (x value), can we use the line to predict the final exam score (predicted y value)? H0: ρ = 0 Ha: ρ ≠ 0 α = 0.05 • The p-value is 0.026 (from LinRegTTest on a calculator or from computer software). • The p-value, 0.026, is less than the significance level of α = 0.05. • Decision: Reject the null hypothesis H0. • Conclusion: There is sufficient evidence to conclude there is a significant linear relationship between the third exam
score (x) and the final exam score (y) because the correlation coefficient is significantly different from zero. Because r is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores. METHOD 2: Using a Table of Critical Values to Make a Decision The 95 Percent Critical Values of the Sample Correlation Coefficient Table (Table 12.9) can be used to give you a good idea of whether the computed value of r is significant. Use it to find the critical values using the degrees of freedom, df = n – 2. The table has already been calculated with α = 0.05. The table tells you the positive critical value, but you should also make that number negative to have two critical values. If r is not between the positive and negative critical values, then the correlation coefficient is significant. If r is significant, then you may use the line for prediction. If r is not significant (between the critical values), you should not use the line to make predictions. Example 12.6 Suppose you computed r = 0.801 using n = 10 data points. The degrees of freedom would be 8 (df = n – 2 = 10 – 2 = 8). Using Table 12.9 with df = 8, we find that the critical value is 0.632. This means the critical values are really ±0.632. Since r = 0.801 and 0.801 > 0.632, r is significant and the line may be used for prediction. If you view this example on a number line, it will help you to see that r is not between the two critical values. Figure 12.11 r is not between –0.632 and 0.632, so r is significant. 12.6 For a given line of best fit, you computed that r = 0.6501 using n = 12 data points, and the critical value found on the table is 0.576. Can the line be used for prediction? Why or why not? Example 12.7 Suppose you computed r = –0.624 with 14 data points, where df = 14 – 2 = 12. The critical values are –0.532 and 0.532. Since –0.624 < –0.532, r is significant and the line can be used for prediction. 710 Chapter 12 \| Linear Regression and Correlation Figure 12.12 r = –0.624 and –0.624 < –0.532. Therefore, r is significant. 12.7 For a given line of best fit, you compute that r = 0.5204 using n = 9 data points, and the critical values are ±0.666. Can the line be used for prediction? Why or why not? Example 12.8 Suppose you computed r = 0.776 and n = 6, with df = 6 -– 2 = 4. The critical values are – 0.811 and 0.811. Since 0.776 is between the two critical values, r is not significant. The line should not be used for prediction. Figure 12.13 –0.811 < r = 0.776 < 0.811. Therefore, r is not significant. 12.8 For a given line of best fit, you compute that r = –0.7204 using n = 8 data points, and the critical value is 0.707. Can the line be used for prediction? Why or why not? Third Exam vs. Final Exam Example: Critical Value Method Consider the third exam/final exam example. The line of best fit is: ŷ = –173.51 + 4.83x, with r = .6631, and there are n = 11 data points. Can the regression line be used for prediction? Given a third exam score (x value), can we use the line to predict the final exam score (predicted y value)? H0: ρ = 0 Ha: ρ ≠ 0 α = 0.05 • Use the 95 Percent Critical Values table for r with df = n – 2 = 11 – 2 = 9. • Using the table with df = 9, we find that the critical value listed is 0.602. Therefore, the critical values are ±0.602. • Since 0.6631 > 0.602, r is significant. • Decision: Reject the null hypothesis. • Conclusion: There is sufficient evidence to conclude there is a significant linear relationship between the third exam score (x) and the final exam score (y) because the correlation coefficient is significantly different from zero. Because r is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 711 Example 12.9 Suppose you computed the following correlation coefficients. Using the table at the end of the chapter, determine whether r is significant and whether the line of best fit associated with each correlation coefficient can be used to predict a y value. If it helps, draw a number line. a. r = –0.567 and the sample size, n, is 19. To solve this problem, first find the degrees of freedom. df = n - 2 = 17. Then, using the table, the critical values are ±0.456. –0.567 < –0.456, or you may say that –0.567 is not between the two critical values. r is significant and may be used for predictions. b. r = 0.708 and the sample size, n, is 9. df = n - 2 = 7 The critical values are ±0.666. 0.708 > 0.666. r is significant and may be used for predictions. c. r = 0.134 and the sample size, n, is 14. df = 14 –- 2 = 12. The critical values are ±0.532. 0.134 is between –0.532 and 0.532. r is not significant and may not be used for predictions. d. r = 0 and the sample size, n, is 5. It doesn’'t matter what the degrees of freedom are because r = 0 will always be between the two critical values, so r is not significant and may not be used for predictions. 12.9 For a given line of best fit, you compute that r = 0 using n = 100 data points. Can the line be used for prediction? Why or why not? Assumptions in Testing the Significance of the Correlation Coefficient Testing the significance of the correlation coefficient requires that certain assumptions about the data be satisfied. The premise of this test is that the data are a sample of observed points taken from a larger population. We have not examined the entire population because it is not possible or feasible to do so. We are examining the sample to draw a conclusion about whether the linear relationship that we see between x and y in the sample data provides strong enough evidence that we can conclude there is a linear relationship between x and y in the population. The regression line equation that we calculate from the sample data gives the best-fit line for our particular sample. We want to use this best-fit line for the sample as an estimate of the best-fit line for the population. Examining the scatter plot and testing the significance of the correlation coefficient helps us determine whether it is appropriate to do this. The assumptions underlying the test of significance are as follows: • There is a linear relationship in the population that models the sample data. Our regression line from the sample is our best estimate of this line in the population. • The y values for any particular x value are normally distributed about the line. This implies there are more y values scattered closer to the line than are scattered farther away. Assumption 1 implies that these normal distributions are centered on the line; the means of these normal distributions of y values lie on the line. • Normal distributions of all the y values have the same shape and spread about the line. • The residual errors are mutually independent (no pattern). 712 Chapter 12 \| Linear Regression and Correlation • The data are produced from a well-designed, random sample or randomized experiment. Figure 12.14 The y values for each x value are normally distributed about the line with the same standard deviation. For each x value, the mean of the y values lies on the regression line. More y values lie near the line than are scattered farther away from the line. 12.4 \| Prediction (Optional) Recall the third exam/final exam example. We found the equation of the best-fit line for the final exam grade as a function of the grade on the third exam. We can now use the least-squares regression line for prediction. Suppose you want to estimate, or predict, the mean final exam score of statistics students who received a 73 on the third exam. The exam scores (x values) range from 65 to 75. Since 73 is between the x values 65 and 75, substitute x = 73 into the equation. Then, We predict that statistics students who earn a grade of 73 on the third exam will earn a grade of 179.08 on the final exam, on average. ^ y = − 173.51 + 4.83(73) = 179.08. Example 12.10 Recall the third exam/final exam example. a. What would you predict the final exam score to be for a student who scored a 66 on the third exam? Solution 12.10 a. 145.27 b. What would you predict the final exam score to be for a student who scored a 90 on the third exam? Solution 12.10 b. The x values in the data are between 65 and 75. 90 is outside the domain of the observed x values in the data (independent variable), so you cannot reliably predict the final exam score for this student. Even though it is possible to enter 90 into the equation for x and calculate a corresponding y value, the y value that you get will not be reliable. To understand how unreliable the prediction can be outside the x values observed in the data, make the substitution x = 90 into the equation: ŷ = –173.51 + 4.83⎛ ⎝90⎞ ⎠ = 261.19. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 713 The final exam score is predicted to be 261.19. The most points that can be awarded for the final exam are 200. 12.10 Data are collected on the relationship between the number of hours per week practicing a musical instrument and scores on a math test. The line of best fit is as follows: ŷ = 72.5 + 2.8x. What would you predict the score on a math test will be for a student who practices a musical instrument for five hours a week? 12.5 \| Outliers In some data sets, there are values (observed data points) called outliers. Outliers are observed data points that are far from the least-squares line. They have large errors, where the error or residual is not very close to the best-fit line. Outliers need to be examined closely. Sometimes, they should not be included in the analysis of the data, like if it is possible that an outlier is a result of incorrect data. Other times, an outlier may hold valuable information about the population under study and should remain included in the data. The key is to examine
carefully what causes a data point to be an outlier. Besides outliers, a sample may contain one or a few points that are called influential points. Influential points are observed data points that are far from the other observed data points in the horizontal direction. These points may have a big effect on the slope of the regression line. To begin to identify an influential point, you can remove it from the data set and determine whether the slope of the regression line is changed significantly. You also want to examine how the correlation coefficient, r, has changed. Sometimes, it is difficult to discern a significant change in slope, so you need to look at how the strength of the linear relationship has changed. Computers and many calculators can be used to identify outliers and influential points. Regression analysis can determine if an outlier is, indeed, an influential point. The new regression will show how omitting the outlier will affect the correlation among the variables, as well as the fit of the line. A graph showing both regression lines helps determine how removing an outlier affects the fit of the model. Identifying Outliers We could guess at outliers by looking at a graph of the scatter plot and best-fit line. However, we would like some guideline regarding how far away a point needs to be to be considered an outlier. As a rough rule of thumb, we can flag as an outlier any point that is located farther than two standard deviations above or below the best-fit line. The standard deviation used is the standard deviation of the residuals or errors. We can do this visually in the scatter plot by drawing an extra pair of lines that are two standard deviations above and below the best-fit line. Any data points outside this extra pair of lines are flagged as potential outliers. Or, we can do this numerically by calculating each residual and comparing it with twice the standard deviation. With regard to the TI-83, 83+, or 84+ calculators, the graphical approach is easier. The graphical procedure is shown first, followed by the numerical calculations. You would generally need to use only one of these methods. Example 12.11 In the third exam/final exam example, you can determine whether there is an outlier. If there is an outlier, as an exercise, delete it and fit the remaining data to a new line. For this example, the new line ought to fit the remaining data better. This means the SSE (sum of the squared errors) should be smaller and the correlation coefficient ought to be closer to 1 or –1. Solution 12.11 714 Chapter 12 \| Linear Regression and Correlation Graphical Identification of Outliers With the TI-83, 83+, or 84+ graphing calculators, it is easy to identify the outliers graphically and visually. If we were to measure the vertical distance from any data point to the corresponding point on the line of best fit and that distance were equal to 2s or more, then we would consider the data point to be too far from the line of best fit. We need to find and graph the lines that are two standard deviations below and above the regression line. Any points that are outside these two lines are outliers. Let’s call these lines Y2 and Y3. As we did with the equation of the regression line and the correlation coefficient, we will use technology to calculate this standard deviation for us. Using the LinRegTTest with these data, scroll down through the output screens to find s = 16.412. Line Y2 = –173.5 + 4.83x – 2(16.4), and line Y3 = –173.5 + 4.83x + 2(16.4), where ŷ = –173.5 + 4.83x is the line of best fit. Y2 and Y3 have the same slope as the line of best fit. Graph the scatter plot with the best-fit line in equation Y1, then enter the two extra lines as Y2 and Y3 in the Y= equation editor. Press ZOOM-9 to get a good view. You will see that the only point that is not between Y2 and Y3 is the point (65, 175). On the calculator screen, it is barely outside these lines, but it is considered an outlier because it is more than two standard deviations away from the best-fit line. The outlier is the student who had a grade of 65 on the third exam and 175 on the final exam. Sometimes a point is so close to the lines used to flag outliers on the graph that it is difficult to tell whether the point is between or outside the lines. On a computer, enlarging the graph may help; on a small calculator screen, zooming in may make the graph clearer. Note that when the graph does not give a clear enough picture, you can use the numerical comparisons to identify outliers. Figure 12.15 12.11 Identify the potential outlier in the scatter plot. The standard deviation of the residuals, or errors, is approximately 8.6. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 715 Figure 12.16 Numerical Identification of Outliers In Table 12.6, the first two columns include the third exam and final exam data. The third column shows the predicted ŷ values calculated from the line of best fit: ŷ = –173.5 + 4.83x. The residuals, or errors, that were mentioned in Section 3 of this chapter have been calculated in the fourth column of the table: Observed y value – predicted y value = y – ŷ. s is the standard deviation of all the y – ŷ = ε values, where n is the total number of data points. If each residual is calculated and squared, and the results are added, we get the SSE. The standard deviation of the residuals is calculated from the SSE as NOTE We divide by (n – 2) because the regression model involves two estimates. s = SSE n − 2 . Rather than calculate the value of s ourselves, we can find s using a computer or calculator. For this example, the calculator function LinRegTTest found s = 16.4 as the standard deviation of the residuals 35; –17; 16; –6; –19; 9; 3; –1; –10; –9; –1. x y ŷ y – ŷ 65 175 140 175 – 140 = 35 67 133 150 133 – 150= –17 71 185 169 185 – 169 = 16 71 163 169 163 – 169 = –6 66 126 145 126 – 145 = –19 75 198 189 198 – 189 = 9 67 153 150 153 – 150 = 3 70 163 164 163 – 164 = –1 Table 12.6 716 Chapter 12 \| Linear Regression and Correlation x y ŷ y – ŷ 71 159 169 159 – 169 = –10 69 151 160 151 – 160 = –9 69 159 160 159 – 160 = –1 Table 12.6 We are looking for all data points for which the residual is greater than 2s = 2(16.4) = 32.8 or less than –32.8. Compare these values with the residuals in column four of the table. The only such data point is the student who had a grade of 65 on the third exam and 175 on the final exam; the residual for this student is 35. How Does the Outlier Affect the Best-Fit Line? Numerically and graphically, we have identified point (65, 175) as an outlier. Recall that recalculation of the least-squares regression line and summary statistics, following deletion of an outlier, may be used to determine whether an outlier is also an influential point. This process also allows you to compare the strength of the correlation of the variables and possible changes in the slope both before and after the omission of any outliers. Compute a new best-fit line and correlation coefficient using the 10 remaining points. On the TI-83, TI-83+, or TI-84+ calculators, delete the outlier from L1 and L2. Using the LinRegTTest, found under Stat and Tests, the new line of best fit and correlation coefficient are the following: ŷ = − 355.19 + 7.39x and r = 0.9121 . The slope is now 7.39, compared to the previous slope of 4.83. This seems significant, but we need to look at the change in r-values as well. The new line shows r = 0.9121 , which indicates a stronger correlation than the original line, with r = 0.6631, since r = 0.9121 is closer to 1. This means the new line is a better fit to the data values. The line can better predict the final exam score given the third exam score. It also means the outlier of (65, 175) was an influential point, since there is a sizeable difference in r-values. We must now decide whether to delete the outlier. If the outlier was recorded erroneously, it should certainly be deleted. Because it produces such a profound effect on the correlation, the new line of best fit allows for better prediction and an overall stronger model. You may use Excel to graph the two least-squares regression lines and compare the slopes and fit of the lines to the data, as shown in Figure 12.17. Figure 12.17 You can see that the second graph shows less deviation from the line of best fit. It is clear that omission of the influential point produced a line of best fit that more closely models the data. Numerical Identification of Outliers: Calculating s and Finding Outliers Manually If you do not have the function LinRegTTest on your calculator, then you must calculate the outlier in the first example by doing the following. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 717 First, square each \|y – ŷ\|. The squares are 352; 172; 162; 62; 192; 92; 32; 12; 102; 92; 12. Then, add (sum) all the \|y – ŷ\| squared terms using the formula 11⎛ ⎝\|yi − ŷi\|⎞ Σ Σ i = 1 i = 1 = 352 + 172 + 162 + 62 + 192 + 92 + 32 + 12 + 102 + 92 + 12 11εi 2 (Recall that yi – ŷi = εi). 2 = ⎠ = 2,440 = SSE. The result, SSE, is the sum of squared errors. Next, calculate s, the standard deviation of all the y – ŷ = ε-values where n = the total number of data points. The calculation is s = SSE n – 2 . For the third exam/final exam example, s = 2440 11 – 2 = 16.47. Next, multiply s by 2: (2)(16.47) = 32.94 32.94 is two standard deviations away from the mean of the y – ŷ values. If we were to measure the vertical distance from any data point to the corresponding point on the line of best fit and that distance is at least 2s, then we would consider the data point to be too far from the line of best fit. We call that point a potential outlier. For the example, if any of the \|y – ŷ\| values are at least 32.94, the corresponding (x, y) data point is a potential outlier. For the third exam/final exam example,
all the \|y – ŷ\| values are less than 31.29 except for the first one, which is 35: 35 > 31.29. That is, \|y – ŷ\| ≥ (2)(s). The point that corresponds to \|y – ŷ\| = 35 is (65, 175). Therefore, the data point (65, 175) is a potential outlier. For this example, we will delete it. (Remember, we do not always delete an outlier.) NOTE When outliers are deleted, the researcher should either record that data were deleted, and why, or the researcher should provide results both with and without the deleted data. If data are erroneous and the correct values are known (e.g., student 1 actually scored a 70 instead of a 65), then this correction can be made to the data. The next step is to compute a new best-fit line using the 10 remaining points. The new line of best fit and the correlation coefficient are ŷ = –355.19 + 7.39x and r = .9121. Example 12.12 Using this new line of best fit (based on the remaining 10 data points in the third exam/final exam example), what would a student who receives a 73 on the third exam expect to receive on the final exam? Is this the same as the prediction made using the original line? Solution 12.12 Using the new line of best fit, ŷ = –355.19 + 7.39(73) = 184.28. A student who scored 73 points on the third exam would expect to earn 184 points on the final exam. The original line predicted that ŷ = –173.51 + 4.83(73) = 179.08, so the prediction using the new line with the 718 Chapter 12 \| Linear Regression and Correlation outlier eliminated differs from the original prediction. 12.12 The data points for the graph from the third exam/final exam example are as follows: (1, 5), (2, 7), (2, 6), (3, 9), (4, 12), (4, 13), (5, 18), (6, 19), (7, 12), and (7, 21). Remove the outlier and recalculate the line of best fit. Find the value of ŷ when x = 10. Example 12.13 The consumer price index (CPI) measures the average change over time in prices paid by urban consumers for consumer goods and services. The CPI affects nearly all Americans because of the many ways it is used. One of its biggest uses is as a measure of inflation. By providing information about price changes in the nation’s economy to government, businesses, and labor forces, the CPI helps them make economic decisions. The president, U.S. Congress, and the Federal Reserve Board use CPI trends to form monetary and fiscal policies. In the following table, x is the year and y is the CPI. x y x y 1915 10.1 1969 36.7 1926 17.7 1975 49.3 1935 13.7 1979 72.6 1940 14.7 1980 82.4 1947 24.1 1986 109.6 1952 26.5 1991 130.7 1964 31.0 1999 166.6 Table 12.7 a. Draw a scatter plot of the data. b. Calculate the least-squares line. Write the equation in the form ŷ = a + bx. c. Draw the line on a scatter plot. d. Find the correlation coefficient. Is it significant? e. What is the average CPI for the year 1990? Solution 12.13 a. See Figure 12.17. b. Using our calculator, ŷ = –3204 + 1.662x is the equation of the line of best fit. c. See Figure 12.17. d. r = 0.8694. The number of data points is n = 14. Use the 95 Percent Critical Values of the Sample Correlation Coefficient table at the end of Chapter 12: In this case, df = 12. The corresponding critical values from the table are ±0.532. Since 0.8694 > 0.532, r is significant. We can use the predicted regression line we found above to make the prediction for x = 1990. e. ŷ = –3204 + 1.662(1990) = 103.4 CPI. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 719 Figure 12.18 NOTE In the example, notice the pattern of the points compared with the line. Although the correlation coefficient is significant, the pattern in the scatter plot indicates that a curve would be a more appropriate model to use than a line. In this example, a statistician would prefer to use other methods to fit a curve to these data, rather than model the data with the line we found. In addition to doing the calculations, it is always important to look at the scatter plot when deciding whether a linear model is appropriate. If you are interested in seeing more years of data, visit the Bureau of Labor Statistics CPI website (ftp://ftp.bls.gov/pub/special.requests/cpi/cpiai.txt). Our data are taken from the column Annual Avg. (third column from the right). For example, you could add more current years of data. Try adding the more recent years: 2004, CPI = 188.9; 2008, CPI = 215.3; and 2011, CPI = 224.9. See how this affects the model. (Check: ŷ = –4436 + 2.295x; r = 0.9018. Is r significant? Is the fit better with the addition of the new points?) 12.13 The following table shows economic development measured in per capita income (PCINC). Year PCINC Year PCINC 1870 1880 1890 1900 1910 340 499 592 757 927 Table 12.8 1920 1,050 1930 1,170 1940 1,364 1950 1,836 1960 2,132 a. What are the independent and dependent variables? b. Draw a scatter plot. c. Use regression to find the line of best fit and the correlation coefficient. d. e. Interpret the significance of the correlation coefficient. Is there a linear relationship between the variables? 720 Chapter 12 \| Linear Regression and Correlation f. Find the coefficient of determination and interpret it. g. What is the slope of the regression equation? What does it mean? h. Use the line of best fit to estimate PCINC for 1900 and for 2000. i. Determine whether there are any outliers. 95 Percent Critical Values of the Sample Correlation Coefficient Table Degrees of Freedom: n – 2 Critical Values: + and – 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 0.997 0.950 0.878 0.811 0.754 0.707 0.666 0.632 0.602 0.576 0.555 0.532 0.514 0.497 0.482 0.468 0.456 0.444 0.433 0.423 0.413 0.404 0.396 0.388 0.381 0.374 0.367 0.361 0.355 Table 12.9 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 721 Degrees of Freedom: n – 2 Critical Values: + and – 30 40 50 60 70 80 90 100 Table 12.9 0.349 0.304 0.273 0.250 0.232 0.217 0.205 0.195 12.6 \| Regression (Distance from School) (Optional) 722 Chapter 12 \| Linear Regression and Correlation 12.1 Regression (Distance From School) Student Learning Outcomes • The student will calculate and construct the line of best fit between two variables. • The student will evaluate the relationship between two variables to determine whether that relationship is significant. Collect the Data Use eight members of your class for the sample. Collect bivariate data (distance an individual lives from school, the cost of supplies for the current term). 1. Complete the table. Distance from School Cost of Supplies This Term Table 12.10 2. Which variable should be the dependent variable and which should be the independent variable? Why? 3. Graph distance vs. cost. Plot the points on the graph. Label both axes with words. Scale both axes. Figure 12.19 Analyze the Data This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 723 Enter your data into a calculator or computer. Write the linear equation, rounding to four decimal places. 1. Calculate the following: a. a = ______ b. b = ______ c. correlation = ______ d. n = ______ e. equation: ŷ = ______ f. Is the correlation significant? Why or why not? (Answer in one to three complete sentences.) 2. Supply an answer for the following scenarios: a. For a person who lives eight miles from campus, predict the total cost of supplies this term. b. For a person who lives 80 miles from campus, predict the total cost of supplies this term. 3. Obtain the graph on a calculator or computer. Sketch the regression line. Figure 12.20 Discussion Questions 1. Answer each question in complete sentences. a. Does the line seem to fit the data? Why? b. What does the correlation imply about the relationship between distance and cost? 2. Are there any outliers? If so, which point is an outlier? 3. Should the outlier, if it exists, be removed? Why or why not? 12.7 \| Regression (Textbook Cost) (Optional) 724 Chapter 12 \| Linear Regression and Correlation 12.2 Regression (Textbook Cost) Student Learning Outcomes • The student will calculate and construct the line of best fit between two variables. • The student will evaluate the relationship between two variables to determine whether that relationship is significant. Collect the Data Survey 10 textbooks. Collect bivariate data (number of pages in a textbook, the cost of the textbook). 1. Complete the table. Number of Pages Cost of Textbook Table 12.11 2. Which variable should be the dependent variable and which should be the independent variable? Why? 3. Graph pages vs. cost. Plot the points on the graph in Analyze the Data. Label both axes with words. Scale both axes. Analyze the Data Enter your data into a calculator or computer. Write the linear equation, rounding to four decimal places. 1. Calculate the following: a. a = ______ b. b = ______ c. correlation = ______ d. n = ______ e. equation: y = ______ f. Is the correlation significant? Why or why not? (Answer in complete sentences.) 2. Supply an answer for the following scenarios: a. For a textbook with 400 pages, predict the cost. b. For a textbook with 600 pages, predict the cost. 3. Obtain the graph on a calculator or computer. Sketch the regression line. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 725 Figure 12.21 Discussion Questions 1. Answer each question in complete sentences. a. Does the line seem to fit the data? Why? b. What does the correlation imply about the relationship between the number of pages and the cost? 2. Are there any outliers? If so, which point is an outlier? 3. Should the outlier, if it exists, be removed? Why or why not? 12.8 \| Regression (Fuel Efficiency) (Optional) 726 Chapter 12 \| Linear Regression and Correlation 12.3 Regression (Fuel Efficiency) Student Learning Outcomes • The student will calculate and constr
uct the line of best fit between two variables. • The student will evaluate the relationship between two variables to determine whether that relationship is significant. Collect the Data Find a reputable source that provides information on total fuel efficiency (in miles per gallon) and weight (in pounds) of new cars with an automatic transmission. You will use these data to determine the relationship, if any, between the fuel efficiency of a car and its weight. 1. Using your random-number generator, select 20 cars randomly from the list and record their weight and fuel efficiency into Table 12.12. Weight Fuel Efficiency Table 12.12 2. Which variable is the dependent variable and which is the independent variable? Why? 3. By hand, draw a scatter plot of weight vs. fuel efficiency. Plot the points on graph paper. Label both axes with words. Scale both axes accurately. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 727 Figure 12.22 Analyze the Data Enter your data into a calculator or computer. Write the linear equation, rounding to four decimal places. 1. Calculate the following: a. a = ______ b. b = ______ c. correlation = ______ d. n = ______ e. equation: ŷ = ______ 2. Obtain a graph of the regression line on a calculator. Sketch the regression line on the same axes as your scatter plot. Discussion Questions 1. Is the correlation significant? Explain how you determined this in complete sentences. 2. 3. Is the relationship a positive one or a negative one? Explain how you can tell and what this means in terms of weight and fuel efficiency. In one or two complete sentences, what is the practical interpretation of the slope of the least-squares line in terms of fuel efficiency and weight? 4. For a car that weighs 4,000 pounds, predict its fuel efficiency. Include units. 5. Can we predict the fuel efficiency of a car that weighs 10,000 pounds using the least-squares line? Explain why or why not. 6. Answer each question in complete sentences. a. Does the line seem to fit the data? Why or why not? b. What does the correlation imply about the relationship between fuel efficiency and weight of a car? Is this what you expected? 7. Are there any outliers? If so, which point is an outlier? 728 Chapter 12 \| Linear Regression and Correlation KEY TERMS coefficient of correlation a measure developed by Karl Pearson during the early 1900s that gives the strength of association between the independent variable and the dependent variable; r = n∑ xy − [∑ x][∑ y] 2 (n∑ x2 − [∑ x] )(n∑ y2 − [∑ y] 2 ) where n is the number of data points The coefficient cannot be more than 1 and less than –1. The closer the coefficient is to ±1, the stronger the evidence of a significant linear relationship between x and y. outlier an observation that does not fit the rest of the data CHAPTER REVIEW 12.1 Linear Equations The most basic type of association is a linear association. This type of relationship can be defined algebraically by the equations used (numerically with actual or predicted data values) or graphically from a plotted curve. Lines are classified as straight curves. Algebraically, a linear equation typically takes the form y = mx + b, where m and b are constants, x is the independent variable, and y is the dependent variable. In a statistical context, a linear equation is written in the form y = a + bx, where a and b are the constants. This form is used to help you distinguish the statistical context from the algebraic context. In the equation y = a + bx, the constant b that multiplies the x variable (b is called a coefficient) is called the slope. The slope describes the rate of change between the independent and dependent variables; in other words, the rate of change describes the change that occurs in the dependent variable as the independent variable is changed. In the equation y = a + bx, the constant a is called the y-intercept. Graphically, the y-intercept is the y-coordinate of the point where the graph of the line crosses the y-axis. At this point, x = 0. The slope of a line is a value that describes the rate of change between the independent and dependent variables. The slope tells us how the dependent variable (y) changes for every one-unit increase in the independent (x) variable, on average. The y-intercept is used to describe the dependent variable when the independent variable equals zero. Graphically, the slope is represented by three line types in elementary statistics. 12.2 The Regression Equation A regression line, or a line of best fit, can be drawn on a scatter plot and used to predict outcomes for the x and y variables in a given data set or sample data. There are several ways to find a regression line, but usually the least-squares regression line is used because it creates a uniform line. Residuals, also called errors, measure the distance from the actual value of y and the estimated value of y. The sum of squared errors, or SSE, when set to its minimum, calculates the points on the line of best fit. Regression lines can be used to predict values within the given set of data but should not be used to make predictions for values outside the set of data. The correlation coefficient, r, measures the strength of the linear association between x and y. The variable r has to be between –1 and +1. When r is positive, x and y tend to increase and decrease together. When r is negative, x increases and y decreases, or the opposite occurs: x decreases and y increases. The coefficient of determination, r2, is equal to the square of the correlation coefficient. When expressed as a percentage, r2 represents the percentage of variation in the dependent variable, y, that can be explained by variation in the independent variable, x, using the regression line. 12.3 Testing the Significance of the Correlation Coefficient (Optional) Linear regression is a procedure for fitting a straight line of the form ŷ = a + bx to data. The conditions for regression are as follows: • Linear: In the population, there is a linear relationship that models the average value of y for different values of x. • Independent: The residuals are assumed to be independent. • Normal: The y values are distributed normally for any value of x. • Equal variance: The standard deviation of the y values is equal for each x value. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 729 • Random: The data are produced from a well-designed random sample or a randomized experiment. The slope b and intercept a of the least-squares line estimate the slope β and intercept α of the population (true) regression line. To estimate the population standard deviation of y (σ) use the standard deviation of the residuals: s = SSE n − 2 . The variable ρ (rho) is the population correlation coefficient. To test the null hypothesis, H0: ρ = hypothesized value, use a linear regression t-test. The most common null hypothesis is H0: ρ = 0, which indicates there is no linear relationship between x and y in the population. The TI-83, 83+, 84, 84+ calculator function LinRegTTest can perform this test (STATS, TESTS, LinRegTTest). 12.4 Prediction (Optional) After determining the presence of a strong correlation coefficient and calculating the line of best fit, you can use the leastsquares regression line to make predictions about your data. 12.5 Outliers To determine whether a point is an outlier, do one of the following: 1. Input the following equations into the TI 83, 83+, 84, or 84+ calculator: y1 = a + bx y2 = a + bx + 2s y3 = a + bx – 2s where s is the standard deviation of the residuals. If any point is above y2 or below y3, then the point is considered to be an outlier. 2. Use the residuals and compare their absolute values to 2s, where s is the standard deviation of the residuals. If the absolute value of any residual is greater than or equal to 2s, then the corresponding point is an outlier. 3. Note: The calculator function LinRegTTest (STATS, TESTS, LinRegTTest) calculates s. FORMULA REVIEW 12.1 Linear Equations where a is the y-intercept and b is the slope. Standard Deviation of the Residuals: y = a + bx, where a is the y-intercept and b is the slope. The variable x is the independent variable and y is the dependent variable. s = SSE n − 2 , where SSE = sum of squared errors, and n = the number of data points. 12.3 Testing the Significance of the Correlation Coefficient (Optional) Least-Squares Line or Line of Best Fit: ŷ = a + bx, PRACTICE 12.1 Linear Equations Use the following information to answer the next three exercises. A vacation resort rents scuba equipment to certified divers. The resort charges an up-front fee of $25 and another fee of $12.50 an hour. 1. What are the dependent and independent variables? 2. Find the equation that expresses the total fee in terms of the number of hours the equipment is rented. 730 Chapter 12 \| Linear Regression and Correlation 3. Graph the equation from Exercise 12.2. Use the following information to answer the next two exercises. A credit card company charges $10 when a payment is late and $5 a day each day the payment remains unpaid. 4. Find the equation that expresses the total fee in terms of the number of days the payment is late. 5. Graph the equation from Exercise 12.4. 6. Is the equation y = 10 + 5x – 3x2 linear? Why or why not? 7. Which of the following equations are linear? a. y = 6x + 8 b. y + 7 = 3x c. y – x = 8x2 d. 4y = 8 8. Does the graph in Figure 12.23 show a linear equation? Why or why not? Figure 12.23 Use the following information to answer the next exercise. Table 12.13 contains real data for the first two decades of flu reporting. Year Number of Flu Cases Diagnosed Number of Flu Deaths Pre-1981 91 1981 1982 1983 1984 1985 1986 1987 1988 1989 319 1,170 3,076 6,240 11,776 19,032 28,564 35,447 42,674 Table 12.13 29 121 453 1,482 3,466 6,878 11,987 16,162 20,868 27
,591 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 731 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 48,634 59,660 78,530 78,834 71,874 68,505 59,347 47,149 38,393 25,174 25,522 25,643 26,464 31,335 36,560 41,055 44,730 49,095 49,456 38,510 20,736 19,005 18,454 17,347 17,402 16,371 Total 802,118 489,093 Table 12.13 9. Use the columns Year and Number of Flu Cases Diagnosed. Why is year the independent variable and number of flu cases diagnosed the dependent variable (instead of the reverse)? Use the following information to answer the next two exercises. A specialty cleaning company charges an equipment fee and an hourly labor fee. A linear equation that expresses the total amount of the fee the company charges for each session is y = 50 + 100x. 10. What are the independent and dependent variables? 11. What is the y-intercept, and what is the slope? Interpret them using complete sentences. Use the following information to answer the next three questions. As a result of erosion, a river shoreline is losing several thousand pounds of soil each year. A linear equation that expresses the total amount of soil lost per year is y = 12,000x. 12. What are the independent and dependent variables? 13. How many pounds of soil does the shoreline lose in a year? 14. What is the y-intercept? Interpret its meaning. Use the following information to answer the next two exercises. The price of a single issue of stock can fluctuate throughout the day. A linear equation that represents the price of stock for Shipment Express is y = 15 – 1.5x, where x is the number of hours passed in an eight-hour day of trading. 15. What are the slope and y-intercept? Interpret their meaning. 16. If you owned this stock, would you want a positive or negative slope? Why? 732 Chapter 12 \| Linear Regression and Correlation 12.2 The Regression Equation 17. Table 12.16 below represents the relationship between the number of hours spent studying and final exam grades. x (number of hours spent studying) y (final exam grades) 3 5 1 2 6 8 4 7 Table 12.14 50 72 45 51 80 96 65 90 Fill in the following chart as a first step in finding the line of best fit, using the median–median approach. Group x (no. of hours spent studying) y (final exam grades) Median x Value Median y Value 1 2 3 Table 12.15 Use the following information to answer the next five exercises. A random sample of 10 professional athletes produced the following data, where x is the number of endorsements the player has and y is the amount of money made, in millions of dollars 12 9 9 3 13 4 10 Table 12.16 18. Draw a scatter plot of the data. 19. Use regression to find the equation for the line of best fit. 20. Draw the line of best fit on the scatter plot. 21. What is the slope of the line of best fit? What does it represent? 22. What is the y-intercept of the line of best fit? What does it represent? 23. What does an r value of zero mean? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 733 24. When n = 2 and r = 1, are the data significant? Explain. 25. When n = 100 and r = –0.89, is there a significant correlation? Explain. 12.3 Testing the Significance of the Correlation Coefficient (Optional) 26. When testing the significance of the correlation coefficient, what is the null hypothesis? 27. When testing the significance of the correlation coefficient, what is the alternative hypothesis? 28. If the level of significance is 0.05 and the p-value is 0.04, what conclusion can you draw? 12.4 Prediction (Optional) Use the following information to answer the next two exercises. An electronics retailer used regression to find a simple model to predict sales growth in the first quarter of the new year (January through March). The model is good for 90 days, where x is the day. The model can be written as ŷ = 101.32 + 2.48x, where ŷ is in thousands of dollars. 29. What would you predict the sales to be on day 60? 30. What would you predict the sales to be on day 90? Use the following information to answer the next three exercises. A landscaping company is hired to mow the grass for several large properties. The total area of the properties is 1,345 acres. The rate at which one person can mow is ŷ = 1350 – 1.2x, where x is the number of hours and ŷ represents the number of acres left to mow. 31. How many acres are left to mow after 20 hours of work? 32. How many acres are left to mow after 100 hours of work? 33. How many hours does it take to mow all the lawns, or when is ŷ = 0? Use the following information to answer the next 14 exercises. Table 12.17 contains real data for the first two decades of flu reporting. Year Number of Flu Cases Diagnosed Number of Flu Deaths Pre-1981 91 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 319 1,170 3,076 6,240 11,776 19,032 28,564 35,447 42,674 48,634 59,660 78,530 78,834 71,874 68,505 59,347 29 121 453 1,482 3,466 6,878 11,987 16,162 20,868 27,591 31,335 36,560 41,055 44,730 49,095 49,456 38,510 Table 12.17 Adults and Adolescents Only, United States 734 Chapter 12 \| Linear Regression and Correlation 1997 1998 1999 2000 2001 2002 47,149 38,393 25,174 25,522 25,643 26,464 Total 802,118 20,736 19,005 18,454 17,347 17,402 16,371 489,093 Table 12.17 Adults and Adolescents Only, United States 34. Graph year versus number of flu cases diagnosed (plot the scatter plot). Do not include pre-1981 data. 35. Perform a linear regression. What is the linear equation? Round to the nearest whole number. Find the following: Write the equations: • Linear equation: __________ • a = ________ • b = ________ • r = ________ • n = ________ 36. Solve. a. When x = 1985, ŷ = _____. b. When x = 1990, ŷ = _____. c. When x = 1970, ŷ = _____. Why doesn’t this answer make sense? 37. Does the line seem to fit the data? Why or why not? 38. What does the correlation imply about the relationship between time (years) and the number of diagnosed flu cases reported in the United States? 39. Plot the two points on the graph. Then, connect the two points to form the regression line. 40. Write the equation: ŷ = ____________. 41. Hand-draw a smooth curve on the graph that shows the flow of the data. 42. Does the line seem to fit the data? Why or why not? 43. Do you think a linear fit is best? Why or why not? 44. What does the correlation imply about the relationship between time (years) and the number of diagnosed flu cases reported in the United States? 45. Graph year vs. number flu cases diagnosed. Do not include pre-1981. Label both axes with words. Scale both axes. 46. Enter your data into your calculator or computer. The pre-1981 data should not be included. Why is that so? Write the linear equation, rounding to four decimal places. 47. Calculate the following: • a = _____ • b = _____ • correlation = _____ • n = _____ 12.5 Outliers 48. Marcus states that all outliers are influential points. Is he correct? Explain. Use the following information to answer the next four exercises. The scatter plot shows the relationship between hours spent studying and exam scores. The line shown is the calculated line of best fit. The correlation coefficient is 0.69. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 735 Figure 12.24 49. Do there appear to be any outliers? 50. A point is removed and the line of best fit is recalculated. The new correlation coefficient is 0.98. Does the point appear to have been an outlier? Why? 51. What effect did the potential outlier have on the line of best fit? 52. Are you more or less confident in the predictive ability of the new line of best fit? 53. The sum of squared errors (SSE) for a data set of 18 numbers is 49. What is the standard deviation? 54. The standard deviation for the SSE for a data set is 9.8. What is the cutoff for the vertical distance that a point can be from the line of best fit to be considered an outlier? HOMEWORK 12.1 Linear Equations 55. For each of the following situations, state the independent variable and the dependent variable. a. A study is done to determine whether elderly drivers are involved in more motor vehicle fatalities than other drivers. The number of fatalities per 100,000 drivers is compared with the age of drivers. Insurance companies base life insurance premiums partially on the age of the applicant. b. A study is done to determine whether the weekly grocery bill changes based on the number of family members. c. d. Utility bills vary according to power consumption. e. A study is done to determine whether a higher education reduces the crime rate in a population. 736 Chapter 12 \| Linear Regression and Correlation 56. Piece-rate systems are widely debated incentive payment plans. In a recent study of loan officer effectiveness, the following piece-rate system was examined: % of goal reached < 80 80 100 120 Incentive n/ a $4,000, with an additional $125 added per percentage point from 81% to 99% $6,500, with an additional $125 added per percentage point from 101% to 119% $9,500, with an additional $125 added per percentage point starting at 121% Table 12.18 If a loan officer makes 95 percent of his or her goal, write the linear function that applies based on the incentive plan table. In context, explain the y-intercept and slope. 12.2 The Regression Equation 57. What is the process through which we can calculate a line that goes through a scatter plot with a linear pattern? 58. Explain what it means when a correlation has an r2 value of .72. 59. Can a coefficient of determination be negative? Why or why not? 60. The table below represents the relationship between SAT scores on the math portion of the test and high school grade point averages (GPAs). Use the median–-median line approach to find the equation for the line of best fit. x (SAT math scores) y (GPAs) 6
24 544 363 373 350 741 262 587 327 364 261 Table 12.19 90 86 70 71 65 98 60 87 62 67 50 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 737 12.4 Prediction (Optional) 61. Recently, the annual numbers of driver deaths per 100,000 people for the selected age groups are as follows: Age (years) Number of Driver Deaths (per 100,000 people) 16–19 20–24 25–34 35–54 55–74 75+ Table 12.20 38 36 24 20 18 28 a. For each age group, pick the midpoint of the interval for the x value. For the 75+ group, use 80. b. Using age as the independent variable and number of driver deaths per 100,000 people as the dependent variable, make a scatter plot of the data. c. Calculate the least-squares (best–fit) line. Put the equation in the form ŷ = a + bx. d. Find the correlation coefficient. Is it significant? e. Predict the number of deaths for ages 40 years and 60 years. f. Based on the given data, is there a linear relationship between age of a driver and driver fatality rate? g. What is the slope of the least-squares (best-fit) line? Interpret the slope. 62. Table 12.21 shows the life expectancy for an individual born in the United States in certain years. Year of Birth Life Expectancy in years 1930 1940 1950 1965 1973 1982 1987 1992 2010 Table 12.21 59.7 62.9 70.2 69.7 71.4 74.5 75 75.7 78.7 a. Decide which variable should be the independent variable and which should be the dependent variable. b. Draw a scatter plot of the ordered pairs. c. Calculate the least-squares line. Put the equation in the form ŷ = a + bx. d. Find the correlation coefficient. Is it significant? e. Find the estimated life expectancy for an individual born in 1950 and for one born in 1982. f. Why aren’t the answers to Part E the same as the values in Table 12.21 that correspond to those years? g. Use the two points in Part E to plot the least-squares line on your graph from Part B. h. Based on the data, is there a linear relationship between the year of birth and life expectancy? i. Are there any outliers in the data? j. Using the least-squares line, find the estimated life expectancy for an individual born in 1850. Does the least- squares line give an accurate estimate for that year? Explain why or why not. k. What is the slope of the least-squares (best-fit) line? Interpret the slope. 738 Chapter 12 \| Linear Regression and Correlation 63. The maximum discount value of the Entertainment® card for the Fine Dining section, 10th edition, for various pages is given in Table 12.22. Page Number Maximum Value ($) 4 14 25 32 43 57 72 85 90 Table 12.22 16 19 15 17 19 15 16 15 17 a. Decide which variable should be the independent variable and which should be the dependent variable. b. Draw a scatter plot of the ordered pairs. c. Calculate the least-squares line. Put the equation in the form ŷ = a + bx. d. Find the correlation coefficient. Is it significant? e. Find the estimated maximum values for the restaurants on page 10 and on page 70. f. Does it appear that the restaurants giving the maximum value are placed in the beginning of the Fine Dining section? How did you arrive at your answer? g. Suppose there are 200 pages of restaurants. What do you estimate to be the maximum value for a restaurant listed on page 200? Is the least-squares line valid for page 200? Why or why not? h. i. What is the slope of the least-squares (best-fit) line? Interpret the slope. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 739 64. Table 12.23 gives the gold medal times for every other Summer Olympics for the women’s 100-meter freestyle in swimming. Year Time in seconds 1912 1924 1932 1952 1960 1968 1976 1984 1992 2000 2008 Table 12.23 82.2 72.4 66.8 66.8 61.2 60.0 55.65 55.92 54.64 53.8 53.1 a. Decide which variable should be the independent variable and which should be the dependent variable. b. Draw a scatter plot of the data. c. Does it appear from inspection that there is a relationship between the variables? Why or why not? d. Calculate the least-squares line. Put the equation in the form ŷ = a + bx. e. Find the correlation coefficient. Is the decrease in times significant? f. Find the estimated gold medal time for 1932. Find the estimated time for 1984. g. Why are the answers from Part F different from the chart values? h. Does it appear that a line is the best way to fit the data? Why or why not? i. Use the least-squares line to estimate the gold medal time for the next Summer Olympics. Do you think your answer is reasonable? Why or why not? 740 65. State Alabama Colorado Hawaii Iowa Maryland Missouri 7 8 6 4 8 8 New Jersey 9 Ohio South Carolina Utah Wisconsin Table 12.24 4 13 4 9 Chapter 12 \| Linear Regression and Correlation No. of Letters in Name Year Entered the Union Rank for Entering the Union Area in square miles 1819 1876 1959 1846 1788 1821 1787 1803 1788 1896 1848 22 38 50 29 7 24 3 17 8 45 30 52,423 104,100 10,932 56,276 12,407 69,709 8,722 44,828 32,008 84,904 65,499 We are interested in whether the number of letters in a state name depends on the year the state entered the Union. a. Decide which variable should be the independent variable and which should be the dependent variable. b. Draw a scatter plot of the data. c. Does it appear from inspection that there is a relationship between the variables? Why or why not? d. Calculate the least-squares line. Put the equation in the form ŷ = a + bx. e. Find the correlation coefficient. What does it imply about the significance of the relationship? f. Find the estimated number of letters (to the nearest integer) a state name would have if it entered the Union in 1900. Find the estimated number of letters a state name would have if it entered the Union in 1940. g. Does it appear that a line is the best way to fit the data? Why or why not? h. Use the least-squares line to estimate the number of letters for a new state that enters the Union this year. Can the least-squares line be used to predict it? Why or why not? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 741 12.5 Outliers 66. Given the information in Table 12.30, which represents the relationship between final exam math grades and final exam history grades, decide whether point (56, 95) is an influential point. Explain how you arrived at your decision. Show all work. x (final exam math grades) y (final exam history grades) 54 56 77 74 63 51 88 72 69 56 Table 12.25 60 68 82 78 69 55 97 77 78 95 742 Chapter 12 \| Linear Regression and Correlation 67. In Table 12.31, the height (sidewalk to roof) of notable tall buildings in America is compared with the number of stories of the building (beginning at street level). Height (in feet) Stories 1,050 428 362 529 790 401 380 1,454 1,127 700 Table 12.26 57 28 26 40 60 22 38 110 100 46 a. Using stories as the independent variable and height as the dependent variable, make a scatter plot of the data. b. Does it appear from inspection that there is a relationship between the variables? c. Calculate the least-squares line. Put the equation in the form ŷ = a + bx. d. Find the correlation coefficient. Is it significant? e. Find the estimated heights for a building that has 32 stories and for a building that has 94 stories. f. Based on the data in Table 12.26, is there a linear relationship between the number of stories in tall buildings and the height of the buildings? g. Are there any outliers in the data? If so, which point(s)? h. What is the estimated height of a building with six stories? Does the least-squares line give an accurate estimate of height? Explain why or why not. i. Based on the least-squares line, adding an extra story is predicted to add about how many feet to a building? j. What is the slope of the least-squares (best-fit) line? Interpret the slope. 68. Ornithologists (scientists who study birds) tag sparrow hawks in 13 different colonies to study their population. They gather data for the percentage of new sparrow hawks in each colony and the percentage of those that have returned from migration. Percent return: 74, 66, 81, 52, 73, 62, 52, 45, 62, 46, 60, 46, 38 Percent new: 5, 6, 8, 11, 12, 15, 16, 17, 18, 18, 19, 20, 20 a. Enter the data into a calculator and make a scatter plot. b. Use the calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from Part A. c. Explain what the slope and y-intercept of the regression line tell us. d. How well does the regression line fit the data? Explain your response. e. Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain. f. An ecologist wants to predict how many birds will join another colony of sparrow hawks to which 70 percent of the adults from the previous year have returned. What is the prediction? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 743 69. The following table shows data on average per capita coffee consumption and death rate from heart disease in a random sample of 10 countries. Yearly Coffee Consumption (liters) 2.5 3.9 2.9 2.4 2.9 0.8 9.1 2.7 0.8 0.7 No. of Deaths from Heart Disease 221 167 131 191 220 297 71 172 211 300 Table 12.27 a. Enter the data into a calculator and make a scatter plot. b. Use the calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from Part A. c. Explain what the slope and y-intercept of the regression line tell us. d. How well does the regression line fit the data? Explain your response. e. Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain. f. Do the data provide convincing evidence that there is a linear relati
onship between the amount of coffee consumed and the heart disease death rate? Carry out an appropriate test at a significance level of 0.05 to help answer this question. 70. The following table consists of one student athlete’s time (in minutes) to swim 2,000 yards and the student’s heart rate (beats per minute) after swimming on a random sample of 10 days. Swim Time Heart Rate 34.12 35.72 34.72 34.05 34.13 35.73 36.17 35.57 35.37 35.57 Table 12.28 144 152 124 140 152 146 128 136 144 148 a. Enter the data into a calculator and make a scatter plot. b. Use the calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from Part A. c. Explain what the slope and y-intercept of the regression line tell us. d. How well does the regression line fit the data? Explain your response. e. Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain. 744 Chapter 12 \| Linear Regression and Correlation 71. A researcher is investigating whether population impacts homicide rate. He uses demographic data from Detroit, Michigan, to compare homicide rates and the population. Population Size Homicide Rate per 100,000 People 558,724 538,584 519,171 500,457 482,418 465,029 448,267 432,109 416,533 401,518 387,046 373,095 359,647 Table 12.29 8.6 8.9 8.52 8.89 13.07 14.57 21.36 28.03 31.49 37.39 46.26 47.24 52.33 a. Use a calculator to construct a scatter plot of the data. What is the independent variable? Why? b. Use the calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot. c. Discuss what the following mean in context: i. The slope of the regression equation ii. The y-intercept of the regression equation iii. The correlation coefficient, r iv. The coefficient of determination, r2 d. Do the data provide convincing evidence that there is a linear relationship between population size and homicide rate? Carry out an appropriate test at a significance level of 0.05 to help answer this question. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 745 72. School Princeton Harvey Mudd CalTech U.S. Naval Academy West Point MIT Lehigh University NYU-Poly Babson College Stanford Table 12.30 Mid-Career Salary (in thousands of U.S. dollars) Yearly Tuition (in U.S. dollars) 137 135 127 122 120 118 118 117 117 114 28,540 40,133 39,900 0 0 42,050 43,220 39,565 40,400 54,506 Use the data in the Table 12.35 to determine the linear regression line equation with the outliers removed. Is there a linear correlation for the data set with outliers removed? Justify your answer. REFERENCES 12.1 Linear Equations Centers for Disease Control and Prevention. (n.d.). Retrieved from https://www.cdc.gov/ National Center for HIV/AIDS, Viral Hepatitis, STD, and TB Prevention. (n.d.). Centers for Disease Control and Prevention. Retrieved from https://www.cdc.gov/nchhstp/default.htm 12.4 Prediction (Optional) Centers for Disease Control and Prevention. (n.d.). Retrieved from https://www.cdc.gov/ National Center for HIV/AIDS, Viral Hepatitis, STD, and TB Prevention. (n.d.). Centers for Disease Control and Prevention. Retrieved from https://www.cdc.gov/nchhstp/default.htm National Center https://www.cdc.gov/nchs/index.htm for Health Statistics. (n.d.). Centers for Disease Control and Prevention. Retrieved from Census U.S. Bureau. motor_vehicle_accidents_and_fatalities.html (n.d.). Retrieved from http://www.census.gov/compendia/statab/cats/transportation/ 12.5 Outliers Committee on Ways and Means, U.S. House of Representatives. (n.d.). Washington, DC: U.S. Department of Health and Human Services. Microsoft Bookshelf. (n.d.). Physician’s Desk Reference Staff. (1990). Physician’s desk reference. Ohio: Medical Economics Company. U.S. Bureau of Labor Statistics. (n.d.). Retrieved from https://www.bls.gov/ 746 Chapter 12 \| Linear Regression and Correlation BRINGING IT TOGETHER: HOMEWORK 73. The average number of people in a family who attended college for various years is given in Table 12.31. Year No. of Family Members Attending College 1969 1973 1975 1979 1983 1988 1991 Table 12.31 4.0 3.6 3.2 3.0 3.0 3.0 2.9 a. Using year as the independent variable and number of family members attending college as the dependent variable, draw a scatter plot of the data. b. Calculate the least-squares line. Put the equation in the form ŷ = a + bx. c. Does the y-intercept, a, have any meaning here? d. Find the correlation coefficient. Is it significant? e. Pick two years between 1969 and 1991 and find the estimated number of family members attending college. f. Based on the data in Table 12.31, is there a linear relationship between the year and the average number of family members attending college? g. Using the least-squares line, estimate the number of family members attending college for 1960 and 1995. Does the least-squares line give an accurate estimate for those years? Explain why or why not. h. Are there any outliers in the data? i. What is the estimated average number of family members attending college for 1986? Does the least-squares line give an accurate estimate for that year? Explain why or why not. j. What is the slope of the least-squares (best-fit) line? Interpret the slope. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 747 74. The percent of female wage and salary workers who are paid hourly rates is given in Table 12.32 for the years 1979 to 1992. Year Percent of Workers Paid Hourly Rates 1979 1980 1981 1982 1983 1984 1985 1986 1987 1990 1992 Table 12.32 61.2 60.7 61.3 61.3 61.8 61.7 61.8 62.0 62.7 62.8 62.9 a. Using year as the independent variable and percent of workers paid hourly rates as the dependent variable, draw a scatter plot of the data. b. Does it appear from inspection that there is a relationship between the variables? Why or why not? c. Does the y-intercept, a, have any meaning here? d. Calculate the least-squares line. Put the equation in the form ŷ = a + bx. e. Find the correlation coefficient. Is it significant? f. Find the estimated percentages for 1991 and 1988. g. Based on the data, is there a linear relationship between the year and the percentage of female wage and salary earners who are paid hourly rates? h. Are there any outliers in the data? i. What is the estimated percentage for the year 2050? Does the least-squares line give an accurate estimate for that year? Explain why or why not. j. What is the slope of the least-squares (best-fit) line? Interpret the slope. Use the following information to answer the next two exercises. The cost of a leading liquid laundry detergent in different sizes is given in Table 12.33. Size (ounces) Cost ($) Cost per Ounce 16 32 64 200 Table 12.33 3.99 4.99 5.99 10.99 748 75. 76. Chapter 12 \| Linear Regression and Correlation a. Using size as the independent variable and cost as the dependent variable, draw a scatter plot. b. Does it appear from inspection that there is a relationship between the variables? Why or why not? c. Calculate the least-squares line. Put the equation in the form ŷ = a + bx. d. Find the correlation coefficient. Is it significant? e. f. g. Does it appear that a line is the best way to fit the data? Why or why not? h. Are there any outliers in the given data? i. If the laundry detergent were sold in a 40 oz. size, what is the estimated cost? If the laundry detergent were sold in a 90 oz. size, what is the estimated cost? Is the least-squares line valid for predicting what a 300 oz. size of the laundry detergent would cost? Why or why not? j. What is the slope of the least-squares (best-fit) line? Interpret the slope. a. Complete Table 12.33 for the cost per ounce of the different sizes of laundry detergent. b. Using size as the independent variable and cost per ounce as the dependent variable, draw a scatter plot of the data. c. Does it appear from inspection that there is a relationship between the variables? Why or why not? d. Calculate the least-squares line. Put the equation in the form ŷ = a + bx. e. Find the correlation coefficient. Is it significant? f. g. h. Does it appear that a line is the best way to fit the data? Why or why not? i. Are there any outliers in the the data? j. If the laundry detergent were sold in a 40 oz. size, what is the estimated cost per ounce? If the laundry detergent were sold in a 90 oz. size, what is the estimated cost per ounce? Is the least-squares line valid for predicting what a 300 oz. size of the laundry detergent would cost per ounce? Why or why not? k. What is the slope of the least-squares (best-fit) line? Interpret the slope. 77. According to a flyer published by Prudential Insurance Company, the costs of approximate probate fees and taxes for selected net taxable estates are as follows: Net Taxable Estate ($) Approximate Probate Fees and Taxes ($) 600,000 750,000 1,000,000 1,500,000 2,000,000 2,500,000 3,000,000 Table 12.34 30,000 92,500 203,000 438,000 688,000 1,037,000 1,350,000 a. Decide which variable should be the independent variable and which should be the dependent variable. b. Draw a scatter plot of the data. c. Does it appear from inspection that there is a relationship between the variables? Why or why not? d. Calculate the least-squares line. Put the equation in the form ŷ = a + bx. e. Find the correlation coefficient. Is it significant? f. Find the estimated total cost for a net taxable estate of $1,000,000. Find the cost for $2,500,000. g. Does it appear that a line is the best way to fit the data? Why or why not? h. Are there any outliers in the data? i. Based on these results, what would be the probate fees and taxes for an estate that does not have any assets? j. What is the slope of the least-squares (best-fit) line? Interpret the slope. This OpenStax book is available fo
r free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 749 78. The following are advertised sale prices of color televisions at Anderson’s: Size (inches) Sale Price ($) 9 20 27 31 35 40 60 Table 12.35 147 197 297 447 1,177 2,177 2,497 a. Decide which variable should be the independent variable and which should be the dependent variable. b. Draw a scatter plot of the data. c. Does it appear from inspection that there is a relationship between the variables? Why or why not? d. Calculate the least-squares line. Put the equation in the form ŷ = a + bx. e. Find the correlation coefficient. Is it significant? f. Find the estimated sale price for a 32-inch television. Find the cost for a 50-inch television. g. Does it appear that a line is the best way to fit the data? Why or why not? h. Are there any outliers in the data? i. What is the slope of the least-squares (best-fit) line? Interpret the slope. 79. Table 12.36 shows the average heights for American boys in 1990. Age (years) Height (centimeters) Birth 2 3 5 7 10 14 Table 12.36 50.8 83.8 91.4 106.6 119.3 137.1 157.5 a. Decide which variable should be the independent variable and which should be the dependent variable. b. Draw a scatter plot of the data. c. Does it appear from inspection that there is a relationship between the variables? Why or why not? d. Calculate the least-squares line. Put the equation in the form ŷ = a + bx. e. Find the correlation coefficient. Is it significant? f. Find the estimated average height for a 1-year-old. Find the estimated average height for an 11-year-old. g. Does it appear that a line is the best way to fit the data? Why or why not? h. Are there any outliers in the data? i. Use the least-squares line to estimate the average height for a 62-year-old man. Do you think that your answer is reasonable? Why or why not? j. What is the slope of the least-squares (best-fit) line? Interpret the slope. 750 80. State Alabama Colorado Hawaii Iowa Maryland Missouri 7 8 6 4 8 8 New Jersey 9 Ohio South Carolina Utah Wisconsin Table 12.37 4 13 4 9 Chapter 12 \| Linear Regression and Correlation No. of Letters in Name Year Entered the Union Rank for Entering the Union Area (square miles) 1819 1876 1959 1846 1788 1821 1787 1803 1788 1896 1848 22 38 50 29 7 24 3 17 8 45 30 52,423 104,100 10,932 56,276 12,407 69,709 8,722 44,828 32,008 84,904 65,499 We are interested in whether there is a relationship between the ranking of a state and the area of the state. a. What are the independent and dependent variables? b. What do you think the scatter plot will look like? Make a scatter plot of the data. c. Does it appear from inspection that there is a relationship between the variables? Why or why not? d. Calculate the least-squares line. Put the equation in the form ŷ = a + bx. e. Find the correlation coefficient. What does it imply about the significance of the relationship? f. Find the estimated areas for Alabama and for Colorado. Are they close to the actual areas? g. Use the two points in Part F to plot the least-squares line on your graph from Part B. h. Does it appear that a line is the best way to fit the data? Why or why not? i. Are there any outliers? j. Use the least-squares line to estimate the area of a new state that enters the Union. Can the least-squares line be used to predict it? Why or why not? k. Delete Hawaii and substitute Alaska for it. Alaska is a state with an area of 656,424 square miles. l. Calculate the new least-squares line. m. Find the estimated area for Alabama. Is it closer to the actual area with this new least-squares line or with the previous one that included Hawaii? Why do you think that’s the case? n. Do you think that, in general, newer states are larger than the original states? SOLUTIONS 1 dependent variable: fee amount independent variable: time This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 751 3 Figure 12.25 5 Figure 12.26 7 y = 6x + 8, 4y = 8, and y + 7 = 3x are all linear equations. 9 The number of flu cases depends on the year. Therefore, year becomes the independent variable and the number of flu cases is the dependent variable. 11 The y-intercept is 50 (a = 50). At the start of the cleaning, the company charges a one-time fee of $50 (this is when x = 0). The slope is 100 (b = 100). For each session, the company charges $100 for each hour they clean. 13 12,000 lb of soil 15 The slope is –1.5 (b = –1.5). This means the stock is losing value at a rate of $1.50 per hour. The y-intercept is $15 (a = 15). This means the price of stock before the trading day was $15. 752 17 Chapter 12 \| Linear Regression and Correlation Group x (no. of hours spent studying) y (final exam grades) Median x value Median y value Table 12.38 19 ŷ = 2.23 + 1.99x 45 50 51 65 72 80 90 96 2 4.5 7 50 68.5 90 21 The slope is 1.99 (b = 1.99). It means that for every endorsement deal a professional player gets, he gets an average of another $1.99 million in pay each year. 23 It means that there is no correlation between the data sets. 25 Yes. There are enough data points and the value of r is strong enough to show there is a strong negative correlation between the data sets. 27 Ha: ρ ≠ 0 29 $250,120 31 1326 acres 33 1125 hours, or when x = 1125 35 Check student solution. 36 a. When x = 1985, ŷ = 25,52. b. When x = 1990, ŷ = 34,275. c. When x = 1970, ŷ = –725. Why doesn’t this answer make sense? The range of x values was 1981 to 2002; the year 1970 is not in this range. The regression equation does not apply, because predicting for the year 1970 is extrapolation, which requires a different process. Also, a negative number does not make sense in this context, when we are predicting flu cases diagnosed. 38 Also, the correlation r = 0.4526. If r is compared with the value in the 95 Percent Critical Values of the Sample Correlation Coefficient Table, because r > 0.423, r is significant, and you would think that the line could be used for prediction. But, the scatter plot indicates otherwise. 39 Check student’ solution. 40 ŷ = 3,448,225 + 1750x 42 There was an increase in flu cases diagnosed until 1993. From 1993 through 2002, the number of flu cases diagnosed declined each year. It is not appropriate to use a linear regression line to fit to the data. 44 Because there is no linear association between year and number of flu cases diagnosed, it is not appropriate to calculate a linear correlation coefficient. When there is a linear association and it is appropriate to calculate a correlation, we cannot say that one variable causes the other variable. 46 We don’t know if the pre-1981 data were collected from a single year. So, we don’t have an accurate x value for this figure. Regression equation: ŷ (number of flu cases) = –3,448,225 + 1749.777 (year). This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 753 Coefficients Intercept –3,448,225 x Variable 1 1,749.777 Table 12.39 47 • a = –3,488,225 • b = 1,750 • correlation = 0.4526 • n = 22 48 No, he is not correct. An outlier is only an influential point if it significantly impacts the slope of the least-squares regression line and the correlation coefficient, r. If omission of this data point from the calculation of the regression line does not show much impact on the slope or r-value, then the outlier is not considered an influential point. For different reasons, it still may be determined that the data point must be omitted from the data set. 49 Yes. There appears to be an outlier at (6, 58). 51 The potential outlier flattened the slope of the line of best fit because it was below the data set. It made the line of best fit less accurate as a predictor for the data. 53 s = 1.75 55 a. b. c. d. e. independent variable: age; dependent variable: fatalities independent variable: number of family members; dependent variable: grocery bill independent variable: age of applicant; dependent variable: insurance premium independent variable: power consumption; dependent variable: utility independent variable: higher education (years); dependent variable: crime rates 58 It means that 72 percent of the variation in the dependent variable (y) can be explained by the variation in the independent variable (x). 754 60 Chapter 12 \| Linear Regression and Correlation x (SAT math scores) y (GPAs) 261 262 327 350 363 364 373 544 587 624 741 Table 12.40 50 60 62 65 70 67 71 86 87 90 98 We must remember to check the order of the y values within each group as well. We notice that the y values in the second group are not in order from the least value to the greatest value; these values thus must be reordered, meaning the median y value for that group is 70. Group x (SAT math scores) y (GPAs) Median x value Median y value 1 2 3 261 262 327 350 363 364 373 544 587 624 741 Table 12.41 50 60 62 65 67 70 71 86 87 90 98 294.5 364 61 70 605.5 88.5 The ordered pairs are (294.5, 61), (364, 70), and (605.5, 88.5). The slope can be calculated using the formula m = y3 − y1 x3 − x1 . Substituting the median x and y values, from the first and third groups gives m = 88.5 − 61 605.5 − 294.5 , which simplifies to m ≈ 0.09. The y-intercept may be found using the formula b = ∑ y − m∑ x 3 . The sum of the median x values is 1264, and the sum of the median y values is 219.5. Substituting these sums and the slope into the formula gives b = 219.5 − 0.09(1264) 3 , which simplifies to b ≈ 35.25. The line of best fit is represented as y = mx + b . Thus, the equation can be written as y = 0.09x + 35.25. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 755 61 b. Check student solution. c. ŷ = 35.5818045 – 0.19182491x d. r = –0.57874 For four degrees of freedom and alpha = 0.05, the LinRegTTest gives a p value of 0.2288, so we do not reject the null hypothesis; there is not a significant
linear relationship between deaths and age. Using the table of critical values for the correlation coefficient, with four degrees of freedom, the critical value is 0.811. The correlation coefficient r = –0.57874 is not less than –0.811, so we do not reject the null hypothesis. f. There is not a linear relationship between the two variables, as evidenced by a p value greater than 0.05. 63 a. We wonder if the better discounts appear earlier in the book, so we select page as x and discount as y. b. Check student solution. c. ŷ = 17.21757 – 0.01412x d. r = – 0.2752 For seven degrees of freedom and alpha = 0.05, LinRegTTest gives a p value = 0.4736, so we do not reject; there is a not a significant linear relationship between page and discount. Using the table of critical values for the correlation coefficient, with seven gives degrees of freedom, the critical value is 0.666. The correlation coefficient xi = –0.2752 is not less than 0.666, so we do not reject. f. There is not a significant linear correlation so it appears there is no relationship between the page and the amount of the discount. As the page number increases by one page, the discount decreases by $0.01412. 65 a. Year is the independent or x variable; the number of letters is the dependent or y variable. b. Check student’s solution. c. No. d. ŷ = 47.03 – 0.0216x e. –0.4280. The r value indicates that there is not a significant correlation between the year the state entered the Union and the number of letters in the name. g. No. The relationship does not appear to be linear; the correlation is not significant. 66 Using LinRegTTest, the output for the original least-squares regression line is y = 26.14 + 0.7539x and r = 0.6657. The output for the new least-squares regression line, after omitting the outlier of (56, 95), is ŷ = 6.36 + 1.0045x and r = 0.9757. The slope of the new line is quite a bit different from the slope of the original least-squares regression line, but the larger change is shown in the r-values, such that the new line has an r-value that has increased to a value that is almost equal to one. Thus, it may be stated that the outlier (56, 95) is also an influential point. 68 a. and b. Check student solution. c. The slope of the regression line is –0.3031 with a y-intercept of 31.93. In context, the y-intercept indicates that when there are no returning sparrow hawks, there will be almost 32 percent new sparrow hawks, which doesn’t make sense, because if there are no returning birds, then the new percentage would have to be 100% (this is an example of why we do not extrapolate). The slope tells us that for each percentage increase in returning birds, the percentage of new birds in the colony decreases by 30.3 percent. d. If we examine r2, we see that only 57.52 percent of the variation in the percentage of new birds is explained by the model and the correlation coefficient, r = –.7584 only indicates a somewhat strong correlation between returning and new percentages. e. The ordered pair (66, 6) generates the largest residual of 6.0. This means that when the observed return percentage is 66 percent, our observed new percentage, 6 percent, is almost 6 percent less than the predicted new value of 11.98 percent. If we remove this data pair, we see only an adjusted slope of –0.2789 and an adjusted intercept of 30.9816. In other words, although these data generate the largest residual, it is not an outlier, nor is the data pair an influential point. f. If there are 70 percent returning birds, we would expect to see y =– 0.2789(70) + 30.9816 = 0.114 or 11.4 percent new birds in the colony. 756 70 a. Check student solution. b. Check student solution. Chapter 12 \| Linear Regression and Correlation c. We have a slope of –1.4946 with a y-intercept of 193.88. The slope, in context, indicates that for each additional minute added to the swim time, the heart rate decreases by 1.5 beats per minute. If the student is not swimming at all, the y-intercept indicates that his heart rate will be 193.88 beats per minute. Although the slope has meaning (the longer it takes to swim 2000 m, the less effort the heart puts out), the y-intercept does not make sense. If the athlete is not swimming (resting), then his heart rate should be very low. d. Because only 1.5 percent of the heart rate variation is explained by this regression equation, we must conclude that this association is not explained with a linear relationship. e. Point (34.72, 124) generates the largest residual: –11.82. This means that our observed heart rate is almost 12 beats less than our predicted rate of 136 beats per minute. When this point is removed, the slope becomes –2.953, with the y-intercept changing to 247.1616. Although the linear association is still very weak, we see that the removed data pair can be considered an influential point in the sense that the y-intercept becomes more meaningful. 72 If we remove the two service academies (the tuition is $0.00), we construct a new regression equation of y = –0.0009x + 160, with a correlation coefficient of 0.71397 and a coefficient of determination of 0.50976. This allows us to say there is a fairly strong linear association between tuition costs and salaries if the service academies are removed from the data set. 73 c. No. The y-intercept would occur at year 0, which doesn’t exist. 74 a. Check student's solution. b. Yes. c. No, the y-intercept would occur at year 0, which doesn’t exist. d. ŷ = −266.8863 + 0.1656x. e. 0.9448, yes. f. 62.8233, 62.3265. g. Yes. h. No, (1987, 62.7). i. 72.5937, no. j. Slope = 0.1656. As the year increases by one, the percent of workers paid hourly rates tends to increase by 0.1656. 76 a. Size (ounces) Cost ($) Cost per ounce 16 32 64 200 Table 12.42 3.99 4.99 5.99 10.99 24.94 15.59 9.36 5.50 b. Check student solution. c. There is a linear relationship for the sizes 16 through 64, but that linear trend does not continue to the 200-oz size. d. ŷ = 20.2368 – 0.0819x e. r = –.8086 f. 40-oz: 16.96 cents/oz This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 757 g. 90-oz: 12.87 cents/oz h. The relationship is not linear; the least-squares line is not appropriate. i. There are no outliers. j. No. You would be extrapolating. The 300-oz size is outside the range of x. k. X = –0.08194. For each additional ounce in size, the cost per ounce decreases by 0.082 cents. 78 a. Size is x, the independent variable, and price is y, the dependent variable. b. Check student solution. c. The relationship does not appear to be linear. d. ŷ = –745.252 + 54.75569x. e. r = .8944 and yes, it is significant. f. 32-inch: $1006.93, 50-inch: $1992.53. g. No, the relationship does not appear to be linear. However, r is significant. h. No, the 60-inch TV. i. For each additional inch, the price increases by $54.76. 758 80 Chapter 12 \| Linear Regression and Correlation a. Rank is the independent variable and area is the dependent variable. b. Check student solution. c. There appears to be a linear relationship, with one outlier. d. ŷ (area) = 24177.06 + 1010.478x e. r = .50047. r is not significant, so there is no relationship between the variables. f. Alabama: 46,407.576 square miles, Colorado: 62,575.224 square miles. g. The Alabama estimate is closer than the Colorado estimate. h. If the outlier is removed, there is a linear relationship. i. There is one outlier (Hawaii). j. rank 51: 75,711.4 square miles, no. k. Alabama Colorado Hawaii Iowa Maryland Missouri New Jersey Ohio 7 8 6 4 8 8 9 4 1819 1876 1959 1846 1788 1821 1787 1803 South Carolina 13 1788 Utah Wisconsin 4 9 1896 1848 Table 12.43 22 38 50 29 7 24 3 17 8 45 30 52,423 104,100 10,932 56,276 12,407 69,709 8,722 44,828 32,008 84,904 65,499 l. ŷ = –87065.3 + 7828.532x. m. Alabama: 85,162.404; the prior estimate was closer. Alaska is an outlier. n. Yes, with the exception of Hawaii. 73 c. No. The y-intercept would occur at year 0, which doesn’t exist. 74 a. Check student's solution. b. Yes. c. No, the y-intercept would occur at year 0, which doesn’t exist. d. ŷ = −266.8863 + 0.1656x. e. 0.9448, yes. f. 62.8233, 62.3265. g. Yes. h. No, (1987, 62.7). i. 72.5937, no. j. Slope = 0.1656. As the year increases by one, the percent of workers paid hourly rates tends to increase by 0.1656. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 \| Linear Regression and Correlation 759 76 a. Size (ounces) Cost ($) Cost per ounce 16 32 64 200 Table 12.44 3.99 4.99 5.99 10.99 24.94 15.59 9.36 5.50 b. Check student solution. c. There is a linear relationship for the sizes 16 through 64, but that linear trend does not continue to the 200-oz size. d. ŷ = 20.2368 – 0.0819x e. r = –.8086 f. 40-oz: 16.96 cents/oz g. 90-oz: 12.87 cents/oz h. The relationship is not linear; the least-squares line is not appropriate. i. There are no outliers. j. No. You would be extrapolating. The 300-oz size is outside the range of x. k. X = –0.08194. For each additional ounce in size, the cost per ounce decreases by 0.082 cents. 78 a. Size is x, the independent variable, and price is y, the dependent variable. b. Check student solution. c. The relationship does not appear to be linear. d. ŷ = –745.252 + 54.75569x. e. r = .8944 and yes, it is significant. f. 32-inch: $1006.93, 50-inch: $1992.53. g. No, the relationship does not appear to be linear. However, r is significant. h. No, the 60-inch TV. i. For each additional inch, the price increases by $54.76. 80 a. Rank is the independent variable and area is the dependent variable. b. Check student solution. c. There appears to be a linear relationship, with one outlier. d. ŷ (area) = 24177.06 + 1010.478x e. r = .50047. r is not significant, so there is no relationship between the variables. f. Alabama: 46,407.576 square miles, Colorado: 62,575.224 square miles. g. The Alabama estimate is closer than the Colorado estimate. h. If the outlier is removed, there is a lin
ear relationship. i. There is one outlier (Hawaii). j. rank 51: 75,711.4 square miles, no. 760 k. Chapter 12 \| Linear Regression and Correlation Alabama Colorado Hawaii Iowa Maryland Missouri New Jersey Ohio 7 8 6 4 8 8 9 4 1819 1876 1959 1846 1788 1821 1787 1803 South Carolina 13 1788 Utah Wisconsin 4 9 1896 1848 Table 12.45 22 38 50 29 7 24 3 17 8 45 30 52,423 104,100 10,932 56,276 12,407 69,709 8,722 44,828 32,008 84,904 65,499 l. ŷ = –87065.3 + 7828.532x. m. Alabama: 85,162.404; the prior estimate was closer. Alaska is an outlier. n. Yes, with the exception of Hawaii. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 761 13 \| F DISTRIBUTION AND ONE-WAY ANOVA Figure 13.1 One-way ANOVA is used to measure information from several groups. Introduction Chapter Objectives By the end of this chapter, the student should be able to do the following: Interpret the F probability distribution as the number of groups and the sample size change • • Discuss two uses for the F distribution: one-way ANOVA and the test of two variances • Conduct and interpret one-way ANOVA • Conduct and interpret hypothesis tests of two variances Many statistical applications in psychology, social science, business administration, and the natural sciences involve several groups. For example, an environmentalist is interested in knowing if the average amount of pollution varies among several bodies of water. A sociologist is interested in knowing if the amount of income a person earns varies according to his or her 762 Chapter 13 \| F Distribution and One-way Anova upbringing. A consumer looking for a new car might compare the average gas mileage of several models. For hypothesis tests comparing averages across more than two groups, statisticians have developed a method called analysis of variance (abbreviated ANOVA). In this chapter, you will study the simplest form of ANOVA called single factor or oneway ANOVA. You will also study the F distribution, used for one-way ANOVA, and the test of two variances. This is a very brief overview of one-way ANOVA. You will study this topic in much greater detail in future statistics courses. One-way ANOVA, as it is presented here, relies heavily on a calculator or computer. 13.1 \| One-Way ANOVA The purpose of a one-way ANOVA test is to determine the existence of a statistically significant difference among several group means. The test uses variances to help determine if the means are equal or not. To perform a one-way ANOVA test, there are five basic assumptions to be fulfilled: • Each population from which a sample is taken is assumed to be normal. • All samples are randomly selected and independent. • The populations are assumed to have equal standard deviations (or variances). • The factor is a categorical variable. • The response is a numerical variable. The Null and Alternative Hypotheses The null hypothesis is that all the group population means are the same. The alternative hypothesis is that at least one pair of means is different. For example, if there are k groups H0: μ1 = μ2 = μ3 = ... = μk Ha: At least two of the group means μ1, μ2, μ3, ..., μk are not equal. That is, μi ≠ μj for some i ≠ j. The graphs, a set of box plots representing the distribution of values with the group means indicated by a horizontal line through the box, help in the understanding of the hypothesis test. In the first graph (red box plots), H0: μ1 = μ2 = μ3 and the three populations have the same distribution if the null hypothesis is true. The variance of the combined data is approximately the same as the variance of each of the populations. If the null hypothesis is false, then the variance of the combined data is larger, which is caused by the different means as shown in the second graph (green box plots). This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 763 Figure 13.2 (a) We fail to reject H0 as it may be true. All the means are about the same; the differences may be due to random variation. (b) We reject H0 as all the means are not the same; the differences are too large to be due to random variation. 13.2 \| The F Distribution and the F Ratio The distribution used for the hypothesis test is a new one. It is called the F distribution, named after Sir Ronald Fisher, an English statistician. The F statistic is a ratio (a fraction). There are two sets of degrees of freedom: one for the numerator and one for the denominator. For example, if F follows an F distribution and the number of degrees of freedom for the numerator is 4, and the number of degrees of freedom for the denominator is 10, then F ~ F4,10. NOTE The F distribution is derived from the Student’s t-distribution. The values of the F distribution are squares of the corresponding values of the t-distribution. One-way ANOVA expands the t-test for comparing more than two groups. The scope of that derivation is beyond the level of this course. It is preferable to use ANOVA when there are more than two groups instead of performing pairwise t-tests because performing multiple tests introduces the likelihood of making a Type 1 error. To calculate the F ratio, two estimates of the variance are made. 1. Variance between samples: an estimate of σ2 that is the variance of the sample means multiplied by n, when the sample sizes are the same. If the samples are different sizes, the variance between samples is weighted to account for the different sample sizes. The variance is also called variation due to treatment or explained variation. 2. Variance within samples: an estimate of σ2 that is the average of the sample variances, also known as a pooled variance. When the sample sizes are different, the variance within samples is weighted. The variance is also called the variation due to error or unexplained variation. • SSbetween = the sum of squares that represents the variation among the different samples • SSwithin = the sum of squares that represents the variation within samples that is due to chance To find a sum of squares mean, add together squared quantities which, in some cases, may be weighted. We used sum of squares to calculate the sample variance and the sample standard deviation in Descriptive Statistics. MS means mean square. MSbetween is the variance between groups, and MSwithin is the variance within groups. 764 Chapter 13 \| F Distribution and One-way Anova Calculation of Sum of Squares and Mean Square • k = the number of different groups • nj = the size of the jth group • sj = the sum of the values in the jth group • n = total number of all the values combined (total sample size: ∑nj) • x = one value: ∑x = ∑sj • Sum of squares of all values from every group combined: ∑x2 • Between group variability: SStotal = ∑x2 – ⎛ ⎝∑ x2⎞ n ⎠ • Total sum of squares: ∑x2 – 2 ⎠ ⎛ ⎝∑ x⎞ n • Explained variation: SS(between) = ∑ (s j)2 n j ⎡ ⎢ ⎣ sum of 2 ⎞ ⎠ ⎛ ⎝∑ s j n ⎤ ⎥ − ⎦ squares representing variation among the different samples • Unexplained variation: sum of squares representing variation within samples due to chance SSwithin = SStotal – SSbetween • dfs for different groups (dfs for the numerator): df = k – 1 • Equation for errors within samples (dfs for the denominator): dfwithin = n – k • Mean square (variance estimate) explained by the different groups: MSbetween = SSbetween d fbetween • Mean square (variance estimate) that is due to chance (unexplained): MSwithin = SSwithin d fwithin MSbetween and MSwithin can be written as follows: • MSbetween = SSbetween d fbetween = SSbetween k − 1 • MSwithin = SSwithin d fwithin = SSwithin n − k The one-way ANOVA test depends on the fact that MSbetween can be influenced by population differences among means of the several groups. Since MSwithin compares values of each group to its own group mean, the fact that group means might be different does not affect MSwithin. The null hypothesis says that all groups are samples from populations having the same normal distribution. The alternate hypothesis says that at least two of the sample groups come from populations with different normal distributions. If the null hypothesis is true, MSbetween and MSwithin should both estimate the same value. NOTE The null hypothesis says that all the group population means are equal. The hypothesis of equal means implies that the populations have the same normal distribution because it is assumed that the populations are normal and that they have equal variances. F Ratio or F Statistic F = MSbetween MSwithin If MSbetween and MSwithin estimate the same value, following the belief that H0 is true, then the F ratio should be approximately equal to 1. Mostly, just sampling errors would contribute to variations away from 1. As it turns out, MSbetween This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 765 consists of the population variance plus a variance produced from the differences between the samples. MSwithin is an estimate of the population variance. Since variances are always positive, if the null hypothesis is false, MSbetween will generally be larger than MSwithin. Then the F ratio will be larger than 1. However, if the population effect is small, it is not unlikely that MSwithin will be larger in a given sample. The previous calculations were done with groups of different sizes. If the groups are the same size, the calculations simplify somewhat and the F ratio can be written as follows: F Ratio formula when the groups are the same size F = 2 n ⋅ s x¯ s2 pooled where • n = the sample size • dfnumerator = k – 1 • dfdenominator = n – k • s2 pooled = the mean of the sample variances (pooled variance) • s x¯ 2 = the variance of the sample means Data is typically put into a table for easy viewing. One-way ANOVA results are often displayed in this manner by computer software. Source of Variation Sum of Squares (SS) Degrees
of Freedom (df) Mean Square (MS) F SS(Factor) SS(Error) SS(Total MS(Factor) = SS(Factor)/(k – 1) F = MS(Factor)/MS(Error) MS(Error) = SS(Error)/(n – k) Factor (Between) Error (Within) Total Table 13.1 Example 13.1 Three different diet plans are to be tested for mean weight loss. The entries in the table are the weight losses for the different plans. The one-way ANOVA results are shown in Table 13.2. Plan 1: n1 = 4 Plan 2: n2 = 3 Plan 3: n3 = 3 5 4.5 4 3 Table 13.2 3.5 7 4.5 8 4 3.5 s1 = 16.5, s2 = 15, s3 = 15.5 Following are the calculations needed to fill in the one-way ANOVA table. The table is used to conduct a hypothesis test. 766 Chapter 13 \| F Distribution and One-way Anova SS(between) = ∑ (s j) ⎞ ⎛ ⎝∑ s j ⎠ n = 2 s1 4 2 s2 3 + + 2 s3 3 (s1 + s2 + s3)2 10 − where n1 = 4, n2 = 3, n3 = 3, and n = n1 + n2 + n3 = 10 = (16.5)2 4 + (15)2 3 + (15.5)2 3 − (16.5 + 15 + 15.5)2 10 SS(between) = 2.2458 ⎝∑ x⎞ n S(total) = ∑ x2 − ⎛ ⎠ 2 = ⎝52 + 4.52 + 42 + 32 + 3.52 + 72 + 4.52 + 82 + 42 + 3.52⎞ ⎛ ⎠ − (5 + 4.5 + 4 + 3 + 3.5 + 7 + 4.5 + 8 + 4 + 3.5)2 10 = 244 − 472 10 = 244 − 220.9 SS(total) = 23.1 SS(within) = SS(total) − SS(between) = 23.1 − 2.2458 SS(within) = 20.8542 One-way ANOVA Table: The formulas for SS(Total), SS(Factor) = SS(Between), and SS(Error) = SS(Within) as shown previously. The same information is provided by the TI calculator hypothesis test function ANOVA in STAT TESTS (syntax is ANOVA[L1, L2, L3] where L1, L2, L3 have the data from Plan 1, Plan 2, Plan 3, respectively). Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F Factor (Between) SS(Factor) = SS(Between) = 2.2458 k – 1 = 3 groups – 1 = 2 SS(Error) = SS(Within) = 20.8542 SS(Total) = 2.2458 + 20.8542 = 23.1 n – k = 10 total data – 3 groups = 7 n – 1 = 10 total data – 1 = 9 Error (Within) Total Table 13.3 F = MS(Factor)/MS(Error) = 1.1229/2.9792 = 0.3769 MS(Factor) = SS(Factor)/(k – 1) = 2.2458/2 = 1.1229 MS(Error) = SS(Error)/(n – k) = 20.8542/7 = 2.9792 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 767 13.1 As part of an experiment to see how different types of soil cover would affect slicing tomato production, Marist College students grew tomato plants under different soil cover conditions. Groups of three plants each had one of the following treatments: • Bare soil • A commercial ground cover • Black plastic • Straw • Compost All plants grew under the same conditions and were the same variety. Students recorded the weight in grams of tomatoes produced by each of the n = 15 plants, as seen in Table 13.4. Bare: n1 = 3 Ground Cover: n2 = 3 Plastic: n3 = 3 Straw: n4 = 3 Compost: n5 = 3 2,625 2,997 4,915 Table 13.4 5,348 5,682 5,482 6,583 8,560 3,830 7,285 6,897 9,230 6,277 7,818 8,677 Create the one-way ANOVA table. The one-way ANOVA hypothesis test is always right-tailed because larger F values are way out in the right tail of the F distribution curve and tend to make us reject H0. Notation The notation for the F distribution is F ~ Fdf(num),df(denom), where df(num) = dfbetween and df(denom) = dfwithin. The mean for the F distribution is μ = d f (denom) d f (denom) – 2 . 13.3 \| Facts About the F Distribution The following are facts about the F distribution: • The curve is not symmetrical but skewed to the right. • There is a different curve for each set of dfs. • The F statistic is greater than or equal to zero. • As the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal. • Other uses for the F distribution include comparing two variances and two-way analysis of variance. Two-way analysis is beyond the scope of this chapter. 768 Chapter 13 \| F Distribution and One-way Anova Figure 13.3 Example 13.2 Let’s return to the slicing tomato exercise in Try It. The means of the tomato yields under the five mulching conditions are represented by μ1, μ2, μ3, μ4, μ5. We will conduct a hypothesis test to determine if all means are the same or at least one is different. Using a significance level of 5 percent, test the null hypothesis that there is no difference in mean yields among the five groups against the alternative hypothesis that at least one mean is different from the rest. Solution 13.2 The null and alternative hypotheses are as follows: H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: μi ≠ μj for some i ≠ j The one-way ANOVA results are shown in Table 13.4 Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F Factor (Between) 36,648,561 5 – 1 = 4 Error (Within) 20,446,726 15 – 5 = 10 Total 57,095,287 15 – 1 = 14 36,648,561 4 20,446,726 10 = 9,162,140 9,162,140 2,044,672.6 = 4.4810 = 2,044,672.6 Table 13.5 Distribution for the test: F4,10 df(num) = 5 – 1 = 4 df(denom) = 15 – 5 = 10 Test statistic: F = 4.4810 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 769 Figure 13.4 Probability statement: p-value = P(F > 4.481) = 0.0248 Compare α and the p-value: α = 0.05, p-value = 0.0248 Make a decision: Since α > p-value, we reject H0. Conclusion: At the 5 percent significance level, we have reasonably strong evidence that differences in mean yields for slicing tomato plants grown under different mulching conditions are unlikely to be due to chance alone. We may conclude that at least some of the mulches led to different mean yields. To find these results on the calculator: Press STAT. Press 1:EDIT. Put the data into the lists L1,L2,L3,L4,L5. Press STAT, arrow over to TESTS, and arrow down to ANOVA. Press ENTER, and then enter (L1,L2,L3,L4,L5). Press ENTER. You will see that the values in the foregoing ANOVA table are easily produced by the calculator, including the test statistic and the p-value of the test. The calculator displays: F = 4.4810 p = 0.0248 (p-value) Factor df = 4 SS = 36648560.9 MS = 9162140.23 Error df = 10 SS = 20446726 MS = 2044672.6 770 Chapter 13 \| F Distribution and One-way Anova 13.2 MRSA, or Staphylococcus aureus, can cause serious bacterial infections in hospital patients. Table 13.6 shows various colony counts from different patients who may or may not have MRSA. The data from the table is plotted in Figure 13.5. Conc = 0.6 Conc = 0.8 Conc = 1.0 Conc = 1.2 Conc = 1.4 9 66 98 Table 13.6 16 93 82 22 147 120 30 199 148 27 168 132 Plot of the data for the different concentrations: Figure 13.5 Test whether the mean numbers of colonies are the same or are different. Construct the ANOVA table by hand or by using a TI-83, 83+, or 84+ calculator, find the p-value, and state your conclusion. Use a 5 percent significance level. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 771 Example 13.3 Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in Table 13.7. Sorority 1 Sorority 2 Sorority 3 Sorority 4 2.17 1.85 2.83 1.69 3.33 2.63 1.77 3.25 1.86 2.21 2.63 3.78 4.00 2.55 2.45 3.79 3.45 3.08 2.26 3.18 Table 13.7 Mean Grades for Four Sororities Using a significance level of 1 percent, is there a difference in mean grades among the sororities? Solution 13.3 Let μ1, μ2, μ3, μ4 be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each five. NOTE This is an example of a balanced design, because each factor (i.e., sorority) has the same number of observations. H0: μ1 = μ2 = μ3 = μ4 Ha: Not all of the means μ1, μ2, μ3, μ4 are equal. Distribution for the test: F3,16 where k = 4 groups and n = 20 samples in total. df(num)= k – 1 = 4 – 1 = 3 df(denom) = n – k = 20 – 4 = 16 Calculate the test statistic: F = 2.23 Graph 772 Chapter 13 \| F Distribution and One-way Anova Figure 13.6 Probability statement: p-value = P(F > 2.23) = 0.1241 Compare α and the p-value: α = 0.01 p-value = 0.1241 α < p-value Make a decision: Since α < p-value, we cannot reject H0. Conclusion: There is not sufficient evidence to conclude that there is a difference among the mean grades for the sororities. Put the data into lists L1, L2, L3, and L4. Press STAT and arrow over to TESTS. Arrow down to F:ANOVA. Press ENTER and enter (L1,L2,L3,L4). The calculator displays the F statistic, the p-value, and the values for the one-way ANOVA table: F = 2.2303 p = 0.1241 (p-value) Factor df = 3 SS = 2.88732 MS = 0.96244 Error df = 16 SS = 6.9044 MS = 0.431525 13.3 Four sports teams took a random sample of players regarding their GPAs for the last year. The results are shown in Table 13.8. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 773 Basketball Baseball Hockey Lacrosse 3.6 2.9 2.5 3.3 3.8 2.1 2.6 3.9 3.1 3.4 4.0 2.0 2.6 3.2 3.2 2.0 3.6 3.9 2.7 2.5 Table 13.8 GPAs for four sports teams Use a significance level of 5 percent and determine if there is a difference in GPA among the teams. 774 Chapter 13 \| F Distribution and One-way Anova Example 13.4 A fourth-grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother’s garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew five plants. At the end of the growing period, each plant was measured, producing the data in inches in Table 13.9. Tommy’s Plants Tara’s Plants Nick’s Plants 24 21 23 30 23 Table 13.9 25 31 23 20 28 23 27 22 30 20 Does it appear that the three
soils in which the bean plants were grown produce the same mean height? Test at a 3 percent level of significance. Solution 13.4 This time, we will perform the calculations that lead to the F' statistic. Notice that each group has the same number of plants, so we will use the formula F' = 2 . n ⋅ s x¯ s2 pooled First, calculate the sample mean and sample variance of each group. Tommy's Plants Tara's Plants Nick's Plants 24.2 11.7 25.4 18.3 24.4 16.3 Sample Mean Sample Variance Table 13.10 Next, calculate the variance of the three group means by calculating the variance of 24.2, 25.4, and 24.4. Variance of the group means = 0.413 = s x¯ 2 , 2 = (5)(0.413) where n = 5 is the sample size (number of plants each child grew). then MSbetween = ns x¯ Calculate the mean of the three sample variances (11.7, 18.3, and 16.3). Mean of the sample variances = 15.433 = s2 then MSwithin = s2 pooled = 15.433. pooled, The F statistic (or F ratio) is F = MSbetween MSwithin = 2 ns x¯ s2 pooled = (5)(0.413) 15.433 = 0.134. The dfs for the numerator = the number of groups – 1 = 3 – 1 = 2. The dfs for the denominator = the total number of samples – the number of groups = 15 – 3 = 12. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 775 The distribution for the test is F2,12 and the F statistic is F = 0.134. The p-value is P(F > 0.134) = 0.8759. Decision: Since α = 0.03 and the p-value = 0.8759, we do not reject H0. Why? Conclusion: With a 3 percent level of significance from the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different. To calculate the p-value: •Press 2nd DISTR, •Arrow down to Fcdf and press ENTER, •Enter 0.134, E99, 2, 12, and •Press ENTER. The p-value is 0.8759. 13.4 Another fourth grader also grew bean plants, but in a jelly-like mass. The heights were (in inches) 24, 28, 25, 30, and 32. Do a one-way ANOVA test on the four groups. Are the heights of the bean plants different? Use the same method as shown in Example 13.4. From the class, create four groups of the same size as follows: men under 22, men at least 22, women under 22, women at least 22. Have each member of each group record the number of states in the United States he or she has visited. Run an ANOVA test to determine if the average number of states visited in the four groups are the same. Test at a 1 percent level of significance. Use one of the solution sheets in Appendix E. 13.4 \| Test of Two Variances Another use of the F distribution is testing two variances. It is often desirable to compare two variances rather than two averages. For instance, college administrators would like two college professors grading exams to have the same variation in their grading. For a lid to fit a container, the variation in the lid and the container should be the same. A supermarket might be interested in the variability of check-out times for two checkers. To perform a F test of two variances, it is important that the following are true: • The populations from which the two samples are drawn are normally distributed. • The two populations are independent of each other. Unlike most other tests in this book, the F test for equality of two variances is very sensitive to deviations from normality. If the two distributions are not normal, the test can give higher p-values than it should, or lower ones, in ways that are unpredictable. Many texts suggest that students not use this test at all, but in the interest of completeness we include it here. Suppose we sample randomly from two independent normal populations. Let σ1 2 and σ2 2 be the population variances and 776 Chapter 13 \| F Distribution and One-way Anova 2 be the sample variances. Let the sample sizes be n1 and n2. Since we are interested in comparing the two sample 2 and s2 s1 variances, we use the F ratio F = ⎡ ⎢(s1)2 (σ1)2 ⎣ ⎡ ⎢(s2)2 (σ2) has the distribution F ~ F(n1 – 1, n2 – 1), where n1 – 1 are the degrees of freedom for the numerator and n2 – 1 are the degrees of freedom for the denominator. If the null hypothesis is σ1 2 = σ2 2 , then the F ratio becomes F = ⎡ ⎢(s1)2 (σ1)2 ⎣ ⎡ ⎢(s2)2 (σ2)s1)2 (s2)2 . NOTE The F ratio could also be (s2)2 (s1)2 . It depends on Ha and on which sample variance is larger. If the two populations have equal variances, then s1 2 and s2 2 are close in value and F = (s1)2 (s2)2 is close to 1. But if the two population variances are very different, s1 variance causes the ratio (s1)2 (s2)2 to be greater than 1. If s1 2 and s2 2 are far apart, then F = (s1)2 (s2)2 is a large number. 2 and s2 2 tend to be very different, too. Choosing s1 2 as the larger sample Therefore, if F is close to 1, the evidence favors the null hypothesis (the two population variances are equal). But if F is much larger than 1, then the evidence is against the null hypothesis. A test of two variances may be left-tailed, right-tailed, or two-tailed. Example 13.5 Two college instructors are interested in whethe there is any variation in the way they grade math exams. They each grade the same set of 30 exams. The first instructor’s grades have a variance of 52.3. The second instructor’s grades have a variance of 89.9. Test the claim that the first instructor’s variance is smaller. In most colleges, it is desirable for the variances of exam grades to be nearly the same among instructors. The level of significance is 10 percent. Solution 13.5 Let 1 and 2 be the subscripts that indicate the first and second instructor, respectively. n1 = n2 = 30. 2 = σ2 H0: σ1 2 and Ha: σ1 2 . 2 < σ2 Calculate the test statistic: By the null hypothesis (σ1 2 = σ2 2 ) , the F statistic is This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 777 F = ⎡ ⎢(s1)2 (σ1)2 ⎣ ⎡ ⎢(s2)2 (σ2)s1)2 (s2)2 = 52.3 89.9 = 0.5818. Distribution for the test: F29,29 where n1 – 1 = 29 and n2 – 1 = 29. Graph: This test is left-tailed. Draw the graph, labeling and shading appropriately. Figure 13.7 Probability statement: p-value = P(F < 0.5818) = 0.0753. Compare α and the p-value: α = 0.10; α > p-value. Make a decision: Since α > p-value, reject H0. Conclusion: With a 10 percent level of significance from the data, there is sufficient evidence to conclude that the variance in grades for the first instructor is smaller. Press STAT and arrow over to TESTS. Arrow down to D:2-SampFTest. Press ENTER. Arrow to Stats (89.9) , and 30. Press ENTER after and press ENTER. For Sx1, n1, Sx2, and n2, enter each. Arrow to σ1: and < σ2. Press ENTER. Arrow down to Calculate and press ENTER. F = 0.5818 and p-value = 0.0753. Do the procedure again and try Draw instead of Calculate. (52.3) , 30, 778 Chapter 13 \| F Distribution and One-way Anova 13.5 The New York Choral Society divides male singers into four categories from highest voices to lowest: Tenor1, Tenor2, Bass1, and Bass2. In the table are heights of the men in the Tenor1 and Bass2 groups. One suspects that taller men will have lower voices, and that the variance of height may go up with the lower voices as well. Do we have good evidence that the variance of the heights of singers in each of these two groups (Tenor1 and Bass2) are different? Tenor1 Bass2 Tenor1 Bass2 Tenor1 Bass2 69 72 71 66 76 74 71 66 68 72 75 67 75 74 72 72 74 72 67 70 65 72 70 68 64 73 66 72 74 70 66 68 75 68 70 72 Table 13.11 68 67 64 67 70 70 69 72 71 74 75 13.5 \| Lab: One-Way ANOVA This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 779 13.1 One-Way ANOVA Student Learning Outcome • The student will conduct a simple one-way ANOVA test involving three variables. Collect the Data 1. Record the price per pound of eight fruits, eight vegetables, and eight breads in your local supermarket. Fruits Vegetables Breads Table 13.12 2. Explain how you could try to collect the data randomly. Analyze the Data and Conduct a Hypothesis Test 1. State the null hypothesis and the alternative hypothesis. 2. Compute the following: a. Fruit i. ii. x¯ = ______ s x = ______ iii. n = ______ b. Vegetables i. ii. x¯ = ______ s x = ______ iii. n = ______ c. Bread i. ii. x¯ = ______ s x = ______ iii. n = ______ 3. Find the following: a. df(num) = ______ 780 Chapter 13 \| F Distribution and One-way Anova b. df(denom) = ______ 4. State the approximate distribution for the test. 5. Test statistic: F = ______ 6. Sketch a graph of this situation. Clearly label and scale the horizontal axis and shade the region(s) corresponding to the p-value. 7. p-value = ______ 8. Test at α = 0.05. State your decision and conclusion. 9. a. Decision: why did you make this decision? b. Conclusion (write a complete sentence): c. Based on the results of your study, is there a need to investigate any of the food groups’s prices? Why or why not? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 781 KEY TERMS analysis of variance also referred to as ANOVA; a method of testing whether the means of three or more populations are equal The method is applicable if • all populations of interest are normally distributed, • • the populations have equal standard deviations, and samples (not necessarily of the same size) are randomly and independently selected from each population. The test statistic for analysis of variance is the F ratio. one-way ANOVA applicable if a method of testing whether the means of three or more populations are equal; the method is • all populations of interest are normally distributed, • • • the populations have equal standard deviations, samples (not necessarily of the same size) are randomly and independently selected from each population, and there is one independent variable and one dependent variable. The test statistic for analysis of variance is the F ratio variance mean of the squared deviations from the mean; the square of the st
andard deviation For a set of data, a deviation can be represented as x – x¯ where x is a value of the data and x¯ is the sample mean. The sample variance is equal to the sum of the squares of the deviations divided by the difference of the sample size and 1. CHAPTER REVIEW 13.1 One-Way ANOVA Analysis of variance extends the comparison of two groups to several, each a level of a categorical variable (factor). Samples from each group are independent and must be randomly selected from normal populations with equal variances. We test the null hypothesis of equal means of the response in every group versus the alternative hypothesis of one or more group means being different from the others. A one-way ANOVA hypothesis test determines if several population means are equal. The distribution for the test is the F distribution with two different degrees of freedom. Assumptions: • Each population from which a sample is taken is assumed to be normal. • All samples are randomly selected and independent. • The populations are assumed to have equal standard deviations (or variances). 13.2 The F Distribution and the F Ratio Analysis of variance compares the means of a response variable for several groups. ANOVA compares the variation within each group to the variation of the mean of each group. The ratio of these two is the F statistic from an F distribution with (number of groups – 1) as the numerator degrees of freedom and (number of observations – number of groups) as the denominator degrees of freedom. These statistics are summarized in the ANOVA table. 13.3 Facts About the F Distribution The graph of the F distribution is always positive and skewed right, though the shape can be mounded or exponential depending on the combination of numerator and denominator degrees of freedom. The F statistic is the ratio of a measure of the variation in the group means to a similar measure of the variation within the groups. If the null hypothesis is correct, then the numerator should be small compared to the denominator. A small F statistic will result, and the area under the F curve to the right will be large, representing a large p-value. When the null hypothesis of equal group means is incorrect, then the numerator should be large compared to the denominator, giving a large F statistic and a small area (small p-value) 782 Chapter 13 \| F Distribution and One-way Anova to the right of the statistic under the F curve. When the data have unequal group sizes (unbalanced data), then techniques from The F Distribution and the F Ratio need to be used for hand calculations. In the case of balanced data, where the groups are the same size, simplified calculations based on group means and variances may be used. In practice, software is usually employed in the analysis. As in any analysis, graphs of various sorts should be used in conjunction with numerical techniques. Always look at your data! 13.4 Test of Two Variances The F test for the equality of two variances rests heavily on the assumption of normal distributions. The test is unreliable if this assumption is not met. If both distributions are normal, then the ratio of the two sample variances is distributed as an F statistic, with numerator and denominator degrees of freedom that are one less than the samples sizes of the corresponding two groups. A test of two variances hypothesis test determines if two variances are the same. The distribution for the hypothesis test is the F distribution with two different degrees of freedom. Assumptions: • The populations from which the two samples are drawn are normally distributed. • The two populations are independent of each other. • k = the number of groups • nj = the size of the jth group • sj = the sum of the values in the jth group • n = the total number of all values (observations) combined • x = one value (one observation) from the data • • 2 = the variance of the sample means s x¯ = the mean of the sample variances (pooled s2 pooled variance) 13.4 Test of Two Variances F has the distribution F ~ F(n1 – 1, n2 – 1) F = 2 s1 2 σ1 2 s2 2 σ2 If σ1 = σ2, then F = s1 s2 2 2 FORMULA REVIEW 13.2 The F Distribution and the F Ratio SSbetween = ∑ (s j) ⎞ ⎠ ⎛ ⎝∑ s j n SStotal = ∑ x2 − 2 ⎞ ⎠ ⎛ ⎝∑ x n SSwithin = SStotal − SSbetween dfbetween = df(num) = k – 1 dfwithin = df(denom) = n – k MSbetween = SSbetween d fbetween MSwithin = SSwithin d fwithin F = MSbetween MSwithin F ratio when the groups are the same size: F = 2 ns x¯ s2 pooled Mean of the F distribution: µ = d f (num) d f (denom) − 1 where This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 783 PRACTICE 13.1 One-Way ANOVA Use the following information to answer the next five exercises. There are five basic assumptions that must be fulfilled to perform a one-way ANOVA test. What are they? 1. Write one assumption. 2. Write another assumption. 3. Write a third assumption. 4. Write a fourth assumption. 5. Write the final assumption. 6. State the null hypothesis for a one-way ANOVA test if there are four groups. 7. State the alternative hypothesis for a one-way ANOVA test if there are three groups. 8. When do you use an ANOVA test? 13.2 The F Distribution and the F Ratio Use the following information to answer the next seven exercises. Groups of men from three different areas of the country are to be tested for mean weight. The entries in Table 13.13 are the weights for the different groups. Group 1 Group 2 Group 3 216 198 240 187 176 202 213 284 228 210 Table 13.13 170 165 182 197 201 9. What is the sum of squares factor? 10. What is the sum of squares error? 11. What is the df for the numerator? 12. What is the df for the denominator? 13. What is the mean square factor? 14. What is the mean square error? 15. What is the F statistic? Use the following information to answer the next eight exercises. Girls from four different soccer teams are to be tested for mean goals scored per game. The entries in Table 13.14 are the goals per game for the different teams. Team 1 Team 2 Team 3 Team Table 13.14 784 Chapter 13 \| F Distribution and One-way Anova Team 1 Team 2 Team 3 Team 4 3 2 4 4 0 0 3 2 Table 13.14 16. What is SSbetween? 17. What is the df for the numerator? 18. What is MSbetween? 19. What is SSwithin? 20. What is the df for the denominator? 21. What is MSwithin? 22. What is the F statistic? 23. Judging by the F statistic, do you think it is likely or unlikely that you will reject the null hypothesis? 13.3 Facts About the F Distribution 24. An F statistic can have what values? 25. What happens to the curves as the degrees of freedom for the numerator and the denominator get larger? Use the following information to answer the next seven exercises. Four basketball teams took a random sample of players regarding how high each player can jump (in inches). The results are shown in Table 13.15. Team 1 Team 2 Team 3 Team 4 Team 5 36 42 51 32 35 38 48 50 39 38 44 46 41 39 40 Table 13.15 26. What is the df(num)? 27. What is the df(denom)? 28. What are the sum of squares and mean squares factors? 29. What are the sum of squares and mean squares errors? 30. What is the F statistic? 31. What is the p-value? 32. At the 5 percent significance level, is there a difference in the mean jump heights among the teams? Use the following information to answer the next seven exercises. A video game developer is testing a new game on three different groups. Each group represents a different target market for the game. The developer collects scores from a random sample from each group. The results are shown in Table 13.16. Group A Group B Group C 101 108 151 149 101 109 Table 13.16 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 785 Group A Group B Group C 98 107 111 160 112 126 198 186 160 Table 13.16 33. What is the df(num)? 34. What is the df(denom)? 35. What are the SSbetween and MSbetween? 36. What are the SSwithin and MSwithin? 37. What is the F Statistic? 38. What is the p-value? 39. At the 10 percent significance level, are the scores among the different groups different? Use the following information to answer the next three exercises. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. Northeast South West Central East 16.3 16.1 16.4 16.5 16.9 16.5 16.4 16.2 16.4 16.5 16.6 16.1 16.2 16.6 16.5 16.4 17.1 17.2 16.6 16.8 x¯ = s2 = ________ ________ ________ ________ ________ ________ ________ ________ ________ ________ Table 13.17 Enter the data into your calculator or computer. 40. p-value = ______ State the decisions and conclusions (in complete sentences) for the following preconceived levels of α. 41. α = 0.05 a. Decision: ____________________________ b. Conclusion: ____________________________ 42. α = 0.01 a. Decision: ____________________________ b. Conclusion: ____________________________ 13.4 Test of Two Variances Use the following information to answer the next two exercises. There are two assumptions that must be true to perform an F test of two variances. 43. Name one assumption that must be true. 786 Chapter 13 \| F Distribution and One-way Anova 44. What is the other assumption that must be true? Use the following information to answer the next seven exercises. Two coworkers commute from the same building. They are interested in whether there is any variation in the time it takes them to drive to work. They each record their times for 20 commutes. The first worker’s times have a variance of 12.1. The second worker’s times have a variance of 16.9. The first worker thinks that he is more consistent with his commute times. Test the claim at the 10 percent level. Assume that
commute times are normally distributed. 45. State the null and alternative hypotheses. 46. What is s1 in this problem? 47. What is s2 in this problem? 48. What is n? 49. What is the F statistic? 50. What is the p-value? 51. Is the claim accurate? Use the following information to answer the next four exercises. Two students are interested in whether there is variation in their test scores for math class. There are 15 total math tests they have taken so far. The first student’s grades have a standard deviation of 38.1. The second student’s grades have a standard deviation of 22.5. The second student thinks his scores are more consistent. 52. State the null and alternative hypotheses. 53. What is the F statistic? 54. What is the p-value? 55. At the 5 percent significance level, do we reject the null hypothesis? Use the following information to answer the next three exercises. Two cyclists are comparing the variances of their overall paces going uphill. Each cyclist records his or her speeds going up 35 hills. The first cyclist has a variance of 23.8, and the second cyclist has a variance of 32.1. The cyclists want to see if their variances are the same or different. Assume that speeds are normally distributed. 56. State the null and alternative hypotheses. 57. What is the F statistic? 58. At the 5 percent significance level, what can we say about the cyclists’ variances? HOMEWORK This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 787 13.1 One-Way ANOVA 59. Three different traffic routes are tested for mean driving time. The entries in the Table 13.18 are the driving times in minutes on the three different routes. Route 1 Route 2 Route 3 30 32 27 35 27 29 28 36 16 41 22 31 Table 13.18 State SSbetween, SSwithin, and the F statistic. 60. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. Northeast South West Central East 16.3 16.1 16.4 16.5 16.9 16.5 16.4 16.2 16.4 16.5 16.6 16.1 16.2 16.6 16.5 16.4 17.1 17.2 16.6 16.8 x¯ = s2 = ________ ________ ________ ________ ________ ________ ________ ________ ________ ________ Table 13.19 State the hypotheses. H0: ____________ Ha: ____________ 13.2 The F Distribution and the F Ratio Use the following information to answer the next three exercises. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. Northeast South West Central East 16.3 16.1 16.4 16.5 16.9 16.5 16.4 16.2 16.4 16.5 16.6 16.1 16.2 16.6 16.5 16.4 17.1 17.2 16.6 16.8 Table 13.20 788 Chapter 13 \| F Distribution and One-way Anova Northeast South West Central East ________ ________ ________ ________ ________ ________ ________ ________ ________ ________ x¯ = s2 = Table 13.20 H0: µ1 = µ2 = µ3 = µ4 = µ5 Hα: At least any two of the group means µ1, µ2, …, µ5 are not equal. 61. degrees of freedom – numerator: df(num) = _________ 62. degrees of freedom – denominator: df(denom) = ________ 63. F statistic = ________ 13.3 Facts About the F Distribution DIRECTIONS Use a solution sheet to conduct the following hypothesis tests. The solution sheet can be found in Appendix E. 64. Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat’s weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again, and the net gain in grams is recorded. Using a significance level of 10 percent, test the hypothesis that the three formulas produce the same mean weight gain. Linda’s Rats (g) Tuan’s Rats (g) Javier’s Rats (g) 43.5 39.4 41.3 46.0 38.2 Table 13.21 47.0 40.5 38.9 46.3 44.2 51.2 40.9 37.9 45.0 48.6 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 789 65. A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are in Table 13.22. Using a 5 percent significance level, test the hypothesis that the three mean commuting mileages are the same. Working-Class Professional (middle incomes) Professional (wealthy) 17.8 26.7 49.4 9.4 65.4 47.1 19.5 51.2 Table 13.22 16.5 17.4 22.0 7.4 9.4 2.1 6.4 13.9 8.5 6.3 4.6 12.6 11.0 28.6 15.4 9.3 Use the following information to answer the next two exercises. Table 13.23 lists the number of pages in four different types of magazines. Home Decorating News Health Computer 172 286 163 205 197 Table 13.23 87 94 123 106 101 82 153 87 103 96 104 136 98 207 146 66. Using a significance level of 5 percent, test the hypothesis that the four magazine types have the same mean length. 67. Eliminate one magazine type that you now feel has a mean length different from the others. Redo the hypothesis test, testing that the remaining three means are statistically the same. Use a new solution sheet. Based on this test, are the mean lengths for the remaining three magazines statistically the same? 790 Chapter 13 \| F Distribution and One-way Anova 68. A researcher wants to know if the mean times (in minutes) that people watch their favorite news station are the same. Suppose that Table 13.24 shows the results of a study. CNN FOX Local 72 37 56 60 51 45 12 18 38 23 35 15 43 68 50 31 22 Table 13.24 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. 69. Are the means for the final exams the same for all statistics class delivery types? Table 13.25 shows the scores on final exams from several randomly selected classes that used the different delivery types. Online Hybrid Face-to-Face 83 73 84 81 72 84 77 80 81 Table 13.25 80 78 84 81 86 79 82 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. 70. Are the mean number of times a month a person eats out the same for whites, blacks, Hispanics, and Asians? Suppose that Table 13.26 shows the results of a study. White Black Hispanic Asian 4 1 5 2 6 8 2 4 6 Table 13.26 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 791 71. Are the mean numbers of daily visitors to a ski resort the same for the three types of snow conditions? Suppose that Table 13.27 shows the results of a study. Powder Machine Made Hard Packed 1,210 1,080 1,537 941 Table 13.27 2,107 1,149 862 1,870 1,528 1,382 2,846 1,638 2,019 1,178 2,233 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. 792 Chapter 13 \| F Distribution and One-way Anova 72. Sanjay made identical paper airplanes out of three different weights of paper: light, medium, and heavy. He made four airplanes from each of the weights and launched them himself across the room. Here are the distances (in meters) that his planes flew. Paper Type/Trial Trial 1 Trial 2 Trial 3 Trial 4 Heavy Medium 5.1 meters 3.1 meters 4.7 meters 5.3 meters 4 meters 3.5 meters 4.5 meters 6.1 meters Light 3.1 meters 3.3 meters 2.1 meters 1.9 meters Table 13.28 Figure 13.8 a. Take a look at the data in the graph. Look at the spread of data for each group (light, medium, heavy). Does it seem reasonable to assume a normal distribution with the same variance for each group? b. Why is this a balanced design? c. Calculate the sample mean and sample standard deviation for each group. d. Does the weight of the paper have an effect on how far the plane will travel? Use a 1 percent level of significance. Complete the test using the method shown in the bean plant example in Example 13.4. ◦ Variance of the group means __________ ◦ MSbetween= ___________ ◦ Mean of the three sample variances ___________ ◦ MSwithin = _____________ ◦ F statistic = ____________ ◦ df(num) = __________, df(denom) = ___________ ◦ Number of groups _______ ◦ Number of observations _______ ◦ p-value = __________ (P(F > _______) = __________) ◦ Graph the p-value. ◦ Decision: _______________________ ◦ Conclusion: _______________________________________________________________ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 793 73. DDT is a pesticide that has been banned from use in the United States and most other areas of the world. It is quite effective but persisted in the environment and over time proved to be harmful to higher-level organisms. Famously, egg shells of eagles and other raptors were believed to be thinner and prone to breakage in the nest because of ingestion of DDT in the food chain of the birds. An experiment was conducted on the number of eggs (fecundity) laid by female fruit flies. There are three groups of flies. One group was bred to be resistant to DDT (the RS group)
. Another was bred to be especially susceptible to DDT (SS). The third group was a control line of nonselected or typical fruit flies (NS). Here are the data: RS SS NS RS SS NS 12.8 38.4 35.4 22.4 23.1 22.6 21.6 32.9 27.4 27.5 29.4 40.4 14.8 48.5 19.3 20.3 16 34.4 23.1 20.9 41.8 38.7 20.1 30.4 34.6 11.6 20.3 26.4 23.3 14.9 19.7 22.3 37.6 23.7 22.9 51.8 22.6 30.2 36.9 26.1 22.5 33.8 29.6 33.4 37.3 29.5 15.1 37.9 416.4 26.7 228.2 38.6 31 29.5 20.3 39 23.4 44.4 16.9 42.4 29.3 12.8 33.7 23.2 16.1 36.6 914.9 14.6 29.2 23.6 10.8 47.4 27.3 12.2 41.7 Table 13.29 The values are the average number of eggs laid daily for each of 75 flies (25 in each group) over the first 14 days of their lives. Using a 1 percent level of significance, are the mean rates of egg selection for the three strains of fruit fly different? If so, in what way? Specifically, the researchers were interested in whether the selectively bred strains were different from the nonselected line, and whether the two selected lines were different from each other. Here is a chart of the three groups: Figure 13.9 794 Chapter 13 \| F Distribution and One-way Anova 74. The data shown is the recorded body temperatures of 130 subjects as estimated from available histograms. Traditionally, we are taught that the normal human body temperature is 98.6 °F. This is not quite correct for everyone. Are the mean temperatures among the four groups different? Calculate 95 percent confidence intervals for the mean body temperature in each group and comment about the confidence intervals. FL FH ML MH FL FH ML MH 96.4 96.8 96.3 96.9 98.4 98.6 98.1 98.6 96.7 97.7 96.7 97 98.7 98.6 98.1 98.6 97.2 97.8 97.1 97.1 98.7 98.6 98.2 98.7 97.2 97.9 97.2 97.1 98.7 98.7 98.2 98.8 97.4 97.6 97.7 97.8 98 98 98 98 97.3 97.4 98.7 98.7 98.2 98.8 97.4 97.5 98.8 98.8 98.2 98.8 97.4 97.6 98.8 98.8 98.3 98.9 97.4 97.7 98.8 98.8 98.4 97.8 98.1 97.5 97.8 98.8 98.9 98.4 97.9 98.3 97.6 97.9 99.2 97.9 98.3 97.6 98 98.3 97.8 98.2 98.4 97.8 98 98 98 99.3 99 99 98.5 98.5 99.2 99.1 98.6 99.5 99.1 98.6 99 99 99 98.2 98.4 97.8 98.3 99.2 98.7 98.2 98.4 97.9 98.4 99.4 99.1 98.2 98.4 98.2 98.5 98.2 98.6 98 98 98 98.4 98.6 98.6 99.9 99.3 100 99.4 100.8 Table 13.30 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 795 13.4 Test of Two Variances 75. Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat’s weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again and the net gain in grams is recorded. Linda’s Rats Tuan’s Rats Javier’s Rats 43.5 39.4 41.3 46.0 38.2 Table 13.31 47.0 40.5 38.9 46.3 44.2 51.2 40.9 37.9 45.0 48.6 Determine whether the variance in weight gain is statistically the same between Javier’s and Linda’s rats. Test at a significance level of 10 percent. 76. A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are as follows. Working-Class Professional (middle incomes) Professional (wealthy) 17.8 26.7 49.4 9.4 65.4 47.1 19.5 51.2 Table 13.32 16.5 17.4 22.0 7.4 9.4 2.1 6.4 13.9 8.5 6.3 4.6 12.6 11.0 28.6 15.4 9.3 Determine whether the variance in mileage driven is statistically the same between the working class and professional (middle income) groups. Use a 5 percent significance level. Use the following information to answer the next two exercises. The following table lists the number of pages in four different types of magazines. Home Decorating News Health Computer 172 286 163 205 Table 13.33 87 94 123 106 82 153 87 103 104 136 98 207 796 Chapter 13 \| F Distribution and One-way Anova Home Decorating News Health Computer 197 101 96 146 Table 13.33 77. Which two magazine types do you think have the same variance in length? 78. Which two magazine types do you think have different variances in length? 79. Is the variance for the amount of money, in dollars, that shoppers spend on Saturdays at the mall the same as the variance for the amount of money that shoppers spend on Sundays at the mall? Suppose that Table 13.34 shows the results of a study. Saturday Sunday Saturday Sunday 75 18 150 94 62 73 Table 13.34 44 58 61 19 99 60 89 62 0 124 50 31 118 137 82 39 127 141 73 80. Are the variances for incomes on the East Coast and the West Coast the same? Suppose that Table 13.35 shows the results of a study. Income is shown in thousands of dollars. Assume that both distributions are normal. Use a level of significance of 0.05. East West 71 126 42 51 44 90 88 38 47 30 82 75 52 115 67 Table 13.35 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 797 81. Thirty men in college were taught a method of finger tapping. They were randomly assigned to three groups of 10, with each receiving one of three doses of caffeine: 0 mg, 100 mg, or 200 mg. This is approximately the amount in zero, one, or two cups of coffee. Two hours after ingesting the caffeine, the men had the rate of finger tapping per minute recorded. The experiment was double blind, so neither the recorders nor the students knew which group they were in. Does caffeine affect the rate of tapping, and if so how? Here are the data: 0 mg 100 mg 200 mg 0 mg 100 mg 200 mg 242 244 247 242 246 248 245 248 247 243 Table 13.36 246 250 248 246 245 245 248 248 244 242 246 247 250 246 244 248 252 250 248 250 82. King Manuel I Komnenos ruled the Byzantine Empire from Constantinople (Istanbul) during the years A.D. 1145–1170. The empire was very powerful during his reign but declined significantly afterward. Coins minted during his era were found in Cyprus, an island in the eastern Mediterranean Sea. Nine coins were from his first coinage, seven from the second, four from the third, and seven from the fourth. These spanned most of his reign. We have data on the silver content of the coins: First Coinage Second Coinage Third Coinage Fourth Coinage 6.9 9.0 6.6 8.1 9.3 9.2 8.6 4.9 5.5 4.6 4.5 5.3 5.6 5.5 5.1 6.2 5.8 5.8 5.9 6.8 6.4 7.0 6.6 7.7 7.2 6.9 6.2 Table 13.37 Did the silver content of the coins change over the course of Manuel’s reign? Here are the means and variances of each coinage. The data are unbalanced. First Second Third Fourth Mean 6.7444 8.2429 4.875 5.6143 Variance 0.2953 1.2095 0.2025 0.1314 Table 13.38 798 Chapter 13 \| F Distribution and One-way Anova 83. The American League and the National League of Major League Baseball are each divided into three divisions: East, Central, and West. Many years, fans talk about some divisions being stronger (having better teams) than other divisions. This may have consequences for the postseason. For instance, in 2012 Tampa Bay won 90 games and did not play in the postseason, while Detroit won only 88 and did play in the postseason. This may have been an oddity, but is there good evidence that in the 2012 season, the American League divisions were significantly different in overall records? Use the following data to test whether the mean number of wins per team in the three American League divisions were the same. Note that the data are not balanced, as two divisions had five teams, while one had only four. Division Team Wins East East East East East NY Yankees Baltimore Tampa Bay Toronto Boston 95 93 90 73 69 Table 13.39 Division Team Wins Central Detroit Central Chicago Sox Central Kansas City Central Cleveland Central Minnesota 88 85 72 68 66 Table 13.40 Division Team Wins West West West West Oakland Texas LA Angels Seattle 94 93 89 75 Table 13.41 REFERENCES 13.2 The F Distribution and the F Ratio Marist College School of Science. (n.d.). Tomato data (Unpublished student research). Marist College School of Science, Poughkeepsie, NY. 13.3 Facts About the F Distribution ESPN. (2012). MLB standings – 2012. Retrieved from http://espn.go.com/mlb/standings/_/year/2012. Hand, D. J. et al. (1994). A Handbook of Small Datasets: Data for Fruitfly Fecundity. London: Chapman & Hall. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 799 Hand, D. J. et al. (1994). A Handbook of Small Datasets. London: Chapman & Hall, p. 50. Hand. A Handbook of Small Datasets. p. 118. Mackowiak, P. A.,Wasserman, S. S., & Levine, M. M. (1992). A critical appraisal of 98.6 degrees F, the upper limit of the normal body temperature, and other legacies of Carl Reinhold August Wunderlich. Journal of the American Medical Association, 268, 1578–1580. Private K–12 school in San Jose, CA. (1994). Data from a fourth grade classroom. 13.4 Test of Two Variances ESPN. (2012). MLB standings – 2012. Retrieved from http://espn.go.com/mlb/standings/_/year/2012/type/vs-division/ order/true. SOLUTIONS 1 Each population from which a sample is taken is assumed to be normal. 3 The populations are assumed to have equal standard deviations (or variances). 5 The response is a numerical value. 7 Ha: At least two of the group means μ1, μ2, μ3 are not equal. 9 4,939.2 11 2 13 2,469.6 15 3.7416 17 3 19 13.2 21 0.825 23 Because a one-way ANOVA test is always right-tailed, a high F statistic corresponds to a low p value, so it is likely that we will reject the null hypothesis. 25 The curves approximate the normal distribution. 27 10 29 SS = 237.33; MS = 23.73 31 0.1614 33 two 35 SS = 5,700.4; MS = 2,850.2 37 3.6101 39 Yes, there is enough evidence to show that the scores among the groups are statistically significant at the 10 percent level. 43 The populations from which the two samples are drawn are normally distributed. 45 H0: σ1 = σ2 Ha: σ1 < σ2 or H0: σ1 47 4.11 2 = σ2 2 Ha: σ1 2 2 < σ2 49 0.7159 51 No, at the 10 p
ercent level of significance, we do not reject the null hypothesis and state that the data do not show that the variation in drive times for the first worker is less than the variation in drive times for the second worker. 53 2.8674 800 Chapter 13 \| F Distribution and One-way Anova 55 Reject the null hypothesis. There is enough evidence to say that the variance of the grades for the first student is higher than the variance in the grades for the second student. 57 0.7414 59 SSbetween = 26 SSwithin = 441 F = 0.2653 62 df(denom) = 15 64 a. H0: µL = µT = µJ b. Ha: at least any two of the means are different c. df(num) = 2; df(denom) = 12 d. F distribution e. 0.67 f. 0.5305 g. Check student’s solution. h. Decision: Do not reject null hypothesis. i. Conclusion: There is insufficient evidence to conclude that the means are different. 67 a. Ha: µc = µn = µh b. At least any two of the magazines have different mean lengths. c. df(num) = 2, df(denom) = 12 d. F distribtuion e. F = 15.28 f. p-value = 0.0005 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: There is sufficient evidence to conclude that the mean lengths of the magazines are different. 69 a. H0: μo = μh = μf b. At least two of the means are different. c. df(n) = 2, df(d) = 13 d. F2,13 e. 0.64 f. 0.5437 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: The mean scores of different class delivery are not different. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 801 71 a. H0: μp = μm = μh b. At least any two of the means are different. c. df(n) = 2, df(d) = 12 d. F2,12 e. 3.13 f. 0.0807 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: There is not sufficient evidence to conclude that the mean numbers of daily visitors are different. 73 The data appear normally distributed from the chart and of similar spread. There do not appear to be any serious outliers, so we may proceed with our ANOVA calculations, to see if we have good evidence of a difference between the three groups. H0: μ1 = μ2 = μ3 Ha: μi ≠ μj some i ≠ j Define μ1, μ2, μ3, as the population mean number of eggs laid by the three groups of fruitflies. F statistic = 8.6657 p-value = 0.0004 Figure 13.10 Decision: Since the p-value is less than the level of significance of 0.01, we reject the null hypothesis. Conclusion: We have good evidence that the average number of eggs laid during the first 14 days of life for these three strains of fruitflies are different. Interestingly, if you perform a two sample t test to compare the RS and NS groups they are significantly different (p = 0.0013). Similarly, SS and NS are significantly different (p = 0.0006). However, the two selected groups, RS and SS are not significantly different (p = 0.5176). Thus we appear to have good evidence that selection either for resistance or for susceptibility involves a reduced rate of egg production (for these specific strains) as compared to flies that were not selected for resistance or susceptibility to DDT. Here, genetic selection has apparently involved a loss of fecundity. 75 a. H0 : σ1 2 2 = σ2 2 2 ≠ σ1 b. Ha : σ1 c. df(num) = 4; df(denom) = 4 d. F4, 4 802 e. 3.00 Chapter 13 \| F Distribution and One-way Anova f. 2(0.1563) = 0.3126. Using the TI-83+/84+ function 2-SampFtest, you get the test statistic as 2.9986 and p-value directly as 0.3127. If you input the lists in a different order, you get a test statistic of 0.3335 but the p-value is the same because this is a two-tailed test. g. Check student's solution. h. Decision: Do not reject the null hypothesis. i. Conclusion: There is insufficient evidence to conclude that the variances are different. 78 The answers may vary. Sample answer: Home decorating magazines and news magazines have different variances. 80 a. H0: = σ1 2 2 = σ2 2 2 ≠ σ1 b. Ha: σ1 c. df(n) = 7, df(d) = 6 d. F7,6 e. 0.8117 f. 0.7825 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: There is not sufficient evidence to conclude that the variances are different. 82 Here is a strip chart of the silver content of the coins: Figure 13.11 While there are differences in spread, it is not unreasonable to use ANOVA techniques. Here is the completed ANOVA table: This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 \| F Distribution and One-way Anova 803 Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F Factor (between) Error (within) Total Table 13.42 37.748 11.015 48.763 4 – 1 = 3 27 – 4 = 23 27 – 1 = 26 12.5825 0.4789 26.272 P(F > 26.272) = 0. Reject the null hypothesis for any alpha. There is sufficient evidence to conclude that the mean silver content among the four coinages are different. From the strip chart, it appears that the first and second coinages had higher silver contents than the third and fourth. 83 Here is a stripchart of the number of wins for the 14 teams in the AL for the 2012 season. Figure 13.12 While the spread seems similar, there may be some question about the normality of the data, given the wide gaps in the middle near the 0.500 mark of 82 games (teams play 162 games each season in MLB). However, one-way ANOVA is robust. Here is the ANOVA table for the data: Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F Factor (between) Error (within) Total Table 13.43 344.16 1,219.55 1,563.71 3 – 1 = 2 14 – 3 = 11 14 – 1 = 13 172.08 110.87 1.5521 P(F > 1.5521) = 0.2548 Since the p-value is so large, there is not good evidence against the null hypothesis of equal means. We decline to reject the null hypothesis. Thus, for 2012, there is not any good evidence of a significant difference in mean number of wins between the divisions of the American League. 804 Chapter 13 \| F Distribution and One-way Anova This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 805 APPENDIX A: APPENDIX A REVIEW EXERCISES (CH 3–13) These review exercises are designed to provide extra practice on concepts learned before a particular chapter. For example, the review exercises for Chapter 3 cover material learned in Chapters 1 and 2. Chapter 3 Use the following information to answer the next six exercises. In a survey of 100 stocks on NASDAQ, the average percent increase for the past year was 9 percent for NASDAQ stocks. 1. The average increase for all NASDAQ stocks is the — A. population B. statistic C. parameter D. sample E. variable 2. All of the NASDAQ stocks are — A. population B. statistics C. parameter D. sample E. variable 3. Nine percent is — A. population B. statistics C. parameter D. sample E. variable 4. The 100 NASDAQ stocks in the survey are — A. population B. statistic C. parameter D. sample E. variable 806 Appendix A 5. The percent increase for one stock in the survey is — A. population B. statistic C. parameter D. sample E. variable 6. Would the data collected by qualitative, quantitative discrete, or quantitative continuous? Use the following information to answer the next two exercises. Thirty people spent two weeks around Mardi Gras in New Orleans. Their two-week weight gain is below. Note—a loss is shown by a negative weight gain. Weight Gain Frequency –2 –1 0 1 4 6 11 Table A1 3 5 2 4 13 2 1 7. Calculate the following values: A. The average weight gain for the two weeks B. The standard deviation C. The first, second, and third quartiles 8. Construct a histogram and box plot of the data. Chapter 4 Use the following information to answer the next two exercises. A recent poll concerning credit cards found that 35 percent of respondents use a credit card that gives them a mile of air travel for every dollar they charge. Thirty percent of the respondents charge more than $2,000 per month. Of those respondents who charge more than $2,000, 80 percent use a credit card that gives them a mile of air travel for every dollar they charge. 9. What is the probability that a randomly selected respondent will spend more than $2,000 and use a credit card that gives them a mile of air travel for every dollar they charge? A. B. C. D. (.30)(.35) (.80)(.35) (.80)(.30) (.80) 10. Are using a credit card that gives a mile of air travel for each dollar spent and charging more than $2,000 per month independent events? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A A. Yes B. No, and they are not mutually exclusive either C. No, but they are mutually exclusive D. Not enough information given to determine the answer 807 11. A sociologist wants to know the opinions of employed adult women about government funding for day care. She obtains a list of 520 members of a local business and professional women’s club and mails a questionnaire to 100 of these women selected at random. Sixty-eight questionnaires are returned. What is the population in this study? A. All employed adult women B. All the members of a local business and professional women’s club C. The 100 women who received the questionnaire D. All employed women with children Use the following information to answer the next two exercises. An article from the San Jose Mercury News was concerned with the racial mix of the 1,500 students at Prospect High School in Saratoga, CA. The table summarizes the results. Male and female values are approximate. Suppose one Prospect High School student is randomly selected. Gender/Ethnic Group White Asian Hispanic Black American Indian Male Female Table A2 400 440 468 132 115 140 35 40 16 14 12. Find the probability that a student is Asian or male. 13. Find the probability that a student
is black given that the student is female. 14. A sample of pounds lost, in a certain month, by individual members of a weight reducing clinic produced the following statistics: • Mean = 5 lbs • Median = 4.5 lbs • Mode = 4 lbs • Standard deviation = 3.8 lbs • First quartile = 2 lbs • Third quartile = 8.5 lbs What is the correct statement? A. One fourth of the members lost exactly two pounds. B. The middle 50 percent of the members lost from two to 8.5 lbs. C. Most people lost 3.5 to 4.5 lbs. D. All of the choices above are correct. 15. What does it mean when a data set has a standard deviation equal to zero? A. All values of the data appear with the same frequency. B. The mean of the data is also zero. C. All of the data have the same value. 808 Appendix A D. There are no data to begin with. 16. Which statement describes the illustration? Figure A1 A. The mean is equal to the median. B. There is no first quartile. C. The lowest data value is the median. D. The median equals Q1 + Q3 2 . 17. According to a recent article in the San Jose Mercury News the average number of babies born with significant hearing loss—deafness—is approximately 2 per 1,000 babies in a healthy baby nursery. The number climbs to an average of 30 per 1,000 babies in an intensive care nursery. Suppose that 1,000 babies from healthy baby nurseries were randomly surveyed. Find the probability that exactly two babies were born deaf. 18. A friend offers you the following deal: For a $10 fee, you may pick an envelope from a box containing 100 seemingly identical envelopes. However, each envelope contains a coupon for a free gift. • Ten of the coupons are for a free gift worth $6. • Eighty of the coupons are for a free gift worth $8. • Six of the coupons are for a free gift worth $12. • Four of the coupons are for a free gift worth $40. Based upon the financial gain or loss over the long run, should you play the game? A. Yes, I expect to come out ahead in money. B. No, I expect to come out behind in money. C. It doesn’t matter. I expect to break even. Use the following information to answer the next four exercises. Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that he/she truly has the flu—and not just a nasty cold—is only about 4 percent. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu. 19. Define the random variable and list its possible values. 20. State the distribution of X. 21. Find the probability that at least four of the 25 patients actually have the flu. 22. On average, for every 25 patients calling in, how many do you expect to have the flu? Use the following information to answer the next two exercises. Different types of writing can sometimes be distinguished by the number of letters in the words used. A student interested in this fact wants to study the number of letters of words used by Tom Clancy in his novels. She opens a Clancy novel at random and records the number of letters of the first 250 words on the page. 23. What kind of data was collected? A. Qualitative B. Quantitative continuous This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 809 C. Quantitative discrete 24. What is the population under study? Chapter 5 Use the following information to answer the next five exercises. A recent study of mothers of junior high school children in Santa Clara County reported that 76 percent of the mothers are employed in paid positions. Of those mothers who are employed, 64 percent work full-time—more than 35 hours per week—and 36 percent work part-time. However, out of all of the mothers in the population, 49 percent work full-time. The population under study is made up of mothers of junior high school children in Santa Clara County. Let E = employed and F = full-time employment. 25. A. Find the percent of all mothers in the population that are not employed. B. Find the percent of mothers in the population that are employed part-time. 26. The type of employment is considered to be what type of data? 27. Find the probability that a randomly selected mother works part-time given that she is employed. 28. Find the probability that a randomly selected person from the population will be employed or work full-time. 29. Being employed and working part-time— A. mutually exclusive events? Why or why not? B. independent events? Why or why not? Use the following additional information to answer the next two exercises. We randomly pick 10 mothers from the above population. We are interested in the number of the mothers that are employed. Let X = number of mothers that are employed. 30. State the distribution for X. 31. Find the probability that at least six are employed. 32. We expect the statistics discussion board to have, on average, 14 questions posted to it per week. We are interested in the number of questions posted to it per day. A. Define X. B. What are the values that the random variable may take on? C. State the distribution for X. D. Find the probability that from 10 to 14—inclusive—questions are posted to the listserv on a randomly picked day. 33. A person invests $1,000 into stock of a company that hopes to go public in one year. The probability that the person will lose all his money after one year, that is, his stock will be worthless, is 35 percent. The probability that the person’s stock will still have a value of $1,000 after one year, that is, no profit and no loss, is 60 percent. The probability that the person’s stock will increase in value by $10,000 after one year, that is, will be worth $11,000, is 5 percent. Find the expected profit after one year. 34. Rachel’s piano cost $3,000. The average cost for a piano is $4,000 with a standard deviation of $2,500. Becca’s guitar cost $550. The average cost for a guitar is $500 with a standard deviation of $200. Matt’s drums cost $600. The average cost for drums is $700 with a standard deviation of $100. Whose cost was lowest when compared to his or her own instrument? 810 Appendix A Figure A2 35. Explain why each statement is either true or false given the box plot in Figure A2. A. Twenty-five percent of the data are at most five. B. There is the same amount of data from 4–5 as there is from 5–7. C. There are no data values of three. D. Fifty percent of the data are four. Using the following information to answer the next two exercises. 64 faculty members were asked the number of cars they owned—including spouse and children’s cars. The results are given in the following graph. Figure A3 36. Find the approximate number of responses that were three. 37. Find the first, second, and third quartiles. Use them to construct a box plot of the data. Use the following information to answer the next three exercises. Table A3 shows data gathered from 15 girls on the Snow Leopard soccer team when they were asked how they liked to wear their hair. Supposed one girl from the team is randomly selected. Hair Style/Hair Color Blond Brown Black Ponytail Plain Table A3 3 2 2 2 5 1 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 811 38. Find the probability that the girl has black hair GIVEN that she wears a ponytail. 39. Find the probability that the girl wears her hair plain OR has brown hair. 40. Find the probability that the girl has blond hair AND that she wears her hair plain. Chapter 6 Use the following information to answer the next two exercises. X ~ U(3, 13) 41. Explain which of the following are false and which are true. A. f(x) = 1 10 , 3 ≤ x ≤ 13 B. There is no mode. C. The median is less than the mean. D. P(x > 10) = P(x ≤ 6) 42. Calculate A. B. C. the mean, the median, and the 65th percentile. Figure A4 43. Which of the following is true for the box plot in Figure A4? A. Twenty-five percent of the data are at most five. B. There is about the same amount of data from 4–5 as there is from 5–7. C. There are no data values of three. D. Fifty percent of the data are four. 44. If P(G\|H) = P(G), then which of the following is correct? A. G and H are mutually exclusive events. B. P(G) = P(H) C. Knowing that H has occurred will affect the chance that G will happen. D. G and H are independent events. 45. If P(J) = .3, P(K) = .63, and J and K are independent events, then explain which are correct and which are incorrect. A. P(J AND K) = 0 B. P(J OR K) = .9 C. P(J OR K) = .72 D. P(J) ≠ P(J\|K) 812 Appendix A 46. On average, five students from each high school class get full scholarships to four-year colleges. Assume that most high school classes have about 500 students. X = the number of students from a high school class that get full scholarships to four-year schools. Which of the following is the distribution of X? A. P(5) B. B(500, 5) C. Exp ⎛ ⎝ ⎞ ⎠ 1 5 D. N ⎛ ⎝5, (.01)(.99) 500 ⎞ ⎠ Chapter 7 Use the following information to answer the next three exercises. Richard’s Furniture Company delivers furniture from 10 a.m. to 2 p.m. continuously and uniformly. We are interested in how long—in hours—past the 10 a.m. start time that individuals wait for their delivery. 47. X ~ ________ A. U(0, 4) B. U(10, 20) C. Exp(2) D. N(2, 1) 48. The average wait time is — A. one hour B. C. D. two hours two and a half hours four hours 49. Suppose that it is now past noon on a delivery day. The probability that a person must wait at least 1.5 more hours is — A. B. C. D. 1 4 1 2 3 4 3 8 50. Given X ~ Exp ⎛ ⎝ ⎞ ⎠ 1 3 A. Find P(x > 1). B. Calculate the minimum value for the upper quartile. C. Find P ⎛ ⎝x = 1 3 ⎞ ⎠ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 813 Appendix A 51. • Forty percent of full-time students took four years to graduate. • Thirty percent of full-time students took five years to graduate. • Twenty percent of full-time students took six years to graduate. • Ten percent of full-time students took seven years to graduate. The expected time for full-time s
tudents to graduate is — A. B. C. D. four years four and a half years five years five and a half years 52. Which of the following distributions is described by the following example? Many people can run a short distance of under two miles, but as the distance increases, fewer people can run that far. A. binomial B. uniform C. exponential D. normal 53. The length of time to brush one’s teeth is generally thought to be exponentially distributed with a mean of 3 4 minutes. Find the probability that a randomly selected person brushes his or her teeth less than 3 4 minutes. A. .5 B. C. D. 3 4 .43 .63 54. Which distribution accurately describes the following situation? The chance that a teenage boy regularly gives his mother a kiss goodnight is about 20 percent. Fourteen teenage boys are randomly surveyed. Let X = the number of teenage boys that regularly give their mother a kiss goodnight. A. B(14,.20) B. P(2.8) C. N(2.8,2.24) D. Exp ⎛ ⎝ ⎞ ⎠ 1 .20 55. A 2008 report on technology use states that approximately 20 percent of U.S. households have never sent an email. Suppose that we select a random sample of fourteen U.S. households. Let X = the number of households in a 2008 sample of 14 households that have never sent an email. A. B(14,.20) B. P(2.8) C. N(2.8,2.24) 814 D. Exp ⎛ ⎝ ⎞ ⎠ 1 .20 Appendix A Chapter 8 Use the following information to answer the next three exercises. Suppose that a sample of 15 randomly chosen people were put on a special weight-loss diet. The amount of weight lost, in pounds, follows an unknown distribution with mean equal to 12 pounds and standard deviation equal to three pounds. Assume that the distribution for the weight loss is normal. 56. To find the probability that the mean amount of weight lost by 15 people is no more than 14 pounds, the random variable should be ________. A. number of people who lost weight on the special weight-loss diet B. C. D. the number of people who were on the diet the mean amount of weight lost by 15 people on the special weight-loss diet the total amount of weight lost by 15 people on the special weight-loss diet 57. Find the probability asked for in Question 56. 58. Find the 90th percentile for the mean amount of weight lost by 15 people. Using the following information to answer the next three exercises. The time of occurrence of the first accident during rushhour traffic at a major intersection is uniformly distributed between the three hour interval 4 p.m. to 7 p.m. Let X = the amount of time—hours—it takes for the first accident to occur. 59. What is the probability that the time of occurrence is within the first half-hour or the last hour of the period from 4 to 7 p.m.? A. It cannot be determined from the information given. B. C. D. 1 6 1 2 1 3 60. The 20th percentile occurs after how many hours? A. B. C. .20 .60 .50 D. 1 61. Assume Ramon has kept track of the times for the first accidents to occur for 40 different days. Let C = the total cumulative time. Then C follows which distribution? A. U(0,3) B. Exp(13) C. N(60, 5.477) D. N(1.5, .01875) 62. Using the information in Question 61, find the probability that the total time for all first accidents to occur is more than 43 hours. Use the following information to answer the next two exercises. The length of time a parent must wait for his children to This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 815 clean their rooms is uniformly distributed in the time interval from one to 15 days. 63. How long must a parent expect to wait for his children to clean their rooms? A. 8 days B. 3 days C. 14 days D. 6 days 64. What is the probability that a parent will wait more than six days given that the parent has already waited more than three days? A. B. C. D. .5174 .0174 .7500 .2143 Use the following information to answer the next five exercises. Twenty percent of the students at a local community college live in within five miles of the campus. Thirty percent of the students at the same community college receive some kind of financial aid. Of those who live within five miles of the campus, 75 percent receive some kind of financial aid. 65. Find the probability that a randomly chosen student at the local community college does not live within five miles of the campus. A. 80 percent B. 20 percent C. 30 percent D. Cannot be determined 66. Find the probability that a randomly chosen student at the local community college lives within five miles of the campus or receives some kind of financial aid. A. 50 percent B. 35 percent C. 27.5 percent D. 75 percent 67. Are living in student housing within five miles of the campus and receiving some kind of financial aid mutually exclusive? A. Yes B. No C. Cannot be determined 68. The interest rate charged on the financial aid is ________ data. A. Quantitative discrete B. Quantitative continuous C. Qualitative discrete D. Qualitative 816 Appendix A 69. The following information is about the students who receive financial aid at the local community college. • 1st quartile = $250 • 2nd quartile = $700 • 3rd quartile = $1,200 These amounts are for the school year. If a sample of 200 students is taken, how many are expected to receive $250 or more? A. 50 B. 250 C. 150 D. Cannot be determined Use the following information to answer the next two exercises. P(A) = .2, P(B) = .3; A and B are independent events. 70. P(A AND B) = — A. B. .5 .6 C. 0 D. .06 71. P(A OR B) = — A. B. C. .56 .5 .44 D. 1 72. If H and D are mutually exclusive events, P(H) = .25, P(D) = .15, then P(H\|D). A. 1 B. 0 C. D. .40 .0375 Chapter 9 73. Rebecca and Matt are 14 year old twins. Matt’s height is two standard deviations below the mean for 14 year old boys’ height. Rebecca’s height is .10 standard deviations above the mean for 14 year old girls’ height. Interpret this. A. Matt is 2.1 inches shorter than Rebecca. B. Rebecca is very tall compared to other 14 year old girls. C. Rebecca is taller than Matt. D. Matt is shorter than the average 14 year old boy. 74. Construct a histogram of the IPO data (see Appendix C). Use the following information to answer the next three exercises. Ninety homeowners were asked the number of estimates they obtained before having their homes fumigated. Let X = the number of estimates. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 817 x Relative Frequency Cumulative Relative Frequency 1 2 4 5 .3 .2 .4 .1 Table A4 75. Complete the cumulative frequency column. 76. Calculate the sample mean (a), the sample standard deviation (b), and the percent of the estimates that fall at or below four (c). 77. Calculate the median, M, the first quartile, Q1, and the third quartile Q3. Then construct a box plot of the data. 78. The middle 50 percent of the data are between ________ and ________. Use the following information to answer the next three exercises. Seventy fifth and sixth graders were asked their favorite dinner. Pizza Hamburgers Spaghetti Fried Shrimp 5th Grader 15 6th Grader 15 6 7 Table A5 9 10 0 8 79. Find the probability that one randomly chosen child is in the 6th grade and prefers fried shrimp. A. B. C. D. 32 70 8 32 8 8 8 70 80. Find the probability that a child does not prefer pizza. A. B. C. 30 70 30 40 40 70 D. 1 81. Find the probability a child is in the fifth grade given that the child prefers spaghetti. A. B. 9 19 9 70 Appendix A 818 C. D. 9 30 19 70 82. A sample of convenience is a random sample. A. True B. False 83. A statistic is a number that is a property of the population. A. True B. False 84. You should always throw out any data that are outliers. A. True B. False 85. Lee bakes pies for a small restaurant in Felton, CA. She generally bakes 20 pies in a day, on average. Of interest is the number of pies she bakes each day. A. Define the random variable X. B. State the distribution for X. C. Find the probability that Lee bakes more than 25 pies in any given day. 86. Six different brands of Italian salad dressing were randomly selected at a supermarket. The grams of fat per serving are 7, 7, 9, 6, 8, and 5. Assume that the underlying distribution is normal. Calculate a 95 percent confidence interval for the population mean grams of fat per serving of Italian salad dressing sold in supermarkets. 87. Given: uniform, exponential, normal distributions. Match each to a statement below. A. mean = median ≠ mode B. mean > median > mode C. mean = median = mode Chapter 10 Use the following information to answer the next three exercises. In a survey at Kirkwood Ski Resort the following information was recorded. 0–10 11–20 21–40 40+ Ski 10 Snowboard 6 12 17 30 12 8 5 Table A6 Suppose that one person from Table A6 was randomly selected. 88. Find the probability that the person was a skier or was age 11–20. 89. Find the probability that the person was a snowboarder given he or she was age 21–40. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 819 90. Explain which of the following are true and which are false. A. Sport and age are independent events. B. Ski and age 11–20 are mutually exclusive events. C. P(Ski AND age 21–40) < P(Ski\|age 21–40) D. P(Snowboard OR age 0–10) < P(Snowboard\|age 0–10) 91. The average length of time a person with a broken leg wears a cast is approximately six weeks. The standard deviation is about three weeks. Thirty people who had recently healed from broken legs were interviewed. State the distribution that most accurately reflects total time to heal for the 30 people. 92. The distribution for X is uniform. What can we say for certain about the distribution for X ¯ when n = 1? ¯ A. The distribution for X ¯ B. The distribution for X ¯ C. The distribution for X ¯ D. The distribution for X is still uniform with the same mean and standard deviation as the distribution for X. is normal with the different mean and a different standard deviation as the distribution for X. is normal with the same mean but a larger standard deviation tha
n the distribution for X. is normal with the same mean but a smaller standard deviation than the distribution for X. 93. The distribution for X is uniform. What can we say for certain about the distribution for ∑ X when n = 50? A. The distribution for ∑ X is still uniform with the same mean and standard deviation as the distribution for X. B. The distribution for ∑ X is normal with the same mean but a larger standard deviation as the distribution for X. C. The distribution for ∑ X is normal with a larger mean and a larger standard deviation than the distribution for X. D. The distribution for ∑ X is normal with the same mean but a smaller standard deviation than the distribution for X. Use the following information to answer the next three exercises. A group of students measured the lengths of all the carrots in a five-pound bag of baby carrots. They calculated the average length of baby carrots to be 2.0 inches with a standard deviation of 0.25 inches. Suppose we randomly survey 16 five-pound bags of baby carrots. 94. State the approximate distribution for X ¯ ~ ________. X ¯ , the distribution for the average lengths of baby carrots in 16 five-pound bags. 95. Explain why we cannot find the probability that one individual randomly chosen carrot is greater than 2.25 inches. is between 2.0 and 2.25 inches. 96. Find the probability that x¯ Use the following information to answer the next three exercises. At the beginning of the term, the amount of time a student waits in line at the campus store is normally distributed with a mean of five minutes and a standard deviation of two minutes. 97. Find the 90th percentile of waiting time in minutes. 98. Find the median waiting time for one student. 99. Find the probability that the average waiting time for 40 students is at least 4.5 minutes. Chapter 11 Use the following information to answer the next the time that owners keep their cars—purchased new—is normally distributed with a mean of seven years and a standard deviation of two years. We are interested in how long an individual keeps his car—purchased new. Our population is people who buy their cars new. four exercises. Suppose that 820 Appendix A 100. Sixty percent of individuals keep their cars at most how many years? 101. Suppose that we randomly survey one person. Find the probability that person keeps his or her car less than 2.5 years. 102. If we are to pick individuals 10 at a time, find the distribution for the mean car length ownership. 103. If we are to pick 10 individuals, find the probability that the sum of their ownership time is more than 55 years. 104. For which distribution is the median not equal to the mean? A. Uniform B. Exponential C. Normal D. Student t 105. Compare the standard normal distribution to the Student’s t distribution, centered at zero. Explain which of the following are true and which are false. A. As the number surveyed increases, the area to the left of –1 for the Student’s t distribution approaches the area for the standard normal distribution. B. As the degrees of freedom decrease, the graph of the Student’s t distribution looks more like the graph of the standard normal distribution. C. If the number surveyed is 15, the normal distribution should never be used. Use the following information to answer the next five exercises. We are interested in the checking account balance of 24-old college students. We randomly survey 16 20-year-old college students. We obtain a sample mean of $640 and a sample standard deviation of $150. Let X = checking account balance of an individual 20-year-old college student. 106. Explain why we cannot determine the distribution of X. 107. If you were to create a confidence interval or perform a hypothesis test for the population mean checking account balance of 20-year-old college students, what distribution would you use? 108. Find the 95 percent confidence interval for the true mean checking account balance of a 20-year-old college student. 109. What type of data is the balance of the checking account considered to be? 110. What type of data is the number of 20-year-olds considered to be? 111. On average, a busy emergency room gets a patient with a shotgun wound about once per week. We are interested in the number of patients with a shotgun wound the emergency room gets per 28 days. A. Define the random variable X. B. State the distribution for X. C. Find the probability that the emergency room gets no patients with shotgun wounds in the next 28 days. Use the following information to answer the next two exercises. The probability that a certain slot machine will pay back money when a quarter is inserted is .30. Assume that each play of the slot machine is independent from each other. A person puts in 15 quarters for 15 plays. 112. Is the expected number of plays of the slot machine that will pay back money greater than, less than, or the same as the median? Explain your answer. 113. Is it likely that exactly eight of the 15 plays would pay back money? Justify your answer numerically. 114. A game is played with the following rules: • It costs $10 to enter. • A fair coin is tossed four times. • • If you do not get four heads or four tails, you lose your $10. If you get four heads or four tails, you get back your $10, plus $30 more. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 821 Over the long run of playing this game, what are your expected earnings? 115. • The mean grade on a math exam in Rachel’s class was 74, with a standard deviation of five. Rachel earned an 80. • The mean grade on a math exam in Becca’s class was 47, with a standard deviation of two. Becca earned a 51. • The mean grade on a math exam in Matt’s class was 70, with a standard deviation of eight. Matt earned an 83. Find whose score was the best, compared to his or her own class. Justify your answer numerically. Use the following information to answer the next two exercises. A random sample of 70 compulsive gamblers were asked the number of days they go to casinos per week. The results are given in the following graph. Figure A5 116. Find the number of responses that were five. 117. Find the mean, standard deviation, the median, the first quartile, the third quartile, and the IQR. 118. Based upon research at De Anza College, it is believed that about 19 percent of the student population speaks a language other than English at home. Suppose that a study was done this year to see if that percent has decreased. Ninetyeight students were randomly surveyed with the following results: Fourteen said that they speak a language other than English at home. A. State an appropriate null hypothesis. B. State an appropriate alternative hypothesis. C. Define the random variable, P′. D. Calculate the test statistic. E. Calculate the p-value. F. At the 5 percent level of decision, what is your decision about the null hypothesis? G. What is the Type I error? H. What is the Type II error? 119. Assume that you are an emergency paramedic called in to rescue victims of an accident. You need to help a patient who is bleeding profusely. The patient is also considered to be a high risk for contracting a blood-borne illness. Assume that the null hypothesis is that the patient does not have the a blood-borne illness. What is a Type I error? 120. It is often said that Californians are more casual than the rest of Americans. Suppose that a survey was done to see if the proportion of Californian professionals that wear jeans to work is greater than the proportion of non-Californian professionals. Fifty of each was surveyed with the following results: Fifteen Californians wear jeans to work and six non- 822 Appendix A Californians wear jeans to work. Let C = Californian professional; NC = non-Californian professional A. State appropriate null and alternate hypotheses. B. Define the random variable. C. Calculate the test statistic and p-value. D. At the 5 percent significance level, what is your decision? E. What is the Type I error? F. What is the Type II error? Use the following information to answer the next two exercises. A group of statistics students have developed a technique that they feel will lower their anxiety level on statistics exams. They measured their anxiety level at the start of the quarter and again at the end of the quarter. Recorded is the paired data in that order: (1,000, 900); (1,200, 1,050); (600, 700); (1,300, 1,100); (1,000, 900); (900, 900). 121. This is a test of (pick the best answer) — A. B. large samples, and independent means small samples, and independent means C. dependent means 122. State the distribution to use for the test. Chapter 12 Use the following information to answer the next two exercises. A recent survey of U.S. teenagers was answered by 720 teenagers, age 15–18. Six percent of teenagers surveyed said they are planning on going to college in another country. We are interested in the true proportion of U.S. teens, ages 15–18, who are planning on going to college in another country. 123. Find the 95 percent confidence interval for the true proportion of U.S. teens, ages 15–19, who are planning to go to college in another country. 124. The report also stated that the results of the survey are accurate to within ±3.7 percent at the 95 percent confidence level. Suppose that a new study is to be done. It is desired to be accurate to within 2 percent of the 95 percent confidence level. What is the minimum number that should be surveyed? 125. Given X ~ Exp ⎛ ⎝ ⎞ ⎠ 1 3 . Sketch the graph that depicts: P(x > 1). Use the following information to answer the next three exercises. The amount of money a customer spends in one trip to the supermarket is known to have an exponential distribution. Suppose the mean amount of money a customer spends in one trip to the supermarket is $72. 126. Find the probability that one customer spends less than $72 in one trip to the supermarket? 127. Suppose five customers pool their money. How much
money altogether would you expect the five customers to spend in one trip to the supermarket in dollars? 128. State the distribution to use if you want to find the probability that the mean amount spent by five customers in one trip to the supermarket is less than $60. Chapter 13 Use the following information to answer the next two exercises. Suppose that the probability of a drought in any independent year is 20 percent. Out of those years in which a drought occurs, the probability of water rationing is 10 percent. However, in any year, the probability of water rationing is 5 percent. 129. What is the probability of both a drought and water rationing occurring? 130. Out of the years with water rationing, find the probability that there is a drought. Use the following information to answer the next three exercises. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 823 Apple Pumpkin Pecan Female 40 Male 20 Table A7 10 30 30 10 131. Suppose that one individual is randomly chosen. Find the probability that the person’s favorite pie is apple or the person is male. 132. Suppose that one male is randomly chosen. Find the probability his favorite pie is pecan. 133. Conduct a hypothesis test to determine if favorite pie type and gender are independent. Use the following information to answer the next two exercises. Let’s say that the probability that an adult watches the news at least once per week is .60. 134. We randomly survey 14 people. On average, how many people do we expect to watch the news at least once per week? 135. We randomly survey 14 people. Of interest is the number that watch the news at least once per week. State the distribution of X. X ~ ________. 136. The following histogram is most likely to be a result of sampling from which distribution? Figure A6 A. Chi-square B. Geometric C. Uniform D. Binomial 137. The ages of De Anza evening students is known to be normally distributed with a population mean of 40 and a population standard deviation of six. A sample of six De Anza evening students reported their ages in years as: 28; 35; 47; 45; 30; 50. Find the probability that the mean of six ages of randomly chosen students is less than 35 years. Hint—Find the sample mean. 138. A math exam was given to all the fifth grade children attending Country School. Two random samples of scores were taken. The null hypothesis is that the mean math scores for boys and girls in fifth grade are the same. Conduct a hypothesis test. 824 Appendix A n ¯ x Boys 55 82 Girls 60 86 Table A8 s2 29 46 139. In a survey of 80 males, 55 had played an organized sport growing up. Of the 70 females surveyed, 25 had played an organized sport growing up. We are interested in whether the proportion for males is higher than the proportion for females. Conduct a hypothesis test. 140. Which of the following is preferable when designing a hypothesis test? A. Maximize α and minimize β B. Minimize α and maximize β C. Maximize α and β D. Minimize α and β Use the following information to answer the next three exercises. One hundred twenty people were surveyed as to their favorite beverage. The results are below. Beverage/Age 0–9 10–19 20–29 30+ Totals Milk Soda Juice Totals Table A9 14 3 7 10 8 12 24 330 6 26 12 44 0 15 7 22 30 52 38 120 141. Are the events of milk and 30+— A. independent events? Justify your answer. B. mutually exclusive events? Justify your answer. 142. Suppose that one person is randomly chosen. Find the probability that person is 10–19 given that he or she prefers juice. 143. Are Preferred Beverage and Age independent events? Conduct a hypothesis test. 144. Given the following histogram, which distribution is the data most likely to come from? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 825 Figure A7 A. Uniform B. Exponential C. Normal D. Chi-square Solutions Chapter 3 1. C Parameter 2. A Population 3. B Statistic 4. D Sample 5. E Variable 6. quantitative continuous 7. A. 2.27 B. 3.04 C. –1, 4, 4 8. Answers will vary. Chapter 4 9. C (.80)(.30) 10. B No, and they are not mutually exclusive either. 11. A All employed adult women 12. .5773 13. .0522 14. B The middle fifty percent of the members lost from 2 to 8.5 lbs. 15. C All of the data have the same value. 16. C The lowest data value is the median. 17. .279 826 Appendix A 18. B No, I expect to come out behind in money. 19. X = the number of patients calling in claiming to have the flu, who actually have the flu. X = 0, 1, 2, …25 20. B(25, .04) 21. .0165 22. 1 23. C Quantitative discrete 24. all words used by Tom Clancy in his novels Chapter 5 25. A. 24 percent B. 27 percent 26. qualitative 27. .36 28. .7636 29. A. no B. no 30. B(10, .76) 31. .9330 32. A. X = the number of questions posted to the statistics listserv per day. B. X = 0, 1, 2,… C. X ~ P(2) D. 0 33. $150 34. Matt 35. A. False B. True C. False D. False 36. 16 37. first quartile: 2 second quartile: 2 third quartile: 3 38. 0.5 39. 7 15 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 827 Appendix A 40. 2 15 Chapter 6 41. A. True B. True C. False – the median and the mean are the same for this symmetric distribution. D. True 42. A. 8 B. 8 C. P(x < k) = 0.65 = (k – 3) ⎛ ⎝ ⎞ ⎠ 1 10 . k = 9.5 43. A. False – 3 4 of the data are at most five. B. True – each quartile has 25 percent of the data. C. False – that is unknown. D. False – 50 percent of the data are four or less. 44. D G and H are independent events. 45. A. False – J and K are independent so they are not mutually exclusive which would imply dependency (meaning P(J AND K) is not 0). B. False – see answer c. C. True – P(J OR K) = P(J) + P(K) – P(J AND K) = P(J) + P(K) – P(J)P(K) = .3 + .6 – (.3)(.6) = .72. Note the P(J AND K) = P(J)P(K) because J and K are independent. D. False – J and K are independent so P(J) = P(J\|K). 46. A P(5) Chapter 7 47. A U(0, 4) 48. B 2 hours 49. A 1 4 50. A. .7165 B. 4.16 C. 0 51. C 5 years 828 52. C exponential 53. .63 54. A B(14, .20) 55. A B(14, .20) Chapter 8 56. C The mean amount of weight lost by 15 people on the special weight-loss diet. Appendix A 57. .9951 58. 12.99 59. C 1 2 60. B .60 61. C N(60, 5.477) 62. .9990 63. A eight days 64. C .7500 65. A 80 percent 66. B 35 percent 67. B no 68. B Quantitative continuous 69. C 150 70. D .06 71. C .44 72. B 0 Chapter 9 73. D Matt is shorter than the average 14 year old boy. 74. Answers will vary. 75. x Relative Frequency Cumulative Relative Frequency 1 2 4 5 .3 .2 .4 .1 Table A10 .3 .2 .4 .1 76. A. 2.8 B. 1.48 C. 90 percent 77. M = 3; Q1 = 1; Q3 = 4 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 829 Appendix A 78. 1 and 4 79. D 8 70 80. C 40 70 81. A 9 19 82. B False 83. B False 84. B False 85. A. X = the number of pies Lee bakes every day. B. P(20) C. .1122 86. CI: (5.25, 8.48) 87. A. uniform B. exponential C. normal Chapter 10 88. 77 100 89. 12 42 90. A. False B. False C. True D. False 91. N(180, 16.43) ¯ 92. A The distribution for X is still uniform with the same mean and standard deviation as the distribution for X. 93. C The distribution for ∑ X is normal with a larger mean and a larger standard deviation than the distribution for X. 94. N ⎛ ⎝2, .25 16 ⎞ ⎠ 95. Answers will vary. 96. .5000 97. 7.6 98. 5 99. .9431 Appendix A 830 Chapter 11 100. 7.5 101. .0122 102. N(7, .63) 103. .9911 104. B exponential 105. A. True B. False C. False 106. Answers will vary. 107. Student’s t with df = 15 108. (560.07, 719.93) 109. quantitative continuous data 110. quantitative discrete data 111. A. X = the number of patients with a shotgun wound the emergency room gets per 28 days. B. P(4) C. .0183 112. greater than 113. no; P(x = 8) = .0348 114. You will lose $5. 115. Becca 116. 14 117. sample mean = 3.2 sample standard deviation = 1.85 median = 3 Q1 = 2 Q3 = 5 IQR = 3 118. d. z = –1.19 e. .1171 f. Do not reject the null hypothesis. 119. We conclude that the patient does have the illness when, in fact, the patient does not. 120. c. z = 2.21; p = .0136 d. Reject the null hypothesis. e. We conclude that the proportion of Californian professionals that wear jeans to work is greater than the proportion of non-Californian professionals when, in fact, it is not greater. f. We cannot conclude that the proportion of Californian professionals that wear jeans to work is greater than the proportion of non-Californian professionals when, in fact, it is greater. 121. C dependent means 122. t5 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 831 Appendix A Chapter 12 123. (.0424, .0770) 124. 2,401 125. Check student's solution. 126. .6321 127. $360 128. N ⎛ ⎝72, ⎞ ⎠ 72 5 Chapter 13 129. .02 130. .40 131. 100 140 132. 10 60 133. p-value = 0; reject the null hypothesis; conclude that they are dependent events 134. 8.4 135. B(14, .60) 136. D Binomial 137. .3669 138. p-value = .0006; reject the null hypothesis; conclude that the averages are not equal 139. p-value = 0; reject the null hypothesis; conclude that the proportion of males is higher 140. minimize α and β 141. A. no B. yes, P(M AND 30+) = 0 142. 12 38 143. no; p-value = 0 144. A uniform References Baran, D. (2010). Twenty percent of Americans have never used email. Retrieved from http://www.webguild.org/20080519/ 20-percent-of-americans-have-never-used-email. Parade Magazine. (n.d.). Retrieved from https://parade.com/. San Jose Mercury News. (n.d.). Retrieved from http://www.mercurynews.com/. 832 Appendix A This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 833 APPENDIX B: APPENDIX B PRACTICE TESTS (1–4) AND FINAL EXAMS Practice Test 1 1.1: Definitions of Statistics, Probability, and Key Terms Use the following information to answer the next three exercises. A grocery store is interested in how much money, on average, their customers spend each visit in the produce department. Using their store records, they draw a sample of 1,000 visi
ts and calculate each customer’s average spending on produce. 1. Identify the population, sample, parameter, statistic, variable, and data for this example. A. population B. sample C. parameter D. statistic E. variable F. data 2. What kind of data is amount of money spent on produce per visit? A. Qualitative B. Quantitative-continuous C. Quantitative-discrete 3. The study finds that the mean amount spent on produce per visit by the customers in the sample is $12.84. This is an example of a A. Population B. Sample C. Parameter D. Statistic E. Variable 1.2: Data, Sampling, and Variation in Data and Sampling Use the following information to answer the next two exercises. A health club is interested in knowing how many times a typical member uses the club in a week. They decide to ask every tenth customer on a specified day to complete a short survey, including information about how many times they have visited the club in the past week. 4. What kind of a sampling design is this? A. Cluster B. Stratified 834 C. Simple random D. Systematic 5. Number of visits per week is what kind of data? A. Qualitative B. Quantitative-continuous C. Quantitative-discrete Appendix B 6. Describe a situation in which you would calculate a parameter, rather than a statistic. 7. The U.S. federal government conducts a survey of high school seniors concerning their plans for future education and employment. One question asks whether they are planning to attend a four-year college or university in the following year. Fifty percent answer yes to this question. That 50 percent is a A. Parameter B. Statistic C. Variable D. Data 8. Imagine that the U.S. federal government had the means to survey all high school seniors in the United States concerning their plans for future education and employment, and found that 50 percent were planning to attend a four-year college or university in the following year. This 50 percent is an example of a A. Parameter B. Dtatistic C. Variable D. Data Use the following information to answer the next three exercises. A survey of a random sample of 100 nurses working at a large hospital asked how many years they had been working in the profession. Their answers are summarized in the following (incomplete) table. 9. Fill in the blanks in the table and round your answers to two decimal places for the Relative Frequency and Cumulative Relative Frequency cells. # of years Frequency Relative Frequency Cumulative Relative Frequency < 5 5–10 > 10 Table B1 25 30 empty 10. What proportion of nurses have five or more years of experience? 11. What proportion of nurses have 10 or fewer years of experience? 12. Describe how you might draw a random sample of 30 students from a lecture class of 200 students. 13. Describe how you might draw a stratified sample of students from a college, where the strata are the students’ class standing (freshman, sophomore, junior, or senior). 14. A manager wants to draw a sample, without replacement, of 30 employees from a workforce of 150. Describe how the chance of being selected will change over the course of drawing the sample. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 835 15. The manager of a department store decides to measure employee satisfaction by selecting four departments at random, and conducting interviews with all the employees in those four departments. What type of survey design is this? A. Cluster B. Stratified C. Simple random D. Systematic 16. A popular American television sports program conducts a poll of viewers to see which team they believe will win the National Football League (NFL) championship this year. Viewers vote by calling a number displayed on the television screen and telling the operator which team they think will win. Do you think that those who participate in this poll are representative of all football fans in America? 17. Two researchers studying vaccination rates independently draw samples of 50 children, aged three–18 months, from a large urban area, and determine if they are up to date on their vaccinations. One researcher finds that 84 percent of the children in her sample are up to date, and the other finds that 86 percent in his sample are up to date. Assuming both followed proper sampling procedures and did their calculations correctly, what is a likely explanation for this discrepancy? 18. A high school increased the length of the school day from 6.5 to 7.5 hours. Students who wished to attend this high school were required to sign contracts pledging to put forth their best effort on their school work and to obey the school rules; if they did not wish to do so, they could attend another high school in the district. At the end of one year, student performance on statewide tests had increased by 10 percentage points over the previous year. Does this prove that a longer school day improves student achievement? 19. You read a newspaper article reporting that eating almonds leads to increased life satisfaction. The study was conducted by the Almond Growers Association, and was based on a randomized survey asking people about their consumption of various foods, including almonds, and also about their satisfaction with different aspects of their life. Does anything about this poll lead you to question its conclusion? 20. Why is non-response a problem in surveys? 1.3: Frequency, Frequency Tables, and Levels of Measurement 21. Compute the mean of the following numbers, and report your answer using one more decimal place than is present in the original data: 14, 5, 18, 23, 6 1.4: Experimental Design and Ethics 22. A psychologist is interested in whether the size of tableware (bowls, plates, etc.) influences how much college students eat. He randomly assigns 100 college students to one of two groups. The first is served a meal using normal-sized tableware, while the second is served the same meal but using tableware that it 20 percent smaller than normal. He records how much food is consumed by each group. Identify the following components of this study. A. population B. sample C. experimental units D. explanatory variable E. F. treatment response variable 23. A researcher analyzes the results of the Scholastic Aptitude Test (SAT) over a five-year period and finds that male students on average score higher on the math section, and female students on average score higher on the verbal section. She concludes that these observed differences in test performance are due to genetic factors. Explain how lurking variables could offer an alternative explanation for the observed differences in test scores. 24. Explain why it would not be possible to use random assignment to study the health effects of exercise. 25. A professor conducts a telephone survey of a city’s population by drawing a sample of numbers from the phone book and having her student assistants call each of the selected numbers once to administer the survey. What are some sources of 836 bias with this survey? Appendix B 26. A professor offers extra credit to students who take part in her research studies. What is an ethical problem with this method of recruiting subjects? 2.1: Stem-and Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs Use the following information to answer the next four exercises. The midterm grades on a chemistry exam, graded on a scale of 0 to 100, were 62, 64, 65, 65, 68, 70, 72, 72, 74, 75, 75, 75, 76, 78, 78, 81, 83, 83, 84, 85, 87, 88, 92, 95, 98, 98, 100, 100, 740 27. Do you see any outliers in this data? If so, how would you address the situation? 28. Construct a stem plot for this data, using only the values in the range zero–100. 29. Describe the distribution of exam scores. 2.2: Histograms, Frequency Polygons, and Time Series Graphs 30. In a class of 35 students, seven students received scores in the 70–79 range. What is the relative frequency of scores in this range? Use the following information to answer the next three exercises. You conduct a poll of 30 students to see how many classes they are taking this term. Your results are 1; 1; 1; 1 2; 2; 2; 2; 2 3; 3; 3; 3; 3; 3; 3; 3 4; 4; 4; 4; 4; 4; 4; 4; 4 5; 5; 5; 5 31. You decide to construct a histogram of this data. What will be the range of your first bar, and what will be the central point? 32. What will be the widths and central points of the other bars? 33. Which bar in this histogram will be the tallest, and what will be its height? 34. You get data from the U.S. Census Bureau on the median household income for your city, and decide to display it graphically. Which is the better choice for this data, a bar graph or a histogram? 35. You collect data on the color of cars driven by students in your statistics class, and want to display this information graphically. Which is the better choice for this data, a bar graph or a histogram? 2.3: Measures of the Location of the Data 36. Your daughter brings home test scores showing that she scored in the 80th percentile in math and the 76th percentile in reading for her grade. Interpret these scores. 37. You have to wait 90 minutes in the emergency room of a hospital before you can see a doctor. You learn that your wait time was in the 82nd percentile of all wait times. Explain what this means, and whether you think it is good or bad. 2.4: Box Plots Use the following information to answer the next three exercises. 1; 1; 2; 3; 4; 4; 5; 5; 6; 7; 7; 8; 9 38. What is the median for this data? 39. What is the first quartile for this data? 40. What is the third quartile for this data? Use the following information to answer the next four exercises. This box plot represents scores on the final exam for a physics class. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 837 Figure B1 41. What is the median for this data, and how do you know? 42. What are the first and third quartiles for this data, and how do you know? 43. What is the interquartile range for this data? 44. What is the range fo
r this data? 2.5: Measures of the Center of the Data 45. In a marathon, the median finishing time was 3:35:04 (three hours, 35 minutes, and four seconds). You finished in 3:34:10. Interpret the meaning of the median time, and discuss your time in relation to it. Use the following information to answer the next three exercises. The values, in thousands of dollars, for houses on a block, are 45; 47; 47.5; 51; 53.5; 125. 46. Calculate the mean for this data. 47. Calculate the median for this data. 48. Which do you think better reflects the average value of the homes on this block? 2.6: Skewness and the Mean, Median, and Mode 49. In a left-skewed distribution, which is greater? A. The mean B. The media C. The mode 50. In a right-skewed distribution, which is greater? A. The mean B. The median C. The mode 51. In a symmetrical distribution, what will be the relationship among the mean, median, and mode? 2.7: Measures of the Spread of the Data Use the following information to answer the next four exercises. 10; 11; 15; 15; 17; 22 52. Compute the mean and standard deviation for this data; use the sample formula for the standard deviation. 53. What number is two standard deviations above the mean of this data? 54. Express the number 13.7 in terms of the mean and standard deviation of this data. 55. In a biology class, the scores on the final exam were normally distributed, with a mean of 85 and a standard deviation of five. Susan got a final exam score of 95. Express her exam result as a z score, and interpret its meaning. 3.1: Terminology Use the following information to answer the next two exercises. You have a jar full of marbles: 50 are red, 25 are blue, and 15 are yellow. Assume you draw one marble at random for each trial and replace it before the next trial. Let P(R) = the probability of drawing a red marble. Let P(B) = the probability of drawing a blue marble. 838 Appendix B Let P(Y) = the probability of drawing a yellow marble. 56. Find P(B). 57. Which is more likely, drawing a red marble or a yellow marble? Justify your answer numerically. Use the following information to answer the next two exercises. The following are probabilities describing a group of college students. Let P(M) = the probability that the student is male Let P(F) = the probability that the student is female Let P(E) = the probability the student is majoring in education Let P(S) = the probability the student is majoring in science 58. Write the symbols for the probability that a student, selected at random, is both female and a science major. 59. Write the symbols for the probability that the student is an education major, given that the student is male. 3.2: Independent and Mutually Exclusive Events 60. Events A and B are independent. If P(A) = 0.3 and P(B) = 0.5, find P(A AND B). 61. C and D are mutually exclusive events. If P(C) = 0.18 and P(D) = 0.03, find P(C OR D). 3.3: Two Basic Rules of Probability 62. In a high school graduating class of 300, 200 students are going to college, 40 are planning to work full-time, and 80 are taking a gap year. Are these events mutually exclusive? Use the following information to answer the next two exercises. An archer hits the center of the target (the bullseye) 70 percent of the time. However, she is a streak shooter, and if she hits the center on one shot, her probability of hitting it on the shot immediately following is 0.85. Written in probability notation P(A) = P(B) = P(hitting the center on one shot) = 0.70 P(B\|A) = P(hitting the center on a second shot, given that she hit it on the first) = 0.85 63. Calculate the probability that she will hit the center of the target on two consecutive shots. 64. Are P(A) and P(B) independent in this example? 3.4: Contingency Tables Use the following information to answer the next three exercises. The following contingency table displays the number of students who report studying at least 15 hours per week, and how many made the honor roll in the past semester. Honor Roll No Honor Roll Total Study at least 15 hours/week Study less than 15 hours/week 125 200 193 Total Table B2 1,000 65. Complete the table. 66. Find P (honor roll\|study at least 15 hours per week). 67. What is the probability a student studies less than 15 hours per week? 68. Are the events study at least 15 hours per week and makes the honor roll independent? Justify your answer numerically. 3.5: Tree and Venn Diagrams 69. At a high school, some students play on the tennis team and some play on the soccer team, but neither plays both tennis and soccer. Draw a Venn diagram illustrating this. 70. At a high school, some students play tennis, some play soccer, and some play both. Draw a Venn diagram illustrating this. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 839 Practice Test 1 Solutions 1.1: Definitions of Statistics, Probability, and Key Terms 1. A. population: all the shopping visits by all the store’s customers B. sample: the 1,000 visits drawn for the study C. parameter: the average expenditure on produce per visit by all the store’s customers D. statistic: the average expenditure on produce per visit by the sample of 1,000 E. variable: the expenditure on produce for each visit F. data: the dollar amounts spent on produce; for instance, $15.40, $11.53, etc. 2. C 3. D 1.2: Data, Sampling, and Variation in Data and Sampling 4. D 5. C 6. Answers will vary. Sample Answer: Any solution in which you use data from the entire population is acceptable. For instance, a professor might calculate the average exam score for her class: Because the scores of all members of the class were used in the calculation, the average is a parameter. 7. B 8. A 9. 10. 0.75 11. 0.55 # of years Frequency Relative Frequency Cumulative Relative Frequency < 5 5–10 > 10 Table B3 25 30 45 0.25 0.30 0.45 0.25 0.55 1 12. Answers will vary. Sample Answer: One possibility is to obtain the class roster and assign each student a number from 1 to 200. Then, use a random number generator or table of random number to generate 30 numbers between 1 and 200, and select the students matching the random numbers. It would also be acceptable to write each student’s name on a card, shuffle them in a box, and draw 30 names at random. 13. One possibility would be to obtain a roster of students enrolled in the college, including the class standing for each student. Then, you would draw a proportionate random sample from within each class. For instance, if 30 percent of the students in the college are freshman, then 30 percent of your sample would be drawn from the freshman class. 14. For the first person picked, the chance of any individual being selected is one in 150. For the second person, it is one in 149, for the third it is one in 148, and so on. For the 30th person selected, the chance of selection is one in 121. 15. A 16. No. There are at least two chances for bias. First, the viewers of this particular program may not be representative of American football fans as a whole. Second, the sample will be self-selected, because people have to make a phone call in order to take part, and those people are probably not representative of the American football fan population as a whole. 840 Appendix B 17. These results (84 percent in one sample, 86 percent in the other) are probably due to sampling variability. Each researcher drew a different sample of children, and you would not expect them to get exactly the same result, although you would expect the results to be similar, as they are in this case. 18. No. The improvement could also be due to self-selection: Only motivated students were willing to sign the contract, and they would have done well even in a school with 6.5 hour days. Because both changes were implemented at the same time, it is not possible to separate out their influence. 19. At least two aspects of this poll are troublesome. The first is that it was conducted by a group who would benefit by the result—almond sales are likely to increase if people believe that eating almonds will make them happier. The second is that this poll found that almond consumption and life satisfaction are correlated, but it does not establish that eating almonds causes satisfaction. It is equally possible, for instance, that people with higher incomes are more likely to eat almonds and are also more satisfied with their lives. 20. You want the sample of people who take part in a survey to be representative of the population from which they are drawn. People who refuse to take part in a survey often have different views than those who do participate, and so even a random sample may produce biased results if a large percentage of those selected refuse to participate in a survey. 1.3: Frequency, Frequency Tables, and Levels of Measurement 21. 13.2 1.4: Experimental Design and Ethics 22. A. population: all college students B. sample: the 100 college students in the study C. experimental units: each individual college student who participated D. explanatory variable: the size of the tableware E. F. treatment: tableware that is 20 percent smaller than normal response variable: the amount of food eaten 23. There are many lurking variables that could influence the observed differences in test scores. Perhaps the boys, on average, have taken more math courses than the girls, and the girls have taken more English classes than the boys. Perhaps the boys have been encouraged by their families and teachers to prepare for a career in math and science, and thus have put more effort into studying math, while the girls have been encouraged to prepare for fields like communication and psychology that are more focused on language use. A study design would have to control for these and other potential lurking variables (anything that could explain the observed difference in test scores, other than the genetic explanation) in order to draw a scientifically sound conclusion about genetic differences. 24. To use random assi
gnment, you would have to be able to assign people to either exercise or not exercise. Because exercise has many beneficial effects, this would not be an ethical experiment. We will study people who chose to exercise and compare them to people who chose not to exercise, and try to control for the other ways those two groups may differ (lurking variables). 25. Sources of bias include the fact that not everyone has a telephone, that cell phone numbers are often not listed in published directories, and that an individual might not be at home at the time of the phone call; all these factors make it likely that the respondents to the survey will not be representative of the population as a whole. 26. Research subjects should not be coerced into participation, and offering extra credit in exchange for participation could be construed as coercion. In addition, this method will result in a volunteer sample, which cannot be assumed to be representative of the population as a whole. 2.1: Stem-and Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs 27. The value 740 is an outlier, because the exams were graded on a scale of zero to 100, and 740 is far outside that range. It may be a data entry error, with the actual score being 74, so the professor should check that exam again to see what the actual score was. 28. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 841 Stem Leaf 10 0 0 Table B4 29. Most scores on this exam were in the range of 70–89, with a few scoring in the 60–69 range, and a few in the 90–100 range. 2.2: Histograms, Frequency Polygons, and Time Series Graphs 30. RF = 7 35 = 0.2 31. The range will be 0.5–1.5, and the central point will be 1. 32. Range 1.5–2.5, central point 2; range 2.5–3.5, central point 3; range 3.5–4.5, central point 4; range 4.5–5.5, central point 5. 33. The bar from 3.5 to 4.5, with a central point of 4, will be tallest; its height will be nine, because there are nine students taking four courses. 34. The histogram is a better choice, because income is a continuous variable. 35. A bar graph is the better choice, because this data is categorical rather than continuous. 2.3: Measures of the Location of the Data 36. Your daughter scored better than 80 percent of the students in her grade on math and better than 76 percent of the students in reading. Both scores are very good, and place her in the upper quartile, but her math score is slightly better in relation to her peers than her reading score. 37. You had an unusually long wait time, which is bad: 82 percent of patients had a shorter wait time than you, and only 18 percent had a longer wait time. 2.4: Box Plots 38. 5 39. 3 40. 7 41. The median is 86, as represented by the vertical line in the box. 42. The first quartile is 80, and the third quartile is 92, as represented by the left and right boundaries of the box. 43. IQR = 92 – 80 = 12 44. Range = 100 – 75 = 25 2.5: Measures of the Center of the Data 45. Half the runners who finished the marathon ran a time faster than 3:35:04, and half ran a time slower than 3:35:04. Your time is faster than the median time, so you did better than more than half of the runners in this race. 46. 61.5, or $61,500 47. 49.25, or $49,250 48. The median, because the mean is distorted by the high value of one house. 2.6: Skewness and the Mean, Median, and Mode 49. C Appendix B 842 50. A 51. They will all be fairly close to one another. 2.7: Measures of the Spread of the Data 52. Mean: 15 Standard deviation: 4.3 μ = 10 + 11 + 15 + 15 + 17 + 22 6 = 15 2 ∑ ⎛ ⎝x − x¯ ⎞ ⎠ n − 1 s = = 94 5 = 4.3 53. 15 + (2)(4.3) = 23.6 54. 13.7 is one standard deviation below the mean of this data, because 15 – 4.3 = 10.7 55. z = 95 − 85 5 = 2.0 Susan’s z score was 2.0, meaning she scored two standard deviations above the class mean for the final exam. 3.1: Terminology 56. P(B) = 25 90 = 0.28 57. Drawing a red marble is more likely. P(R) = 50 80 P(Y) = 15 80 = 0.19 = 0.62 58. P(F AND S) 59. P(E\|M) 3.2: Independent and Mutually Exclusive Events 60. P(A AND B) = (0.3)(0.5) = 0.15 61. P(C OR D) = 0.18 + 0.03 = 0.21 3.3: Two Basic Rules of Probability 62. No, they cannot be mutually exclusive, because they add up to more than 300. Therefore, some students must fit into two or more categories (e.g., both going to college and working full time). 63. P(A and B) = (P(B\|A))(P(A)) = (0.85)(0.70) = 0.595 64. No. If they were independent, P(B) would be the same as P(B\|A). We know this is not the case, because P(B) = 0.70 and P(B\|A) = 0.85. 3.4: Contingency Tables 65. Honor roll No honor roll Total Study at least 15 hours/week 482 Study less than 15 hours/week 125 Total Table B5 607 200 193 393 682 318 1,000 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 843 66. P(honor roll\|study at least 15 hours word per week) = 482 1,000 = 0.482 67. P(study less than 15 hours word per week) = 125 + 193 1,000 = 0.318 68. Let P(S) = study at least 15 hours per week Let P(H) = make the honor roll From the table, P(S) = 0.682, P(H) = 0.607, and P(S AND H) = 0.482. If P(S) and P(H) were independent, then P(S AND H) would equal (P(S))(P(H)). However, (P(S))(P(H)) = (0.682)(0.607) = 0.414, while P(S AND H) = 0.482. Therefore, P(S) and P(H) are not independent. 3.5: Tree and Venn Diagrams 69. Figure B2 70. Figure B3 Practice Test 2 4.1: Probability Distribution Function (PDF) for a Discrete Random Variable Use the following information to answer the next five exercises. You conduct a survey among a random sample of students at a particular university. The data collected includes their major, the number of classes they took the previous semester, and the amount of money they spent on books purchased for classes in the previous semester. 844 Appendix B 1. If X = student’s major, then what is the domain of X? 2. If Y = the number of classes taken in the previous semester, what is the domain of Y? 3. If Z = the amount of money spent on books in the previous semester, what is the domain of Z? 4. Why are X, Y, and Z in the previous example random variables? 5. After collecting data, you find that, for one case, z = –7. Is this a possible value for Z? 6. What are the two essential characteristics of a discrete probability distribution? Use this discrete probability distribution represented in this table to answer the following six questions. The university library records the number of books checked out by each patron over the course of one day, with the following result: x P(x) 0 1 2 3 4 0.20 0.45 0.20 0.10 0.05 Table B6 7. Define the random variable X for this example. 8. What is P(x > 2)? 9. What is the probability a patron will check out at least one book? 10. What is the probability a patron will take out no more than three books? 11. If the table listed P(x) as 0.15, how would you know that there was a mistake? 12. What is the average number of books taken out by a patron? 4.2: Mean or Expected Value and Standard Deviation Use the following information to answer the next four exercises. Three jobs are open in a company: one in the accounting department, one in the human resources department, and one in the sales department. The accounting job receives 30 applicants, and the human resources and sales department 60 applicants. 13. If X = the number of applications for a job, use this information to fill in Table B7. x P(x) xP(x) Table B7 14. What is the mean number of applicants? 15. What is the PDF for X? 16. Add a fourth column to the table, for (x – μ)2P(x). 17. What is the standard deviation of X? 4.3: Binomial Distribution 18. In a binomial experiment, if p = 0.65, what does q equal? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 845 19. What are the required characteristics of a binomial experiment? 20. Joe conducts an experiment to see how many times he has to flip a coin before he gets four heads in a row. Does this qualify as a binomial experiment? Use the following information to answer the next three exercises. In a particular community, 65 percent of households include at least one person who has graduated from college. You randomly sample 100 households in this community. Let X = the number of households including at least one college graduate. 21. Describe the probability distribution of X. 22. What is the mean of X? 23. What is the standard deviation of X? Use the following information to answer the next four exercises. Joe is the star of his school’s baseball team. His batting average is 0.400, meaning that for every 10 times he comes to bat (an at-bat), four of those times he gets a hit. You decide to track his batting performance for his next 20 at-bats. 24. Define the random variable X in this experiment. 25. Assuming Joe’s probability of getting a hit is independent and identical across all 20 at-bats, describe the distribution of X. 26. Given this information, what number of hits do you predict Joe will get? 27. What is the standard deviation of X? 4.4: Geometric Distribution 28. What are the three major characteristics of a geometric experiment? 29. You decide to conduct a geometric experiment by flipping a coin until it comes up heads. This takes five trials. Represent the outcomes of this trial, using H for heads and T for tails. 30. You are conducting a geometric experiment by drawing cards from a normal 52-card pack, with replacement, until you draw the Queen of Hearts. What is the domain of X for this experiment? 31. You are conducting a geometric experiment by drawing cards from a normal 52-card deck, without replacement, until you draw a red card. What is the domain of X for this experiment? Use the following information to answer the next three exercises. In a particular university, 27 percent of students are engineering majors. You decide to select students at random until you choose one that is an engineering major. Let X = the number of students you select until you find one that is an engineering m
ajor. 32. What is the probability distribution of X? 33. What is the mean of X? 34. What is the standard deviation of X? 4.5: Hypergeometric Distribution 35. You draw a random sample of 10 students to participate in a survey, from a group of 30, consisting of 16 boys and 14 girls. You are interested in the probability that seven of the students chosen will be boys. Does this qualify as a hypergeometric experiment? List the conditions and whether or not they are met. 36. You draw five cards, without replacement, from a normal 52-card deck of playing cards, and are interested in the probability that two of the cards are spades. What are the group of interest, size of the group of interest, and sample size for this example? 4.6: Poisson Distribution 37. What are the key characteristics of the Poisson distribution? Use the following information to answer the next three exercises. The number of drivers to arrive at a toll booth in an hour can be modeled by the Poisson distribution. 38. If X = the number of drivers, and the average numbers of drivers per hour is four, how would you express this distribution? 39. What is the domain of X? 40. What are the mean and standard deviation of X? 846 Appendix B 5.1: Continuous Probability Functions 41. You conduct a survey of students to see how many books they purchased the previous semester, the total amount they paid for those books, the number they sold after the semester was over, and the amount of money they received for the books they sold. Which variables in this survey are discrete, and which are continuous? 42. With continuous random variables, we never calculate the probability that X has a particular value, but we always speak in terms of the probability that X has a value within a particular range. Why is this? 43. For a continuous random variable, why are P(x < c) and P(x ≤ c) equivalent statements? 44. For a continuous probability function, P(x < 5) = 0.35. What is P(x > 5), and how do you know? 45. Describe how you would draw the continuous probability distribution described by the function f (x) = 1 10 for 0 ≤ x ≤ 10. What type of a distribution is this? 46. For the continuous probability distribution described by the function f (x) = 1 10 for 0 ≤ x ≤ 10. what is the P(0 < x < 4)? 5.2: The Uniform Distribution 47. For the continuous probability distribution described by the function f (x) = 1 10 for 0 ≤ x ≤ 10, what is the P(2 < x < 5)? Use the following information to answer the next four exercises. The number of minutes that a patient waits at a medical clinic to see a doctor is represented by a uniform distribution between zero and 30 minutes, inclusive. 48. If X equals the number of minutes a person waits, what is the distribution of X? 49. Write the probability density function for this distribution. 50. What is the mean and standard deviation for waiting time? 51. What is the probability that a patient waits less than 10 minutes? 5.3: The Exponential Distribution 52. The distribution of the variable X, representing the average time to failure for an automobile battery, can be written as X ~ Exp(m). Describe this distribution in words. 53. If the value of m for an exponential distribution is 10, what are the mean and standard deviation for the distribution? 54. Write the probability density function for a variable distributed as X ~ Exp(0.2). 6.1: The Standard Normal Distribution 55. Translate this statement about the distribution of a random variable X into words: X ~ (100, 15). 56. If the variable X has the standard normal distribution, express this symbolically. Use the following information for the next six exercises. According to the World Health Organization, distribution of height in centimeters for girls aged five years and zero months has the distribution X ~ N(109, 4.5). 57. What is the z score for a height of 112 inches? 58. What is the z score for a height of 100 centimeters? 59. Find the z score for a height of 105 centimeters and explain what that means in the context of the population. 60. What height corresponds to a z score of 1.5 in this population? 61. Using the empirical rule, we expect about 68 percent of the values in a normal distribution to lie within one standard deviation above or below the mean. What does this mean, in terms of a specific range of values, for this distribution? 62. Using the empirical rule, about what percentage of heights in this distribution do you expect to be between 95.5 cm and 122.5 cm? 6.2: Using the Normal Distribution Use the following information to answer the next four exercises. The distributor of raffle tickets claims that 20 percent of the tickets are winners. You draw a sample of 500 tickets to test this proposition. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 847 63. Can you use the normal approximation to the binomial for your calculations? Why or why not. 64. What are the expected mean and standard deviation for your sample, assuming the distributor’s claim is true? 65. What is the probability that your sample will have a mean greater than 100? 66. If the z score for your sample result is –2, explain what this means, using the empirical rule. 7.1: The Central Limit Theorem for Sample Means (Averages) 67. What does the central limit theorem state with regard to the distribution of sample means? 68. The distribution of results from flipping a fair coin is uniform: Heads and tails are equally likely on any flip, and over a large number of trials, you expect about the same number of heads and tails. Yet if you conduct a study by flipping 30 coins and recording the number of heads, and repeat this 100 times, the distribution of the mean number of heads will be approximately normal. How is this possible? 69. The mean of a normally-distributed population is 50, and the standard deviation is four. If you draw 100 samples of size 40 from this population, describe what you would expect to see in terms of the sampling distribution of the sample mean. 70. X is a random variable with a mean of 25 and a standard deviation of two. Write the distribution for the sample mean of samples of size 100 drawn from this population. 71. Your friend is doing an experiment drawing samples of size 50 from a population with a mean of 117 and a standard deviation of 16. This sample size is large enough to allow use of the central limit theorem, so he says the standard deviation of the sampling distribution of sample means will also be 16. Explain why this is wrong, and calculate the correct value. 72. You are reading a research article that refers to the standard error of the mean. What does this mean, and how is it calculated? Use the following information to answer the next six exercises. You repeatedly draw samples of n = 100 from a population with a mean of 75 and a standard deviation of 4.5. 73. What is the expected distribution of the sample means? 74. One of your friends tries to convince you that the standard error of the mean should be 4.5. Explain what error your friend made. 75. What is the z score for a sample mean of 76? 76. What is the z score for a sample mean of 74.7? 77. What sample mean corresponds to a z score of 1.5? 78. If you decrease the sample size to 50, will the standard error of the mean be smaller or larger? What would be its value? Use the following information to answer the next two questions. We use the empirical rule to analyze data for samples of size 60 drawn from a population with a mean of 70 and a standard deviation of 9. 79. What range of values would you expect to include 68 percent of the sample means? 80. If you increased the sample size to 100, what range would you expect to contain 68 percent of the sample means, applying the empirical rule? 7.2: The Central Limit Theorem for Sums 81. How does the central limit theorem apply to sums of random variables? 82. Explain how the rules applying the central limit theorem to sample means, and to sums of a random variable, are similar. 83. If you repeatedly draw samples of size 50 from a population with a mean of 80 and a standard deviation of four, and calculate the sum of each sample, what is the expected distribution of these sums? Use the following information to answer the next four exercises. You draw one sample of size 40 from a population with a mean of 125 and a standard deviation of seven. 84. Compute the sum. What is the probability that the sum for your sample will be less than 5,000? 85. If you drew samples of this size repeatedly, computing the sum each time, what range of values would you expect to contain 95 percent of the sample sums? 86. What value is one standard deviation below the mean? 87. What value corresponds to a z score of 2.2? 848 Appendix B 7.3: Using the Central Limit Theorem 88. What does the law of large numbers say about the relationship between the sample mean and the population mean? 89. Applying the law of large numbers, which sample mean would you expect to be closer to the population mean: a sample of size 10 or a sample of size 100? Use this information for the next three questions. A manufacturer makes screws with a mean diameter of 0.15 cm (centimeters) and a range of 0.10 cm to 0.20 cm; within that range, the distribution is uniform. 90. If X = the diameter of one screw, what is the distribution of X? 91. Suppose you repeatedly draw samples of size 100 and calculate their mean. Applying the central limit theorem, what is the distribution of these sample means? 92. Suppose you repeatedly draw samples of 60 and calculate their sum. Applying the central limit theorem, what is the distribution of these sample sums? Practice Test 2 Solutions Probability Distribution Function (PDF) for a Discrete Random Variable 1. The domain of X = {English, Mathematics, . . .}, i.e., a list of all the majors offered at the university, plus undeclared. 2. The domain of Y = {0, 1, 2, . . .}; i.e., the integers from zero to the upper limit of classes allowed by the university
. 3. The domain of Z = any amount of money from zero upwards. 4. Because they can take any value within their domain, and their value for any particular case is not known until the survey is completed. 5. No, because the domain of Z includes only positive numbers (you cannot spend a negative amount of money). Possibly the value –7 is a data entry error, or a special code to indicate that the student did not answer the question. 6. The probabilities must sum to 1.0, and the probabilities of each event must be between 0 and 1, inclusive. 7. Let X = the number of books checked out by a patron. 8. P(x > 2) = 0.10 + 0.05 = 0.15 9. P(x ≥ 0) = 1 – 0.20 = 0.80 10. P(x ≤ 3) = 1 – 0.05 = 0.95 11. The probabilities would sum to 1.10, and the total probability in a distribution must always equal 1.0. 12. x¯ = 0(0.20) + 1(0.45) + 2(0.20) + 3(0.10) + 4(0.05) = 1.35 Mean or Expected Value and Standard Deviation 13. x P(x) xP(x) 30 0.33 9.90 40 0.33 13.20 60 0.33 19.80 Table B8 14. x¯ = 9.90 + 13.20 + 19.80 = 42.90 15. P(x = 30) = 0.33 P(x = 40) = 0.33 P(x = 60) = 0.33 16. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 849 x P(x) xP(x) (x – μ)2P(x) 30 0.33 9.90 (30 – 42.90)2(0.33) = 54.91 40 0.33 13.20 (40 – 42.90)2(0.33) = 2.78 60 0.33 19.90 (60 – 42.90)2(0.33) = 96.49 Table B9 17. σ x = 54.91 + 2.78 + 96.49 = 12.42 Binomial Distribution 18. q = 1 – 0.65 = 0.35 19. 1. There are a fixed number of trials. 2. There are only two possible outcomes, and they add up to one. 3. The trials are independent and conducted under identical conditions. 20. No, because there are not a fixed number of trials 21. X ~ B(100, 0.65) 22. μ = np = 100(0.65) = 65 23. σ x = npq = 100(0.65)(0.35) = 4.77 24. X = Joe gets a hit in one at-bat (in one occasion of his coming to bat) 25. X ~ B(20, 0.4) 26. μ = np = 20(0.4) = 8 27. σ x = npq = 20(0.40)(0.60) = 2.19 4.4: Geometric Distribution 28. 1. A series of Bernoulli trials are conducted until one is a success, and then the experiment stops. 2. At least one trial is conducted, but there is no upper limit to the number of trials. 3. The probability of success or failure is the same for each trial. 29. T T T T H 30. The domain of X = {1, 2, 3, 4, 5, . . . n}. Because you are drawing with replacement, there is no upper bound to the number of draws that may be necessary. 31. The domain of X = {1, 2, 3, 4, 5, 6, 7, 8., 9, 10, 11, 12, . . . 27}. Because you are drawing without replacement, and 26 of the 52 cards are red, you have to draw a red card within the first 17 draws. 32. X ~ G(0.24) 33. μ = 1 p = 1 0.27 = 3.70 34. σ = 1 − p p2 = 1 − 0.27 0.272 = 3.16 4.5: Hypergeometric Distribution 35. Yes, because you are sampling from a population composed of two groups (boys and girls), have a group of interest (boys), and are sampling without replacement (hence, the probabilities change with each pick, and you are not performing Bernoulli trials). 850 Appendix B 36. The group of interest is the cards that are spades, the size of the group of interest is 13, and the sample size is five. 4.6: Poisson Distribution 37. A Poisson distribution models the number of events occurring in a fixed interval of time or space, when the events are independent and the average rate of the events is known. 38. X ~ P(4) 39. The domain of X = {0, 1, 2, 3, . . .}; i.e., any integer from 0 upwards. 40.1: Continuous Probability Functions 41. The discrete variables are the number of books purchased, and the number of books sold after the end of the semester. The continuous variables are the amount of money spent for the books, and the amount of money received when they were sold. 42. Because for a continuous random variable, P(x = c) = 0, where c is any single value. Instead, we calculate P(c < x < d); i.e., the probability that the value of x is between the values c and d. 43. Because P(x = c) = 0 for any continuous random variable. 44. P(x > 5) = 1 – 0.35 = 0.65, because the total probability of a continuous probability function is always 1. 45. This is a uniform probability distribution. You would draw it as a rectangle with the vertical sides at 0 and 20, and the horizontal sides at 1 10 and 0. ⎛ 46. P(0 < x < 4) = (4 − 0) ⎝ ⎞ ⎠ = 0.4 1 10 5.2: The Uniform Distribution ⎛ 47. P(2 < x < 5) = (5 − 2) ⎝ ⎞ ⎠ = 0.3 1 10 48. X ~ U(0, 15) 49. f (x) = 1 b − a for (a ≤ x ≤ b) so f (x) = 1 30 for (0 ≤ x ≤ 30) 50. μ = a + b 2 = 0 + 30 5 = 15.0 σ = (b − a)2 12 = (30 − 0)2 12 = 8.66 ⎛ 51. P(x < 10) = (10) ⎝ ⎞ ⎠ = 0.33 1 30 5.3: The Exponential Distribution 52. X has an exponential distribution with decay parameter m and mean and standard deviation 1 m . In this distribution, there will be relatively large numbers of small values, with values becoming less common as they become larger. 53. μ = σ = 1 m = 1 10 54. f(x) = 0.2e–0.2x where x ≥ 0. = 0.1 6.1: The Standard Normal Distribution 55. The random variable X has a normal distribution with a mean of 100 and a standard deviation of 15. 56. X ~ N(0,1) This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 851 57. z = 58 so z = 112 − 109 4.5 = 0.67 so z = 100 − 109 4.5 = − 2.00 59. z = 105 − 109 4.5 = −0.89 This girl is shorter than average for her age, by 0.89 standard deviations. 60. 109 + (1.5)(4.5) = 115.75 cm 61. We expect about 68 percent of the heights of girls aged five years and zero months to be between 104.5 cm and 113.5 cm. 62. We expect 99.7 percent of the heights in this distribution to be between 95.5 cm and 122.5 cm, because that range represents the values three standard deviations above and below the mean. 6.2: Using the Normal Distribution 63. Yes, because both np and nq are greater than five. np = (500)(0.20) = 100 and nq = 500(0.80) = 400 64. μ = np = (500)(0.20) = 100 σ = npq = 500(0.20)(0.80) = 8.94 65. Fifty percent, because in a normal distribution, half the values lie above the mean. 66. The results of our sample were two standard deviations below the mean, suggesting it is unlikely that 20 percent of the raffle tickets are winners, as claimed by the distributor, and that the true percentage of winners is lower. Applying the Empirical Rule, if that claim were true, we would expect to see a result this far below the mean only about 2.5 percent of the time. 7.1: The Central Limit Theorem for Sample Means (Averages) 67. The central limit theorem states that if samples of sufficient size are drawn from a population, the distribution of sample means will be normal, even if the distribution of the population is not normal. 68. The sample size of 30 is sufficiently large in this example to apply the central limit theorem. This theorem states that, for samples of sufficient size drawn from a population, the sampling distribution of the sample mean will approach normality, regardless of the distribution of the population from which the samples were drawn. 69. You would not expect each sample to have a mean of 50, because of sampling variability. However, you would expect the sampling distribution of the sample means to cluster around 50, with an approximately normal distribution, so that values close to 50 are more common than values further removed from 50. ¯ 70. X ¯ ∼ N(25, 0.2) because X ∼ N ⎛ ⎝μ x, σ x n ⎞ ⎠ 71. The standard deviation of the sampling distribution of the sample means can be calculated using the formula ⎛ ⎝ σ x n ⎞ ⎠ , which in this case is ⎛ ⎝ ⎞ ⎠ 16 50 is therefore 2.26. . The correct value for the standard deviation of the sampling distribution of the sample means 72. The standard error of the mean is another name for the standard deviation of the sampling distribution of the sample ⎞ mean. Given samples of size n drawn from a population with standard deviation σx, the standard error of the mean is ⎛ ⎠ . ⎝ σ x n 73. X ~ N(75, 0.45) 74. Your friend forgot to divide the standard deviation by the square root of n. 75. z = x¯ − μ x σ x = 76 − 75 4.5 = 2.2 852 Appendix B 76. z = x¯ − μ x σ x = 74.7 − 75 4.5 = −0.67 77. 75 + (1.5)(0.45) = 75.675 78. The standard error of the mean will be larger, because you will be dividing by a smaller number. The standard error of the mean for samples of size n = 50 is ⎛ ⎝ = 0.64 σ x n ⎞ ⎠ = 4.5 50 79. You would expect this range to include values up to one standard deviation above or below the mean of the sample means. In this case: 70 + 9 60 = 68.84 so you would expect 68 percent of the sample means to be between 68.84 and = 71.16 and 70 − 9 60 71.16. 80. 70 + 9 = 70.9 and 70 − 9 = 69.1 so you would expect 68 percent of the sample means to be between 69.1 100 and 70.9. Note that this is a narrower interval due to the increased sample size. 100 7.2: The Central Limit Theorem for Sums 81. For a random variable X, the random variable ΣX will tend to become normally distributed as the size n of the samples used to compute the sum increases. 82. Both rules state that the distribution of a quantity (the mean or the sum) calculated on samples drawn from a population will tend to have a normal distribution as the sample size increases, regardless of the distribution of population from which the samples are drawn. 83. ΣX ∼ N ⎛ ⎝nμ x, ( n)(σ x)⎞ ⎠ so ΣX ∼ N(4,000, 28.3) 84. The probability is 0.50, because 5,000 is the mean of the sampling distribution of sums of size 40 from this population. Sums of random variables computed from a sample of sufficient size are normally distributed, and in a normal distribution, half the values lie below the mean. 85. Using the empirical rule, you would expect 95 percent of the values to be within two standard deviations of the mean. Using the formula for the standard deviation is for a sample sum ( n)(σ x) = ⎛ ⎠(7) = 44.3, so you would expect 95 percent of the values to be between 5,000 + (2)(44.3) and 5,000 – (2)(44.3), or between 4,911.4 and 588.6. ⎝ 40⎞ 86. μ − ( n)(σ x) = 5,000 − ⎛ ⎝ 40⎞ ⎠(7) = 4,955.7 87. 5,000 + (2.2)⎛ ⎝ 40⎞ ⎠(7) = 5097.4 7.3: Using the Central Limit Theorem 88. The law of large numbers says that, as sa
mple size increases, the sample mean tends to get nearer and nearer to the population mean. 89. You would expect the mean from a sample of size 100 to be nearer to the population mean, because the law of large numbers says that, as sample size increases, the sample mean tends to approach the population mean. 90. X ~ N(0.10, 0.20) ¯ 91. X ∼ N ⎛ ⎝μ x, σ x n ⎞ ⎠ and the standard deviation of a uniform distribution is b − a 12 . In this example, the standard deviation of the distribution is b − a 12 = 0.10 12 = 0.03 ¯ so X ∼ N(0.15, 0.003) 92. ΣX ∼ N((n)(μ x), ( n)(σ x)) so ΣX ∼ N(9.0, 0.23) This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 853 Practice Test 3 8.1: Confidence Interval, Single Population Mean, Population Standard Deviation Known, Normal Use the following information to answer the next seven exercises. You draw a sample of size 30 from a normally distributed population with a standard deviation of four. 1. What is the standard error of the sample mean in this scenario, rounded to two decimal places? 2. What is the distribution of the sample mean? 3. If you want to construct a two-sided 95 percent confidence interval, how much probability will be in each tail of the distribution? 4. What is the appropriate z score and error bound or margin of error (EBM) for a 95 percent confidence interval for this data? 5. Rounding to two decimal places, what is the 95 percent confidence interval if the sample mean is 41? 6. What is the 90 percent confidence interval if the sample mean is 41? Round to two decimal places 7. Suppose the sample size in this study had been 50, rather than 30. What would the 95 percent confidence interval be if the sample mean is 41? Round your answer to two decimal places. 8. For any given data set and sampling situation, which would you expect to be wider: a 95 percent confidence interval or a 99 percent confidence interval? 8.2: Confidence Interval, Single Population Mean, Standard Deviation Unknown, Student’s t 9. Comparing graphs of the standard normal distribution (z distribution) and a t distribution with 15 degrees of freedom (df), how do they differ? 10. Comparing graphs of the standard normal distribution (z distribution) and a t distribution with 15 degrees of freedom (df), how are they similar? Use the following information to answer the next five exercises. Body temperature is known to be distributed normally among healthy adults. Because you do not know the population standard deviation, you use the t distribution to study body temperature. You collect data from a random sample of 20 healthy adults and find that your sample temperatures have a mean of 98.4 and a sample standard deviation of 0.3 (both in degrees Fahrenheit). 11. What are the degrees of freedom (df) for this study? 12. For a two-tailed 95 percent confidence interval, what is the appropriate t value to use in the formula? 13. What is the 95 percent confidence interval? 14. What is the 99 percent confidence interval? Round to two decimal places. 15. Suppose your sample size had been 30 rather than 20. What would the 95 percent confidence interval be then? Round to two decimal places 8.3: Confidence Interval for a Population Proportion Use this information to answer the next four exercises. You conduct a poll of 500 randomly selected city residents, asking them if they own an automobile. Of the respondents, 280 say they own an automobile, and 220 say they do not. 16. Find the sample proportion and sample standard deviation for this data. 17. What is the 95 percent two-sided confidence interval? Round to four decimal places. 18. Calculate the 90 percent confidence interval. Round to four decimal places. 19. Calculate the 99 percent confidence interval. Round to four decimal places. Use the following information to answer the next three exercises. You are planning to conduct a poll of community members aged 65 and older, to determine how many own mobile phones. You want to produce an estimate whose 95 percent confidence interval will be within four percentage points (plus or minus) of the true population proportion. Use an estimated population proportion of 0.5. 20. What sample size do you need? 854 Appendix B 21. Suppose you knew from prior research that the population proportion was 0.6. What sample size would you need? 22. Suppose you wanted a 95 percent confidence interval within three percentage points of the population. Assume the population proportion is 0.5. What sample size do you need? 9.1: Null and Alternate Hypotheses 23. In your state, 58 percent of registered voters in a community are registered as republicans. You want to conduct a study to see if this also holds up in your community. State the null and alternative hypotheses to test this. 24. You believe that at least 58 percent of registered voters in a community are registered as republicans. State the null and alternative hypotheses to test this. 25. The mean household value in a city is $268,000. You believe that the mean household value in a particular neighborhood is lower than the city average. Write the null and alternative hypotheses to test this. 26. State the appropriate alternative hypothesis to this null hypothesis: H0: μ = 107 27. State the appropriate alternative hypothesis to this null hypothesis: H0: p < 0.25 9.2: Outcomes and the Type I and Type II Errors 28. If you reject H0 when H0 is correct, what type of error is this? 29. If you fail to reject H0 when H0 is false, what type of error is this? 30. What is the relationship between the Type II error and the power of a test? 31. A new blood test is being developed to screen patients for cancer. Positive results are followed up by a more accurate (and expensive) test. It is assumed that the patient does not have cancer. Describe the null hypothesis and the Type I and Type II errors for this situation, and explain which type of error is more serious. 32. Explain in words what it means that a screening test for TB has an α level of 0.10. The null hypothesis is that the patient does not have TB. 33. Explain in words what it means that a screening test for TB has a β level of 0.20. The null hypothesis is that the patient does not have TB. 34. Explain in words what it means that a screening test for TB has a power of 0.80. 9.3: Distribution Needed for Hypothesis Testing 35. If you are conducting a hypothesis test of a single population mean, and you do not know the population variance, what test will you use if the sample size is 10 and the population is normal? 36. If you are conducting a hypothesis test of a single population mean, and you know the population variance, what test will you use? 37. If you are conducting a hypothesis test of a single population proportion, with np and nq greater than or equal to five, what test will you use, and with what parameters? 38. Published information indicates that, on average, college students spend less than 20 hours studying per week. You draw a sample of 25 students from your college and find the sample mean to be 18.5 hours, with a standard deviation of 1.5 hours. What distribution will you use to test whether study habits at your college are the same as the national average, and why? 39. A published study says that 95 percent of American children are vaccinated against a disease, with a standard deviation of 1.5 percent. You draw a sample of 100 children from your community and check their vaccination records to see if the vaccination rate in your community is the same as the national average. What distribution will you use for this test, and why? 9.4: Rare Events, the Sample, Decision, and Conclusion 40. You are conducting a study with an α level of 0.05. If you get a result with a p-value of 0.07, what will be your decision? 41. You are conducting a study with α = 0.01. If you get a result with a p-value of 0.006, what will be your decision? Use the following information to answer the next five exercises. According to the World Health Organization, the average height of a one-year-old child is 29”. You believe children with a particular disease are smaller than average, so you draw a sample of 20 children with this disease and find a mean height of 27.5” and a sample standard deviation of 1.5”. 42. What are the null and alternative hypotheses for this study? 43. What distribution will you use to test your hypothesis, and why? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 855 44. What is the test statistic and the p-value? 45. Based on your sample results, what is your decision? 46. Suppose the mean for your sample was 25. Redo the calculations and describe what your decision would be. 9.5: Additional Information and Full Hypothesis Test Examples 47. You conduct a study using α = 0.05. What is the level of significance for this study? 48. You conduct a study, based on a sample drawn from a normally distributed population with a known variance, with the following hypotheses: H0: μ = 35.5 Ha: μ ≠ 35.5 Will you conduct a one-tailed or two-tailed test? 49. You conduct a study, based on a sample drawn from a normally distributed population with a known variance, with the following hypotheses: H0: μ ≥ 35.5 Ha: μ < 35.5 Will you conduct a one-tailed or two-tailed test? Use the following information to answer the next three exercises. Nationally, 80 percent of adults own an automobile. You are interested in whether the same proportion in your community own cars. You draw a sample of 100 and find that 75 percent own cars. 50. What are the null and alternative hypotheses for this study? 51. What test will you use, and why? 10.1: Comparing Two Independent Population Means with Unknown Population Standard Deviations 52. You conduct a poll of political opinions, interviewing both members of 50 married couples. Are the groups in this study independent or matched? 53. You are testing a new drug to treat insomnia. You randomly assign 80 volunteer subjects to either th

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