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SUMMARIZING DATA GUIDED PRACTICE 2.10 The observation 64 thousand, a suspected outlier, was found to be an accurate observation. What would such an observation suggest about the nature of character counts in emails?6 GUIDED PRACTICE 2.11 Consider a data set that consists of the following numbers: 12, 12, 12, 12, 12, 13, 13, 14, 14, 15, 19. Which graph would better illustrate the data: a stem-and-leaf plot or a dot plot? Explain.7 2.1.3 Histograms Stem-and-leaf plots and dot plots are ideal for displaying data from small samples because they show the exact values of the observations and how frequently they occur. However, they are impractical for larger samples. For larger samples, rather than showing the frequency of every value, we prefer to think of the value as belonging to a bin. For example, in the email50 data set, we create a table of counts for the number of cases with character counts between 0 and 5,000, then the number of cases between 5,000 and 10,000, and so on. Such a table, shown in Figure 2.7, is called a frequency table. Bins usually include the observations that fall on their left (lower) boundary and exclude observations that fall on their right (upper) boundary. This is called left inclusive. For example, 5 (i.e. 5000) would be counted in the 5-10 bin, not in the 0-5 bin. These binned counts are plotted as bars in Figure 2.8 into what is called a histogram or frequency histogram, which resembles the stacked dot plot shown in Figure 2.6. Characters (in thousands) Count 0-5 5-10 10-15 15-20 20-25 25-30 19 12 6 2 3 5 · · · · · · 55-60 60-65 0 1 Figure 2.7: The counts for the binned num char data. Figure 2.8: A histogram of num char. This histogram uses bins or class intervals of width 5. Explore this histogram and dozens of histograms using American Community Survey data on Tableau Public . 6That occasionally there may be very long emails. 7Because all the values begin with 1, there would be only one stem (or two in a split stem-and-leaf). This would not provide a good sense of the distribution. For example, the gap between 15 and 19 would not be visually apparent. A dot plot would be better here. Number of Characters (in thousands)Frequency01020304050600510152005101520253035404550556065 2.1. EXAMINING NUMERICAL DATA 65 GUIDED PRACTICE 2.12 What can you see in the dot plot and stem-and-leaf plot that you cannot see in the frequency histogram?8 DRAWING HISTOGRAMS 1. The variable is always placed on the horizontal axis. Before drawing the histogram, label both axes and draw a scale for each. 2. Draw bars such that the height of the bar is the frequency of that bin and the width of the bar corresponds to the bin width. Histograms provide a view of the data density. Higher bars represent where the data are relatively more common. For instance, there are many more emails between 0 and 10,000 characters than emails between 10,000 and 20,000 in the data set. The bars make it easy to see how the density of the data changes relative to the number of characters. EXAMPLE 2.13 How many emails had fewer than 10 thousand characters? The height of the bars corresponds to frequency. There were 19 cases from 0 to less than 5 thousand and 12 cases from 5 thousand to less than 10 thousand, so there were 19 + 12 = 31 emails with fewer than 10 thousand characters. EXAMPLE 2.14 Approximately how many emails had fewer than 1 thousand characters? Based just on this histogram, we cannot know the exact answer to this question. We only know that 19 emails had between 0 and 5 thousand characters. If the number of emails is evenly distribution on this interval, then we can estimate that approximately 19/5 ≈ 4 emails fell in the range between 0 and 1 thousand. EXAMPLE 2.15 What percent of the emails had 10 thousand or more characters? From the ﬁrst example, we know that 31 emails had fewer than 10 thousand characters. Since there are 50 emails in total, there must be 19 emails that have 10 thousand or more characters. To ﬁnd the percent, compute 19/50 = 0.38 = 38%. 8Character counts for individual emails. 66 CHAPTER 2. SUMMARIZING DATA Sometimes questions such as the ones above can be answered more easily with a cumulative frequency histogram. This type of histogram shows cumulative, or total, frequency achieved by each bin, rather than the frequency in that particular bin. Characters (in thousands) Cumulative Frequency 0-5 5-10 10-15 15-20 20-25 25-30 30-35 · · · 55-60 60-65 19 31 37 39 42 47 47 · · · 49 50 Figure 2.9: The cumulative frequencies for the binned num char data. Figure 2.10: A cumulative frequency histogram of num char. This histogram uses bins or class intervals of width 5. Compare frequency, relative frequency, cumula. tive frequency, and cumulative relative frequency histograms on Tableau Public EXAMPLE 2.16 How many of the emails had fewer than 20 thousand characters? By tracing the height of the 15-20 thousand bin over to the vertical axis, we can see that it has a height just under 40 on the cumulative frequency scale. Therefore, we estimate that ≈39 of the emails had fewer than 30 thousand characters. Note that, unlike with a regular frequency histogram, we do not add up the height of the bars in a cumulative frequency histogram because each bar already represents a cumulative sum. EXAMPLE 2.17 Using the cumulative frequency table and histogram, how many of the emails had 10-15 thousand characters? To answer this question, we do a subtraction. 37 emails had less than or equal to 10-15 thousand characters and 31 emails had less than or equal to 5-10 thousand characters, so 6 emails must have had 10-15 thousand characters. Number of Characters (in thousands)Cumulative Frequency01020304050600102030405005101520253035404550556065 2.1. EXAMINING NUMERICAL DATA 67 EXAMPLE 2.18 Approximately 25 of the emails had fewer than how many characters? This time we are given a cumulative frequency, so we start at 25 on the vertical axis and trace it across to see which bin it hits. It hits the 5-10 thousand bin, so 25 of the emails had fewer than a value somewhere between 5 and 10 thousand characters. Knowing that 25 of the emails had fewer than a value between 5 and 10 thousand characters is useful information, but it is even more useful if we know what percent of the total 25 represents. Knowing that there were 50 total emails tells us that 25/50 = 0.5 = 50% of the emails had fewer than a value between 5 and 10 thousand characters. When we want to know what fraction or percent of the data meet a certain criteria, we use relative frequency instead of frequency. Relative frequency is a fancy term for percent or proportion. It tells us how large a number is relative to the total. Just as we constructed a frequency table, frequency histogram, and cumulative frequency histogram, we can construct a relative frequency table, relative frequency histogram, and cumulative relative frequency histogram. GUIDED PRACTICE 2.19 How will the shape of the relative frequency histograms diﬀer from the frequency histograms?9 PAY CLOSE ATTENTION TO THE VERTICAL AXIS OF A HISTOGRAM We can misinterpret a histogram if we forget to check whether the vertical axis represents frequency, relative frequency, cumulative frequency, or cumulative relative frequency. 2.1.4 Describing Shape Frequency and relative frequency histograms are especially convenient for describing the shape of the data distribution. Figure 2.8 shows that most emails have a relatively small number of characters, while fewer emails have a very large number of characters. When data trail oﬀ to the right in this way and have a longer right tail, the shape is said to be right skewed.10 Data sets with the reverse characteristic – a long, thin tail to the left – are said to be left skewed. We also say that such a distribution has a long left tail. Data sets that show roughly equal trailing oﬀ in both directions are called symmetric. LONG TAILS TO IDENTIFY SKEW When data trail oﬀ in one direction, the distribution has a long tail. If a distribution has a long left tail, it is left skewed. If a distribution has a long right tail, it is right skewed. GUIDED PRACTICE 2.20 Take a look at the dot plot in Figure 2.6. Can you see the skew in the data? Is it easier to see the skew in the frequency histogram, the dot plot, or the stem-and-leaf plot?11 9The shape will remain exactly the same. Changing from frequency to relative frequency involves dividing all the frequencies by the same number, so only the vertical scale (the numbers on the y-axis) change. 10Other ways to describe data that are right skewed: skewed to the right, skewed to the high end, or skewed to the positive end. 11The skew is visible in all three plots. However, it is not easily visible in the cumulative frequency histogram. 68 CHAPTER 2. SUMMARIZING DATA GUIDED PRACTICE 2.21 Would you expect the distribution of number of pets per household to be right skewed, left skewed, or approximately symmetric? Explain.12 In addition to looking at whether a distribution is skewed or symmetric, histograms, stem-andleaf plots, and dot plots can be used to identify modes. A mode is represented by a prominent peak in the distribution.13 There is only one prominent peak in the histogram of num char. Figure 2.11 shows histograms that have one, two, or three prominent peaks. Such distributions are called unimodal, bimodal, and multimodal, respectively. Any distribution with more than 2 prominent peaks is called multimodal. Notice that in Figure 2.8 there was one prominent peak in the unimodal distribution with a second less prominent peak that was not counted since it only diﬀers from its neighboring bins by a few observations. Figure 2.11: Counting only prominent peaks, the distributions are (left to right) unimodal, bimodal, and multimodal. GUIDED PRACTICE 2.22 Height measurements of young students and adult teachers at a K-3 elementary school were taken. How many modes would you anticipate in this height data set?14 LOOKING FOR MODES
Looking for modes isn’t about ﬁnding a clear and correct answer about the number of modes in a distribution, which is why prominent is not rigorously deﬁned in this book. The important part of this examination is to better understand your data and how it might be structured. 12We suspect most households would have 0, 1, or 2 pets but that a smaller number of households will have 3, 4, 5, or more pets, so there will be greater density over the small numbers, suggesting the distribution will have a long right tail and be right skewed. 13Another deﬁnition of mode, which is not typically used in statistics, is the value with the most occurrences. It is common to have no observations with the same value in a data set, which makes this other deﬁnition useless for many real data sets. 14There might be two height groups visible in the data set: one of the students and one of the adults. That is, the data are probably bimodal. 048121605101504812162005101504812162005101520 2.1. EXAMINING NUMERICAL DATA 69 2.1.5 Descriptive versus inferential statistics Finally, we note that the graphical summaries of this section and the numerical summaries of the next section fall into the realm of descriptive statistics. Descriptive statistics is about describing or summarizing data; it does not attribute properties of the data to a larger population. Inferential statistics, on the other hand, uses samples to generalize or to infer something about a larger population. We will have to wait until Chapter 5 to enter the exciting world of inferential statistics. Section summary • A scatterplot is a bivariate display illustrating the relationship between two numerical variables. The observations must be paired, which is to say that they correspond to the same case or individual. The linear association between two variables can be positive or negative, or there can be no association. Positive association means that larger values of the ﬁrst variable are associated with larger values of the second variable. Negative association means that larger values of the ﬁrst variable are associated with smaller values of the second variable. Additionally, the association can follow a linear trend or a curved (nonlinear) trend. • When looking at a univariate display, researchers want to understand the distribution of the variable. The term distribution refers to the values that a variable takes and the frequency of those values. When looking at a distribution, note the presence of clusters, gaps, and outliers. • Distributions may be symmetric or they may have a long tail. If a distribution has a long left tail (with greater density over the higher numbers), it is left skewed. If a distribution has a long right tail (with greater density over the smaller numbers), it is right skewed. • Distributions may be unimodal, bimodal, or multimodal. • Two graphs that are useful for showing the distribution of a small number of observations are the stem-and-leaf plot and dot plot. These graphs are ideal for displaying data from small samples because they show the exact values of the observations and how frequently they occur. However, they are impractical for larger data sets. • For larger data sets it is common to use a frequency histogram or a relative frequency histogram to display the distribution of a variable. This requires choosing bins of an appropriate width. • To see cumulative amounts, use a cumulative frequency histogram. A cumulative rel- ative frequency histogram is ideal for showing percentile. • Descriptive statistics describes or summarizes data, while inferential statistics uses sam- ples to generalize or infer something about a larger population. 70 CHAPTER 2. SUMMARIZING DATA Exercises 2.1 ACS, Part I. Each year, the US Census Bureau surveys about 3.5 million households with The American Community Survey (ACS). Data collected from the ACS have been crucial in government and policy decisions, helping to determine the allocation of federal and state funds each year. Some of the questions asked on the survey are about their income, age (in years), and gender. The table below contains this information for a random sample of 20 respondents to the 2012 ACS.15 female Income Age Gender 53,000 1600 70,000 12,800 1,200 30,000 4,500 20,000 25,000 42,000 28 male 18 54 male 22 male 18 34 male 21 male 28 29 33 male female female female 1 2 3 4 5 6 7 8 9 10 Income Age Gender female 34 female 55 33 female 41 male 47 30 male 61 male 50 24 19 male 670 29,000 44,000 48,000 30,000 60,000 108,000 5,800 50,000 11,000 female female female 11 12 13 14 15 16 17 18 19 20 (a) Create a scatterplot of income vs. age, and describe the relationship between these two variables. (b) Now create two scatterplots: one for income vs. age for males and another for females. (c) How, if at all, do the relationships between income and age diﬀer for males and females? 2.2 MLB stats. A baseball team’s success in a season is usually measured by their number of wins. In order to win, the team has to have scored more points (runs) than their opponent in any given game. As such, number of runs is often a good proxy for the success of the team. The table below shows number of runs, home runs, and batting averages for a random sample of 10 teams in the 2014 Major League Baseball season.16 Team 1 Baltimore 2 Boston 3 Cincinnati 4 Cleveland 5 Detroit 6 Houston 7 Minnesota 8 NY Yankees 9 Pittsburgh 10 San Francisco Runs Home runs Batting avg. 0.256 211 0.244 123 0.238 131 0.253 142 0.277 155 0.242 163 0.254 128 0.245 147 0.259 156 0.255 132 705 634 595 669 757 629 715 633 682 665 (a) Draw a scatterplot of runs vs. home runs. (b) Draw a scatterplot of runs vs. batting averages. (c) Are home runs or batting averages more strongly associated with number of runs? Explain your rea- soning. 15United States Census Bureau. Summary File. 2012 American Community Survey. U.S. Census Bureau’s American Community Survey Oﬃce, 2013. Web. 16ESPN: MLB Team Stats - 2014. 2.1. EXAMINING NUMERICAL DATA 71 2.3 Fiber in your cereal . The Cereal FACTS report provides information on nutrition content of cereals as well as who they are targeted for (adults, children, families). We have selected a random sample of 20 cereals from the data provided in this report. Shown below are the ﬁber contents (percentage of ﬁber per gram of cereal) for these cereals.17 Brand 1 Pebbles Fruity 2 Rice Krispies Treats 3 Pebbles Cocoa 4 Pebbles Marshmallow 5 Frosted Rice Krispies 6 Rice Krispies 7 Trix 8 Honey Comb 9 Rice Krispies Gluten Free 10 Frosted Flakes Fiber % 0.0% 0.0% 0.0% 0.0% 0.0% 3.0% 3.1% 3.1% 3.3% 3.3% Brand 11 Cinnamon Toast Crunch 12 Reese’s Puﬀs 13 Cheerios Honey Nut 14 Lucky Charms 15 Pebbles Boulders Chocolate PB 16 Corn Pops 17 Frosted Flakes Reduced Sugar 18 Cliﬀord Crunch 19 Apple Jacks 20 Dora the Explorer Fiber % 3.3% 3.4% 7.1% 7.4% 7.4% 9.4% 10.0% 10.0% 10.7% 11.1% (a) Create a stem and leaf plot of the distribution of the ﬁber content of these cereals. (b) Create a dot plot of the ﬁber content of these cereals. (c) Create a histogram and a relative frequency histogram of the ﬁber content of these cereals. (d) What percent of cereals contain more than 7% ﬁber? 2.4 Sugar in your cereal. The Cereal FACTS report from Exercise 2.3 also provides information on sugar content of cereals. We have selected a random sample of 20 cereals from the data provided in this report. Shown below are the sugar contents (percentage of sugar per gram of cereal) for these cereals. Brand 1 Rice Krispies Gluten Free 2 Rice Krispies 3 Dora the Explorer 4 Frosted Flakes Red. Sugar 5 Cliﬀord Crunch 6 Rice Krispies Treats 7 Pebbles Boulders Choc. PB 8 Cinnamon Toast Crunch 9 Trix 10 Honey Comb Sugar % 3% 12% 22% 27% 27% 30% 30% 30% 31% 31% Brand 11 Corn Pops 12 Cheerios Honey Nut 13 Reese’s Puﬀs 14 Pebbles Fruity 15 Pebbles Cocoa 16 Lucky Charms 17 Frosted Flakes 18 Pebbles Marshmallow 19 Frosted Rice Krispies 20 Apple Jacks Sugar % 31% 32% 34% 37% 37% 37% 37% 37% 40% 43% (a) Create a stem and leaf plot of the distribution of the sugar content of these cereals. (b) Create a dot plot of the sugar content of these cereals. (c) Create a histogram and a relative frequency histogram of the sugar content of these cereals. (d) What percent of cereals contain more than 30% sugar? 17JL Harris et al. “Cereal FACTS 2012: Limited progress in the nutrition quality and marketing of children’s cereals”. In: Rudd Center for Food Policy & Obesity. 12 (2012). 72 CHAPTER 2. SUMMARIZING DATA 2.5 Mammal life spans. Data were collected on life spans (in years) and gestation lengths (in days) for 62 mammals. A scatterplot of life span versus length of gestation is shown below.18 (a) What type of an association is apparent between life span and length of gestation? (b) What type of an association would you expect to see if the axes of the plot were reversed, i.e. if we plotted length of gestation versus life span? (c) Are life span and length of gestation inde- pendent? Explain your reasoning. 2.6 Associations. Indicate which of the plots show (a) a positive association, (b) a negative association, or (c) no association. Also determine if the positive and negative associations are linear or nonlinear. Each part may refer to more than one plot. 18T. Allison and D.V. Cicchetti. “Sleep in mammals: ecological and constitutional correlates”. In: Arch. Hydrobiol 75 (1975), p. 442. lllllllllllllllllllllllllllllllllllllllllllllllllllllllGestation (days)Life Span (years)02004006000255075100(1)(2)(3)(4) 2.2. NUMERICAL SUMMARIES AND BOX PLOTS 73 2.2 Numerical summaries and box plots What are the diﬀerent ways to measure the center of a distribution, and why is there more than one way to measure the center? How do you know if a value is “far” from the center? What does it mean to an outlier? We will continue with the email50 data set and investigate multiple quantitative summarizes for numerical data. Learning objectives 1. Calculate, interpret, and compare the two measures of
center (mean and median) and the three measures of spread (standard deviation, interquartile range, and range). 2. Understand how the shape of a distribution aﬀects the relationship between the mean and the median. 3. Identify and apply the two rules of thumb for identify outliers (one involving standard deviation and mean and the other involving Q1 and Q3). 4. Describe the distribution a numerical variable with respect to center, spread, and shape, noting the presence of outliers. 5. Find the 5 number summary and IQR, and draw a box plot with outliers shown. 6. Understand the eﬀect changing units has on each of the summary quantities. 7. Use quartiles, percentiles, and Z-scores to measure the relative position of a data point within the data set. 8. Compare the distribution of a numerical variable using dot plots / histograms with the same scale, back-to-back stem-and-leaf plots, or parallel box plots. Compare the distributions with respect to center, spread, shape, and outliers. 2.2.1 Measures of center In the previous section, we saw that modes can occur anywhere in a data set. Therefore, mode is not a measure of center. We understand the term center intuitively, but quantifying what is the center can be a little more challenging. This is because there are diﬀerent deﬁnitions of center. Here we will focus on the two most common: the mean and median. The mean, sometimes called the average, is a common way to measure the center of a distribution of data. To ﬁnd the mean number of characters in the 50 emails, we add up all the character counts and divide by the number of emails. For computational convenience, the number of characters is listed in the thousands and rounded to the ﬁrst decimal. ¯x = 21.7 + 7.0 + · · · + 15.8 50 = 11.6 The sample mean is often labeled ¯x. The letter x is being used as a generic placeholder for the variable of interest, num char, and the bar on the x communicates that the average number of characters in the 50 emails was 11,600. 74 CHAPTER 2. SUMMARIZING DATA MEAN The sample mean of a numerical variable is computed as the sum of all of the observations divided by the number of observations: ¯x = 1 n xi = xi n = x1 + x2 + · · · + xn n where is the capital Greek letter sigma and xi means take the sum of all the individual x values. x1, x2, . . . , xn represent the n observed values. GUIDED PRACTICE 2.23 Examine Equations (2.23) and (2.23) above. What does x1 correspond to? And x2? What does xi represent?19 GUIDED PRACTICE 2.24 What was n in this sample of emails?20 The email50 data set represents a sample from a larger population of emails that were received in January and March. We could compute a mean for this population in the same way as the sample mean, however, the population mean has a special label: µ. The symbol µ is the Greek letter mu and represents the average of all observations in the population. Sometimes a subscript, such as x, is used to represent which variable the population mean refers to, e.g. µx. EXAMPLE 2.25 The average number of characters across all emails can be estimated using the sample data. Based on the sample of 50 emails, what would be a reasonable estimate of µx, the mean number of characters in all emails in the email data set? (Recall that email50 is a sample from email.) The sample mean, 11,600, may provide a reasonable estimate of µx. While this number will not be perfect, it provides a point estimate of the population mean. In Chapter 5 and beyond, we will develop tools to characterize the reliability of point estimates, and we will ﬁnd that point estimates based on larger samples tend to be more reliable than those based on smaller samples. EXAMPLE 2.26 We might like to compute the average income per person in the US. To do so, we might ﬁrst think to take the mean of the per capita incomes across the 3,142 counties in the county data set. What would be a better approach? The county data set is special in that each county actually represents many individual people. If we were to simply average across the income variable, we would be treating counties with 5,000 and 5,000,000 residents equally in the calculations. Instead, we should compute the total income for each county, add up all the counties’ totals, and then divide by the number of people in all the counties. If we completed these steps with the county data, we would ﬁnd that the per capita income for the US is $27,348.43. Had we computed the simple mean of per capita income across counties, the result would have been just $22,504.70! Example 2.26 used what is called a weighted mean, which will not be a key topic in this textbook. However, we have provided an online supplement on weighted means for interested readers: www.openintro.org/go?id=stat extra weighted mean 19x1 corresponds to the number of characters in the ﬁrst email in the sample (21.7, in thousands), x2 to the number of characters in the second email (7.0, in thousands), and xi corresponds to the number of characters in the ith email in the data set. 20The sample size was n = 50. 2.2. NUMERICAL SUMMARIES AND BOX PLOTS 75 The median provides another measure of center. The median splits an ordered data set in half. There are 50 character counts in the email50 data set (an even number) so the data are perfectly split into two groups of 25. We take the median in this case to be the average of the two middle observations: (6,768 + 7,012)/2 = 6,890. When there are an odd number of observations, there will be exactly one observation that splits the data into two halves, and in this case that observation is the median (no average needed). MEDIAN: THE NUMBER IN THE MIDDLE In an ordered data set, the median is the observation right in the middle. If there are an even number of observations, the median is the average of the two middle values. Graphically, we can think of the mean as the balancing point. The median is the value such that 50% of the area is to the left of it and 50% of the area is to the right of it. Figure 2.12: A histogram of num char with its mean and median shown. EXAMPLE 2.27 Based on the data, why is the mean greater than the median in this data set? Consider the three largest values of 42 thousand, 43 thousand, and 64 thousand. These values drag up the mean because they substantially increase the sum (the total). However, they do not drag up the median because their magnitude does not change the location of the middle value. THE MEAN FOLLOWS THE TAIL In a right skewed distribution, the mean is greater than the median. In a left skewed distribution, the mean is less than the median. In a symmetric distribution, the mean and median are approximately equal. GUIDED PRACTICE 2.28 Consider the distribution of individual income in the United States. Which is greater: the mean or median? Why?21 21Because a small percent of individuals earn extremely large amounts of money while the majority earn a modest amount, the distribution is skewed to the right. Therefore, the mean is greater than the median. Number of Characters (in thousands)Frequency010203040506070051015200510152025303540455055606570MeanMedian 76 CHAPTER 2. SUMMARIZING DATA 2.2.2 Standard deviation as a measure of spread The U.S. Census Bureau reported that in 2019, the median family income was $80,944 and the mean family income was $108,587. Is a family income of $60,000 far from the mean or somewhat close to the mean? In order to answer this question, it is not enough to know the center of the data set and its range (maximum value - minimum value). We must know about the variability of the data set within that range. Low variability or small spread means that the values tend to be more clustered together. High variability or large spread means that the values tend to be far apart. EXAMPLE 2.29 Is it possible for two data sets to have the same range but diﬀerent spread? If so, give an example. If not, explain why not. Yes. An example is: 1, 1, 1, 1, 1, 9, 9, 9, 9, 9 and 1, 5, 5, 5, 5, 5, 5, 5, 5, 5, 9. The ﬁrst data set has a larger spread because values tend to be farther away from each other while in the second data set values are clustered together at the mean. Here, we introduce the standard deviation as a measure of spread. Though its formula is a bit tedious to calculate by hand, the standard deviation is very useful in data analysis and roughly describes how far away, on average, the observations are from the mean. We call the distance of an observation from its mean its deviation. Below are the deviations for the 1st, 2nd, 3rd, and 50th observations in the num char variable. For computational convenience, the number of characters is listed in the thousands and rounded to the ﬁrst decimal. x1 − ¯x = 21.7 − 11.6 = 10.1 x2 − ¯x = 7.0 − 11.6 = −4.6 x3 − ¯x = 0.6 − 11.6 = −11.0 ... x50 − ¯x = 15.8 − 11.6 = 4.2 If we square these deviations and then take an average, the result is about equal to the sample variance, denoted by s2: s2 = = 10.12 + (−4.6)2 + (−11.0)2 + · · · + 4.22 50 − 1 102.01 + 21.16 + 121.00 + · · · + 17.64 49 = 172.44 We divide by n − 1, rather than dividing by n, when computing the variance; you need not worry about this mathematical nuance for the material in this textbook. Notice that squaring the deviations does two things. First, it makes large values much larger, seen by comparing 10.12, (−4.6)2, (−11.0)2, and 4.22. Second, it gets rid of any negative signs. The standard deviation is deﬁned as the square root of the variance: √ s = 172.44 = 13.13 The standard deviation of the number of characters in an email is about 13.13 thousand. A subscript of x may be added to the variance and standard deviation, i.e. s2 x and sx, as a reminder that these are the variance and standard deviation of the observations represented by x1, x2, ..., xn. The x subscript is usually omitted when it is clear which data the variance or standard deviation is referencing. 2.2. NUMERICAL SUMMARIES AND BOX PLOTS 77 CALCULATING THE STANDARD DEVIATION The standard d
eviation is the square root of the variance. It is roughly the “typical” distance of the observations from the mean. sX = 1 n − 1 (xi − ¯x)2 The variance is useful for mathematical reasons, but the standard deviation is easier to interpret because it has the same units as the data set. The units for variance will be the units squared (e.g. meters2). Formulas and methods used to compute the variance and standard deviation for a population are similar to those used for a sample.22 However, like the mean, the population values have special symbols: σ2 for the variance and σ for the standard deviation. The symbol σ is the Greek letter sigma. THINKING ABOUT THE STANDARD DEVIATION It is useful to think of the standard deviation as the “typical” or “average” distance that observations fall from the mean. EXAMPLE 2.30 Earlier, we reported that the mean family income in the U.S. in 2019 was $108,587. Estimating the standard deviation of income as approximately $50,000, is a family income of $60,000 far from the mean or relatively close to the mean? Because $60,000 is less that one standard deviation from the mean, it is relatively close to the mean. If the value were more than 2 standard deviations away from the mean, we would consider it far from the mean. In the next section, we encounter a bell-shaped distribution known as the normal distribution. The empirical rule tells us that for nearly normal distributions, about 68% of the data will be within one standard deviation of the mean, about 95% will be within two standard deviations of the mean, and about 99.7% will be within three standard deviations of the mean. However, as seen in Figures 2.13 and 2.14, these percentages generally do not hold if the distribution is not bell-shaped. Figure 2.13: In the num char data, 40 of the 50 emails (80%) are within 1 standard deviation of the mean, and 47 of the 50 emails (9 4%) are within 2 standard deviations. The empirical rule does not hold well for skewed data, as shown in this example. 22The only diﬀerence is that the population variance has a division by n instead of n − 1. Number of Characters (in thousands)llllllllllllllllllllllllllllllllllllllllllllllllll010203040506070 78 CHAPTER 2. SUMMARIZING DATA Figure 2.14: Three very diﬀerent population distributions with the same mean µ = 0 and standard deviation σ = 1. GUIDED PRACTICE 2.31 On page 67, the concept of shape of a distribution was introduced. A good description of the shape of a distribution should include modality and whether the distribution is symmetric or skewed to one side. Using Figure 2.14 as an example, explain why such a description is important.23 When describing any distribution, comment on the three important characteristics of center, spread, and shape. Also note any especially unusual cases. EXAMPLE 2.32 In the data’s context (the number of characters in emails), describe the distribution of the num char variable shown in the histogram below. The distribution of email character counts is unimodal and very strongly skewed to the right. Many of the counts fall near the mean at 11,600, and most fall within one standard deviation (13,130) of the mean. There is one exceptionally long email with about 65,000 characters. In this chapter we use standard deviation as a descriptive statistic to describe the variability in a given data set. In Chapter 4 we will use standard deviation to assess how close a sample mean is likely to be to the population mean. 23Figure 2.14 shows three distributions that look quite diﬀerent, but all have the same mean, variance, and standard deviation. Using modality, we can distinguish between the ﬁrst plot (bimodal) and the last two (unimodal). Using skewness, we can distinguish between the last plot (right skewed) and the ﬁrst two. While a graph tells a more complete story, we can use modality and shape (symmetry/skew) to characterize basic information about a distribution. −3−2−10123−3−2−10123−3−2−10123Number of Characters (in thousands)Frequency01020304050600510152005101520253035404550556065 2.2. NUMERICAL SUMMARIES AND BOX PLOTS 79 2.2.3 Z-scores Knowing how many standard deviations a value is from the mean is often more useful than simply knowing how far a value is from the mean. EXAMPLE 2.33 Previously, we saw that the mean family income in the U.S. in 2019 was $108,587. Let’s round this to $100,000 and estimate the standard deviation of income as $50,000. Using these estimates, how many standard deviations above the mean is a family income of $200,000? The value $200,000 is $100,000 above the mean. $100,000 is 2 standard deviations above the mean. This can be found by doing 200, 000 − 100, 000 50, 000 = 2 The number of standard deviations a value is above or below the mean is known as the Z-score. A Z-score has no units, and therefore is sometimes also called standard units. THE Z-SCORE The Z-score of an observation is the number of standard deviations it falls above or below the mean. We compute the Z-score for an observation x that follows a distribution with mean µ and standard deviation σ using Z = x − µ σ Observations above the mean always have positive Z-scores, while those below the mean always have negative Z-scores. If an observation is equal to the mean, then the Z-score is 0. EXAMPLE 2.34 Head lengths of brushtail possums have a mean of 92.6 mm and standard deviation 3.6 mm. Compute the Z-scores for possums with head lengths of 95.4 mm and 85.8 mm. For x1 = 95.4 mm: For x2 = 85.8 mm: Z1 = x1 − µ σ = 95.4 − 92.6 3.6 = 0.78 Z2 = 85.8 − 92.6 3.6 = −1.89 We can use Z-scores to roughly identify which observations are more unusual than others. An observation x1 is said to be more unusual than another observation x2 if the absolute value of its Zscore is larger than the absolute value of the other observation’s Z-score: \|Z1\| > \|Z2\|. This technique is especially insightful when a distribution is symmetric. 80 CHAPTER 2. SUMMARIZING DATA GUIDED PRACTICE 2.35 Which of the observations in Example 2.34 is more unusual?24 GUIDED PRACTICE 2.36 Let X represent a random variable from a distribution with µ = 3 and σ = 2, and suppose we observe x = 5.19. (a) Find the Z-score of x. (b) Interpret the Z-score.25 Because Z-scores have no units, they are useful for comparing distance to the mean for distri- butions that have diﬀerent standard deviations or diﬀerent units. EXAMPLE 2.37 The average daily high temperature in June in LA is 77◦F with a standard deviation of 5◦F. The average daily high temperature in June in Iceland is 13◦C with a standard deviation of 3◦C. Which would be considered more unusual: an 83◦F day in June in LA or a 19◦C day in June in Iceland? Both values are 6◦ above the mean. However, they are not the same number of standard deviations above the mean. 83 is (83−77)/5 = 1.2 standard deviations above the mean, while 19 is (19−13)/3 = 2 standard deviations above the mean. Therefore, a 19◦C day in June in Iceland would be more unusual than an 83◦F day in June in LA. 2.2.4 Box plots and quartiles A box plot summarizes a data set using ﬁve summary statistics while also plotting unusual observations, called outliers. Figure 2.15 provides a box plot of the num char variable from the email50 data set. The ﬁve summary statistics used in a box plot are known as the ﬁve-number summary, which consists of the minimum, the maximum, and the three quartiles (Q1, Q2, Q3) of the data set being studied. Q2 represents the second quartile, which is equivalent to the 50th percentile (i.e. the median). Previously, we saw that Q2 (the median) for the email50 data set was the average of the two middle values: 6,768+7,012 = 6,890. Q1 represents the ﬁrst quartile, which is the 25th percentile, and is the median of the smaller half of the data set. There are 25 values in the lower half of the data set, so Q1 is the middle value: 2,454 characters. Q3 represents the third quartile, or 75th percentile, and is the median of the larger half of the data set: 15,829 characters. 2 We calculate the variability in the data using the range of the middle 50% of the data: Q3−Q1 = 13,375. This quantity is called the interquartile range (IQR, for short). It, like the standard deviation, is a measure of variability or spread in data. The more variable the data, the larger the standard deviation and IQR tend to be. 24Because the absolute value of Z-score for the second observation (x2 = 85.8 mm → Z2 = −1.89) is larger than that of the ﬁrst (x1 = 95.4 mm → Z1 = 0.78), the second observation has a more unusual head length. 25(a) Its Z-score is given by Z = x−µ σ = 5.19−3 2 = 2.19/2 = 1.095. (b) The observation x is 1.095 standard deviations above the mean. We know it must be above the mean since Z is positive. 2.2. NUMERICAL SUMMARIES AND BOX PLOTS 81 Figure 2.15: A labeled box plot for the number of characters in 50 emails. The median (6,890) splits the data into the bottom 50% and the top 50%. Explore dozens of boxplots with histograms using American Community Survey data on Tableau Public . INTERQUARTILE RANGE (IQR) The IQR is the length of the box in a box plot. It is computed as where Q1 and Q3 are the 25th and 75th percentiles. IQR = Q3 − Q1 OUTLIERS IN THE CONTEXT OF A BOX PLOT When in the context of a box plot, deﬁne an outlier as an observation that is more than 1.5 × IQR above Q3 or 1.5 × IQR below Q1. Such points are marked using a dot or asterisk in a box plot. To build a box plot, draw an axis (vertical or horizontal) and draw a scale. Draw a dark line denoting Q2, the median. Next, draw a line at Q1 and at Q3. Connect the Q1 and Q3 lines to form a rectangle. The width of the rectangle corresponds to the IQR and the middle 50% of the data is in this interval. Extending out from the rectangle, the whiskers attempt to capture all of the data remaining outside of the box, except outliers. In Figure 2.15, the upper whisker does not extend to the last three points, which are beyond Q3 + 1.5 × IQR and are outliers, so it extends only to the last p
oint below this limit.26 The lower whisker stops at the lowest value, 33, since there are no additional data to reach. Outliers are each marked with a dot or asterisk. In a sense, the box is like the body of the box plot and the whiskers are like its arms trying to reach the rest of the data. 26You might wonder, isn’t the choice of 1.5×IQR for deﬁning an outlier arbitrary? It is! In practical data analyses, we tend to avoid a strict deﬁnition since what is an unusual observation is highly dependent on the context of the data. Number of Characters (in thousands)010203040506070lower whiskerQ1 (first quartile)Q2 (median)Q3 (third quartile)upper whiskermax whisker reachoutliers 82 CHAPTER 2. SUMMARIZING DATA EXAMPLE 2.38 Compare the box plot to the graphs previously discussed: stem-and-leaf plot, dot plot, frequency and relative frequency histogram. What can we learn more easily from a box plot? What can we learn more easily from the other graphs? It is easier to immediately identify the quartiles from a box plot. The box plot also more prominently highlights outliers. However, a box plot, unlike the other graphs, does not show the distribution of the data. For example, we cannot generally identify modes using a box plot. EXAMPLE 2.39 Is it possible to identify skew from the box plot? Yes. Looking at the lower and upper whiskers of this box plot, we see that the lower 25% of the data is squished into a shorter distance than the upper 25% of the data, implying that there is greater density in the low values and a tail trailing to the upper values. This box plot is right skewed. GUIDED PRACTICE 2.40 True or false: there is more data between the median and Q3 than between Q1 and the median.27 EXAMPLE 2.41 Consider the following ordered data set. 5 5 9 10 15 16 30 40 80 Find the 5 number summary and identify how small or large a value would need to be to be considered an outlier. Are there any outliers in this data set? There are nine numbers in this data set. Because n is odd, the median is the middle number: 15. When ﬁnding Q1, we ﬁnd the median of the lower half of the data, which in this case includes 4 numbers (we do not include the 15 as belonging to either half of the data set). Q1 then is the average of 5 and 9, which is Q1 = 7, and Q3 is the average of 30 and 40, so Q3 = 35. The min is 5 and the max is 80. To see how small a number needs to be to be an outlier on the low end we do: On the high end we need: Q1 − 1.5 × IQR = Q1 − 1.5 × (Q3 − Q1) = 7 − 1.5 × (35 − 7) = −35 Q3 + 1.5 × IQR = Q3 + 1.5 × (Q3 − Q1) = 35 + 1.5 × (35 − 7) = 77 There are no numbers less than -41, so there are no outliers on the low end. The observation at 80 is greater than 77, so 80 is an outlier on the high end. 27False. Since Q1 is the 25th percentile and the median is the 50th percentile, 25% of the data fall between Q1 and the median. Similarly, 25% of the data fall between Q2 and the median. The distance between the median and Q3 is larger because that 25% of the data is more spread out. 2.2. NUMERICAL SUMMARIES AND BOX PLOTS 83 2.2.5 Technology: summarizing 1-variable statistics Online calculators such as Desmos or a handheld calculator can be used to calculate summary statistics. More advanced statistical software packages include R (which was used for most of the graphs in this text), Python, SAS, and STATA. Get started quickly with this Desmos 1-VarStats Calculator (available at openintro.org/ahss/desmos). Calculator instructions TI-83/84: ENTERING DATA The ﬁrst step in summarizing data or making a graph is to enter the data set into a list. Use STAT, Edit. 1. Press STAT. 2. Choose 1:Edit. 3. Enter data into L1 or another list. CASIO FX-9750GII: ENTERING DATA 1. Navigate to STAT (MENU button, then hit the 2 button or select STAT). 2. Optional: use the left or right arrows to select a particular list. 3. Enter each numerical value and hit EXE. 84 CHAPTER 2. SUMMARIZING DATA TI-84: CALCULATING SUMMARY STATISTICS Use the STAT, CALC, 1-Var Stats command to ﬁnd summary statistics such as mean, standard deviation, and quartiles. 1. Enter the data as described previously. 2. Press STAT. 3. Right arrow to CALC. 4. Choose 1:1-Var Stats. 5. Enter L1 (i.e. 2ND 1) for List. If the data is in a list other than L1, type the name of that list. 6. Leave FreqList blank. 7. Choose Calculate and hit ENTER. TI-83: Do steps 1-4, then type L1 (i.e. 2nd 1) or the list’s name and hit ENTER. Calculating the summary statistics will return the following information. down arrow to see all of the summary statistics. It will be necessary to hit the ¯x Σx Σx2 Sx σx Mean Sum of all the data values Sum of all the squared data values Sample standard deviation Population standard deviation Sample size or # of data points n minX Minimum Q1 Med Median maxX Maximum First quartile TI-83/84: DRAWING A BOX PLOT 1. Enter the data to be graphed as described previously. 2. Hit 2ND Y= (i.e. STAT PLOT). 3. Hit ENTER (to choose the ﬁrst plot). 4. Hit ENTER to choose ON. 5. Down arrow and then right arrow three times to select box plot with outliers. 6. Down arrow again and make Xlist: L1 and Freq: 1. 7. Choose ZOOM and then 9:ZoomStat to get a good viewing window. TI-83/84: WHAT TO DO IF YOU CANNOT FIND L1 OR ANOTHER LIST Restore lists L1-L6 using the following steps: 1. Press STAT. 2. Choose 5:SetUpEditor. 3. Hit ENTER. CASIO FX-9750GII: DRAWING A BOX PLOT AND 1-VARIABLE STATISTICS 1. Navigate to STAT (MENU, then hit 2) and enter the data into a list. 2. Go to GRPH (F1). 3. Next go to SET (F6) to set the graphing parameters. 4. To use the 2nd or 3rd graph instead of GPH1, select F2 or F3. 5. Move down to Graph Type and select the (F6) option to see more graphing options, then select Box (F2). 6. If XList does not show the list where you entered the data, hit LIST (F1) and enter the correct list number. 7. Leave Frequency at 1. 8. For Outliers, choose On (F1). 9. Hit EXE and then choose the graph where you set the parameters F1 (most common), F2, or F3. 10. If desired, explore 1-variable statistics by selecting 1-Var (F1). 2.2. NUMERICAL SUMMARIES AND BOX PLOTS 85 CASIO FX-9750GII: DELETING A DATA LIST 1. Navigate to STAT (MENU, then hit 2). 2. Use the arrow buttons to navigate to the list you would like to delete. 3. Select (F6) to see more options. 4. Select DEL-A (F4) and then F1 to conﬁrm. GUIDED PRACTICE 2.42 Enter the following 10 data points into a calculator. Find the summary statistics and make a box plot of the data.28 5, 8, 1, 19, 3, 1, 11, 18, 20, 5 GUIDED PRACTICE 2.43 Use the email50 data set at openintro.org/data and Screen 2 of this Desmos 1-Var Stats Calculator to summarize the num char variable (number of characters in an email).29 28The summary statistics should be ¯x = 9.1, Sx = 7.48, Q1 = 3, etc. Using a TI, the boxplot looks like this: 29Remember, the Desmos Calculators and Activities in this book can be found at openintro.org/ahss/desmos. Down the email50 CSV ﬁle and open it. Copy and paste the num char column into the Desmos calculator, replacing the data currently in x1. Adjust window as needed and you should get the following: 86 CHAPTER 2. SUMMARIZING DATA 2.2.6 Outliers and robust statistics RULES OF THUMB FOR IDENTIFYING OUTLIERS There are two rules of thumb for identifying outliers: • More than 1.5× IQR below Q1 or above Q3 • More than 2 standard deviations above or below the mean. Both are important for the AP exam. In practice, consider these to be only rough guidelines. GUIDED PRACTICE 2.44 For the email50 data set,Q1 = 2,536 and Q3 = 15, 411. ¯x = 11,600 and s = 13,130. What values would be considered an outlier on the low end using each rule?30 GUIDED PRACTICE 2.45 Because there are no negative values in this data set, there can be no outliers on the low end. What does the fact that there are outliers on the high end but not on the low end suggestion?31 How are the sample statistics of the num char data set aﬀected by the observation, 64,401? What would have happened if this email wasn’t observed? What would happen to these summary statistics if the observation at 64,401 had been even larger, say 150,000? These scenarios are plotted alongside the original data in Figure 2.16, and sample statistics are computed under each scenario in Figure 2.17. Figure 2.16: Dot plots of the original character count data and two modiﬁed data sets. 30 Q1 − 1.5 × IQR = 2536 − 1.5 × (15411 − 2536) = −16, 749.5, so values less than -16,749.5 would be considered an outlier using the ﬁrst rule of thumb. Using the second rule of thumb, a value less than ¯x−2×s = 11, 600−2×13, 130 = −14, 660 would be considered an outlier. Note tht these are just rules of thumb and yield diﬀerent values. 31It suggests that the distribution has a right hand tail, that is, that it is right skewed. llllllllllllllllllllllllllllllllllllllllllllllllll Number of Characters (in thousands)050100150OriginallllllllllllllllllllllllllllllllllllllllllllllllllDrop 64,401llllllllllllllllllllllllllllllllllllllllllllllllll64,401 to 150,000 2.2. NUMERICAL SUMMARIES AND BOX PLOTS 87 scenario original num char data drop 64,401 observation move 64,401 to 150,000 robust median 6,890 6,768 6,890 IQR 12,875 11,702 12,875 not robust s ¯x 13,130 11,600 10,798 10,521 22,434 13,310 Figure 2.17: A comparison of how the median, IQR, mean (¯x), and standard deviation (s) change when extreme observations are present. GUIDED PRACTICE 2.46 (a) Which is more aﬀected by extreme observations, the mean or median? Figure 2.17 may be helpful. (b) Is the standard deviation or IQR more aﬀected by extreme observations?32 The median and IQR are called robust estimates because extreme observations have little eﬀect on their values. The mean and standard deviation are much more aﬀected by changes in extreme observations. EXAMPLE 2.47 The median and IQR do not change much under the three scenarios in Figure 2.17. Why might this be the case? Since there are no large gaps between observations around the three quartiles, adding,
deleting, or changing one value, no matter how extreme that value, will have little eﬀect on their values. GUIDED PRACTICE 2.48 The distribution of vehicle prices tends to be right skewed, with a few luxury and sports cars lingering out into the right tail. If you were searching for a new car and cared about price, should you be more interested in the mean or median price of vehicles sold, assuming you are in the market for a regular car?33 32(a) Mean is aﬀected more. (b) Standard deviation is aﬀected more. Complete explanations are provided in the material following Guided Practice 2.46. 33Buyers of a “regular car” should be concerned about the median price. High-end car sales can drastically inﬂate the mean price while the median will be more robust to the inﬂuence of those sales. 88 CHAPTER 2. SUMMARIZING DATA 2.2.7 Linear transformations of data EXAMPLE 2.49 Begin with the following list: 1, 1, 5, 5. Multiply all of the numbers by 10. What happens to the mean? What happens to the standard deviation? How do these compare to the mean and the standard deviation of the original list? The original list has a mean of 3 and a standard deviation of 2. The new list: 10, 10, 50, 50 has a mean of 30 with a standard deviation of 20. Because all of the values were multiplied by 10, both the mean and the standard deviation were multiplied by 10. 34 EXAMPLE 2.50 Start with the following list: 1, 1, 5, 5. Multiply all of the numbers by -0.5. What happens to the mean? What happens to the standard deviation? How do these compare to the mean and the standard deviation of the original list? The new list: -0.5, -0.5, -2.5, -2.5 has a mean of -1.5 with a standard deviation of 1. Because all of the values were multiplied by -0.5, the mean was multiplied by -0.5. Multiplying all of the values by a negative ﬂipped the sign of numbers, which aﬀects the location of the center, but not the spread. Multiplying all of the values by -0.5 multiplied the standard deviation by +0.5 since the standard deviation cannot be negative. EXAMPLE 2.51 Again, start with the following list: 1, 1, 5, 5. Add 100 to every entry. How do the new mean and standard deviation compare to the original mean and standard deviation? The new list is: 101, 101, 105, 105. The new mean of 103 is 100 greater than the original mean of 3. The new standard deviation of 2 is the same as the original standard deviation of 2. Adding a constant to every entry shifted the values, but did not stretch them. Suppose that a researcher is looking at a list of 500 temperatures recorded in Celsius (C). The mean of the temperatures listed is given as 27◦C with a standard deviation of 3◦C. Because she is not familiar with the Celsius scale, she would like to convert these summary statistics into Fahrenheit (F). To convert from Celsius to Fahrenheit, we use the following conversion: xF = 9 5 xC + 32 Fortunately, she does not need to convert each of the 500 temperatures to Fahrenheit and then recalculate the mean and the standard deviation. The unit conversion above is a linear transformation of the following form, where a = 9/5 and b = 32: aX + b Using the examples as a guide, we can solve this temperature-conversion problem. The mean was 27◦C and the standard deviation was 3◦C. To convert to Fahrenheit, we multiply all of the values by 9/5, which multiplies both the mean and the standard deviation by 9/5. Then we add 32 to all of the values which adds 32 to the mean but does not change the standard deviation further. ¯xF = ¯xC + 32 (27) + 32 9 5 5 9 = 80.6 = σF = σC 9 5 9 5 = 5.4 = (3) 34Here, the population standard deviation was used in the calculation. These properties can be proven mathemat- ically using properties of sigma (summation). 2.2. NUMERICAL SUMMARIES AND BOX PLOTS 89 Figure 2.18: 500 temperatures shown in both Celsius and Fahrenheit. ADDING SHIFTS THE VALUES, MULTIPLYING STRETCHES OR CONTRACTS THEM Adding a constant to every value in a data set shifts the mean but does not aﬀect the standard deviation. Multiplying the values in a data set by a constant will change the mean and the standard deviation by the same multiple, except that the standard deviation will always remain positive. EXAMPLE 2.52 Consider the temperature example. How would converting from Celsuis to Fahrenheit aﬀect the median? The IQR? The median is aﬀected in the same way as the mean and the IQR is aﬀected in the same way as the standard deviation. To get the new median, multiply the old median by 9/5 and add 32. The IQR is computed by subtracting Q1 from Q3. While Q1 and Q3 are each aﬀected in the same way as the median, the additional 32 added to each will cancel when we take Q3 − Q1. That is, the IQR will be increase by a factor of 9/5 but will be unaﬀected by the addition of 32. For a more mathematical explanation of the IQR calculation, see the footnote.35 2.2.8 Comparing numerical data across groups Some of the more interesting investigations can be considered by examining numerical data across groups. The methods required here aren’t really new. All that is required is to make a numerical plot for each group. To make a direct comparison between two groups, create a pair of dot plots or a pair of histograms drawn using the same scales. It is also common to use back-to-back stem-and-leaf plots, parallel box plots, and hollow histograms, the three of which are explored here. We will take a look again at the county data set and compare the median household income for counties that gained population from 2010 to 2017 versus counties that had no gain. While we might like to make a causal connection here, remember that these are observational data and so such an interpretation would be, at best, half-baked. There were 1,454 counties where the population increased from 2010 to 2017, and there were 1,672 counties with no gain (all but one were a loss). A random sample of 100 counties from the ﬁrst group and 50 from the second group are shown in Figure 2.19 to give a better sense of some of the raw median income data. 35New IQR = 9 5 Q3 + 32 − 9 5 Q1 + 32 = 9 5 (Q3 − Q1) = 9 5 × (old IQR). 204060801000204060CelsiusFahrenheitFrequencyTemperature 90 CHAPTER 2. SUMMARIZING DATA Median Income for 150 Counties, in $1000s Population Gain 43.6 51.8 63.3 34.1 46.3 40.6 34.7 51.4 57.6 54.6 49.1 46.5 63 47.6 49 44.7 55.5 42.2 40.7 52.1 45.5 82.2 46.3 54 56.5 69.2 55 57.2 38.9 49.6 55.9 45.6 71.7 38.6 61.5 48.1 60.3 52.8 43.6 62.4 42.9 62 48.4 46.4 44.1 50.9 53.7 39.1 39 35.3 52.7 51.1 56.4 49.8 49.1 39.7 44.1 52.2 46 40.5 39.9 50 56 77.5 57.8 38.8 100.2 63 38.2 44.6 40.6 51.1 80.8 75.2 51.9 61 53.8 53.1 63 46.6 74.2 63.2 50.4 57.2 42.6 45.7 41.9 51.7 51 49.4 51.3 45.1 46.4 48.6 56.7 38.9 34.6 60 42.6 37.1 No Population Gain 50.7 60.3 48.3 40.3 40.4 39.3 45.9 47.2 57 46.1 41.5 42.3 46.4 51.7 44.9 50.5 51.8 29.1 34.9 30.9 27 61.1 51.8 40.7 45.7 34.7 43.4 37 21 33.1 45.7 31.9 52 45.7 38.7 56.3 21.7 43 56.2 48.9 34.5 44.5 43.2 42.1 50.9 41.3 35.4 33.6 56.5 43.4 Figure 2.19: In this table, median household income (in $1000s) from a random sample of 100 counties that had population gains are shown on the left. Median incomes from a random sample of 50 counties that had no population gain are shown on the right. Population: Gain Population: No Gain \| 2 \|12 \| 2 \|79 4\| 3 \|1234 99999987555\| 3 \|5555799 444433322111000\| 4 \|00111223333 999998887666666665555\| 4 \|55666666789 444333222221111110000\| 5 \|1112222 887776666555\| 5 \|6677 33333222100\| 6 \|01 9\| 6 \| 42\| 7 \| 85\| 7 \| 21\| 8 \| Legend: 2 \|1 = 21,000 median income Figure 2.20: Back-to-back stem-and-leaf plot for median income, split by whether the count had a population gain or no gain. 2.2. NUMERICAL SUMMARIES AND BOX PLOTS 91 Figure 2.21: Side-by-side box plot (left panel) and hollow histograms (right panel) for med hh income, where the counties are split by whether or not there was a population gain from 2010 to 2017. Explore this data set on Tableau Public . The side-by-side box plot is a traditional tool for comparing across groups. An example is shown in the left panel of Figure 2.21, where there are two box plots, one for each group, placed into one plotting window and drawn on the same scale. Another useful plotting method uses hollow histograms to compare numerical data across groups. These are just the outlines of histograms of each group put on the same plot, as shown in the right panel of Figure 2.21. GUIDED PRACTICE 2.53 Use the plots in Figure 2.21 to compare the incomes for counties across the two groups. What do you notice about the approximate center of each group? What do you notice about the variability between groups? Is the shape relatively consistent between groups? How many prominent modes are there for each group?36 COMPARING DISTRIBUTIONS When comparing distributions, compare them with respect to center, spread, and shape as well as any unusual observations. Such descriptions should be in context. GUIDED PRACTICE 2.54 What components of each plot in Figure 2.21 do you ﬁnd most useful?37 36Answers may vary a little. The counties with population gains tend to have higher income (median of about $45,000) versus counties without a gain (median of about $40,000). The variability is also slightly larger for the population gain group. This is evident in the IQR, which is about 50% bigger in the gain group. Both distributions show slight to moderate right skew and are unimodal. The box plots indicate there are many observations far above the median in each group, though we should anticipate that many observations will fall beyond the whiskers when examining any data set that contain more than a couple hundred data points. 37Answers will vary. The parallel box plots are especially useful for comparing centers and spreads, while the hollow histograms are more useful for seeing distribution shape, skew, and groups of anomalies. Change in PopulationGainNo Gain$20k$40k$60k$80k$100k$120kMedian Household IncomeMedian Household Income$20k$40k$60k$80k$100kGainNo Gain 9
2 CHAPTER 2. SUMMARIZING DATA GUIDED PRACTICE 2.55 Do these graphs tell us about any association between income for the two groups?38 Looking at an association is diﬀerent than comparing distributions. When comparing distributions, we are interested in questions such as, “Which distribution has a greater average?” and “How do the shapes of the distribution diﬀer?” The number of elements in each data set need not be the same (e.g. height of women and height of men). When we look at association, we are interested in whether there is a positive, negative, or no association between the variables. This requires two data sets of equal length that are essentially paired (e.g. height and weight of individuals). COMPARING DISTRIBUTIONS VERSUS LOOKING AT ASSOCIATION We compare two distributions with respect to center, spread, and shape. To compare the distributions visually, we use 2 single-variable graphs, such as two histograms, two dot plots, parallel box plots, or a back-to-back stem-and-leaf. When looking at association, we look for a positive, negative, or no relationship between the variables. To see association visually, we require a scatterplot. 2.2.9 Mapping data (special topic) The county data set oﬀers many numerical variables that we could plot using dot plots, scatterplots, or box plots, but these miss the true nature of the data. Rather, when we encounter geographic data, we should create an intensity map, where colors are used to show higher and lower values of a variable. Figures 2.22 and 2.23 shows intensity maps for poverty rate in percent (poverty), unemployment rate (unemployment rate), homeownership rate in percent (homeownership), and median household income (median hh income). The color key indicates which colors correspond to which values. The intensity maps are not generally very helpful for getting precise values in any given county, but they are very helpful for seeing geographic trends and generating interesting research questions or hypotheses. EXAMPLE 2.56 What interesting features are evident in the poverty and unemployment rate intensity maps? Poverty rates are evidently higher in a few locations. Notably, the deep south shows higher poverty rates, as does much of Arizona and New Mexico. High poverty rates are evident in the Mississippi ﬂood plains a little north of New Orleans and also in a large section of Kentucky. The unemployment rate follows similar trends, and we can see correspondence between the two variables. In fact, it makes sense for higher rates of unemployment to be closely related to poverty rates. One observation that stand out when comparing the two maps: the poverty rate is much higher than the unemployment rate, meaning while many people may be working, they are not making enough to break out of poverty. GUIDED PRACTICE 2.57 What interesting features are evident in the median hh income intensity map in Figure 2.23(b)?39 38No, to see association we require a scatterplot. Moreover, these data are not paired, so the discussion of association does not make sense here. 39Note: answers will vary. There is some correspondence between high earning and metropolitan areas, where we can see darker spots (higher median household income), though there are several exceptions. You might look for large cities you are familiar with and try to spot them on the map as dark spots. 2.2. NUMERICAL SUMMARIES AND BOX PLOTS 93 (a) (b) Figure 2.22: (a) Intensity map of poverty rate (percent). (b) Intensity map of the unemployment rate (percent). Explore dozens of intensity maps using American Community Survey data on Tableau Public . 2%14%>25%Poverty2%4%>7%Unemployment Rate 94 CHAPTER 2. SUMMARIZING DATA (a) (b) Figure 2.23: (a) Intensity map of homeownership rate (percent). (b) Intensity map of median household income ($1000s). Explore dozens of intensity maps using American Community Survey data on Tableau Public . <55%73%91%Homeownership Rate$19$47>$75Median Household Income 2.2. NUMERICAL SUMMARIES AND BOX PLOTS 95 Section summary • In this section we looked at univariate summaries, including two measures of center and three mea- sures of spread. • When summarizing or comparing distributions, always comment on center, spread, and shape. Also, mention outliers or gaps if applicable. Put descriptions in context, that is, identify the variable(s) being summarized by name and include relevant units. Remember: Center, Spread, and Shape! In context! • Mean and median are measures of center. (A common mistake is to report mode as a measure of center. However, a mode can appear anywhere in a distribution.) – The mean is the sum of all the observations divided by the number of observations, n. ¯x = 1 n xi = xi n = x1+x2+...+xn n – In an ordered data set, the median is the middle number when n is odd. When n is even, the median is the average of the two middle numbers. • Because large values exert more “pull” on the mean, large values on the high end tend to increase the mean more than they increase the median. In a right skewed distribution, therefore, the mean is greater than the median. Analogously, in a left skewed distribution, the mean is less than the median. Remember: The mean follows the tail! The skew is the tail! • Standard deviation (SD) and Interquartile range (IQR) are measures of spread. SD measures the typical spread from the mean, whereas IQR measures the spread of the middle 50% of the data. – To calculate the standard deviation, subtract the average from each value, square all those diﬀerences, add them up, divide by n − 1, then take the square root. Note: The standard deviation is the square root of the variance. sX = 1 n−1 (xi − ¯x)2 – The IQR is the diﬀerence between the third quartile Q3 and the ﬁrst quartile Q1. IQR = Q3 − Q1 • Range is also sometimes used as a measure of spread. The range of a data set is deﬁned as the diﬀerence between the maximum value and the minimum value, i.e. max − min. • Outliers are observations that are extreme relative to the rest of the data. Two rules of thumb for identifying observations as outliers are: – more than 2 standard deviations above or below the mean – more than 1.5 × IQR below Q1 or above Q3 • Mean and SD are sensitive to outliers. Median and IQR are more robust and less sensitive to outliers. • A Z-score represents the number of standard deviations a value in a data set is above or below the mean. To calculate a Z-score use: Z = x−mean SD . • Z-scores do not depend on units. When looking at distributions with diﬀerent units or diﬀerent standard deviations, Z-scores are useful for comparing how far values are away from the mean (relative to the distribution of the data). • Linear transformations of data. Adding a constant to every value in a data set shifts the mean but does not aﬀect the standard deviation. Multiplying the values in a data set by a constant will multiply the mean and the standard deviation by that constant, except that the standard deviation must always remain positive. • Box plots do not show the distribution of a data set in the way that histograms do. Rather, they provide a visual depiction of the 5-number summary, which consists of: min, Q1, Q2, Q3, max. While a box plot does not indicate modes, it can show skew and outliers. 96 CHAPTER 2. SUMMARIZING DATA Exercises 2.7 Smoking habits of UK residents, Part I. A survey was conducted to study the smoking habits of UK residents. The histograms below display the distributions of the number of cigarettes smoked on weekdays and weekends, and they exclude data from people who identiﬁed themselves as non-smokers. Describe the two distributions and compare them.40 2.8 Stats scores, Part I. Below are the ﬁnal exam scores of twenty introductory statistics students. 79, 83, 57, 82, 94, 83, 72, 74, 73, 71, 66, 89, 78, 81, 78, 81, 88, 69, 77, 79 Draw a histogram of these data and describe the distribution. 2.9 Smoking habits of UK residents, Part II. A random sample of 5 smokers from the data set discussed in Exercise 2.7 is provided below. gender Female Male Female Female Female age maritalStatus 51 24 33 17 76 Married Single Married Single Widowed grossIncome £2,600 to £5,200 £10,400 to £15,600 £10,400 to £15,600 £5,200 to £10,400 £5,200 to £10,400 smoke Yes Yes Yes Yes Yes amtWeekends 20 cig/day 20 cig/day 20 cig/day 20 cig/day 20 cig/day amtWeekdays 20 cig/day 15 cig/day 10 cig/day 15 cig/day 20 cig/day (a) Find the mean amount of cigarettes smoked on weekdays and weekends by these 5 respondents. (b) Find the standard deviation of the amount of cigarettes smoked on weekdays and on weekends by these 5 respondents. Is the variability higher on weekends or on weekdays? 2.10 Factory defective rate. A factory quality control manager decides to investigate the percentage of defective items produced each day. Within a given work week (Monday through Friday) the percentage of defective items produced was 2%, 1.4%, 4%, 3%, 2.2%. (a) Calculate the mean for these data. (b) Calculate the standard deviation for these data, showing each step in detail. 40National STEM Centre, Large Datasets from stats4schools. Amount Weekends0102030405060050100Amount Weekdays0102030405060050 2.2. NUMERICAL SUMMARIES AND BOX PLOTS 97 2.11 Days off at a mining plant. Workers at a particular mining site receive an average of 35 days paid vacation, which is lower than the national average. The manager of this plant is under pressure from a local union to increase the amount of paid time oﬀ. However, he does not want to give more days oﬀ to the workers because that would be costly. Instead he decides he should ﬁre 10 employees in such a way as to raise the average number of days oﬀ that are reported by his employees. In order to achieve this goal, should he ﬁre employees who have the most number of days oﬀ, least number of days oﬀ, or those who have about the average number of days oﬀ? 2.12 Medians and IQRs. For each part, compare distributions (1) and (2) based on their medians and IQRs. You do not need t
o calculate these statistics; simply state how the medians and IQRs compare. Make sure to explain your reasoning. (a) (1) 3, 5, 6, 7, 9 (2) 3, 5, 6, 7, 20 (b) (1) 3, 5, 6, 7, 9 (2) 3, 5, 7, 8, 9 (c) (1) 1, 2, 3, 4, 5 (2) 6, 7, 8, 9, 10 (d) (1) 0, 10, 50, 60, 100 (2) 0, 100, 500, 600, 1000 2.13 Means and SDs. For each part, compare distributions (1) and (2) based on their means and standard deviations. You do not need to calculate these statistics; simply state how the means and the standard deviations compare. Make sure to explain your reasoning. Hint: It may be useful to sketch dot plots of the distributions. (a) (1) 3, 5, 5, 5, 8, 11, 11, 11, 13 (2) 3, 5, 5, 5, 8, 11, 11, 11, 20 (c) (1) 0, 2, 4, 6, 8, 10 (2) 20, 22, 24, 26, 28, 30 (b) (1) -20, 0, 0, 0, 15, 25, 30, 30 (2) -40, 0, 0, 0, 15, 25, 30, 30 (d) (1) 100, 200, 300, 400, 500 (2) 0, 50, 300, 550, 600 2.14 Mix-and-match. Describe the distribution in the histograms below and match them to the box plots. (a)506070(b)050100(c)0246(1)0246(2)55606570(3)020406080100 98 CHAPTER 2. SUMMARIZING DATA 2.15 Air quality. Daily air quality is measured by the air quality index (AQI) reported by the Environmental Protection Agency. This index reports the pollution level and what associated health eﬀects might be a concern. The index is calculated for ﬁve major air pollutants regulated by the Clean Air Act and takes values from 0 to 300, where a higher value indicates lower air quality. AQI was reported for a sample of 91 days in 2011 in Durham, NC. The relative frequency histogram below shows the distribution of the AQI values on these days.41 (a) Estimate the median AQI value of this sample. (b) Would you expect the mean AQI value of this sample to be higher or lower than the median? Explain your reasoning. (c) Estimate Q1, Q3, and IQR for the distribution. (d) Would any of the days in this sample be considered to have an unusually low or high AQI? Explain your reasoning. 2.16 Median vs. mean. Estimate the median for the 400 observations shown in the histogram, and note whether you expect the mean to be higher or lower than the median. 2.17 Histograms vs. box plots. Compare the two plots below. What characteristics of the distribution are apparent in the histogram and not in the box plot? What characteristics are apparent in the box plot but not in the histogram? 2.18 Facebook friends. Facebook data indicate that 50% of Facebook users have 100 or more friends, and that the average friend count of users is 190. What do these ﬁndings suggest about the shape of the distribution of number of friends of Facebook users?42 41US Environmental Protection Agency, AirData, 2011. 42Lars Backstrom. “Anatomy of Facebook”. In: Facebook Data Team’s Notes (2011). Daily AQI10203040506000.050.10.150.2405060708090100020406080510152025050100150200510152025 2.2. NUMERICAL SUMMARIES AND BOX PLOTS 99 2.19 Distributions and appropriate statistics, Part I. For each of the following, state whether you expect the distribution to be symmetric, right skewed, or left skewed. Also specify whether the mean or median would best represent a typical observation in the data, and whether the variability of observations would be best represented using the standard deviation or IQR. Explain your reasoning. (a) Number of pets per household. (b) Distance to work, i.e. number of miles between work and home. (c) Heights of adult males. 2.20 Distributions and appropriate statistics, Part II. For each of the following, state whether you expect the distribution to be symmetric, right skewed, or left skewed. Also specify whether the mean or median would best represent a typical observation in the data, and whether the variability of observations would be best represented using the standard deviation or IQR. Explain your reasoning. (a) Housing prices in a country where 25% of the houses cost below $350,000, 50% of the houses cost below $450,000, 75% of the houses cost below $1,000,000 and there are a meaningful number of houses that cost more than $6,000,000. (b) Housing prices in a country where 25% of the houses cost below $300,000, 50% of the houses cost below $600,000, 75% of the houses cost below $900,000 and very few houses that cost more than $1,200,000. (c) Number of alcoholic drinks consumed by college students in a given week. Assume that most of these students don’t drink since they are under 21 years old, and only a few drink excessively. (d) Annual salaries of the employees at a Fortune 500 company where only a few high level executives earn much higher salaries than all the other employees. 2.21 Income at the coffee shop. The ﬁrst histogram below shows the distribution of the yearly incomes of 40 patrons at a college coﬀee shop. Suppose two new people walk into the coﬀee shop: one making $225,000 and the other $250,000. The second histogram shows the new income distribution. Summary statistics are also provided. n Min. 1st Qu. Median Mean 3rd Qu. Max. SD (1) 40 60,680 63,620 65,240 65,090 66,160 69,890 2,122 (2) 42 60,680 63,710 65,350 73,300 66,540 250,000 37,321 (a) Would the mean or the median best represent what we might think of as a typical income for the 42 patrons at this coﬀee shop? What does this say about the robustness of the two measures? (b) Would the standard deviation or the IQR best represent the amount of variability in the incomes of the 42 patrons at this coﬀee shop? What does this say about the robustness of the two measures? 2.22 Midrange. The midrange of a distribution is deﬁned as the average of the maximum and the minimum of that distribution. Is this statistic robust to outliers and extreme skew? Explain your reasoning (1)$60k$62.5k$65k$67.5k$70k04812(2)$60k$110k$160k$210k$260k04812 100 CHAPTER 2. SUMMARIZING DATA 2.23 Commute times. The US census collects data on time it takes Americans to commute to work, among many other variables. The histogram below shows the distribution of average commute times in 3,142 US counties in 2010. Also shown below is a spatial intensity map of the same data. (a) Describe the numerical distribution for commute times. (b) Describe the spatial distribution of commuting times using the map provided. 2.24 Hispanic/Latine population. The US census collects data on race and ethnicity of Americans, among many other variables. The histogram below shows the distribution of the percentage of the population that is Hispanic/Latine in 3,142 counties in the US in 2017. (a) Describe the distribution of percent of population that is Hispanic/Latine for counties in the US. (b) What features of the distribution of the Hispanic/Latine population in US counties are apparent in the map but not in the histogram? What features are apparent in the histogram but not the map? Mean work travel (in min)102030400100200419>33Hispanic/Latinx %0204060801000500100015002000020>40 2.3. NORMAL DISTRIBUTION 101 2.3 Normal distribution What proportion of adults have systolic blood pressure above 140? What is the probability of getting more than 250 heads in 400 tosses of a fair coin? If the average weight of a piece of carry-on luggage is 11 pounds, what is the probability that 200 random carry on pieces will weigh more than 2500 pounds? If 55% of a population supports a certain candidate, what is the probability that she will have less than 50% support in a random sample of size 200? There is one distribution that can help us answer all of these questions. Can you guess what it is? That’s right – it’s the normal distribution. Learning objectives 1. Use Z-scores and the standard normal model to approximate a distribution where appropriate. 2. Find probabilities and percentiles using the normal approximation. 3. Find the value that corresponds to a given percentile when the distribution is approximately normal. 2.3.1 Normal distribution model Among all the distributions we see in practice, one is overwhelmingly the most common. The symmetric, unimodal, bell curve is ubiquitous throughout statistics. Indeed it is so common, that people often know it as the normal curve or normal distribution.43 A normal curve is shown in Figure 2.24. Figure 2.24: A normal curve. The normal distribution always describes a symmetric, unimodal, bell-shaped curve. However, these curves can look diﬀerent depending on the details of the model. Speciﬁcally, the normal distribution model can be adjusted using two parameters: mean and standard deviation. As you can probably guess, changing the mean shifts the bell curve to the left or right, while changing the standard deviation stretches or constricts the curve. 43It is also introduced as the Gaussian distribution after Frederic Gauss, the ﬁrst person to formalize its mathe- matical expression. 102 CHAPTER 2. SUMMARIZING DATA Figure 2.25 shows the normal distribution with mean 0 and standard deviation 1 in the left panel and the normal distributions with mean 19 and standard deviation 4 in the right panel. Figure 2.26 shows these distributions on the same axis. Figure 2.25: Both curves represent the normal distribution. However, they diﬀer in their center and spread. Figure 2.26: The normal distributions shown in Figure 2.25 but plotted together and on the same scale. Because the mean and standard deviation describe a normal distribution exactly, they are called the distribution’s parameters. The normal distribution with mean µ = 0 and standard deviation σ = 1 is called the standard normal distribution.. NORMAL DISTRIBUTION FACTS Many variables are nearly normal, but none are exactly normal. The normal distribution, while never perfect, provides very close approximations for a variety of scenarios. We will use it to model data as well as probability distributions. −3−2−10123Y71115192327310102030 2.3. NORMAL DISTRIBUTION 103 2.3.2 Using the normal distribution to approximate empirical distributions We often want to put data onto a standardized scale, which can make comparisons more reasonable. EXAMPLE 2.58 Figure 2.27 shows the mean and standard deviation fo
r total scores on the SAT and ACT. The distribution of SAT and ACT scores are both nearly normal. Suppose Ann scored 1300 on her SAT and Tom scored 24 on his ACT. Who performed better? As we saw in section 2.2.3, we can use Z-scores to compare observations from diﬀerent distributions. Using Ann’s SAT score, 1300 , along with the SAT mean and SD, we can ﬁnd Ann’s Z-score. ZAnn = xAnn − µSAT σSAT = 1300 − 1100 200 = 1 Similarly, using Tom’s ACT score, 24, along with the ACT mean and SD we can ﬁnd his Z-score. ZT om = xTom − µACT σACT = 24 − 21 6 = 0.5 Because Ann’s score was 1 standard deviation above the mean, while Tom’s score was 0.5 standard deviations above the mean, we can say that Ann did better than Tom. Mean SD SAT ACT 21 1100 6 200 Figure 2.27: Mean and standard deviation for the SAT and ACT. Assuming that both the SAT and ACT distributions are nearly normally distributed, what percent of test takers scored lower than Ann? What percent scored lower than Tom? To answer these question exactly, we would need all of the data. However, if we use the information that SAT and ACT distributions are nearly normal, we can estimate these percents. Figure 2.28 shows these distributions modeled with a normal curve. If we can ﬁnd the percent of the normal curve that is to the left of Ann’s score, we could use that percent as our estimate of the percent of the data points that are smaller than Ann’s score. We call this process normal approximation. The steps are: 1. First verify that the distribution can be reasonably modeled with a normal distribution. 2. Convert value or values of interest to Z-scores. 3. Find the relevant area/percent under the standard normal curve. We use the area/percent that we ﬁnd from the normal curve as our estimate of the desired percent. Figure 2.28: Ann’s and Tom’s scores shown with the distributions of SAT and ACT scores. X700900110013001500Ann915212733Tom 104 CHAPTER 2. SUMMARIZING DATA 2.3.3 Finding areas under the normal curve It’s very useful in statistics to be able to identify areas of distributions, especially tail areas. For instance, what percent of people have an SAT score below Ann’s score of 1300? This is the same as Ann’s percentile. We previously determined that a score of 1300 corresponds to a Z-score of 1 and that SAT scores are approximately normally distributed. We can visualize such a tail area by sketching a normal curve and shading everything below Z = 1 as shown in Figure 2.29. Figure 2.29: The area to the left of the Z-score represents the percentile of the observation. There are many techniques for ﬁnding this area, and we’ll discuss three of the options. 1. The most common approach in practice is to use statistical software. For example, in the program R, we could ﬁnd the area shown in Figure 2.29 using the following command, which takes in the Z-score of 1 and returns the lower tail area: .....> pnorm(1) .....[1] 0.8413447 Using the online Desmos calculator, we could do: normaldist( ), check the “Find Cumulative Probability (CDF)” box and set Max to 1. According to these calculation, the area shaded that is below Z = 1 is 0.841, so we estimate that 84.1% of SAT test takers score below 1300 and that Ann is at the 84th percentile. There are many other software options, such as Python or SAS; even spreadsheet programs such as Excel and Google Sheets support these calculations. 2. A common strategy in classrooms is to use a graphing calculator, such as a TI or Casio calculaInstructions for ﬁnding areas of a normal distribution using these calculators are provided in tor. Section 2.3.7. 3. The last option for ﬁnding tail areas is to use what’s called a probability table; these are occasionally used in classrooms but rarely in practice. Appendix C.2 contains such a table and a guide for how to use it. We will solve normal distribution problems in this section by always ﬁrst ﬁnding the Z-score. The reason is that we will encounter close parallels called test statistics beginning in Chapter 5; these are, in many instances, an equivalent of a Z-score. Readers may ﬁnd it helpful to familiarize themselves with one of the options above before continuing on to the applications that follow. −3−2−10123 2.3. NORMAL DISTRIBUTION 105 2.3.4 Normal probability examples Combined SAT scores are approximated well by a normal model with mean 1100 and standard deviation 200. EXAMPLE 2.59 What is the probability that a randomly selected SAT taker scores at least 1190 on the SAT? The probability that a randomly selected SAT taker scores at least 1190 on the SAT is equivalent to the proportion of all SAT takers that score at least 1190 on the SAT. First, always draw and label a picture of the normal distribution. (Drawings need not be exact to be useful.) We are interested in the probability that a randomly selected score will be above 1190, so we shade this upper tail: The picture shows the mean and the values at 2 standard deviations above and below the mean. The simplest way to ﬁnd the shaded area under the curve makes use of the Z-score of the cutoﬀ value. With µ = 1100, σ = 200, and the cutoﬀ value x = 1190, the Z-score is computed as Z = x − µ σ = 1190 − 1100 200 = 90 200 = 0.45 Next, we want to ﬁnd the area under the normal curve to right of Z = 0.45. Using technology, we ﬁnd P (Z > 0.45) = 0.3264. The probability that a randomly selected score is at least 1190 on the SAT is 0.3264. ALWAYS DRAW A PICTURE FIRST, AND FIND THE Z-SCORE SECOND For any normal probability situation, always always always draw and label the normal curve and shade the area of interest ﬁrst. The picture will provide an estimate of the probability. After drawing a ﬁgure to represent the situation, identify the Z-score for the observation of interest. GUIDED PRACTICE 2.60 If the probability that a randomly selected score is at least 1190 is 0.3264, what is the probability that the score is less than 1190? Draw the normal curve representing this exercise, shading the lower region instead of the upper one.44 44We found the probability in Example 2.59: 0.6736. A picture for this exercise is represented by the shaded area below “0.6736” in Example 2.59. 70011001500 106 CHAPTER 2. SUMMARIZING DATA EXAMPLE 2.61 Edward earned a 1030 on his SAT. What is his percentile? First, a picture is needed. Edward’s percentile is the proportion of people who do not get as high as a 1030. These are the scores to the left of 1030. Identifying the mean µ = 1100, the standard deviation σ = 200, and the cutoﬀ for the tail area x = 1030 makes it easy to compute the Z-score: Z = x − µ σ = 1030 − 1100 200 = −0.35 Using technology we ﬁnd that P (Z < −0.35) = 0.3632. Edward is at the 36th percentile. GUIDED PRACTICE 2.62 Use the results of Example 2.61 to compute the proportion of SAT takers who did better than Edward. Also draw a new picture.45 The last several problems have focused on ﬁnding the probability or percentile for a particular obser- vation. It is also possible to identify the value corresponding to a particular percentile. 45If Edward did better than 36% of SAT takers, then about 64% must have done better than him. 7001100150070011001500 2.3. NORMAL DISTRIBUTION 107 EXAMPLE 2.63 Carlos believes he can get into his preferred college if he scores at least in the 80th percentile on the SAT. What score should he aim for? Here, we are given a percentile rather than a Z-score, so we work backwards. As always, ﬁrst draw the picture. We want to ﬁnd the observation that corresponds to the 80th percentile. First, we ﬁnd the Z-score associated with the 80th percentile. Using technology, we ﬁnd that P (Z < 0.84) = 0.80. In any normal distribution, a value with a Z-score of 0.84 will be at the 80th percentile. Once we have the Z-score, we work backwards to ﬁnd x. Z = x − µ σ 0.84 = x − 1100 200 0.84 × 200 + 1100 = x x = 1268 The 80th percentile on the SAT corresponds to a score of 1268. GUIDED PRACTICE 2.64 Imani scored at the 72nd percentile on the SAT. What was her SAT score?46 IF THE DATA ARE NOT NEARLY NORMAL, DON’T USE THE NORMAL APPROXIMATION Before using the normal approximation method, verify that the data or distribution is approximately normal. If it is not, the normal approximation will give incorrect results. Also remember that all answers based on normal approximations are in fact approximations and are not exact. Finally, we should observe that it is possible for a normal random variable to fall 4, 5, or even more standard deviations from the mean. The probability of being further than 4 standard deviations from the mean is about 1-in-15,000. For 5 and 6 standard deviations, it is about 1-in-2 million and 1-in-500 million, respectively. However, while the tails of the normal distribution extend inﬁnitely in either direction, our data sets are ﬁnite and normal approximation in the extreme tails is unlikely to be very accurate, even for bell-shaped data sets. 46First, draw a picture! The closest percentile in the table to 0.72 is 0.7190, which corresponds to Z = 0.58. Next, set up the Z-score formula and solve for x: 0.58 = x−1100 200 → x = 1216. Imani scored 1216. 5007009001100130015001700 108 CHAPTER 2. SUMMARIZING DATA 2.3.5 68-95-99.7 rule Here, we present a useful rule of thumb for the probability of falling within 1, 2, and 3 standard deviations of the mean in the normal distribution. The 68-95-99.7 rule, also known as the empirical rule, will be useful in a wide range of practical settings, especially when trying to make a quick estimate without a calculator or Z-table. Figure 2.30: Probabilities for falling within 1, 2, and 3 standard deviations of the mean in a normal distribution. GUIDED PRACTICE 2.65 Use the Z-table to conﬁrm that about 68%, 95%, and 99.7% of observations fall within 1, 2, and 3, standard deviations of the mean in the normal distribution, respectively. For instance, ﬁrst ﬁnd the area that falls between Z = −1 and Z = 1, which should have an area of about 0.68. Similarly there should be an area of a
bout 0.95 between Z = −2 and Z = 2.47 It is possible for a normal random variable to fall 4, 5, or even more standard deviations from the mean. However, these occurrences are very rare if the data are nearly normal. The probability of being further than 4 standard deviations from the mean is about 1-in-15,000. For 5 and 6 standard deviations, it is about 1-in-2 million and 1-in-500 million, respectively. GUIDED PRACTICE 2.66 SAT scores closely follow the normal model with mean µ = 1100 and standard deviation σ = 200. (a) About what percent of test takers score 700 to 1500? (b) What percent score between 1100 and 1500?48 47First draw the pictures. To ﬁnd the area between Z = −1 and Z = 1, use the normal probability table to determine the areas below Z = −1 and above Z = 1. Next verify the area between Z = −1 and Z = 1 is about 0.68. Repeat this for Z = −2 to Z = 2 and also for Z = −3 to Z = 3. 48(a) 700 and 1500 represent two standard deviations above and below the mean, which means about 95% of test takers will score between 700 and 1500. (b) Since the normal model is symmetric, then half of the test takers from part (a) ( 95% 2 = 47.5% of all test takers) will score 700 to 1500 while 47.5% score between 1100 and 1500. m-3sm-2sm-smm+sm+2sm+3s99.7%95%68% 2.3. NORMAL DISTRIBUTION 109 2.3.6 Evaluating the normal approximation (special topic) It is important to remember normality is always an approximation. Testing the appropriateness of the normal assumption is a key step in many data analyses. The distribution of heights of US males is well approximated by the normal model. We are interested in proceeding under the assumption that the data are normally distributed, but ﬁrst we must check to see if this is reasonable. There are two visual methods for checking the assumption of normality that can be implemented and interpreted quickly. The ﬁrst is a simple histogram with the best ﬁtting normal curve overlaid on the plot, as shown in the left panel of Figure 2.31. The sample mean ¯x and standard deviation s are used as the parameters of the best ﬁtting normal curve. The closer this curve ﬁts the histogram, the more reasonable the normal model assumption. Another more common method is examining a normal probability plot,49 shown in the right panel of Figure 2.31. The closer the points are to a perfect straight line, the more conﬁdent we can be that the data follow the normal model. Figure 2.31: A sample of 100 male heights. The observations are rounded to the nearest whole inch, explaining why the points appear to jump in increments in the normal probability plot. EXAMPLE 2.67 Consider all NBA players from the 2018-2019 season presented in Figure 2.32. Based on the graphs, are NBA player heights normally distributed? We ﬁrst create a histogram and normal probability plot of the NBA player heights. The histogram in the left panel is slightly left skewed, which contrasts with the symmetric normal distribution. The points in the normal probability plot do not appear to closely follow a straight line but show what appears to be a “wave”. NBA player heights do not appear to come from a normal distribution. GUIDED PRACTICE 2.68 Figure 2.33 shows normal probability plots for two distributions that are skewed. One distribution is skewed to the low end (left skewed) and the other to the high end (right skewed). Which is which?50 49Also commonly called a quantile-quantile plot. 50Examine where the points fall along the vertical axis. In the ﬁrst plot, most points are near the low end with fewer observations scattered along the high end; this describes a distribution that is right skewed. The second plot shows the opposite features, and this distribution is left skewed. Male Heights (inches)6065707580llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllTheoretical QuantilesMale Heights (inches)−2−1012657075 110 CHAPTER 2. SUMMARIZING DATA Figure 2.32: Histogram and normal probability plot for the NBA heights from the 2018-2019 season. Figure 2.33: Normal probability plots for Guided Practice 2.68. 2.3.7 Technology: ﬁnding normal probabilities Get started quickly with a Desmos Normal Calculator that we’ve put together (visit openintro.org/ahss/desmos). Height (inches)70758085llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllTheoretical Quantilesobserved−3−2−1012370758085lllllllllllllllllllllllllObservedTheoretical quantiles−2−1012012llllllllllllllllllllllllllllllllllllllllllllllllllObservedTheoretical quantiles−2−101251015 2.3. NORMAL DISTRIBUTION 111 TI-84: FINDING AREA UNDER THE NORMAL CURVE Use 2ND VARS, normalcdf to ﬁnd an area/proportion/probability between two Z-scores or to the left or right of a Z-score. 1. Choose 2ND VARS (i.e. DISTR). 2. Choose 2:normalcdf. 3. Enter the lower (left) Z-score and the upper (right) Z-score. • If ﬁnding just a lower tail area, set lower to -5. • If ﬁnding just an upper tail area, set upper to 5. 4. Leave µ as 0 and σ as 1. 5. Down arrow, choose Paste, and hit ENTER. TI-83: Do steps 1-2, then enter the lower bound and upper bound separated by a comma, e.g. normalcdf(2, 5), and hit ENTER. CASIO FX-9750GII: FINDING AREA UNDER THE NORMAL CURVE 1. Navigate to STAT (MENU, then hit 2). 2. Select DIST (F5), then NORM (F1), and then Ncd (F2). 3. If needed, set Data to Variable (Var option, which is F2). 4. Enter the Lower Z-score and the Upper Z-score. Set σ to 1 and µ to 0. • If ﬁnding just a lower tail area, set Lower to -5. • For an upper tail area, set Upper to 5. 5. Hit EXE, which will return the area probability (p) along with the Z-scores for the lower and upper bounds. GUIDED PRACTICE 2.69 Use a calculator or software to conﬁrm that about 68%, 95%, and 99.7% of observations fall within 1, 2, and 3, standard deviations of the mean in the normal distribution, respectively.51 GUIDED PRACTICE 2.70 Find the area under the normal curve between -1.5 and 1.5. 52 EXAMPLE 2.71 Use a calculator to determine what percentile corresponds to a Z-score of 1.5 for a normal distribution.53 To ﬁnd an area under the normal curve using a calculator, ﬁrst identify a lower bound and an upper bound. We want all of the area to the left of 1.5, so the lower bound should be -∞. However, the area under the curve is negligible when Z is smaller than -5, so we will use -5 as the lower bound. Using a lower bound of -5 and an upper bound of 1.5, we get P (Z < 1.5) = 0.933. 51To ﬁnd the area between Z = −1 and Z = 1, let lower bound be -1 and upper bound be 1. We ﬁnd that P (−1 < Z < 1) = 0.6827. Similarly, P (−2 < Z < 2) = 0.9545 and P (−3 < Z < 3) = 0.9973. 52Lower bound is -1.5 and upper bound is 1.5. The area under the normal curve between -1.5 and 1.5 = P (−1.5 < Z < 1.5) = 0.866. Note that is not simply the average of 0.6827 and 0.9545, as the normal curve is not a rectangle. 53normalcdf gives the result without drawing the graph. To draw the graph, do 2nd VARS, DRAW, 1:ShadeNorm. However, beware of errors caused by other plots that might interfere with this plot. −3−2−10123 112 CHAPTER 2. SUMMARIZING DATA GUIDED PRACTICE 2.72 Find the area under the normal curve to right of Z = 2. 54 TI-84: FIND A Z-SCORE THAT CORRESPONDS TO A PERCENTILE Use 2ND VARS, invNorm to ﬁnd the Z-score that corresponds to a given percentile. 1. Choose 2ND VARS (i.e. DISTR). 2. Choose 3:invNorm. 3. Let Area be the percentile as a decimal (the area to the left of desired Z-score). 4. Leave µ as 0 and σ as 1. 5. Down arrow, choose Paste, and hit ENTER. TI-83: Do steps 1-2, then enter the percentile as a decimal, e.g. invNorm(.40), then hit ENTER. CASIO FX-9750GII: FIND A Z-SCORE THAT CORRESPONDS TO A PERCENTILE 1. Navigate to STAT (MENU, then hit 2). 2. Select DIST (F5), then NORM (F1), and then InvN (F3). 3. If needed, set Data to Variable (Var option, which is F2). 4. Decide which tail area to use (Tail), the tail area (Area), and then enter the σ and µ values. 5. Hit EXE. EXAMPLE 2.73 Use a calculator to ﬁnd the Z-score that corresponds to the 40th percentile. Letting area be 0.40, a calculator gives -0.253. This means that Z = −0.253 corresponds to the 40th percentile, that is, P (Z < −0.253) = 0.40. GUIDED PRACTICE 2.74 Find the Z-score such that 20 percent of the area is to the right of that Z-score.55 54Now we want to shade to the right. Therefore our lower bound will be 2 and the upper bound will be +5 (or a number bigger than 5) to get P (Z > 2) = 0.023. 55If 20% of the area is the right, then 80% of the area is to the left. Letting area be 0.80, we get Z = 0.841. 2.3. NORMAL DISTRIBUTION 113 Section summary • A Z-score represents the number of standard deviations a value in a data set is above or below the mean. To calculate a Z-score use: Z = x−mean SD . • The normal distribution is the most commonly used distribution in Statistics. Many distribution are approximately normal, but none are exactly normal. • The empirical rule (68-95-99.7 Rule) comes from the normal distribution. The closer a distribution is to normal, the better this rule will hold. • It is often useful to use the standard normal distribution, which has mean 0 and SD 1, to approximate a discrete histogram. There are two common types of normal approximation problems, and for each a key step is to ﬁnd a Z-score. A: Find the percent or probability of a value greater/less than a given x-value. 1. Verify that the distribution of interest is approximately normal. 2. Calculate the Z-score. Use the provided population mean and SD to standardize the given x-value. 3. Use a calculator function
(e.g. normcdf on a TI) or other technology to ﬁnd the area under the normal curve to the right/left of this Z-score; this is the estimate for the percent/probability. B: Find the x-value that corresponds to a given percentile. 1. Verify that the distribution of interest is approximately normal. 2. Find the Z-score that corresponds to the given percentile (using, for example, invNorm on a TI). 3. Use the Z-score along with the given mean and SD to solve for the x -value. 114 CHAPTER 2. SUMMARIZING DATA Exercises 2.25 Area under the curve, Part I. What percent of a standard normal distribution N (µ = 0, σ = 1) is found in each region? Be sure to draw a graph. (a) Z < −1.35 (b) Z > 1.48 (c) −0.4 < Z < 1.5 (d) \|Z\| > 2 2.26 Area under the curve, Part II. What percent of a standard normal distribution N (µ = 0, σ = 1) is found in each region? Be sure to draw a graph. (a) Z > −1.13 (b) Z < 0.18 (c) Z > 8 (d) \|Z\| < 0.5 2.27 GRE scores, Part I. Sophia who took the Graduate Record Examination (GRE) scored 160 on the Verbal Reasoning section and 157 on the Quantitative Reasoning section. The mean score for Verbal Reasoning section for all test takers was 151 with a standard deviation of 7, and the mean score for the Quantitative Reasoning was 153 with a standard deviation of 7.67. Suppose that both distributions are nearly normal. (a) What is Sophia’s Z-score on the Verbal Reasoning section? On the Quantitative Reasoning section? Draw a standard normal distribution curve and mark these two Z-scores. (b) What do these Z-scores tell you? (c) Relative to others, which section did she do better on? (d) Find her percentile scores for the two exams. (e) What percent of the test takers did better than her on the Verbal Reasoning section? On the Quantitative Reasoning section? (f) Explain why simply comparing raw scores from the two sections could lead to an incorrect conclusion as to which section a student did better on. (g) If the distributions of the scores on these exams are not nearly normal, would your answers to parts (b) - (e) change? Explain your reasoning. 2.28 Triathlon times, Part I. In triathlons, it is common for racers to be placed into age and gender groups. Friends Leo and Mary both completed the Hermosa Beach Triathlon, where Leo competed in the Men, Ages 30 - 34 group while Mary competed in the Women, Ages 25 - 29 group. Leo completed the race in 1:22:28 (4948 seconds), while Mary completed the race in 1:31:53 (5513 seconds). Obviously Leo ﬁnished faster, but they are curious about how they did within their respective groups. Can you help them? Here is some information on the performance of their groups: • The ﬁnishing times of the Men, Ages 30 - 34 group has a mean of 4313 seconds with a standard deviation of 583 seconds. • The ﬁnishing times of the Women, Ages 25 - 29 group has a mean of 5261 seconds with a standard deviation of 807 seconds. • The distributions of ﬁnishing times for both groups are approximately Normal. Remember: a better performance corresponds to a faster ﬁnish. (a) What are the Z-scores for Leo’s and Mary’s ﬁnishing times? What do these Z-scores tell you? (b) Did Leo or Mary rank better in their respective groups? Explain your reasoning. (c) What percent of the triathletes did Leo ﬁnish faster than in his group? (d) What percent of the triathletes did Mary ﬁnish faster than in her group? (e) If the distributions of ﬁnishing times are not nearly normal, would your answers to parts (a) - (d) change? Explain your reasoning. 2.29 GRE scores, Part II. In Exercise 2.27 we saw two distributions for GRE scores: N (µ = 151, σ = 7) for the verbal part of the exam and N (µ = 153, σ = 7.67) for the quantitative part. Use this information to compute each of the following: (a) The score of a student who scored in the 80th percentile on the Quantitative Reasoning section. (b) The score of a student who scored worse than 70% of the test takers in the Verbal Reasoning section. 2.3. NORMAL DISTRIBUTION 115 2.30 Triathlon times, Part II. In Exercise 2.28 we saw two distributions for triathlon times: N (µ = 4313, σ = 583) for Men, Ages 30 - 34 and N (µ = 5261, σ = 807) for the Women, Ages 25 - 29 group. Times are listed in seconds. Use this information to compute each of the following: (a) The cutoﬀ time for the fastest 5% of athletes in the men’s group, i.e. those who took the shortest 5% of time to ﬁnish. (b) The cutoﬀ time for the slowest 10% of athletes in the women’s group. 2.31 LA weather, Part I. deviation of 5◦F. Suppose that the temperatures in June closely follow a normal distribution. (a) What is the probability of observing an 83◦F temperature or higher in LA during a randomly chosen The average daily high temperature in June in LA is 77◦F with a standard day in June? (b) How cool are the coldest 10% of the days (days with lowest high temperature) during June in LA? 2.32 CAPM. The Capital Asset Pricing Model (CAPM) is a ﬁnancial model that assumes returns on a portfolio are normally distributed. Suppose a portfolio has an average annual return of 14.7% (i.e. an average gain of 14.7%) with a standard deviation of 33%. A return of 0% means the value of the portfolio doesn’t change, a negative return means that the portfolio loses money, and a positive return means that the portfolio gains money. (a) What percent of years does this portfolio lose money, i.e. have a return less than 0%? (b) What is the cutoﬀ for the highest 15% of annual returns with this portfolio? 2.33 LA weather, Part II. Exercise 2.31 states that average daily high temperature in June in LA is 77◦F with a standard deviation of 5◦F, and it can be assumed that they to follow a normal distribution. We use the following equation to convert ◦F (Fahrenheit) to ◦C (Celsius): C = (F − 32) × 5 9 . (a) What is the probability of observing a 28◦C (which roughly corresponds to 83◦F) temperature or higher in June in LA? Calculate using the ◦C model from part (a). (b) Did you get the same answer or diﬀerent answers in part (b) of this question and part (a) of Exercise 2.31? Are you surprised? Explain. (c) Estimate the IQR of the temperatures (in ◦C) in June in LA. 2.34 Find the SD. Cholesterol levels for women aged 20 to 34 follow an approximately normal distribution with mean 185 milligrams per deciliter (mg/dl). Women with cholesterol levels above 220 mg/dl are considered to have high cholesterol and about 18.5% of women fall into this category. What is the standard deviation of the distribution of cholesterol levels for women aged 20 to 34? 2.35 Scores on stats ﬁnal, Part I. Below are ﬁnal exam scores of 20 Introductory Statistics students. 1 57, 2 66, 3 69, 4 71, 5 72, 6 73, 7 74, 8 77, 9 78, 10 78, 11 79, 12 79, 13 81, 14 81, 15 82, 16 83, 17 83, 18 88, 19 89, 20 94 The mean score is 77.7 points. with a standard deviation of 8.44 points. Use this information to determine if the scores approximately follow the 68-95-99.7% Rule. 2.36 Heights of female college students, Part I. Below are heights of 25 female college students. 1 54, 2 55, 3 56, 4 56, 5 57, 6 58, 7 58, 8 59, 9 60, 10 60, 11 60, 12 61, 13 61, 14 62, 15 62, 16 63, 17 63, 18 63, 19 64, 20 65, 21 65, 22 67, 23 67, 24 69, 25 73 The mean height is 61.52 inches with a standard deviation of 4.58 inches. Use this information to determine if the heights approximately follow the 68-95-99.7% Rule. 116 CHAPTER 2. SUMMARIZING DATA 2.4 Considering categorical data How do we visualize and summarize categorical data? In this section, we will introduce tables and other basic tools for categorical data that are used throughout this book and will answer the following questions: • Based on the loan50 data, is there an assocation between the categorical variables of homeownership and application type (individual, joint)? • Using the email50 data, does email type provide any useful value in classifying email as spam or not spam? Learning objectives 1. Use a one-way table and a bar chart to summarize a categorical variable. Use counts (frequency) or proportions (relative frequency). 2. Compare distributions of a categorical variable using a two-way table and a side-by-side bar chart, segmented bar chart, or mosaic plot. 3. Calculate marginal and joint frequencies for two-way tables. 2.4.1 Contingency tables and bar charts Figure 2.34 summarizes two variables: app type and homeownership. A table that summarizes data for two categorical variables in this way is called a contingency table. Each value in the table represents the number of times a particular combination of variable outcomes occurred. For example, the value 3496 corresponds to the number of loans in the data set where the borrower rents their home and the application type was by an individual. Row and column totals are also included. The row totals provide the total counts across each row (e.g. 3496 + 3839 + 1170 = 8505), and column totals are total counts down each column. We can also create a table that shows only the overall percentages or proportions for each combination of categories, or we can create a table for a single variable, such as the one shown in Figure 2.35 for the homeownership variable. app type individual joint Total homeownership rent mortgage 3496 362 3858 3839 950 4789 own 1170 183 1353 Total 8505 1495 10000 Figure 2.34: A contingency table for app type and homeownership. homeownership Count 3858 rent 4789 mortgage 1353 own 10000 Total Figure 2.35: A table summarizing the frequencies of each value for the homeownership variable. A bar chart (also called bar plot or bar graph) is a common way to display a single categorical variable. The left panel of Figure 2.36 shows a bar chart for the homeownership variable. In the right panel, the counts are converted into proportions, showing the proportion of observations that are in each level (e.g. 3858/10000 = 0.3858 for rent). 2.4. CONSIDERING CATEGORICAL DATA 117 Figure 2.36: Two bar charts of number. The left panel shows the counts, and the right panel shows the proport
ions in each group. 2.4.2 Row and column proportions Sometimes it is useful to understand the fractional breakdown of one variable in another, and we can modify our contingency table to provide such a view. Figure 2.37 shows the row proportions for Figure 2.34, which are computed as the counts divided by their row totals. The value 3496 at the intersection of individual and rent is replaced by 3496/8505 = 0.411, i.e. 3496 divided by its row total, 8505. So what does 0.411 represent? It corresponds to the proportion of individual applicants who rent. individual joint Total rent mortgage 0.451 0.635 0.479 0.411 0.242 0.386 own Total 1.000 1.000 1.000 0.138 0.122 0.135 Figure 2.37: A contingency table with row proportions for the app type and homeownership variables. The row total is oﬀ by 0.001 for the joint row due to a rounding error. A contingency table of the column proportions is computed in a similar way, where each column proportion is computed as the count divided by the corresponding column total. Figure 2.38 shows such a table, and here the value 0.906 indicates that 90.6% of renters applied as individuals for the loan. This rate is higher compared to loans from people with mortgages (80.2%) or who own their home (85.1%). Because these rates vary between the three levels of homeownership (rent, mortgage, own), this provides evidence that the app type and homeownership variables are associated. individual joint Total rent mortgage 0.802 0.198 1.000 0.906 0.094 1.000 own Total 0.851 0.150 1.000 0.865 0.135 1.000 Figure 2.38: A contingency table with column proportions for the app type and homeownership variables. The total for the last column is oﬀ by 0.001 due to a rounding error. We could also have checked for an association between app type and homeownership in Figure 2.37 using row proportions. When comparing these row proportions, we would look down columns to see if the fraction of loans where the borrower rents, has a mortgage, or owns varied across the individual to joint application types. rentmortgageownFrequency01000200030004000Homeownershiprentmortgageown0.00.10.20.30.4ProportionHomeownership 118 CHAPTER 2. SUMMARIZING DATA GUIDED PRACTICE 2.75 (a) What does 0.451 represent in Figure 2.37? (b) What does 0.802 represent in Figure 2.38?56 GUIDED PRACTICE 2.76 (a) What does 0.122 at the intersection of joint and own represent in Figure 2.37? (b) What does 0.135 represent in the Figure 2.38?57 EXAMPLE 2.77 Data scientists use statistics to ﬁlter spam from incoming email messages. By noting speciﬁc characteristics of an email, a data scientist may be able to classify some emails as spam or not spam with high accuracy. One such characteristic is whether the email contains no numbers, small numbers, or big numbers. Another characteristic is the email format, which indicates whether or not an email has any HTML content, such as bolded text. We’ll focus on email format and spam status using the email data set, and these variables are summarized in a contingency table in Figure 2.39. Which would be more helpful to someone hoping to classify email as spam or regular email for this table: row or column proportions? A data scientist would be interested in how the proportion of spam changes within each email format. This corresponds to column proportions: the proportion of spam in plain text emails and the proportion of spam in HTML emails. If we generate the column proportions, we can see that a higher fraction of plain text emails are spam (209/1195 = 17.5%) than compared to HTML emails (158/2726 = 5.8%). This information on its own is insuﬃcient to classify an email as spam or not spam, as over 80% of plain text emails are not spam. Yet, when we carefully combine this information with many other characteristics, we stand a reasonable chance of being able to classify some emails as spam or not spam with conﬁdence. text HTML Total 367 158 209 spam 3554 2568 not spam 986 3921 2726 1195 Total Figure 2.39: A contingency table for spam and format. Example 2.77 points out that row and column proportions are not equivalent. Before settling on one form for a table, it is important to consider each to ensure that the most useful table is constructed. However, sometimes it simply isn’t clear which, if either, is more useful. EXAMPLE 2.78 Look back to Tables 2.37 and 2.38. Are there any obvious scenarios where one might be more useful than the other? None that we thought were obvious! What is distinct about app type and homeownership vs the email example is that these two variables don’t have a clear explanatory-response variable relationship that we might hypothesize (see Section 1.3.3 for these terms). Usually it is most useful to “condition” on the explanatory variable. For instance, in the email example, the email format was seen as a possible explanatory variable of whether the message was spam, so we would ﬁnd it more interesting to compute the relative frequencies (proportions) for each email format. 56(a) 0.451 represents the proportion of individual applicants who have a mortgage. (b) 0.802 represents the fraction of applicants with mortgages who applied as individuals. 57(a) 0.122 represents the fraction of joint borrowers who own their home. (b) 0.135 represents the home-owning borrowers who had a joint application for the loan. 2.4. CONSIDERING CATEGORICAL DATA 119 2.4.3 Using a bar chart with two variables Contingency tables using row or column proportions are especially useful for examining how two categorical variables are related. Segmented bar charts provide a way to visualize the information in these tables. A segmented bar chart, or stacked bar chart, is a graphical display of contingency table information. For example, a segmented bar chart representing Figure 2.38 is shown in Figure 2.40(a), where we have ﬁrst created a bar chart using the homeownership variable and then divided each group by the levels of app type. One related visualization to the segmented bar chart is the side-by-side bar chart, where an example is shown in Figure 2.40(b). For the last type of bar chart we introduce, the column proportions for the app type and homeownership contingency table have been translated into a standardized segmented bar chart in Figure 2.40(c). This type of visualization is helpful in understanding the fraction of individual or joint loan applications for borrowers in each level of homeownership. Additionally, since the proportions of joint and individual vary across the groups, we can conclude that the two variables are associated. (a) (b) (c) (d) (a) segmented bar chart for homeownership, where the counts Figure 2.40: (b) Side-by-side bar chart for have been further broken down by app type. homeownership and app type. (c) Standardized version of the segmented bar chart. (d) Standardized side-by-side bar chart. See these bar charts on Tableau Public . rentmortgageown01000200030004000jointindividualFrequencyrentmortgageown01000200030004000jointindividualFrequencyrentmortgageown0.00.20.40.60.81.0jointindividualProportionrentmortgageown0.00.20.40.60.81.0Proportionjointindividual 120 CHAPTER 2. SUMMARIZING DATA EXAMPLE 2.79 Examine the four bar charts in Figure 2.40. When is the segmented, side-by-side, standardized segmented bar chart, or standardized side-by-side the most useful? The segmented bar chart is most useful when it’s reasonable to assign one variable as the explanatory variable and the other variable as the response, since we are eﬀectively grouping by one variable ﬁrst and then breaking it down by the others. Side-by-side bar charts are more agnostic in their display about which variable, if any, represents the explanatory and which the response variable. It is also easy to discern the number of cases in of the six diﬀerent group combinations. However, one downside is that it tends to require more horizontal space; the narrowness of Figure 2.40(b) makes the plot feel a bit cramped. Additionally, when two groups are of very diﬀerent sizes, as we see in the own group relative to either of the other two groups, it is diﬃcult to discern if there is an association between the variables. The standardized segmented bar chart is helpful if the primary variable in the segmented bar chart is relatively imbalanced, e.g. the own category has only a third of the observations in the mortgage category, making the simple segmented bar chart less useful for checking for an association. The major downside of the standardized version is that we lose all sense of how many cases each of the bars represents. The last plot is a standardized side-by-side bar chart. It shows the joint and individual groups as proportions within each level of homeownership, and it oﬀers similar beneﬁts and tradeoﬀs to the standardized version of the stacked bar plot. 2.4.4 Mosaic plots A mosaic plot is a visualization technique suitable for contingency tables that resembles a standardized segmented bar chart with the beneﬁt that we still see the relative group sizes of the primary variable as well. To get started in creating our ﬁrst mosaic plot, we’ll break a square into columns for each category of the homeownership variable, with the result shown in Figure 2.41(a). Each column represents a level of homeownership, and the column widths correspond to the proportion of loans in each of those categories. For instance, there are fewer loans where the borrower is an owner than where the borrower has a mortgage. In general, mosaic plots use box areas to represent the number of cases in each category. (a) (b) Figure 2.41: (a) The one-variable mosaic plot for homeownership. (b) Two-variable mosaic plot for both homeownership and app type. To create a completed mosaic plot, the single-variable mosaic plot is further divided into pieces in Figure 2.41(b) using the app type variable. Each column is split proportional to the number of loans from individual and joint borrowers. For example, the second column represents loans where the borro
wer has a mortgage, and it was divided into individual loans (upper) and joint loans (lower). As another example, the bottom segment of the third column represents loans where the borrower owns their home and applied jointly, while the upper segment of this column represents borrowers who are homeowners and ﬁled individually. We can again use this plot to see that the homeownership and app type variables are associated, since some columns are divided in diﬀerent vertical locations than others, which was the same technique used for checking an association in the standardized segmented bar chart. rentmortgageownrentmortgageownjointindiv. 2.4. CONSIDERING CATEGORICAL DATA 121 In Figure 2.42, we chose to ﬁrst split by the homeowner status of the borrower. However, we could have instead ﬁrst split by the application type, as in Figure 2.42. Like with the bar charts, it’s common to use the explanatory variable to represent the ﬁrst split in a mosaic plot, and then for the response to break up each level of the explanatory variable, if these labels are reasonable to attach to the variables under consideration. Figure 2.42: Mosaic plot where loans are grouped by the homeownership variable after they’ve been divided into the individual and joint application types. 2.4.5 The only pie chart you will see in this book A pie chart is shown in Figure 2.43 alongside a bar chart representing the same information. Pie charts can be useful for giving a high-level overview to show how a set of cases break down. However, it is also diﬃcult to decipher details in a pie chart. For example, it takes a couple seconds longer to recognize that there are more loans where the borrower has a mortgage than rent when looking at the pie chart, while this detail is very obvious in the bar chart. While pie charts can be useful, we prefer bar charts for their ease in comparing groups. Figure 2.43: A pie chart and bar chart of homeownership. Compare multiple ways of summarizing a single categorical variable on Tableau Public . indiv.jointrentmortgageownrentmortgageownrentmortgageownHomeownership01000200030004000Frequency 122 CHAPTER 2. SUMMARIZING DATA Section summary • Categorical variables, unlike numerical variables, are simply summarized by counts (how many) and proportions. These are referred to as frequency and relative frequency, respectively. • When summarizing one categorical variable, a one-way frequency table is useful. For summarizing two categorical variables and their relationship, use a two-way frequency table (also known as a contingency table). • To graphically summarize a single categorical variable, use a bar chart. To summarize and compare two categorical variables, use a side-by-side bar chart, a segmented bar chart, or a mosaic plot. • Pie charts are another option for summarizing categorical data, but they are more diﬃcult to read and bar charts are generally a better option. 2.4. CONSIDERING CATEGORICAL DATA 123 Exercises 2.37 Antibiotic use in children. The bar plot and the pie chart below show the distribution of pre-existing medical conditions of children involved in a study on the optimal duration of antibiotic use in treatment of tracheitis, which is an upper respiratory infection. (a) What features are apparent in the bar plot but not in the pie chart? (b) What features are apparent in the pie chart but not in the bar plot? (c) Which graph would you prefer to use for displaying these categorical data? 2.38 Views on immigration. 910 randomly sampled registered voters from Tampa, FL were asked if they thought undocumented workers in the US should be (i) allowed to keep their jobs and apply for US citizenship, (ii) allowed to keep their jobs as temporary guest workers but not allowed to apply for US citizenship, or (iii) lose their jobs and have to leave the country. The results of the survey by political ideology are shown below.58 Political ideology Conservative Moderate Response (i) Apply for citizenship (ii) Guest worker (iii) Leave the country (iv) Not sure Total 57 121 179 15 372 120 113 126 4 363 Liberal Total 278 262 350 20 910 101 28 45 1 175 (a) What percent of these Tampa, FL voters identify themselves as conservatives? (b) What percent of these Tampa, FL voters are in favor of the citizenship option? (c) What percent of these Tampa, FL voters identify themselves as conservatives and are in favor of the citizenship option? (d) What percent of these Tampa, FL voters who identify themselves as conservatives are also in favor of the citizenship option? What percent of moderates share this view? What percent of liberals share this view? (e) Do political ideology and views on immigration appear to be independent? Explain your reasoning. 58SurveyUSA, News Poll #18927, data collected Jan 27-29, 2012. GastrointestinalImmunocompromisedGenetic/metabolicNeuromuscularTraumaRespiratoryCardiovascularPrematurityRelative frequency0.000.100.200.30GastrointestinalImmunocompromisedGenetic/metabolicNeuromuscularTraumaRespiratoryCardiovascularPrematurity 124 CHAPTER 2. SUMMARIZING DATA 2.39 Views on the DREAM Act. A random sample of registered voters from Tampa, FL were asked if they support the DREAM Act, a proposed law which would provide a path to citizenship for people brought illegally to the US as children. The survey also collected information on the political ideology of the respondents. Based on the mosaic plot shown below, do views on the DREAM Act and political ideology appear to be independent? Explain your reasoning.59 2.40 Raise taxes. A random sample of registered voters nationally were asked whether they think it’s better to raise taxes on the rich or raise taxes on the poor. The survey also collected information on the political party aﬃliation of the respondents. Based on the mosaic plot shown below, do views on raising taxes and political aﬃliation appear to be independent? Explain your reasoning.60 59SurveyUSA, News Poll #18927, data collected Jan 27-29, 2012. 60Public Policy Polling, Americans on College Degrees, Classic Literature, the Seasons, and More, data collected Feb 20-22, 2015. ConservativeModerateLiberalSupportNot supportNot sureDemocratRepublicanIndep / OtherRaise taxes on the richRaise taxes on the poorNot sure 2.5. CASE STUDY: MALARIA VACCINE (SPECIAL TOPIC) 125 2.5 Case study: malaria vaccine (special topic) How large does an observed diﬀerence need to be for it to provide convincing evidence that something real is going on, something beyond random variation? Answering this question requires the tools that we will encounter in the later chapters on probability and inference. However, this is such an interesting and important question, and we’ll also address it here using simulation. This section can be covered now or in tandem with Chapter 5: Foundations for Inference. Learning objectives 1. Recognize that an observed diﬀerence in sample statistics may be due to random chance and that we use hypothesis testing to determine if this diﬀerence statistically signiﬁcant (i.e. too large to be attributed to random chance). 2. Set up competing hypotheses and use the results of a simulation to evaluate the degree of support the data provide against the null hypothesis and for the alternative hypothesis. 2.5.1 Variability within data EXAMPLE 2.80 Suppose your professor splits the students in class into two groups: students on the left and students on the right. If ˆpL and ˆpR represent the proportion of students who own an Apple product on the left and right, respectively, would you be surprised if ˆpL did not exactly equal ˆpR ? While the proportions would probably be close to each other, it would be unusual for them to be exactly the same. We would probably observe a small diﬀerence due to chance. GUIDED PRACTICE 2.81 If we don’t think the side of the room a person sits on in class is related to whether the person owns an Apple product, what assumption are we making about the relationship between these two variables?61 We consider a study on a new malaria vaccine called PfSPZ. In this study, volunteer patients were randomized into one of two experiment groups: 14 patients received an experimental vaccine or 6 patients received a placebo vaccine. Nineteen weeks later, all 20 patients were exposed to a drug-sensitive malaria parasite strain; the motivation of using a drug-sensitive strain of parasite here is for ethical considerations, allowing any infections to be treated eﬀectively. The results are summarized in Figure 2.44, where 9 of the 14 treatment patients remained free of signs of infection while all of the 6 patients in the control group patients showed some baseline signs of infection. 61We would be assuming that these two variables are independent. 126 CHAPTER 2. SUMMARIZING DATA treatment outcome vaccine placebo Total infection 5 6 11 no infection Total 14 6 20 9 0 9 Figure 2.44: Summary results for the malaria vaccine experiment. GUIDED PRACTICE 2.82 Is this an observational study or an experiment? What implications does the study type have on what can be inferred from the results?62 In this study, a smaller proportion of patients who received the vaccine showed signs of an infection (35.7% versus 100%). However, the sample is very small, and it is unclear whether the diﬀerence provides convincing evidence that the vaccine is eﬀective. EXAMPLE 2.83 Data scientists are sometimes called upon to evaluate the strength of evidence. When looking at the rates of infection for patients in the two groups in this study, what comes to mind as we try to determine whether the data show convincing evidence of a real diﬀerence? The observed infection rates (35.7% for the treatment group versus 100% for the control group) suggest the vaccine may be eﬀective. However, we cannot be sure if the observed diﬀerence represents the vaccine’s eﬃcacy or is just from random chance. Generally there is a little bit of ﬂuctuation in sample data, and we wouldn’t expect the sample proportions to be exactly equal, even if
the truth was that the infection rates were independent of getting the vaccine. Additionally, with such small samples, perhaps it’s common to observe such large diﬀerences when we randomly split a group due to chance alone! Example 2.83 is a reminder that the observed outcomes in the data sample may not perfectly reﬂect the true relationships between variables since there is random noise. While the observed diﬀerence in rates of infection is large, the sample size for the study is small, making it unclear if this observed diﬀerence represents eﬃcacy of the vaccine or whether it is simply due to chance. We label these two competing claims, H0 and HA, which are spoken as “H-nought” and “H-A”: H0: Independence model. The variables treatment and outcome are independent. They have no relationship, and the observed diﬀerence between the proportion of patients who developed an infection in the two groups, 64.3%, was due to chance. HA: Alternative model. The variables are not independent. The diﬀerence in infection rates of 64.3% was not due to chance, and vaccine aﬀected the rate of infection. What would it mean if the independence model, which says the vaccine had no inﬂuence on the rate of infection, is true? It would mean 11 patients were going to develop an infection no matter which group they were randomized into, and 9 patients would not develop an infection no matter which group they were randomized into. That is, if the vaccine did not aﬀect the rate of infection, the diﬀerence in the infection rates was due to chance alone in how the patients were randomized. Now consider the alternative model: infection rates were inﬂuenced by whether a patient received the vaccine or not. If this was true, and especially if this inﬂuence was substantial, we would expect to see some diﬀerence in the infection rates of patients in the groups. We choose between these two competing claims by assessing if the data conﬂict so much with H0 that the independence model cannot be deemed reasonable. If this is the case, and the data support HA, then we will reject the notion of independence and conclude there was discrimination. 62The study is an experiment, as patients were randomly assigned an experiment group. Since this is an experiment, the results can be used to evaluate a causal relationship between the malaria vaccine and whether patients showed signs of an infection. 2.5. CASE STUDY: MALARIA VACCINE (SPECIAL TOPIC) 127 2.5.2 Simulating the study We’re going to implement simulations, where we will pretend we know that the malaria vaccine being tested does not work. Ultimately, we want to understand if the large diﬀerence we observed is common in these simulations. If it is common, then maybe the diﬀerence we observed was purely due to chance. If it is very uncommon, then the possibility that the vaccine was helpful seems more plausible. Figure 2.44 shows that 11 patients developed infections and 9 did not. For our simulation, we will suppose the infections were independent of the vaccine and we were able to rewind back to when the researchers randomized the patients in the study. If we happened to randomize the patients diﬀerently, we may get a diﬀerent result in this hypothetical world where the vaccine doesn’t inﬂuence the infection. Let’s complete another randomization using a simulation. In this simulation, we take 20 notecards to represent the 20 patients, where we write down “infection” on 11 cards and “no infection” on 9 cards. In this hypothetical world, we believe each patient that got an infection was going to get it regardless of which group they were in, so let’s see what happens if we randomly assign the patients to the treatment and control groups again. We thoroughly shuﬄe the notecards and deal 14 into a vaccine pile and 6 into a placebo pile. Finally, we tabulate the results, which are shown in Figure 2.45. outcome treatment (simulated) vaccine placebo Total infection 7 4 11 no infection Total 14 6 20 7 2 9 Figure 2.45: Simulation results, where any diﬀerence in infection rates is purely due to chance. GUIDED PRACTICE 2.84 What is the diﬀerence in infection rates between the two simulated groups in Figure 2.45? How does this compare to the observed 64.3% diﬀerence in the actual data?63 2.5.3 Checking for independence We computed one possible diﬀerence under the independence model in Guided Practice 2.84, which represents one diﬀerence due to chance. While in this ﬁrst simulation, we physically dealt out notecards to represent the patients, it is more eﬃcient to perform this simulation using a computer. Repeating the simulation on a computer, we get another diﬀerence due to chance: And another: 2 6 3 6 − 9 14 − 8 14 = −0.310 = −0.071 And so on until we repeat the simulation enough times that we have a good idea of what represents the distribution of diﬀerences from chance alone. Figure 2.46 shows a stacked plot of the diﬀerences found from 100 simulations, where each dot represents a simulated diﬀerence between the infection rates (control rate minus treatment rate). Note that the distribution of these simulated diﬀerences is centered around 0. We simulated these diﬀerences assuming that the independence model was true, and under this condition, we expect the diﬀerence to be near zero with some random ﬂuctuation, where near is pretty generous in this case since the sample sizes are so small in this study. 634/6 − 7/14 = 0.167 or about 16.7% in favor of the vaccine. This diﬀerence due to chance is much smaller than the diﬀerence observed in the actual groups. 128 CHAPTER 2. SUMMARIZING DATA Figure 2.46: A stacked dot plot of diﬀerences from 100 simulations produced under the independence model, H0, where in these simulations infections are unaﬀected by the vaccine. Two of the 100 simulations had a diﬀerence of at least 64.3%, the diﬀerence observed in the study. EXAMPLE 2.85 Given the results of the simulation shown in Figure 2.46, about how often would you expect to observe a result as large as 64.3% if H0 were true? Because a result this large happened 2 times out the 100 simulations, we would expect such a large value only 2% of the time if H0 were true. There are two possible interpretations of the results of the study: H0 Independence model. The vaccine has no eﬀect on infection rate, and we just happened to observe a rare event. HA Alternative model. The vaccine has an eﬀect on infection rate, and the diﬀerence we observed was actually due to the vaccine being eﬀective at combatting malaria, which explains the large diﬀerence of 64.3%. Based on the simulations, we have two options. (1) We conclude that the study results do not provide strong enough evidence against the independence model, meaning we do not conclude that the vaccine had an eﬀect in this clinical setting. (2) We conclude the evidence is suﬃciently strong to reject H0, and we assert that the vaccine was useful. Is 2% small enough to make us reject the independence model? That depends on how much evidence we require. The smaller that probability is, the more evidence it provides against H0. Later, we will see that researchers often use a cutoﬀ of 5%, though it can depend upon the situation. Using the 5% cutoﬀ, we would reject the independence model in favor of the alternative. That is, we are concluding the data provide strong evidence that the vaccine provides some protection against malaria in this clinical setting. When there is strong enough evidence that the result points to a real diﬀerence and is not simply due to random variation, we call the result statistically signiﬁcant. One ﬁeld of statistics, statistical inference, is built on evaluating whether such diﬀerences are due to chance. In statistical inference, data scientists evaluate which model is most reasonable given the data. Errors do occur, just like rare events, and we might choose the wrong model. While we do not always choose correctly, statistical inference gives us tools to control and evaluate how often these errors occur. In Chapter 5, we give a formal introduction to the problem of model selection. We spend the next two chapters building a foundation of probability and theory necessary to make that discussion rigorous. llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllDifference in Infection Rates−0.6−0.4−0.20.00.20.40.60.8 2.5. CASE STUDY: MALARIA VACCINE (SPECIAL TOPIC) 129 Exercises 2.41 Side effects of Avandia. Rosiglitazone is the active ingredient in the controversial type 2 diabetes medicine Avandia and has been linked to an increased risk of serious cardiovascular problems such as stroke, heart failure, and death. A common alternative treatment is pioglitazone, the active ingredient in a diabetes medicine called Actos. In a nationwide retrospective observational study of 227,571 Medicare beneﬁciaries aged 65 years or older, it was found that 2,593 of the 67,593 patients using rosiglitazone and 5,386 of the 159,978 using pioglitazone had serious cardiovascular problems. These data are summarized in the contingency table below.64 Treatment Rosiglitazone Pioglitazone Total Cardiovascular problems Yes 2,593 5,386 7,979 No 65,000 154,592 219,592 Total 67,593 159,978 227,571 (a) Determine if each of the following statements is true or false. If false, explain why. Be careful: The In such cases, the statement reasoning may be wrong even if the statement’s conclusion is correct. should be considered false. i. Since more patients on pioglitazone had cardiovascular problems (5,386 vs. 2,593), we can conclude that the rate of cardiovascular problems for those on a pioglitazone treatment is higher. ii. The data suggest that diabetic patients who are taking rosiglitazone are more likely to have cardiovascular problems since the rate of incidence was (2,593 / 67,593 = 0.038) 3.8% for patients on this treatment, while it was only (5,386 / 159,978 = 0.034) 3.4% for patients on pioglitazone. iii. The fact that the rate of incid
ence is higher for the rosiglitazone group proves that rosiglitazone causes serious cardiovascular problems. iv. Based on the information provided so far, we cannot tell if the diﬀerence between the rates of incidences is due to a relationship between the two variables or due to chance. (b) What proportion of all patients had cardiovascular problems? (c) If the type of treatment and having cardiovascular problems were independent, about how many patients in the rosiglitazone group would we expect to have had cardiovascular problems? (d) We can investigate the relationship between outcome and treatment in this study using a randomization technique. While in reality we would carry out the simulations required for randomization using statistical software, suppose we actually simulate using index cards. In order to simulate from the independence model, which states that the outcomes were independent of the treatment, we write whether or not each patient had a cardiovascular problem on cards, shuﬄed all the cards together, then deal them into two groups of size 67,593 and 159,978. We repeat this simulation 1,000 times and each time record the number of people in the rosiglitazone group who had cardiovascular problems. Use the relative frequency histogram of these counts to answer (i)-(iii). i. What are the claims being tested? ii. Compared to the number calculated in part (c), which would provide more support for the alternative hypothesis, more or fewer patients with cardiovascular problems in the rosiglitazone group? iii. What do the simulation results suggest about the relationship between taking rosiglitazone and having cardiovascular problems in diabetic patients? 64D.J. Graham et al. “Risk of acute myocardial infarction, stroke, heart failure, and death in elderly Medicare patients treated with rosiglitazone or pioglitazone”. In: JAMA 304.4 (2010), p. 411. issn: 0098-7484. Simulated rosiglitazone cardiovascular events22502350245000.10.2 130 CHAPTER 2. SUMMARIZING DATA 2.42 Heart transplants. The Stanford University Heart Transplant Study was conducted to determine whether an experimental heart transplant program increased lifespan. Each patient entering the program was designated an oﬃcial heart transplant candidate, meaning that he was gravely ill and would most likely beneﬁt from a new heart. Some patients got a transplant and some did not. The variable transplant indicates which group the patients were in; patients in the treatment group got a transplant and those in the control group did not. Of the 34 patients in the control group, 30 died. Of the 69 people in the treatment group, 45 died. Another variable called survived was used to indicate whether or not the patient was alive at the end of the study.65 (a) Based on the mosaic plot, is survival independent of whether or not the patient got a transplant? Explain your reasoning. (b) What do the box plots below suggest about the eﬃcacy (eﬀectiveness) of the heart transplant treatment. (c) What proportion of patients in the treatment group and what proportion of patients in the control group died? (d) One approach for investigating whether or not the treatment is eﬀective is to use a randomization technique. i. What are the claims being tested? ii. The paragraph below describes the set up for such approach, if we were to do it without using statistical software. Fill in the blanks with a number or phrase, whichever is appropriate. cards representing patients who were alive at the end of cards representing patients who were not. Then, We write alive on the study, and dead on we shuﬄe these cards and split them into two groups: one group of size representing control. We representing treatment, and another group of size calculate the diﬀerence between the proportion of dead cards in the treatment and control groups (treatment - control) and record this value. We repeat this 100 times to build a . Lastly, we calculate the fraction of simulations distribution centered at . If this fraction is low, where the simulated diﬀerences in proportions are we conclude that it is unlikely to have observed such an outcome by chance and that the null hypothesis should be rejected in favor of the alternative. iii. What do the simulation results shown below suggest about the eﬀectiveness of the transplant pro- gram? 65B. Turnbull et al. “Survivorship of Heart Transplant Data”. In: Journal of the American Statistical Association 69 (1974), pp. 74–80. controltreatmentalivedeadSurvival Time (days)controltreatment050010001500simulated differences in proportions−0.25−0.15−0.050.050.150.25llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll 2.5. CASE STUDY: MALARIA VACCINE (SPECIAL TOPIC) 131 Chapter highlights A raw data matrix/table may have thousands of rows. The data need to be summarized in order to makes sense of all the information. In this chapter, we looked at ways to summarize data graphically, numerically, and verbally. Categorical data • A single categorical variable is summarized with counts or proportionsproportion in a one-way table. A bar chart is used to show the frequency or relative frequency of the categories that the variable takes on. • Two categorical variables can be summarized in a two-way table and with a side-by-side bar chart or a segmented bar chart. Numerical data • When looking at a single numerical variable, we try to understand the distribution of the variable. The distribution of a variable can be represented with a frequency table and with a graph, such as a stem-and-leaf plot or dot plot for small data sets, or a histogram for larger data sets. If only a summary is desired, a box plot may be used. • The distribution of a variable can be described and summarized with center (mean or median), spread (SD or IQR), and shape (right skewed, left skewed, approximately symmetric). • Z-scores and percentiles are useful for identifying a data point’s relative position within a data set. • When a distribution is nearly normal, we can use the empirical rule (68-95-99.7 rule), and we can use a normal model to approximate the histogram. • Outliers are values that appear extreme relative to the rest of the data. Investigating outliers can provide insight into properties of the data or may reveal data collection/entry errors. • When comparing the distribution of two variables, use two dot plots, two histograms, a back-to- back stem-and-leaf, or parallel box plots. • To look at the association between two numerical variables, use a scatterplot. Graphs and numbers can summarize data, but they alone are insuﬃcient. It is the role of the researcher or data scientist to ask questions, to use these tools to identify patterns and departure from patterns, and to make sense of this in the context of the data. Strong writing skills are critical for being able to communicate the results to a wider audience. 132 CHAPTER 2. SUMMARIZING DATA Chapter exercises In a class of 25 students, 24 of them took an exam in class and 1 student took 2.43 Make-up exam. a make-up exam the following day. The professor graded the ﬁrst batch of 24 exams and found an average score of 74 points with a standard deviation of 8.9 points. The student who took the make-up the following day scored 64 points on the exam. (a) Does the new student’s score increase or decrease the average score? (b) What is the new average? (c) Does the new student’s score increase or decrease the standard deviation of the scores? 2.44 Infant mortality. The infant mortality rate is deﬁned as the number of infant deaths per 1,000 live births. This rate is often used as an indicator of the level of health in a country. The relative frequency histogram below shows the distribution of estimated infant death rates for 224 countries for which such data were available in 2014.66 (a) Estimate Q1, the median, and Q3 from the histogram. (b) Would you expect the mean of this data set to be smaller or larger than the median? Explain your reasoning. 2.45 TV watchers. Students in an AP Statistics class were asked how many hours of television they watch per week (including online streaming). This sample yielded an average of 4.71 hours, with a standard deviation of 4.18 hours. Is the distribution of number of hours students watch television weekly symmetric? If not, what shape would you expect this distribution to have? Explain your reasoning. 2.46 A new statistic. The statistic median can be used as a measure of skewness. Suppose we have a distribution where all observations are greater than 0, xi > 0. What is the expected shape of the distribution under the following conditions? Explain your reasoning. ¯x (a) (b) (c) ¯x ¯x median = 1 median < 1 median > 1 ¯x 2.47 Oscar winners. The ﬁrst Oscar
awards for best actor and best actress were given out in 1929. The histograms below show the age distribution for all of the best actor and best actress winners from 1929 to 2018. Summary statistics for these distributions are also provided. Compare the distributions of ages of best actor and actress winners.67 Mean SD n Best Actress 36.2 11.9 92 Mean SD n Best Actor 43.8 8.83 92 66CIA Factbook, Country Comparisons, 2014. 67Oscar winners from 1929 – 2012, data up to 2009 from the Journal of Statistics Education data archive and more current data from wikipedia.org. Infant Mortality (per 1000 Live Births)Fraction of Countries02040608010012000.10.20.30.4Best actorBest actress204060800102030405001020304050Age (in years) 2.5. CASE STUDY: MALARIA VACCINE (SPECIAL TOPIC) 133 2.48 Exam scores. The average on a history exam (scored out of 100 points) was 85, with a standard deviation of 15. Is the distribution of the scores on this exam symmetric? If not, what shape would you expect this distribution to have? Explain your reasoning. 2.49 Stats scores. Below are the ﬁnal exam scores of twenty introductory statistics students. 57, 66, 69, 71, 72, 73, 74, 77, 78, 78, 79, 79, 81, 81, 82, 83, 83, 88, 89, 94 Create a box plot of the distribution of these scores. The ﬁve number summary provided below may be useful. Min Q1 Q2 (Median) Q3 Max 57 72.5 78.5 82.5 94 2.50 Marathon winners. The histogram and box plots below show the distribution of ﬁnishing times for male and female winners of the New York Marathon between 1970 and 1999. (a) What features of the distribution are apparent in the histogram and not the box plot? What features are apparent in the box plot but not in the histogram? (b) What may be the reason for the bimodal distribution? Explain. (c) Compare the distribution of marathon times for men and women based on the box plot shown below. (d) The time series plot shown below is another way to look at these data. Describe what is visible in this plot but not in the others. 2.02.42.83.201020Marathon times2.02.42.83.22.02.42.83.2WomenMenllllllllllllllllllllllllllllllMarathon times19701975198019851990199520002.02.42.83.2lWomenMen 134 CHAPTER 2. SUMMARIZING DATA 2.51 Birth weight. In a large study of birth weight of newborns, the weights of 23,419 newborn boys were recorded. The distribution of weights was approximately normal with a mean of 7.44 lbs (3376 grams) and a standard deviation of 1.33 lbs (603 grams). The government classiﬁes a newborn as having low birth weight if the weight is less than 5.5 pounds.68 (a) What percent of these newborns had a low birth weight? (b) Approximately what percent of these babies weighed greater than 10 pounds? (c) Approximately how many of these newborns weighed greater than 10 pounds? (d) How much would a newborn have to weigh in order to be at the 90th percentile among this group? 2.52 Auto insurance premiums. Suppose a newspaper article states that the distribution of auto insurance premiums for residents of California is approximately normal with a mean of $1,650. The article also states that 25% of California residents pay more than $1,800. (a) What is the Z-score that corresponds to the top 25% (or the 75th percentile) of the standard normal distribution? (b) What is the mean insurance cost? What is the cutoﬀ for the 75th percentile? (c) Identify the standard deviation of insurance premiums in California. 2.53 Speeding on the I-5, Part I. The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour.69 (a) What percent of passenger vehicles travel slower than 80 miles/hour? (b) What percent of passenger vehicles travel between 60 and 80 miles/hour? (c) How fast do the fastest 5% of passenger vehicles travel? (d) The speed limit on this stretch of the I-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5. 2.54 Heights of 10 year olds, Part I. Heights of 10 year olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches. (a) What is the probability that a randomly chosen 10 year old is shorter than 48 inches? (b) What is the probability that a randomly chosen 10 year old is between 60 and 65 inches? (c) If the tallest 10% of the class is considered “very tall”, what is the height cutoﬀ for “very tall”? 68www.biomedcentral.com/1471-2393/8/5 69S. Johnson and D. Murray. “Empirical Analysis of Truck and Automobile Speeds on Rural Interstates: Impact of Posted Speed Limits”. In: Transportation Research Board 89th Annual Meeting. 2010. 135 Chapter 3 Probability and probability distributions 3.1 Deﬁning probability 3.2 Conditional probability 3.3 Simulations 3.4 Random variables 3.5 Geometric distribution 3.6 Binomial distribution 136 Probability forms a foundation of statistics, and you’re probably already aware of many of the ideas. However, formalization of the concepts is new for most. This chapter aims to introduce probability concepts through examples that will be familiar to most people. For videos, slides, and other resources, please visit www.openintro.org/ahss 3.1. DEFINING PROBABILITY 137 3.1 Deﬁning probability What is the probability of rolling an even number on a die? Of getting 5 heads in row when tossing a coin? Of drawing a Heart or an Ace from a deck of cards? The study of probability is fun and interesting in its own right, but it also forms the foundation for statistical models and inferential procedures, many of which we will investigate in subsequent chapters. Learning objectives 1. Describe the long-run relative frequency interpretation of probability and understand its rela- tionship to the “Law of Large Numbers”. 2. Use Venn diagrams to represent events and their probabilities and to visualize the complement, union, and intersection of events. 3. Use the General Addition Rule to ﬁnd the probability that at least one of several events occurs. 4. Understand when events are disjoint (mutually exclusive) and how that simpliﬁes the General Addition Rule. 5. Apply the Multiplication Rule for ﬁnding the joint probability of independent events. 3.1.1 Introductory examples EXAMPLE 3.1 A “die”, the singular of dice, is a cube with six faces numbered 1, 2, 3, 4, 5, and 6. What is the chance of getting 1 when rolling a die? If the die is fair, then the chance of a 1 is as good as the chance of any other number. Since there are six outcomes, the chance must be 1-in-6 or, equivalently, 1/6. EXAMPLE 3.2 What is the chance of getting a 1 or 2 in the next roll? 1 and 2 constitute two of the six equally likely possible outcomes, so the chance of getting one of these two outcomes must be 2/6 = 1/3. EXAMPLE 3.3 What is the chance of getting either 1, 2, 3, 4, 5, or 6 on the next roll? 100%. The outcome must be one of these numbers. EXAMPLE 3.4 What is the chance of not rolling a 2? Since the chance of rolling a 2 is 1/6 or 16.¯6%, the chance of not rolling a 2 must be 100% − 16.¯6% = 83.¯3% or 5/6. Alternatively, we could have noticed that not rolling a 2 is the same as getting a 1, 3, 4, 5, or 6, which makes up ﬁve of the six equally likely outcomes and has probability 5/6. 138 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS EXAMPLE 3.5 Consider rolling two dice. If 1/6th of the time the ﬁrst die is a 1 and 1/6th of those times the second die is a 1, what is the chance of getting two 1s? If 16.¯6% of the time the ﬁrst die is a 1 and 1/6th of those times the second die is also a 1, then the chance that both dice are 1 is (1/6) × (1/6) or 1/36. 3.1.2 Probability We use probability to build tools to describe and understand apparent randomness. We often frame probability in terms of a random process giving rise to an outcome. Roll a die → 1, 2, 3, 4, 5, or 6 Flip a coin → H or T Rolling a die or ﬂipping a coin is a seemingly random process and each gives rise to an outcome. PROBABILITY The probability of an outcome is the proportion of times the outcome would occur if we observed the random process an inﬁnite number of times. Probability is deﬁned as a proportion, and it always takes values between 0 and 1 (inclusively). It may also be displayed as a percentage between 0% and 100%. Probability can be illustrated by rolling a die many times. Consider the event “roll a 1”. The relative frequency of an event is the proportion of times the event occurs out of the number of trials. Let ˆpn be the proportion of outcomes that are 1 after the ﬁrst n rolls. As the number of rolls increases, ˆpn (the relative frequency of rolls) will converge to the probability of rolling a 1, p = 1/6. Figure 3.1 shows this convergence for 100,000 die rolls. The tendency of ˆpn to stabilize around p, that is, the tendency of the relative frequency to stabilize around the true probability, is described by the Law of Large Numbers. Figure 3.1: The fraction of die rolls that are 1 at each stage in a simulation. The relative frequency tends to get closer to the probability 1/6 ≈ 0.167 as the number of rolls increases. LAW OF LARGE NUMBERS As more observations are collected, the observed proportion ˆpn of occurrences with a particular outcome after n trials converges to the true probability p of that outcome. n (number of rolls)1101001,00010,000100,0000.000.050.100.150.200.250.30p^n 3.1. DEFINING PROBABILITY 139 Occasionally the proportion will veer oﬀ from the probability and appear to defy the Law of Large Numbers, as ˆpn does many times in Figure 3.1. However, these deviations become smaller as the number of rolls increases. Above we write p as the probability of rolling a 1. We can also write this probability as P (rolling a 1) As we become more comfortable with this notation, we will abbreviate it further. For instance, if it is clear that the process is “rolling a die”, we could abbreviate P (rolling a 1) as P (1). GUIDED PRACTICE 3.6 R
andom processes include rolling a die and ﬂipping a coin. (a) Think of another random process. (b) Describe all the possible outcomes of that process. For instance, rolling a die is a random process with potential outcomes 1, 2, ..., 6. 1 What we think of as random processes are not necessarily random, but they may just be too diﬃcult to understand exactly. The fourth example in the footnote solution to Guided Practice 3.6 suggests a roommate’s behavior is a random process. However, even if a roommate’s behavior is not truly random, modeling her behavior as a random process can still be useful. MODELING A PROCESS AS RANDOM It can be helpful to model a process as random even if it is not truly random. 3.1.3 Disjoint or mutually exclusive outcomes Two outcomes are called disjoint or mutually exclusive if they cannot both happen in the same trial. For instance, if we roll a die, the outcomes 1 and 2 are disjoint since they cannot both occur on a single roll. On the other hand, the outcomes 1 and “rolling an odd number” are not disjoint since both occur if the outcome of the roll is a 1. The terms disjoint and mutually exclusive are equivalent and interchangeable. Calculating the probability of disjoint outcomes is easy. When rolling a die, the outcomes 1 and 2 are disjoint, and we compute the probability that one of these outcomes will occur by adding their separate probabilities: P (1 or 2) = P (1) + P (2) = 1/6 + 1/6 = 1/3 What about the probability of rolling a 1, 2, 3, 4, 5, or 6? Here again, all of the outcomes are disjoint so we add the probabilities: P (1 or 2 or 3 or 4 or 5 or 6) = P (1) + P (2) + P (3) + P (4) + P (5) + P (6) = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1. The Addition Rule guarantees the accuracy of this approach when the outcomes are disjoint. 1Here are four examples. (i) Whether someone gets sick in the next month or not is an apparently random process with outcomes sick and not. (ii) We can generate a random process by randomly picking a person and measuring that person’s height. The outcome of this process will be a positive number. (iii) Whether the stock market goes up or down next week is a seemingly random process with possible outcomes up, down, and no change. Alternatively, we could have used the percent change in the stock market as a numerical outcome. (iv) Whether your roommate cleans her dishes tonight probably seems like a random process with possible outcomes cleans dishes and leaves dishes. 140 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS ADDITION RULE OF DISJOINT OUTCOMES If A1 and A2 represent two disjoint outcomes, then the probability that one of them occurs is given by P (A1 or A2) = P (A1) + P (A2) If there are many disjoint outcomes A1, ..., Ak, then the probability that one of these outcomes will occur is P (A1) + P (A2) + · · · + P (Ak) GUIDED PRACTICE 3.7 We are interested in the probability of rolling a 1, 4, or 5. (a) Explain why the outcomes 1, 4, and 5 are disjoint. (b) Apply the Addition Rule for disjoint outcomes to determine P (1 or 4 or 5).2 GUIDED PRACTICE 3.8 In the email data set in Chapter 2, the number variable described whether no number (labeled none), only one or more small numbers (small), or whether at least one big number appeared in an email (big). Of the 3,921 emails, 549 had no numbers, 2,827 had only one or more small numbers, and 545 had at least one big number. (a) Are the outcomes none, small, and big disjoint? (b) Determine the proportion of emails with value small and big separately. (c) Use the Addition Rule for disjoint outcomes to compute the probability a randomly selected email from the data set has a number in it, small or big.3 Statisticians rarely work with individual outcomes and instead consider sets or collections of outcomes. Let A represent the event where a die roll results in 1 or 2 and B represent the event that the die roll is a 4 or a 6. We write A as the set of outcomes {1, 2} and B = {4, 6}. These sets are commonly called events. Because A and B have no elements in common, they are disjoint events. A and B are represented in Figure 3.2. Figure 3.2: Three events, A, B, and D, consist of outcomes from rolling a die. A and B are disjoint since they do not have any outcomes in common. The Addition Rule applies to both disjoint outcomes and disjoint events. The probability that one of the disjoint events A or B occurs is the sum of the separate probabilities: P (A or B) = P (A) + P (B) = 1/3 + 1/3 = 2/3 2(a) The random process is a die roll, and at most one of these outcomes can come up. This means they are disjoint outcomes. (b) P (1 or 4 or 5) = P (1) + P (4) + P (5) = 1 6 = 1 3(a) Yes. Each email is categorized in only one level of number. (b) Small: 2827 small or big) = P (small) + P (big) = 0.721 + 0.139 = 0.860. 3921 = 0.721. Big: 545 3921 = 0.139. (c) 123456ABD 3.1. DEFINING PROBABILITY 141 GUIDED PRACTICE 3.9 (a) Verify the probability of event A, P (A), is 1/3 using the Addition Rule. (b) Do the same for event B.4 GUIDED PRACTICE 3.10 (a) Using Figure 3.2 as a reference, what outcomes are represented by event D? (b) Are events B and D disjoint? (c) Are events A and D disjoint?5 GUIDED PRACTICE 3.11 In Guided Practice 3.10, you conﬁrmed B and D from Figure 3.2 are disjoint. Compute the probability that either event B or event D occurs.6 3.1.4 Probabilities when events are not disjoint Let’s consider calculations for two events that are not disjoint in the context of a regular deck of 52 cards, represented in Figure 3.3. If you are unfamiliar with the cards in a regular deck, please see the footnote.7 2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ J♣ Q♣ K♣ A♣ 2♦ 3♦ 4♦ 5♦ 6♦ 7♦ 8♦ 9♦ 10♦ J♦ Q♦ K♦ A♦ 2♥ 3♥ 4♥ 5♥ 6♥ 7♥ 8♥ 9♥ 10♥ J♥ Q♥ K♥ A♥ 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠ A♠ Figure 3.3: Representations of the 52 unique cards in a deck. GUIDED PRACTICE 3.12 (a) What is the probability that a randomly selected card is a diamond? (b) What is the probability that a randomly selected card is a face card?8 Venn diagrams are useful when outcomes can be categorized as “in” or “out” for two or three variables, attributes, or random processes. The Venn diagram in Figure 3.4 uses a circle to represent diamonds and another to represent face cards. If a card is both a diamond and a face card, it falls into the intersection of the circles. If it is a diamond but not a face card, it will be in part of the left circle that is not in the right circle (and so on). The total number of cards that are diamonds is given by the total number of cards in the diamonds circle: 10 + 3 = 13. The probabilities are also shown (e.g. 10/52 = 0.1923). 4(a) P (A) = P (1 or 2) = P (1) + P (2) = 1 5(a) Outcomes 2 and 3. (b) Yes, events B and D are disjoint because they share no outcomes. (c) The events A 3 . (b) Similarly, P (B) = 1/3 and D share an outcome in common, 2, and so are not disjoint. 6Since B and D are disjoint events, use the Addition Rule: P (B or D) = P (B) + P (D) = 1 7The 52 cards are split into four suits: ♣ (club), ♦ (diamond), ♥ (heart), ♠ (spade). Each suit has its 13 cards labeled: 2, 3, ..., 10, J (jack), Q (queen), K (king), and A (ace). Thus, each card is a unique combination of a suit and a label, e.g. 4♥ and J♣. The 12 cards represented by the jacks, queens, and kings are called face cards. The cards that are ♦ or ♥ are typically colored red while the other two suits are typically colored black(a) There are 52 cards and 13 diamonds. If the cards are thoroughly shuﬄed, each card has an equal chance 52 = 0.250. (b) Likewise, of being drawn, so the probability that a randomly selected card is a diamond is P (♦) = 13 there are 12 face cards, so P (face card) = 12 52 = 3 13 = 0.231. 142 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Figure 3.4: A Venn diagram for diamonds and face cards. GUIDED PRACTICE 3.13 Using the Venn diagram, verify P (face card) = 12/52 = 3/13.9 Let A represent the event that a randomly selected card is a diamond and B represent the event that it is a face card. How do we compute P (A or B)? Events A and B are not disjoint – the cards J♦, Q♦, and K♦ fall into both categories – so we cannot use the Addition Rule for disjoint events. Instead we use the Venn diagram. We start by adding the probabilities of the two events: P (A) + P (B) = P (♦) + P (face card) = 13/52 + 12/52 However, the three cards that are in both events were counted twice, once in each probability. We must correct this double counting: P (A or B) = P (♦) + P (face card) = P (♦) + P (face card) − P (♦ and face card) = 13/52 + 12/52 − 3/52 = 22/52 = 11/26 Equation (3.14) is an example of the General Addition Rule. GENERAL ADDITION RULE If A and B are any two events, disjoint or not, then the probability that A or B will occur is P (A or B) = P (A) + P (B) − P (A and B) where P (A and B) is the probability that both events occur. SYMBOLIC NOTATION FOR “AND” AND “OR” The symbol ∩ means intersection and is equivalent to “and”. The symbol ∪ means union and is equivalent to “or”. It is common to see the General Addition Rule written as P (A ∪ B) = P (A) + P (B) − P (A ∩ B) “OR” IS INCLUSIVE When we write, “or” in statistics, we mean “and/or” unless we explicitly state otherwise. Thus, A or B occurs means A, B, or both A and B occur. This is equivalent to at least one of A or B occurring. 9The Venn diagram shows face cards split up into “face card but not ♦” and “face card and ♦”. Since these 52 = 3 13 . correspond to disjoint events, P (face card) is found by adding the two corresponding probabilities: 3 52 = 12 52 + 9 10390.19230.05770.1731There are also30 cards that areneither diamondsnor face cardsDiamonds, 0.2500Face cards, 0.2308 3.1. DEFINING PROBABILITY 143 GUIDED PRACTICE 3.14 (a) If A and B are disjoint, describe why this implies P (A and B) = 0. (b) Using part (a), verify that the General Addition Rule simpliﬁes to the simpler Addition Rule for disjoint events if A and B are disjoint.10 GUIDED PRACTICE 3.15 In the email data set with 3,921 e
mails, 367 were spam, 2,827 contained some small numbers but no big numbers, and 168 had both characteristics. Create a Venn diagram for this setup.11 GUIDED PRACTICE 3.16 (a) Use your Venn diagram from Guided Practice 3.15 to determine the probability a randomly drawn email from the email data set is spam and had small numbers (but not big numbers). (b) What is the probability that the email had either of these attributes?12 3.1.5 Complement of an event Rolling a die produces a value in the set {1, 2, 3, 4, 5, 6}. This set of all possible outcomes is called the sample space (S) for rolling a die. We often use the sample space to examine the scenario where an event does not occur. Let D = {2, 3} represent the event that the outcome of a die roll is 2 or 3. Then the complement represents all outcomes in our sample space that are not in D, which is denoted by Dc = {1, 4, 5, 6}. That is, Dc is the set of all possible outcomes not already included in D. Figure 3.5 shows the relationship between D, Dc, and the sample space S. Figure 3.5: Event D = {2, 3} and its complement, Dc = {1, 4, 5, 6}. S represents the sample space, which is the set of all possible events. GUIDED PRACTICE 3.17 (a) Compute P (Dc) = P (rolling a 1, 4, 5, or 6). (b) What is P (D) + P (Dc)?13 10(a) If A and B are disjoint, A and B can never occur simultaneously. (b) If A and B are disjoint, then the last term of Equation (3.14) is 0 (see part (a)) and we are left with the Addition Rule for disjoint events. 11Both the counts and corresponding probabilities (e.g. 2659/3921 = 0.678) are shown. Notice that the number of emails represented in the left circle corresponds to 2659 + 168 = 2827, and the number represented in the right circle is 168 + 199 = 367. 12(a) The solution is represented by the intersection of the two circles: 0.043. (b) This is the sum of the three disjoint probabilities shown in the circles: 0.678 + 0.043 + 0.051 = 0.772. 13(a) The outcomes are disjoint and each has probability 1/6, so the total probability is 4/6 = 2/3. (b) We can also see that P (D) = 1 6 + 1 6 = 1/3. Since D and Dc are disjoint, P (D) + P (Dc) = 1. 145623DDCSsmall numbers and no big numbersspam26591681990.6780.0430.051Other emails: 3921−2659−168−199 = 895(0.228) 144 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS GUIDED PRACTICE 3.18 Events A = {1, 2} and B = {4, 6} are shown in Figure 3.2 on page 140. (a) Write out what Ac and Bc represent. (b) Compute P (Ac) and P (Bc). (c) Compute P (A) + P (Ac) and P (B) + P (Bc).14 An event A together with its complement Ac comprise the entire sample space. Because of this we can say that P (A) + P (Ac) = 1. COMPLEMENT The complement of event A is denoted Ac, and Ac represents all outcomes not in A. A and Ac are mathematically related: P (A) + P (Ac) = 1, i.e. P (A) = 1 − P (Ac) In simple examples, computing A or Ac is feasible in a few steps. However, using the complement can save a lot of time as problems grow in complexity. GUIDED PRACTICE 3.19 A die is rolled 10 times. (a) What is the complement of getting at least one 6 in 10 rolls of the die? (b) What is the complement of getting at most three 6’s in 10 rolls of the die?15 3.1.6 Independence Just as variables and observations can be independent, random processes can be independent, too. Two processes are independent if knowing the outcome of one provides no useful information about the outcome of the other. For instance, ﬂipping a coin and rolling a die are two independent processes – knowing the coin was heads does not help determine the outcome of a die roll. On the other hand, stock prices usually move up or down together, so they are not independent. Example 3.5 provides a basic example of two independent processes: rolling two dice. We want to determine the probability that both will be 1. Suppose one of the dice is red and the other white. If the outcome of the red die is a 1, it provides no information about the outcome of the white die. We ﬁrst encountered this same question in Example 3.5 (page 138), where we calculated the probability using the following reasoning: 1/6th of the time the red die is a 1, and 1/6th of those times the white die will also be 1. This is illustrated in Figure 3.6. Because the rolls are independent, the probabilities of the corresponding outcomes can be multiplied to get the ﬁnal answer: (1/6) × (1/6) = 1/36. This can be generalized to many independent processes. Figure 3.6: 1/6th of the time, the ﬁrst roll is a 1. Then 1/6th of those times, the second roll will also be a 1. 14Brief solutions: (a) Ac = {3, 4, 5, 6} and Bc = {1, 2, 3, 5}. (b) Noting that each outcome is disjoint, add the individual outcome probabilities to get P (Ac) = 2/3 and P (Bc) = 2/3. (c) A and Ac are disjoint, and the same is true of B and Bc. Therefore, P (A) + P (Ac) = 1 and P (B) + P (Bc) = 1. 15(a) The complement of getting at least one 6 in ten rolls of a die is getting zero 6’s in the 10 rolls. (b) The complement of getting at most three 6’s in 10 rolls is getting four, ﬁve, ..., nine, or ten 6’s in 10 rolls. All rolls1/6th of the firstrolls are a 1.1/6th of those times wherethe first roll is a 1 thesecond roll is also a 1. 3.1. DEFINING PROBABILITY 145 EXAMPLE 3.20 What if there was also a blue die independent of the other two? What is the probability of rolling the three dice and getting all 1s? The same logic applies from Example 3.5. If 1/36th of the time the white and red dice are both 1, then 1/6th of those times the blue die will also be 1, so multiply: P (white = 1 and red = 1 and blue = 1) = P (white = 1) × P (red = 1) × P (blue = 1) = (1/6) × (1/6) × (1/6) = 1/216 Examples 3.5 and 3.20 illustrate what is called the Multiplication Rule for independent pro- cesses. MULTIPLICATION RULE FOR INDEPENDENT PROCESSES If A and B represent events from two diﬀerent and independent processes, then the probability that both A and B occur can be calculated as the product of their separate probabilities: P (A and B) = P (A) × P (B) Similarly, if there are k events A1, ..., Ak from k independent processes, then the probability they all occur is P (A1) × P (A2) × · · · × P (Ak) GUIDED PRACTICE 3.21 About 9% of people are left-handed. Suppose 2 people are selected at random from the U.S. population. Because the sample size of 2 is very small relative to the population, it is reasonable to assume these two people are independent. (a) What is the probability that both are left-handed? (b) What is the probability that both are right-handed?16 GUIDED PRACTICE 3.22 Suppose 5 people are selected at random.17 (a) What is the probability that all are right-handed? (b) What is the probability that all are left-handed? (c) What is the probability that not all of the people are right-handed? 16(a) The probability the ﬁrst person is left-handed is 0.09, which is the same for the second person. We apply the Multiplication Rule for independent processes to determine the probability that both will be left-handed: 0.09×0.09 = 0.0081. (b) It is reasonable to assume the proportion of people who are ambidextrous (both right- and left-handed) is nearly 0, which results in P (right-handed) = 1 − 0.09 = 0.91. Using the same reasoning as in part (a), the probability that both will be right-handed is 0.91 × 0.91 = 0.8281. 17(a) The abbreviations RH and LH are used for right-handed and left-handed, respectively. Since each are indepen- dent, we apply the Multiplication Rule for independent processes: P (all ﬁve are RH) = P (ﬁrst = RH, second = RH, ..., ﬁfth = RH) = P (ﬁrst = RH) × P (second = RH) × · · · × P (ﬁfth = RH) = 0.91 × 0.91 × 0.91 × 0.91 × 0.91 = 0.624 (b) Using the same reasoning as in (a), 0.09 × 0.09 × 0.09 × 0.09 × 0.09 = 0.0000059 (c) Use the complement, P (all ﬁve are RH), to answer this question: P (not all RH) = 1 − P (all RH) = 1 − 0.624 = 0.376 146 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Suppose the variables handedness and gender are independent, i.e. knowing someone’s gender provides no useful information about their handedness and vice-versa. Then we can compute whether a randomly selected person is right-handed and female18 using the Multiplication Rule: P (right-handed and female) = P (right-handed) × P (female) = 0.91 × 0.50 = 0.455 GUIDED PRACTICE 3.23 Three people are selected at random.19 (a) What is the probability that the ﬁrst person is male and right-handed? (b) What is the probability that the ﬁrst two people are male and right-handed?. (c) What is the probability that the third person is female and left-handed? (d) What is the probability that the ﬁrst two people are male and right-handed and the third person is female and left-handed? Sometimes we wonder if one outcome provides useful information about another outcome. The question we are asking is, are the occurrences of the two events independent? We say that two events A and B are independent if they satisfy Equation (3.21). EXAMPLE 3.24 If we shuﬄe up a deck of cards and draw one, is the event that the card is a heart independent of the event that the card is an ace? The probability the card is a heart is 1/4 and the probability that it is an ace is 1/13. The probability the card is the ace of hearts is 1/52. We check whether Equation 3.21 is satisﬁed: P (♥) × P (ace) = 1 4 × 1 13 = 1 52 = P (♥ and ace) Because the equation holds, the event that the card is a heart and the event that the card is an ace are independent events. 18The actual proportion of the U.S. population that is female is about 50%, and so we use 0.5 for the probability of sampling a woman. However, this probability does diﬀer in other countries. 19Brief answers are provided. (a) This can be written in probability notation as P (a randomly selected person is male and right-handed) = 0.455. (b) 0.207. (c) 0.045. (d) 0.0093. 3.1. DEFINING PROBABILITY 147 Section summary • When an outcome depends upon a chance process, we can deﬁne the probability of the outcome as the proportion of times it would occur if we re
peated the process an inﬁnite number of times. Also, even when an outcome is not truly random, modeling it with probability can be useful. • The Law of Large Numbers states that the relative frequency, or proportion of times an outcome occurs after n repetitions, stabilizes around the true probability as n gets large. • The probability of an event is always between 0 and 1, inclusive. • The probability of an event and the probability of its complement add up to 1. Sometime we use P (A) = 1 − P (not A) when P (not A) is easier to calculate than P (A). • A and B are disjoint, i.e. mutually exclusive, if they cannot happen together. In this case, the events do not overlap and P (A and B) = 0. • In the special case where A and B are disjoint events: P (A or B) = P (A) + P (B). • When A and B are not disjoint, adding P (A) and P (B) will overestimate P (A or B) because the overlap of A and B will be added twice. Therefore, when A and B are not disjoint, use the General Addition Rule: P (A or B) = P (A) + P (B) − P (A and B).20 • To ﬁnd the probability that at least one of several events occurs, use a special case of the rule of complements: P (at least one) = 1 − P (none). • When only considering two events, the probability that one or the other happens is equal to the probability that at least one of the two events happens. When dealing with more than two events, the General Addition Rule becomes very complicated. Instead, to ﬁnd the probability that A or B or C occurs, ﬁnd the probability that none of them occur and subtract that value from 1. • Two events are independent when the occurrence of one does not change the likelihood of the other. • In the special case where A and B are independent: P (A and B) = P (A) × P (B). 20Often written: P (A ∪ B) = P (A) + P (B) − P (A ∩ B). 148 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Exercises 3.1 True or false. Determine if the statements below are true or false, and explain your reasoning. (a) If a fair coin is tossed many times and the last eight tosses are all heads, then the chance that the next toss will be heads is somewhat less than 50%. (b) Drawing a face card (jack, queen, or king) and drawing a red card from a full deck of playing cards are mutually exclusive events. (c) Drawing a face card and drawing an ace from a full deck of playing cards are mutually exclusive events. 3.2 Roulette wheel. The game of roulette involves spinning a wheel with 38 slots: 18 red, 18 black, and 2 green. A ball is spun onto the wheel and will eventually land in a slot, where each slot has an equal chance of capturing the ball. (a) You watch a roulette wheel spin 3 consecutive times and the ball lands on a red slot each time. What is the probability that the ball will land on a red slot on the next spin? (b) You watch a roulette wheel spin 300 consecutive times and the ball lands on a red slot each time. What is the probability that the ball will land on a red slot on the next spin? (c) Are you equally conﬁdent of your answers to parts (a) and (b)? Why or why not? Photo by H˚akan Dahlstr¨om (http://ﬂic.kr/p/93fEzp) CC BY 2.0 license Below are four versions of the same game. Your archnemesis gets to 3.3 Four games, one winner. pick the version of the game, and then you get to choose how many times to ﬂip a coin: 10 times or 100 times. Identify how many coin ﬂips you should choose for each version of the game. It costs $1 to play each game. Explain your reasoning. (a) If the proportion of heads is larger than 0.60, you win $1. (b) If the proportion of heads is larger than 0.40, you win $1. (c) If the proportion of heads is between 0.40 and 0.60, you win $1. (d) If the proportion of heads is smaller than 0.30, you win $1. 3.4 Backgammon. Backgammon is a board game for two players in which the playing pieces are moved according to the roll of two dice. Players win by removing all of their pieces from the board, so it is usually good to roll high numbers. You are playing backgammon with a friend and you roll two 6s in your ﬁrst roll and two 6s in your second roll. Your friend rolls two 3s in his ﬁrst roll and again in his second row. Your friend claims that you are cheating, because rolling double 6s twice in a row is very unlikely. Using probability, show that your rolls were just as likely as his. 3.5 Coin ﬂips. If you ﬂip a fair coin 10 times, what is the probability of (a) getting all tails? (b) getting all heads? (c) getting at least one tails? 3.6 Dice rolls. If you roll a pair of fair dice, what is the probability of (a) getting a sum of 1? (b) getting a sum of 5? (c) getting a sum of 12? 3.1. DEFINING PROBABILITY 149 A Pew Research survey asked 2,373 randomly sampled registered voters their 3.7 Swing voters. political aﬃliation (Republican, Democrat, or Independent) and whether or not they identify as swing voters. 35% of respondents identiﬁed as Independent, 23% identiﬁed as swing voters, and 11% identiﬁed as both.21 (a) Are being Independent and being a swing voter disjoint, i.e. mutually exclusive? (b) Draw a Venn diagram summarizing the variables and their associated probabilities. (c) What percent of voters are Independent but not swing voters? (d) What percent of voters are Independent or swing voters? (e) What percent of voters are neither Independent nor swing voters? (f) Is the event that someone is a swing voter independent of the event that someone is a political Indepen- dent? 3.8 Poverty and language. The American Community Survey is an ongoing survey that provides data every year to give communities the current information they need to plan investments and services. The 2010 American Community Survey estimates that 14.6% of Americans live below the poverty line, 20.7% speak a language other than English (foreign language) at home, and 4.2% fall into both categories.22 (a) Are living below the poverty line and speaking a foreign language at home disjoint? (b) Draw a Venn diagram summarizing the variables and their associated probabilities. (c) What percent of Americans live below the poverty line and only speak English at home? (d) What percent of Americans live below the poverty line or speak a foreign language at home? (e) What percent of Americans live above the poverty line and only speak English at home? (f) Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home? 3.9 Disjoint vs. independent. In parts (a) and (b), identify whether the events are disjoint, independent, or neither (events cannot be both disjoint and independent). (a) You and a randomly selected student from your class both earn A’s in this course. (b) You and your class study partner both earn A’s in this course. (c) If two events can occur at the same time, must they be dependent? In a multiple choice exam, there are 5 questions and 4 choices for each 3.10 Guessing on an exam. question (a, b, c, d). Nancy has not studied for the exam at all and decides to randomly guess the answers. What is the probability that: (a) the ﬁrst question she gets right is the 5th question? (b) she gets all of the questions right? (c) she gets at least one question right? 21Pew Research Center, With Voters Focused on Economy, Obama Lead Narrows, data collected between April 4-15, 2012. 22U.S. Census Bureau, 2010 American Community Survey 1-Year Estimates, Characteristics of People by Language Spoken at Home. 150 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.11 Educational attainment. The family college data set contains a sample of 792 cases with two variables, teen and parents, and is summarized below.23 The teen variable is either college or not, where the college label means the teen went to college immediately after high school. The parents variable takes the value degree if at least one parent of the teenager completed a college degree. teen college not Total parents degree 231 49 280 not 214 298 512 Total 445 347 792 Table 3.7: Contingency table summarizing the family college data set. (a) For a randomly selected case, what is the probability that a parent completed a college degree? (b) For a randomly selected case, what is the probability that the teen went to college immediately after high school? (c) For a randomly selected case, what is the probability that a parent completed a college degree and teen went to college immediately after high school? (d) Is P(a parent completed college degree and teen went to college immediately after high school) = P(parent completed college degree) x P(teen went to college immediately after high school)? Explain why this is or is not the case. 3.12 School absences. Data collected at elementary schools in DeKalb County, GA suggest that each year roughly 25% of students miss exactly one day of school, 15% miss 2 days, and 28% miss 3 or more days due to sickness.24 (a) What is the probability that a student chosen at random doesn’t miss any days of school due to sickness this year? (b) What is the probability that a student chosen at random misses no more than one day? (c) What is the probability that a student chosen at random misses at least one day? (d) If a parent has two kids at a DeKalb County elementary school, what is the probability that neither kid will miss any school? Note any assumption you must make to answer this question. (e) If a parent has two kids at a DeKalb County elementary school, what is the probability that both kids will miss some school, i.e. at least one day? Note any assumption you make. (f) If you made an assumption in part (d) or (e), do you think it was reasonable? If you didn’t make any assumptions, double check your earlier answers. 23A simulated data set based on real population summaries at nces.ed.gov/pubs2001/2001126.pdf. 24S.S. Mizan et al. “Absence, Extended Absence, and Repeat Tardiness Related to Asthma Status among Elemen- tary School Children”. In: Journal of Asthma 48.3 (2011), pp. 228–234. 3.2. CONDITIONAL PROBABILITY 151 3.2 Conditional pr
obability In this section we will use conditional probabilities to answer the following questions: • What is the likelihood that a machine learning algorithm will misclassify a photo as being about fashion if it is not actually about fashion? • How much more likely are children to attend college whose parents attended college than children whose parents did not attend college? • Given that a person receives a positive test result for a disease, what is the probability that the person actually has the disease? Learning objectives 1. Understand conditional probability and how to calculate it. 2. Calculate joint and conditional probabilities based on a two-way table. 3. Use the General Multiplication Rule to ﬁnd the probability of joint events. 4. Determine whether two events are independent and whether they are mutually exclusive, based on the deﬁnitions of those terms. 5. Draw a tree diagram with at least two branches to organize possible outcomes and their probabilities. Understand that the second branch represents conditional probabilities. 6. Use the tree diagram or Bayes’ Theorem to solve “inverted” conditional probabilities. 152 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.2.1 Exploring probabilities with a contingency table The photo classify data set represents a sample of 1822 photos from a photo sharing website. Data scientists have been working to improve a classiﬁer for whether the photo is about fashion or not, and these 659 photos represent a test for their classiﬁer. Each photo gets two classiﬁcations: the ﬁrst is called mach learn and gives a classiﬁcation from a machine learning (ML) system of either pred fashion or pred not. Each of these 1822 photos have also been classiﬁed carefully by a team of people, which we take to be the source of truth; this variable is called truth and takes values fashion and not. Figure 3.8 summarizes the results. mach learn pred fashion pred not Total truth fashion 197 112 309 not 22 1491 1513 Total 219 1603 1822 Figure 3.8: Contingency table summarizing the photo classify data set. Figure 3.9: A Venn diagram using boxes for the photo classify data set. EXAMPLE 3.25 If a photo is actually about fashion, what is the chance the ML classiﬁer correctly identiﬁed the photo as being about fashion? We can estimate this probability using the data. Of the 309 fashion photos, the ML algorithm correctly classiﬁed 197 of the photos: P (mach learn is pred fashion given truth is fashion) = 197 309 = 0.638 EXAMPLE 3.26 We sample a photo from the data set and learn the ML algorithm predicted this photo was not about fashion. What is the probability that it was incorrect and the photo is about fashion? If the ML classiﬁer suggests a photo is not about fashion, then it comes from the second row in the data set. Of these 1603 photos, 112 were actually about fashion: P (truth is fashion given mach learn is pred not) = 112 1603 = 0.070 ML Predicts FashionFashion Photos0.110.010.06Neither: 0.82 3.2. CONDITIONAL PROBABILITY 153 3.2.2 Marginal and joint probabilities Figure 3.8 includes row and column totals for each variable separately in the photo classify data set. These totals represent marginal probabilities for the sample, which are the probabilities based on a single variable without regard to any other variables. For instance, a probability based solely on the mach learn variable is a marginal probability: P (mach learn is pred fashion) = 219 1822 = 0.12 A probability of outcomes for two or more variables or processes is called a joint probability: P (mach learn is pred fashion and truth is fashion) = 197 1822 = 0.11 It is common to substitute a comma for “and” in a joint probability, although using either the word “and” or a comma is acceptable: P (mach learn is pred fashion, truth is fashion) means the same thing as P (mach learn is pred fashion and truth is fashion) MARGINAL AND JOINT PROBABILITIES If a probability is based on a single variable, it is a marginal probability. The probability of outcomes for two or more variables or processes is called a joint probability. We use table proportions to summarize joint probabilities for the photo classify sample. These proportions are computed by dividing each count in Figure 3.8 by the table’s total, 1822, to obtain the proportions in Figure 3.10. The joint probability distribution of the mach learn and truth variables is shown in Figure 3.11. mach learn: pred fashion mach learn: pred not Total truth: fashion 0.1081 0.0615 0.1696 truth: not 0.0121 0.8183 0.8304 Total 0.1202 0.8798 1.00 Figure 3.10: Probability table summarizing the photo classify data set. Joint outcome mach learn is pred fashion and truth is fashion mach learn is pred fashion and truth is not mach learn is pred not and truth is fashion mach learn is pred not and truth is not Total Probability 0.1081 0.0121 0.0615 0.8183 1.0000 Figure 3.11: Joint probability distribution for the photo classify data set. GUIDED PRACTICE 3.27 Verify Figure 3.11 represents a probability distribution: events are disjoint, all probabilities are non-negative, and the probabilities sum to 1.25 25Each of the four outcome combination are disjoint, all probabilities are indeed non-negative, and the sum of the probabilities is 0.1081 + 0.0121 + 0.0615 + 0.8183 = 1.00. 154 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS We can compute marginal probabilities using joint probabilities in simple cases. For example, the probability that a randomly selected photo from the data set is about fashion is found by summing the outcomes in which truth takes value fashion: P (truth is fashion) = P (mach learn is pred fashion and truth is fashion) + P (mach learn is pred not and truth is fashion) = 0.1081 + 0.0615 = 0.1696 3.2.3 Deﬁning conditional probability The ML classiﬁer predicts whether a photo is about fashion, even if it is not perfect. We would like to better understand how to use information from a variable like mach learn to improve our probability estimation of a second variable, which in this example is truth. The probability that a random photo from the data set is about fashion is about 0.17. If we knew the machine learning classiﬁer predicted the photo was about fashion, could we get a better estimate of the probability the photo is actually about fashion? Absolutely. To do so, we limit our view to only those 219 cases where the ML classiﬁer predicted that the photo was about fashion and look at the fraction where the photo was actually about fashion: P (truth is fashion given mach learn is pred fashion) = 197 219 = 0.900 We call this a conditional probability because we computed the probability under a condition: the ML classiﬁer prediction said the photo was about fashion. There are two parts to a conditional probability, the outcome of interest and the condition. It is useful to think of the condition as information we know to be true, and this information usually can be described as a known outcome or event. We generally separate the text inside our probability notation into the outcome of interest and the condition with a vertical bar: P (truth is fashion given mach learn is pred fashion) = P (truth is fashion \| mach learn is pred fashion) = 197 219 = 0.900 The vertical bar “\|” is read as given. In the last equation, we computed the probability a photo was about fashion based on the condition that the ML algorithm predicted it was about fashion as a fraction: P (truth is fashion \| mach learn is pred fashion) = = # cases where truth is fashion and mach learn is pred fashion # cases where mach learn is pred fashion 197 219 = 0.900 We considered only those cases that met the condition, mach learn is pred fashion, and then we computed the ratio of those cases that satisﬁed our outcome of interest, photo was actually about fashion. Frequently, marginal and joint probabilities are provided instead of count data. For example, disease rates are commonly listed in percentages rather than in a count format. We would like to be able to compute conditional probabilities even when no counts are available, and we use the last equation as a template to understand this technique. 3.2. CONDITIONAL PROBABILITY 155 We considered only those cases that satisﬁed the condition, where the ML algorithm predicted fashion. Of these cases, the conditional probability was the fraction representing the outcome of interest, that the photo was about fashion. Suppose we were provided only the information in Figure 3.10, i.e. only probability data. Then if we took a sample of 1000 photos, we would anticipate about 12.0% or 0.120 × 1000 = 120 would be predicted to be about fashion (mach learn is pred fashion). Similarly, we would expect about 10.8% or 0.108 × 1000 = 108 to meet both the information criteria and represent our outcome of interest. Then the conditional probability can be computed as P (truth is fashion \| mach learn is pred fashion) = = # (truth is fashion and mach learn is pred fashion) # (mach learn is pred fashion) 108 120 = 0.108 0.120 = 0.90 Here we are examining exactly the fraction of two probabilities, 0.108 and 0.120, which we can write as P (truth is fashion and mach learn is pred fashion) and P (mach learn is pred fashion). The fraction of these probabilities is an example of the general formula for conditional probability. CONDITIONAL PROBABILITY The conditional probability of the outcome of interest A given condition B is computed as the following: P (A\|B) = P (A and B) P (B) GUIDED PRACTICE 3.28 (a) Write out the following statement in conditional probability notation: “The probability that the ML prediction was correct, if the photo was about fashion”. Here the condition is now based on the photo’s truth status, not the ML algorithm. (b) Determine the probability from part (a). Figure 3.10 on page 153 may be helpful.26 GUIDED PRACTICE 3.29 (a) Determine the probability that the algorithm is incorrect if it is known the photo is about fashion. (b) Using the answers from part (a
) and Guided Practice 3.28(b), compute P (mach learn is pred fashion \| truth is fashion) + P (mach learn is pred not \| truth is fashion) (c) Provide an intuitive argument to explain why the sum in (b) is 1.27 26(a) If the photo is about fashion and the ML algorithm prediction was correct, then the ML algorithm my have a value of pred fashion: P (mach learn is pred fashion \| truth is fashion) for The equation conditional ﬁnd (b) P (mach learn is pred fashion and truth is fashion) = 0.1081 and P (truth is not) = 0.1696. Then the ratio represents the conditional probability: 0.1081/0.1696 = 0.6374. 27(a) This probability is P (mach learn is pred not, truth is fashion) 0.1696 = 0.3626. (b) The total equals 1. (c) Under the condition the photo is about fashion, the ML algorithm must have either predicted it was about fashion or predicted it was not about fashion. The complement still works for conditional probabilities, provided the probabilities are conditioned on the same information. P (truth is fashion) probability = 0.0615 indicates should ﬁrst we 156 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.2.4 Smallpox in Boston, 1721 The smallpox data set provides a sample of 6,224 individuals from the year 1721 who were exposed to smallpox in Boston.28 Doctors at the time believed that inoculation, which involves exposing a person to the disease in a controlled form, could reduce the likelihood of death. Each case represents one person with two variables: inoculated and result. The variable inoculated takes two levels: yes or no, indicating whether the person was inoculated or not. The variable result has outcomes lived or died. These data are summarized in Tables 3.12 and 3.13. result lived died Total inoculated yes 238 6 244 5136 844 5980 no Total 5374 850 6224 Figure 3.12: Contingency table for the smallpox data set. result lived died Total inoculated yes 0.0382 0.0010 0.0392 no 0.8252 0.1356 0.9608 Total 0.8634 0.1366 1.0000 Figure 3.13: Table proportions for the smallpox data, computed by dividing each count by the table total, 6224. GUIDED PRACTICE 3.30 Write out, in formal notation, the probability a randomly selected person who was not inoculated died from smallpox, and ﬁnd this probability.29 GUIDED PRACTICE 3.31 Determine the probability that an inoculated person died from smallpox. How does this result compare with the result of Guided Practice 3.30?30 GUIDED PRACTICE 3.32 The people of Boston self-selected whether or not to be inoculated. (a) Is this study observational or was this an experiment? (b) Can we infer any causal connection using these data? (c) What are some potential confounding variables that might inﬂuence whether someone lived or died and also aﬀect whether that person was inoculated?31 28Fenner F. 1988. Smallpox and Its Eradication (History of International Public Health, No. 6). Geneva: World Health Organization. ISBN 92-4-156110-6. 29P (result = died \| not inoculated) = P (result = died and not inoculated) 30P (died \| inoculated) = P (died and inoculated) P (not inoculated) = 0.0010 0.9608 = 0.1411. 0.0392 = 0.0255. The death rate for individuals who were = 0.1356 P (inoculated) inoculated is only about 1 in 40 while the death rate is about 1 in 7 for those who were not inoculated. 31Brief answers: (a) Observational. (b) No, we cannot infer causation from this observational study. (c) Accessi- bility to the latest and best medical care, so income may play a role. There are other valid answers for part (c). 3.2. CONDITIONAL PROBABILITY 157 3.2.5 General multiplication rule Section 3.1.6 introduced the Multiplication Rule for independent processes. Here we provide the General Multiplication Rule for events that might not be independent. GENERAL MULTIPLICATION RULE If A and B represent two outcomes or events, then P (A and B) = P (A\|B) × P (B) For the term P (A\|B), it is useful to think of A as the outcome of interest and B as the condition. This General Multiplication Rule is simply a rearrangement of the deﬁnition for conditional probability. EXAMPLE 3.33 Consider the smallpox data set. Suppose we are given only two pieces of information: 96.08% of residents were not inoculated, and 85.88% of the residents who were not inoculated ended up surviving. How could we compute the probability that a resident was not inoculated and lived? We will compute our answer using the General Multiplication Rule and then verify it using Figure 3.13. We want to determine and we are given that P (lived and not inoculated) P (lived \| not inoculated) = 0.8588 P (not inoculated) = 0.9608 Among the 96.08% of people who were not inoculated, 85.88% survived: P (lived and not inoculated) = 0.8588 × 0.9608 = 0.8251 This is equivalent to the General Multiplication Rule. We can conﬁrm this probability in Figure 3.13 at the intersection of no and lived (with a small rounding error). GUIDED PRACTICE 3.34 Use P (inoculated) = 0.0392 and P (lived \| inoculated) = 0.9754 to determine the probability that a person was both inoculated and lived.32 GUIDED PRACTICE 3.35 If 97.54% of the inoculated people lived, what proportion of inoculated people must have died?33 GUIDED PRACTICE 3.36 Based on the probabilities computed above, does it appear that inoculation is eﬀective at reducing the risk of death from smallpox?34 32The answer is 0.0382, which can be veriﬁed using Figure 3.13. 33There were only two possible outcomes: lived or died. This means that 100% - 97.54% = 2.46% of the people who were inoculated died. 34The samples are large relative to the diﬀerence in death rates for the “inoculated” and “not inoculated” groups, so it seems there is an association between inoculated and outcome. However, as noted in the solution to Guided Practice 3.32, this is an observational study and we cannot be sure if there is a causal connection. (Further research has shown that inoculation is eﬀective at reducing death rates.) 158 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.2.6 Sampling without replacement EXAMPLE 3.37 Professors sometimes select a student at random to answer a question. If each student has an equal chance of being selected and there are 15 people in your class, what is the chance that she will pick you for the next question? If there are 15 people to ask and none are skipping class, then the probability is 1/15, or about 0.067. EXAMPLE 3.38 If the professor asks 3 questions, what is the probability that you will not be selected? Assume that she will not pick the same person twice in a given lecture. For the ﬁrst question, she will pick someone else with probability 14/15. When she asks the second question, she only has 14 people who have not yet been asked. Thus, if you were not picked on the ﬁrst question, the probability you are again not picked is 13/14. Similarly, the probability you are again not picked on the third question is 12/13, and the probability of not being picked for any of the three questions is P (not picked in 3 questions) = P (Q1 = not picked, Q2 = not picked, Q3 = not picked.) = 14 15 × 13 14 × 12 13 = 12 15 = 0.80 GUIDED PRACTICE 3.39 What rule permitted us to multiply the probabilities in Example 3.38?35 EXAMPLE 3.40 Suppose the professor randomly picks without regard to who she already selected, i.e. students can be picked more than once. What is the probability that you will not be picked for any of the three questions? Each pick is independent, and the probability of not being picked for any individual question is 14/15. Thus, we can use the Multiplication Rule for independent processes. P (not picked in 3 questions) = P (Q1 = not picked, Q2 = not picked, Q3 = not picked.) = 14 15 × 14 15 × 14 15 = 0.813 You have a slightly higher chance of not being picked compared to when she picked a new person for each question. However, you now may be picked more than once. 35The three probabilities we computed were actually one marginal probability, P (Q1=not picked), and two condi- tional probabilities: P (Q2 = not picked \| Q1 = not picked) P (Q3 = not picked \| Q1 = not picked, Q2 = not picked) Using the General Multiplication Rule, the product of these three probabilities is the probability of not being picked in 3 questions. 3.2. CONDITIONAL PROBABILITY 159 GUIDED PRACTICE 3.41 Under the setup of Example 3.40, what is the probability of being picked to answer all three questions?36 If we sample from a small population without replacement, we no longer have independence between our observations. In Example 3.38, the probability of not being picked for the second question was conditioned on the event that you were not picked for the ﬁrst question. In Example 3.40, the professor sampled her students with replacement: she repeatedly sampled the entire class without regard to who she already picked. GUIDED PRACTICE 3.42 Your department is holding a raﬄe. They sell 30 tickets and oﬀer seven prizes. (a) They place the tickets in a hat and draw one for each prize. The tickets are sampled without replacement, i.e. the selected tickets are not placed back in the hat. What is the probability of winning a prize if you buy one ticket? (b) What if the tickets are sampled with replacement?37 GUIDED PRACTICE 3.43 Compare your answers in Guided Practice 3.42. How much inﬂuence does the sampling method have on your chances of winning a prize?38 Had we repeated Guided Practice 3.42 with 300 tickets instead of 30, we would have found something interesting: the results would be nearly identical. The probability would be 0.0233 without replacement and 0.0231 with replacement. SAMPLING WITHOUT REPLACEMENT When the sample size is only a small fraction of the population (under 10%), observations can be considered independent even when sampling without replacement. 36P (being picked to answer all three questions) = 1 15 37(a) First determine the probability of not winning. The tickets are sampled without replacement, which means the probability you do not win on the ﬁrst draw is 29/30, 28/29 for the second, ..
., and 23/24 for the seventh. The probability you win no prize is the product of these separate probabilities: 23/30. That is, the probability of winning a prize is 1 − 23/30 = 7/30 = 0.233. (b) When the tickets are sampled with replacement, there are seven independent draws. Again we ﬁrst ﬁnd the probability of not winning a prize: (29/30)7 = 0.789. Thus, the probability of winning (at least) one prize when drawing with replacement is 0.211. = 0.00030. 3 38There is about a 10% larger chance of winning a prize when using sampling without replacement. However, at most one prize may be won under this sampling procedure. 160 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.2.7 Independence considerations in conditional probability If two processes are independent, then knowing the outcome of one should provide no informa- tion about the other. We can show this is mathematically true using conditional probabilities. GUIDED PRACTICE 3.44 Let X and Y represent the outcomes of rolling two dice. (a) What is the probability that the ﬁrst die, X, is 1? (b) What is the probability that both X and Y are 1? (c) Use the formula for conditional probability to compute P (Y = 1 \|X = 1). (d) What is P (Y = 1)? Is this diﬀerent from the answer from part (c)? Explain.39 We can show in Guided Practice 3.44(c) that the conditioning information has no inﬂuence by using the Multiplication Rule for independence processes: P (Y = 1\|X = 1) = = P (Y = 1 and X = 1) P (X = 1) P (Y = 1) × P (X = 1) P (X = 1) = P (Y = 1) GUIDED PRACTICE 3.45 Ron is watching a roulette table in a casino and notices that the last ﬁve outcomes were black. He ﬁgures that the chances of getting black six times in a row is very small (about 1/64) and puts his paycheck on red. What is wrong with his reasoning?40 3.2.8 Checking for independent and mutually exclusive events If A and B are independent events, then the probability of A being true is unchanged if B is true. Mathematically, this is written as P (A\|B) = P (A) The General Multiplication Rule states that P (A and B) equals P (A\|B) × P (B). If A and B are independent events, we can replace P (A\|B) with P (A) and the following multiplication rule applies: P (A and B) = P (A) × P (B) CHECKING WHETHER TWO EVENTS ARE INDEPENDENT When checking whether two events A and B are independent, verify one of the following equations holds (there is no need to check both equations): P (A\|B) = P (A) P (A and B) = P (A) × P (B) If the equation that is checked holds true (the left and right sides are equal), A and B are independent. If the equation does not hold, then A and B are dependent. 39Brief solutions: (a) 1/6. (b) 1/36. (c) P (Y = 1 and X= 1) 1/6 = 1/6. (d) The probability is the same as in part (c): P (Y = 1) = 1/6. The probability that Y = 1 was unchanged by knowledge about X, which makes sense as X and Y are independent. = 1/36 P (X= 1) 40He has forgotten that the next roulette spin is independent of the previous spins. Casinos do employ this practice; they post the last several outcomes of many betting games to trick unsuspecting gamblers into believing the odds are in their favor. This is called the gambler’s fallacy. 3.2. CONDITIONAL PROBABILITY 161 EXAMPLE 3.46 Are teenager college attendance and parent college degrees independent or dependent? Figure 3.14 may be helpful. We’ll use the ﬁrst equation above to check for independence. If the teen and parents variables are independent, it must be true that P (teen college \| parent degree) = P (teen college) Using Figure 3.14, we check whether equality holds in this equation. P (teen college \| parent degree) ?= P (teen college) 0.83 = 0.56 The value 0.83 came from a probability calculation using Figure 3.14: 231 280 ≈ 0.83. Because the sides are not equal, teenager college attendance and parent degree are dependent. That is, we estimate the probability a teenager attended college to be higher if we know that one of the teen’s parents has a college degree. teen college not Total parents degree 231 49 280 not 214 298 512 Total 445 347 792 Figure 3.14: Contingency table summarizing the family college data set. GUIDED PRACTICE 3.47 Use the second equation in the box above to show that teenager college attendance and parent college degrees are dependent.41 If A and B are mutually exclusive events, then A and B cannot occur at the same time. Mathematically, this is written as P (A and B) = 0 The General Addition Rule states that P (A or B) equals P (A) + P (B) − P (A and B). If A and B are mutually exclusive events, we can replace P (A and B) with 0 and the following addition rule applies: P (A or B) = P (A) + P (B) 41We check for equality in the following equation: P (teen college, parent degree) ? = P (teen college) × P (parent degree) 231 792 = 0.292 = 445 792 × 280 792 = 0.199 These terms are not equal, which conﬁrms what we learned in Example 3.46: teenager college attendance and parent college degrees are dependent. 162 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS CHECKING WHETHER TWO EVENTS ARE MUTUALLY EXCLUSIVE (DISJOINT) If A and B are mutually exclusive events, then they cannot occur at the same time. If asked to determine if events A and B are mutually exclusive, verify one of the following equations holds (there is no need to check both equations): P (A and B) = 0 P (A or B) = P (A) + P (B) If the equation that is checked holds true (the left and right sides are equal), A and B are mutually exclusive. If the equation does not hold, then A and B are not mutually exclusive. EXAMPLE 3.48 Are teen college attendance and parent college degrees mutually exclusive? Looking in the table, we see that there are 231 instances where both the teenager attended college and parents have a degree, indicating the probability of both events occurring is greater than 0. Since we have found an example where both of these events happen together, these two events are not mutually exclusive. We could more formally show this by computing the probability both events occur at the same time: P (teen college, parent degree) = 231 792 = 0 Since this probability is not zero, teenager college attendance and parent college degrees are not mutually exclusive. MUTUALLY EXCLUSIVE AND INDEPENDENT ARE DIFFERENT If two events are mutually exclusive, then if one is true, the other cannot be true. This implies the two events are in some way connected, meaning they must be dependent. If two events are independent, then if one occurs, it is still possible for the other to occur, meaning the events are not mutually exclusive. DEPENDENT EVENTS NEED NOT BE MUTUALLY EXCLUSIVE. If two events are dependent, we cannot simply conclude they are mutually exclusive. For example, the college attendance of teenagers and a college degree by one of their parents are dependent, but those events are not mutually exclusive. 3.2. CONDITIONAL PROBABILITY 163 3.2.9 Tree diagrams Tree diagrams are a tool to organize outcomes and probabilities around the structure of the data. They are most useful when two or more processes occur in a sequence and each process is conditioned on its predecessors. The smallpox data ﬁt this description. We see the population as split by inoculation: yes and no. Following this split, survival rates were observed for each group. This structure is reﬂected in the tree diagram shown in Figure 3.15. The ﬁrst branch for inoculation is said to be the primary branch while the other branches are secondary. Figure 3.15: A tree diagram of the smallpox data set. Tree diagrams are annotated with marginal and conditional probabilities, as shown in Figure 3.15. This tree diagram splits the smallpox data by inoculation into the yes and no groups with respective marginal probabilities 0.0392 and 0.9608. The secondary branches are conditioned on the ﬁrst, so we assign conditional probabilities to these branches. For example, the top branch in Figure 3.15 is the probability that lived conditioned on the information that inoculated. We may (and usually do) construct joint probabilities at the end of each branch in our tree by multiplying the numbers we come across as we move from left to right. These joint probabilities are computed using the General Multiplication Rule: P (inoculated and lived) = P (inoculated) × P (lived \| inoculated) = 0.0392 × 0.9754 = 0.0382 EXAMPLE 3.49 What is the probability that a randomly selected person who was inoculated died? This is equivalent to P (died \| inoculated). This conditional probability can be found in the second branch as 0.0246. InoculatedResultyes, 0.0392lived, 0.97540.03920.9754 = 0.03824died, 0.02460.03920.0246 = 0.00096no, 0.9608lived, 0.85890.96080.8589 = 0.82523died, 0.14110.96080.1411 = 0.13557 164 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS EXAMPLE 3.50 What is the probability that a randomly selected person lived? There are two ways that a person could have lived: be inoculated and live OR not be inoculated and live. To ﬁnd this probability, we sum the two disjoint probabilities: P (lived) = 0.0392 × 0.9745 + 0.9608 × 0.8589 = 0.03824 + 0.82523 = 0.86347 GUIDED PRACTICE 3.51 After an introductory statistics course, 78% of students can successfully construct tree diagrams. Of those who can construct tree diagrams, 97% passed, while only 57% of those students who could not construct tree diagrams passed. (a) Organize this information into a tree diagram. (b) What is the probability that a student who was able to construct tree diagrams did not pass? (c) What is the probability that a randomly selected student was able to successfully construct tree diagrams and passed? (d) What is the probability that a randomly selected student passed? 42 3.2.10 Bayes’ Theorem In many instances, we are given a conditional probability of the form P (statement about variable 1 \| statement about variable 2) but we would really like to know the inverted conditional probability: P (statement about variable 2 \| statement about variable 1) For example, inst
ead of wanting to know P (lived \| inoculated), we might want to know P (inoculated \| lived). This is more challenging because it cannot be read directly from the tree diagram. In these instances we use Bayes’ Theorem. Let’s begin by looking at a new example. \| able to construct 42(a) The tree diagram is shown to the right. (b) P (not pass tree diagram) = 0.03. (c) P (able to construct tree diagrams and passed) = P (able to construct tree diagrams) × P (passed \| able to construct tree diagrams) = 0.78 × 0.97 = 0.7566. (d) P (passed) = 0.7566 + 0.1254 = 0.8820. Able to constructtree diagramsPass classyes, 0.78pass, 0.970.780.97 = 0.7566fail, 0.030.780.03 = 0.0234no, 0.22pass, 0.570.220.57 = 0.1254fail, 0.430.220.43 = 0.0946 3.2. CONDITIONAL PROBABILITY 165 EXAMPLE 3.52 In Canada, about 0.35% of women over 40 will develop breast cancer in any given year. A common screening test for cancer is the mammogram, but this test is not perfect. In about 11% of patients with breast cancer, the test gives a false negative: it indicates a woman does not have breast cancer when she does have breast cancer. Similarly, the test gives a false positive in 7% of patients who do not have breast cancer: it indicates these patients have breast cancer when they actually do not. If we tested a random woman over 40 for breast cancer using a mammogram and the test came back positive – that is, the test suggested the patient has cancer – what is the probability that the patient actually has breast cancer? We are given suﬃcient information to quickly compute the probability of testing positive if a woman has breast cancer (1.00 − 0.11 = 0.89). However, we seek the inverted probability of cancer given a positive test result: P (has BC \| mammogram+) Here, “has BC” is an abbreviation for the patient actually having breast cancer, and “mammogram+” means the mammogram screening was positive, which in this case means the test suggests the patient has breast cancer. (Watch out for the non-intuitive medical language: a positive test result suggests the possible presence of cancer in a mammogram screening.) We can use the conditional probability formula from the previous section: P (A\|B) = P (A and B) . Our conditional probability can be found as follows: P (B) P (has BC \| mammogram+) = P (has BC and mammogram+) P (mammogram+) The probability that a mammogram is positive is as follows. P (mammogram+) = P (has BC and mammogram+) + P (no BC and mammogram+) A tree diagram is useful for identifying each probability and is shown in Figure 3.16. Using the tree diagram, we ﬁnd that P (has BC \| mammogram+) = = = P (has BC and mammogram+) P (has BC and mammogram+) + P (no BC and mammogram+) 0.0035(0.89) 0.0035(0.89) + 0.9965(0.07) 0.00312 0.07288 ≈ 0.0428 That is, even if a patient has a positive mammogram screening, there is still only a 4% chance that she has breast cancer. Example 3.52 highlights why doctors often run more tests regardless of a ﬁrst positive test result. When a medical condition is rare, a single positive test isn’t generally deﬁnitive. 166 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Figure 3.16: Tree diagram for Example 3.52, computing the probability a random patient who tests positive on a mammogram actually has breast cancer. Consider again the last equation of Example 3.52. Using the tree diagram, we can see that the numerator (the top of the fraction) is equal to the following product: P (has BC and mammogram+) = P (mammogram+\| has BC)P (has BC) The denominator – the probability the screening was positive – is equal to the sum of probabilities for each positive screening scenario: P (mammogram+) = P (mammogram+ and no BC) + P (mammogram+ and has BC) In the example, each of the probabilities on the right side was broken down into a product of a conditional probability and marginal probability using the tree diagram. P (mammogram+) = P (mammogram+ and no BC) + P (mammogram+ and has BC) = P (mammogram+\| no BC)P (no BC) + P (mammogram+\| has BC)P (has BC) We can see an application of Bayes’ Theorem by substituting the resulting probability expressions into the numerator and denominator of the original conditional probability. P (has BC\| mammogram+) = P (mammogram+\| has BC)P (has BC) P (mammogram+\| no BC)P (no BC) + P (mammogram+\| has BC)P (has BC) TruthMammogramcancer, 0.0035positive, 0.890.00350.89 = 0.00312negative, 0.110.00350.11 = 0.00038no cancer, 0.9965positive, 0.070.99650.07 = 0.06976negative, 0.930.99650.93 = 0.92675 3.2. CONDITIONAL PROBABILITY 167 BAYES’ THEOREM: INVERTING PROBABILITIES Consider the following conditional probability for variable 1 and variable 2: P (outcome A1 of variable 1\| outcome B of variable 2) Bayes’ Theorem states that this conditional probability can be identiﬁed as the following fraction: P (B\|A1)P (A1) P (B\|A1)P (A1) + P (B\|A2)P (A2) + · · · + P (B\|Ak)P (Ak) where A2, A3, ..., and Ak represent all other possible outcomes of the ﬁrst variable. Bayes’ Theorem is just a generalization of what we have done using tree diagrams. The formula need not be memorized, since it can always be derived using a tree diagram: • The numerator identiﬁes the probability of getting both A1 and B. • The denominator is the overall probability of getting B. Traverse each branch of the tree diagram that ends with event B. Add up the required products. GUIDED PRACTICE 3.53 Jose visits campus every Thursday evening. However, some days the parking garage is full, often due to college events. There are academic events on 35% of evenings, sporting events on 20% of evenings, and no events on 45% of evenings. When there is an academic event, the garage ﬁlls up about 25% of the time, and it ﬁlls up 70% of evenings with sporting events. On evenings when there are no events, it only ﬁlls up about 5% of the time. If Jose comes to campus and ﬁnds the garage full, what is the probability that there is a sporting event? Use a tree diagram to solve this problem. The tree diagram, with three primary branches, shown to the right. We want is P (sporting event\|garage full) = = P (sporting event and garage full) P (garage full) 0.14 0.0875 + 0.14 + 0.0225 = 0.56. If the garage is full, there is a 56% probability that there is a sporting event. The last several exercises oﬀered a way to update our belief about whether there is a sporting event, academic event, or no event going on at the school based on the information that the parking lot was full. This strategy of updating beliefs using Bayes’ Theorem is actually the foundation of an entire section of statistics called Bayesian statistics. While Bayesian statistics is very important and useful, we will not have time to cover it in this book. EventGarage fullAcademic, 0.35Full, 0.250.350.25 = 0.0875Spaces Available, 0.750.350.75 = 0.2625Sporting, 0.20Full, 0.70.20.7 = 0.14Spaces Available, 0.30.20.3 = 0.06None, 0.45Full, 0.050.450.05 = 0.0225Spaces Available, 0.950.450.95 = 0.4275 168 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Section summary • A conditional probability can be written as P (A\|B) and is read, “Probability of A given B”. P (A\|B) is the probability of A, given that B has occurred. In a conditional probability, we are given some information. In an unconditional probability, such as P (A), we are not given any information. • Sometimes P (A\|B) can be deduced. For example, when drawing without replacement from a deck of cards, P (2nd draw is an Ace \| 1st draw was an Ace) = 3 51 . When this is not the case, as when working with a table or a Venn diagram, one must use the conditional probability rule P (A\|B) = P (A and B) . P (B) • In the last section, we saw that two events are independent when the outcome of one has no eﬀect on the outcome of the other. When A and B are independent, P (A\|B) = P (A). • When A and B are dependent, ﬁnd the probability of A and B using the General Multi- plication Rule: P (A and B) = P (A\|B) × P (B). • In the special case where A and B are independent, P (A and B) = P (A) × P (B). • If A and B are mutually exclusive, they must be dependent, since the occurrence of one of them changes the probability that the other occurs to 0. • When sampling without replacement, such as drawing cards from a deck, make sure to use conditional probabilities when solving and problems. • Sometimes, the conditional probability P (B\|A) may be known, but we are interested in the “inverted” probability P (A\|B). Bayes’ Theorem helps us solve such conditional probabilities that cannot be easily answered. However, rather than memorize Bayes’ Theorem, one can generally draw a tree diagram and apply the conditional probability rule P (A\|B) = P (A and B) . P (B) 3.2. CONDITIONAL PROBABILITY 169 Exercises 3.13 Joint and conditional probabilities. P(A) = 0.3, P(B) = 0.7 (a) Can you compute P(A and B) if you only know P(A) and P(B)? (b) Assuming that events A and B arise from independent random processes, i. what is P(A and B)? ii. what is P(A or B)? iii. what is P(A\|B)? (c) If we are given that P(A and B) = 0.1, are the random variables giving rise to events A and B indepen- dent? (d) If we are given that P(A and B) = 0.1, what is P(A\|B)? 3.14 PB & J. Suppose 80% of people like peanut butter, 89% like jelly, and 78% like both. Given that a randomly sampled person likes peanut butter, what’s the probability that he also likes jelly? A Pew Research poll asked 1,306 Americans “From what you’ve read and 3.15 Global warming. heard, is there solid evidence that the average temperature on earth has been getting warmer over the past few decades, or not?”. The table below shows the distribution of responses by party and ideology, where the counts have been replaced with relative frequencies.43 Response Not Earth is warming warming Conservative Republican Party and Mod/Lib Republican Mod/Cons Democrat Ideology Liberal Democrat Total 0.11 0.06 0.25 0.18 0.60 0.20 0.06 0.07 0.01 0.34 Don’t Know Refuse 0.02 0.01 0.02 0.01 0.06 Total 0.33 0.13 0.34 0.20 1.00 (a) Are
believing that the earth is warming and being a liberal Democrat mutually exclusive? (b) What is the probability that a randomly chosen respondent believes the earth is warming or is a liberal Democrat? (c) What is the probability that a randomly chosen respondent believes the earth is warming given that he is a liberal Democrat? (d) What is the probability that a randomly chosen respondent believes the earth is warming given that he is a conservative Republican? (e) Does it appear that whether or not a respondent believes the earth is warming is independent of their party and ideology? Explain your reasoning. (f) What is the probability that a randomly chosen respondent is a moderate/liberal Republican given that he does not believe that the earth is warming? 43Pew Research Center, Majority of Republicans No Longer See Evidence of Global Warming, data collected on October 27, 2010. 170 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.16 Health coverage, relative frequencies. The Behavioral Risk Factor Surveillance System (BRFSS) is an annual telephone survey designed to identify risk factors in the adult population and report emerging health trends. The following table displays the distribution of health status of respondents to this survey (excellent, very good, good, fair, poor) and whether or not they have health insurance. Health Coverage No Yes Total Excellent Very good 0.0364 0.3123 0.3486 0.0230 0.2099 0.2329 Health Status Good 0.0427 0.2410 0.2838 Fair 0.0192 0.0817 0.1009 Poor 0.0050 0.0289 0.0338 Total 0.1262 0.8738 1.0000 (a) Are being in excellent health and having health coverage mutually exclusive? (b) What is the probability that a randomly chosen individual has excellent health? (c) What is the probability that a randomly chosen individual has excellent health given that he has health coverage? (d) What is the probability that a randomly chosen individual has excellent health given that he doesn’t have health coverage? (e) Do having excellent health and having health coverage appear to be independent? 3.17 Marbles in an urn. Imagine you have an urn containing 5 red, 3 blue, and 2 orange marbles in it. (a) What is the probability that the ﬁrst marble you draw is blue? (b) Suppose you drew a blue marble in the ﬁrst draw. If drawing with replacement, what is the probability of drawing a blue marble in the second draw? (c) Suppose you instead drew an orange marble in the ﬁrst draw. If drawing with replacement, what is the probability of drawing a blue marble in the second draw? (d) If drawing with replacement, what is the probability of drawing two blue marbles in a row? (e) When drawing with replacement, are the draws independent? Explain. 3.18 Socks in a drawer. In your sock drawer you have 4 blue, 5 gray, and 3 black socks. Half asleep one morning you grab 2 socks at random and put them on. Find the probability you end up wearing (a) 2 blue socks (b) no gray socks (c) at least 1 black sock (d) a green sock (e) matching socks 3.19 Chips in a bag. Imagine you have a bag containing 5 red, 3 blue, and 2 orange chips. (a) Suppose you draw a chip and it is blue. If drawing without replacement, what is the probability the next is also blue? (b) Suppose you draw a chip and it is orange, and then you draw a second chip without replacement. What is the probability this second chip is blue? (c) If drawing without replacement, what is the probability of drawing two blue chips in a row? (d) When drawing without replacement, are the draws independent? Explain. 3.2. CONDITIONAL PROBABILITY 171 3.20 Books on a bookshelf. The table below shows the distribution of books on a bookcase based on whether they are nonﬁction or ﬁction and hardcover or paperback. Format Hardcover Paperback Total Type Fiction Nonﬁction Total 13 15 28 59 8 67 72 23 95 (a) Find the probability of drawing a hardcover book ﬁrst then a paperback ﬁction book second when drawing without replacement. (b) Determine the probability of drawing a ﬁction book ﬁrst and then a hardcover book second, when drawing without replacement. (c) Calculate the probability of the scenario in part (b), except this time complete the calculations under the scenario where the ﬁrst book is placed back on the bookcase before randomly drawing the second book. (d) The ﬁnal answers to parts (b) and (c) are very similar. Explain why this is the case. In a classroom with 24 students, 7 students are wearing jeans, 4 are wearing 3.21 Student outﬁts. shorts, 8 are wearing skirts, and the rest are wearing leggings. If we randomly select 3 students without replacement, what is the probability that one of the selected students is wearing leggings and the other two are wearing jeans? Note that these are mutually exclusive clothing options. 3.22 The birthday problem. Suppose we pick three people at random. For each of the following questions, ignore the special case where someone might be born on February 29th, and assume that births are evenly distributed throughout the year. (a) What is the probability that the ﬁrst two people share a birthday? (b) What is the probability that at least two people share a birthday? 3.23 Drawing box plots. After an introductory statistics course, 80% of students can successfully construct box plots. Of those who can construct box plots, 86% passed, while only 65% of those students who could not construct box plots passed. (a) Construct a tree diagram of this scenario. (b) Calculate the probability that a student is able to construct a box plot if it is known that he passed. 3.24 Predisposition for thrombosis. A genetic test is used to determine if people have a predisposition for thrombosis, which is the formation of a blood clot inside a blood vessel that obstructs the ﬂow of blood It is believed that 3% of people actually have this predisposition. The through the circulatory system. genetic test is 99% accurate if a person actually has the predisposition, meaning that the probability of a positive test result when a person actually has the predisposition is 0.99. The test is 98% accurate if a person does not have the predisposition. What is the probability that a randomly selected person who tests positive for the predisposition by the test actually has the predisposition? Lupus is a medical phenomenon where antibodies that are supposed to attack 3.25 It’s never lupus. foreign cells to prevent infections instead see plasma proteins as foreign bodies, leading to a high risk of blood clotting. It is believed that 2% of the population suﬀer from this disease. The test is 98% accurate if a person actually has the disease. The test is 74% accurate if a person does not have the disease. There is a line from the Fox television show House that is often used after a patient tests positive for lupus: “It’s never lupus.” Do you think there is truth to this statement? Use appropriate probabilities to support your answer. 3.26 Exit poll. Edison Research gathered exit poll results from several sources for the Wisconsin recall election of Scott Walker. They found that 53% of the respondents voted in favor of Scott Walker. Additionally, they estimated that of those who did vote in favor for Scott Walker, 37% had a college degree, while 44% of those who voted against Scott Walker had a college degree. Suppose we randomly sampled a person who participated in the exit poll and found that he had a college degree. What is the probability that he voted in favor of Scott Walker?44 44New York Times, Wisconsin recall exit polls. 172 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.3 Simulations What is the probability of getting a sum greater than 16 in three rolls of a die? Finding all possible combinations that satisfy this would be tedious, but we could conduct a physical simulation or a computer simulation to estimate this probability. With modern computing power, simulations have become an important and powerful tool for data scientists. In this section, we will look at the concepts that underlie simulations. Learning objectives 1. Understand the purpose of a simulation and recognize the application of the long-run relative frequency interpretation of probability. 2. Understand how random digit tables work and how to assign digits to outcomes. 3. Be able to repeat a simulation a set number of trials or until a condition is true, and use the results to estimate the probability of interest. 3.3.1 Setting up and carrying out simulations In the previous section we saw how to apply the binomial formula to ﬁnd the probability of exactly x successes in n independent trials when a success has probability p. Sometimes we have a problem we want to solve but we don’t know the appropriate formula, or even worse, a formula may not exist. In this case, one common approach is to estimate the probability using simulations. You may already be familiar with simulations. Want to know the probability of rolling a sum of 7 with a pair of dice? Roll a pair of dice many, many, many times and see what proportion of times the sum is 7. The more times you roll the pair of dice, the better the estimate will tend to be. Of course, such experiments can be time consuming or even infeasible. In this section, we consider simulations using random numbers. Random numbers (or technically, psuedo-random numbers) can be produced using a calculator or computer. Random digits are produced such that each digit, 0-9, is equally likely to come up in each spot. You’ll ﬁnd that occasionally we may have the same number in a row – sometimes multiple times – but in the long run, each digit should appear 1/10th of the time. Row 1 2 3 4 5 6 7 8 9 10 Column 1-5 43087 63432 19025 85117 16285 94342 61099 37537 04510 27217 6-10 41864 72132 83056 16706 56280 18473 14136 58839 16172 12151 11-15 51009 40269 62511 31083 01494 50845 39052 56876 90838 52645 16-20 39689 56103 52598 24816 90240 77757 50235 02960 15210 96218 Figure 3.17: Random number table. A full page of random numbers may be found in Appendix
C.1 on page 509. 3.3. SIMULATIONS 173 EXAMPLE 3.54 Mika’s favorite brand of cereal is running a special where 20% of the cereal boxes contain a prize. Mika really wants that prize. If her mother buys 6 boxes of the cereal over the next few months, what is the probability Mika will get a prize? To solve this problem using simulation, we need to be able to assign digits to outcomes. Each box should have a 20% chance of having a prize and an 80% chance of not having a prize. Therefore, a valid assignment would be: 0, 1 → prize 2-9 → no prize Of the ten possible digits (0, 1, 2, ..., 8, 9 ), two of them, i.e. 20% of them, correspond to winning a prize, which exactly matches the odds that a cereal box contains a prize. In Mika’s simulation, one trial will consist of 6 boxes of cereal, and therefore a trial will require six digits (each digit will correspond to one box of cereal). We will repeat the simulation for 20 trials. Therefore we will need 20 sets of 6 digits. Let’s begin on row 1 of the random digit table, shown in Figure 3.17. If a trial consisted of 5 digits, we could use the ﬁrst 5 digits going across: 43087. Because here a trial consists of 6 digits, it may be easier to read down the table, rather than read across. We will let trial 1 consist of the ﬁrst 6 digits in column 1 (461819 ), trial 2 consist of the ﬁrst 6 digits in column 2 (339564 ), etc. For this simulation, we will end up using the ﬁrst 6 rows of each of the 20 columns. In trial 1, there are two 1 ’s, so we record that as a success; in this trial there were actually two prizes. In trial 2 there were no 0 ’s or 1 ’s, therefore we do not record this as a success. In trial 3 there were three prizes, so we record this as a success. The rest of this exercise is left as a Guided Practice problem for you to complete. GUIDED PRACTICE 3.55 Finish the simulation above and report the estimate for the probability that Mika will get a prize if her mother buys 6 boxes of cereal where each one has a 20% chance of containing a prize.45 GUIDED PRACTICE 3.56 In the previous example, the probability that a box of cereal contains a prize is 20%. The question presented is equivalent to asking, what is the probability of getting at least one prize in six randomly selected boxes of cereal. This probability question can be solved explicitly using the method of complements. Find this probability. How does the estimate arrived at by simulation compare to this probability?46 45The trials that contain at least one 0 or 1 and therefore are successes are trials: 1, 3, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, and 20. There were 17 successes among the 20 trials, so our estimate of the probability based on this simulation is 17/20 = 0.85. 46The true probability is given by 1 − P (no prizes in six boxes) = 1 − 0.86 = 0.74. The estimate arrived at by simulation was 11% too high. Note: We only repeated the simulation 20 times. If we had repeated it 1000 times, we would (very likely) have gotten an estimate closer to the true probability. 174 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS We can also use simulations to estimate quantities other than probabilities. Consider the following example. EXAMPLE 3.57 Let’s say that instead of buying exactly 6 boxes of cereal, Mika’s mother agrees to buy boxes of this cereal until she ﬁnds one with a prize. On average, how many boxes of cereal would one have to buy until one gets a prize? For this question, we can use the same digit assignment. However, our stopping rule is diﬀerent. Each trial may require a diﬀerent number of digits. For each trial, the stopping rule is: look at digits until we encounter a 0 or a 1. Then, record how many digits/boxes of cereal it took. Repeat the simulation for 20 trials, and then average the numbers from each trial. Let’s begin again at row 1. We can read across or down, depending upon what is most convenient. Since there are 20 columns and we want 20 trials, we will read down the columns. Starting at column 1, we count how many digits (boxes of cereal) we encounter until we reach a 0 or 1 (which represent a prize). For trial 1 we see 461, so we record 3. For trial 2 we see 3395641, so we record 7. For trial 3, we see 0, so we record 1. The rest of this exercise is left as a Guided Practice problem for you to complete. GUIDED PRACTICE 3.58 Finish the simulation above and report your estimate for the average number of boxes of cereal one would have to buy until encountering a prize, where the probability of a prize in each box is 20%.47 EXAMPLE 3.59 Now, consider a case where the probability of interest is not 20%, but rather 28%. Which digits should correspond to success and which to failure? This example is more complicated because with only 10 digits, there is no way to select exactly 28% of them. Therefore, each observation will have to consist of two digits. We can use two digits at a time and assign pairs of digits as follows: 00-27 → success 28-99 → failure GUIDED PRACTICE 3.60 Assume the probability of winning a particular casino game is 45%. We want to carry out a simulation to estimate the probability that we will win at least 5 times in 10 plays. We will use 30 trials of the simulation. Assign digits to outcomes. Also, how many total digits will we require to run this simulation?48 47For the 20 trials, the number of digits we see until we encounter a 0 or 1 is: 3,7,1,4,9, 4,1,2,4,5, 5,1,1,1,3, 8,5,2,2,6. Now we take the average of these 20 numbers to get 74/20 = 3.7. 48One possible assignment is: 00-44 → win and 45-99 → lose. Each trial requires 10 pairs of digits, so we will need 30 sets of 10 pairs of digits for a total of 30 × 10 × 2 = 600 digits. 3.3. SIMULATIONS 175 GUIDED PRACTICE 3.61 Assume carnival spinner has 7 slots. We want to carry out a simulation to estimate the probability that we will win at least 10 times in 60 plays. Repeat 100 trials of the simulation. Assign digits to outcomes. Also, how many total digits will we require to run this simulation?49 Does anyone perform simulations like this? Sort of. Simulations are used a lot in statistics, and these often require the same principles covered in this section to properly set up those simulations. The diﬀerence is in implementation after the setup. Rather than use a random number table, a data scientist will write a program that uses a pseudo-random number generator in a computer to run the simulations very quickly – often times millions of trials each second, which provides much more accurate estimates than running a couple dozen trials by hand. Section summary • When a probability is diﬃcult to determine via a formula, one can set up a simulation to estimate the probability. • The relative frequency theory of probability and the Law of Large Numbers are the mathematical underpinning of simulations. A larger number of trials should tend to produce better estimates. • The ﬁrst step to setting up a simulation is to assign digits to represent outcomes. This should be done in such a way as to give the event of interest the correct probability. Then, using a random number table, calculator, or computer, generate random digits (outcomes). Repeat this a speciﬁed number of trials or until a given stopping rule. When this is ﬁnished, count up how many times the event happened and divide that by the number of trials to get the estimate of the probability. 49Note that 1/7 = 0.142857... This makes it tricky to assign digits to outcomes. The best approach here would be to exclude some of the digits from the simulation. We can assign 0 to success and 1-6 to failure. This corresponds to a 1/7 chance of getting a success. If we encounter a 7, 8, or 9, we will just skip over it. Because we don’t know how many 7, 8, or 9 ’s we will encounter, we do not know how many total digits we will end up using for the simulation. (If you want a challenge, try to estimate the total number of digits you would need.) 176 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Exercises 3.27 Smog check, Part I . Suppose 16% of cars fail pollution tests (smog checks) in California. We would like to estimate the probability that an entire ﬂeet of seven cars would pass using a simulation. We assume each car is independent. We only want to know if the entire ﬂeet passed, i.e. none of the cars failed. What is wrong with each of the following simulations to represent whether an entire (simulated) ﬂeet passed? (a) Flip a coin seven times where each toss represents a car. A head means the car passed and a tail means it failed. If all cars passed, we report PASS for the ﬂeet. If at least one car failed, we report FAIL. (b) Read across a random number table starting at line 5. If a number is a 0 or 1, let it represent a failed car. Otherwise the car passes. We report PASS if all cars passed and FAIL otherwise. (c) Read across a random number table, looking at two digits for each simulated car. If a pair is in the range [00-16], then the corresponding car failed. If it is in [17-99], the car passed. We report PASS if all cars passed and FAIL otherwise. 3.28 Left-handed. Studies suggest that approximately 10% of the world population is left-handed. Use ten simulations to answer each of the following questions. For each question, describe your simulation scheme clearly. (a) What is the probability that at least one out of eight people are left-handed? (b) On average, how many people would you have to sample until the ﬁrst person who is left-handed? (c) On average, how many left-handed people would you expect to ﬁnd among a random sample of six people? 3.29 Smog check, Part II . Consider the ﬂeet of seven cars in Exercise 3.27. Remember that 16% of cars fail pollution tests (smog checks) in California, and that we assume each car is independent. (a) Write out how to calculate the probability of the ﬂeet failing, i.e. at least one of the cars in the ﬂeet failing, via simulation. (b) Simulate 5 ﬂeets. Based on these simulations, estimate the probability at l
east one car will fail in a ﬂeet. (c) Compute the probability at least one car fails in a ﬂeet of seven. 3.30 To catch a thief. Suppose that at a retail store, 1/5th of all employees steal some amount of merchandise. The stores would like to put an end to this practice, and one idea is to use lie detector tests to catch and ﬁre thieves. However, there is a problem: lie detectors are not 100% accurate. Suppose it is known that a lie detector has a failure rate of 25%. A thief will slip by the test 25% of the time and an honest employee will only pass 75% of the time. (a) Describe how you would simulate whether an employee is honest or is a thief using a random number table. Write your simulation very carefully so someone else can read it and follow the directions exactly. (b) Using a random number table, simulate 20 employees working at this store and determine if they are honest or not. Make sure to record the random digits assigned to each employee as you will refer back to these in part (c). (c) Determine the result of the lie detector test for each simulated employee from part (b) using a new simulation scheme. (d) How many of these employees are “honest and passed” and how many are “honest and failed”? (e) How many of these employees are “thief and passed” and how many are “thief and failed”? (f) Suppose the management decided to ﬁre everyone who failed the lie detector test. What percent of ﬁred employees were honest? What percent of not ﬁred employees were thieves? 3.4. RANDOM VARIABLES 177 3.4 Random variables The chance of landing on single number in the game of roulette is 1/38 and the pay is 35:1. The chance of landing on Red is 18/38 and the pay is 1:1. Which game has the higher expected value? The higher standard deviation of expected winnings? How do we interpret these quantities in this context? If you were to play each game 20 times, what would the distribution of possible outcomes look like? In this section, we deﬁne and summarize random variables such as this, and we look at some of their properties. Learning objectives 1. Deﬁne a probability distribution and what makes a distribution a valid probability distribution. 2. Summarize a discrete probability distribution graphically using a histogram and verbally with respect to center, spread, and shape. 3. Calculate and interpret the mean (expected value) and standard deviation of a random variable. 4. Calculate the mean and standard deviation of a transformed random variable. 5. Calculate the mean of the sum or diﬀerence of random variables. 6. Calculate the standard deviation of the sum or diﬀerence of random variables when those variables are independent. 3.4.1 Introduction to expected value EXAMPLE 3.62 Two books are assigned for a statistics class: a textbook and its corresponding study guide. The university bookstore determined 20% of enrolled students do not buy either book, 55% buy the textbook only, and 25% buy both books, and these percentages are relatively constant from one term to another. If there are 100 students enrolled, how many books should the bookstore expect to sell to this class? Around 20 students will not buy either book (0 books total), about 55 will buy one book (55 books total), and approximately 25 will buy two books (totaling 50 books for these 25 students). The bookstore should expect to sell about 105 books for this class. GUIDED PRACTICE 3.63 Would you be surprised if the bookstore sold slightly more or less than 105 books?50 50If they sell a little more or a little less, this should not be a surprise. Hopefully Chapter 2 helped make clear that there is natural variability in observed data. For example, if we would ﬂip a coin 100 times, it will not usually come up heads exactly half the time, but it will probably be close. 178 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS EXAMPLE 3.64 The textbook costs $137 and the study guide $33. How much revenue should the bookstore expect from this class of 100 students? About 55 students will just buy a textbook, providing revenue of $137 × 55 = $7, 535 The roughly 25 students who buy both the textbook and the study guide would pay a total of ($137 + $33) × 25 = $170 × 25 = $4, 250 Thus, the bookstore should expect to generate about $7, 535 + $4, 250 = $11, 785 from these 100 students for this one class. However, there might be some sampling variability so the actual amount may diﬀer by a little bit. Figure 3.18: Probability distribution for the bookstore’s revenue from one student. The triangle represents the average revenue per student. EXAMPLE 3.65 What is the average revenue per student for this course? The expected total revenue is $11,785, and there are 100 students. Therefore the expected revenue per student is $11, 785/100 = $117.85. 3.4.2 Probability distributions A probability distribution is a table of all disjoint outcomes and their associated probabili- ties. Figure 3.19 shows the probability distribution for the sum of two dice. RULES FOR PROBABILITY DISTRIBUTIONS A probability distribution is a list of the possible outcomes with corresponding probabilities that satisﬁes three rules: 1. The outcomes listed must be disjoint. 2. Each probability must be between 0 and 1. 3. The probabilities must total 1. cost (dollars)01371700.00.10.20.30.40.5probability 3.4. RANDOM VARIABLES 179 GUIDED PRACTICE 3.66 Figure 3.20 suggests three distributions for household income in the United States. Only one is correct. Which one must it be? What is wrong with the other two?51 Dice sum Probability 2 1 36 3 2 36 4 3 36 5 4 36 6 5 36 7 6 36 8 5 36 9 4 36 10 3 36 11 2 36 12 1 36 Figure 3.19: Probability distribution for the sum of two dice. Income range ($1000s) (a) (b) (c) 0-25 0.18 0.38 0.28 25-50 0.39 -0.27 0.27 50-100 0.33 0.52 0.29 100+ 0.16 0.37 0.16 Figure 3.20: Proposed distributions of US household incomes (Guided Practice 3.66). Chapter 2 emphasized the importance of plotting data to provide quick summaries. Probability distributions can also be summarized in a histogram or bar plot. The probability distribution for the sum of two dice is shown in Figure 3.19 and its histogram is plotted in Figure 3.21. The distribution of US household incomes is shown in Figure 3.22 as a bar plot. The presence of the 100+ category makes it diﬃcult to represent it with a regular histogram. Figure 3.21: A histogram for the probability distribution of the sum of two dice. In these bar plots, the bar heights represent the probabilities of outcomes. If the outcomes are numerical and discrete, it is usually (visually) convenient to make a histogram, as in the case of the sum of two dice. Another example of plotting the bars at their respective locations is shown in Figure 3.18. 51The probabilities of (a) do not sum to 1. The second probability in (b) is negative. This leaves (c), which sure enough satisﬁes the requirements of a distribution. One of the three was said to be the actual distribution of US household incomes, so it must be (c). Dice sum234567891011120.000.050.100.15Probability 180 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Figure 3.22: A bar graph for the probability distribution of US household income. Because it is artiﬁcially separated into four unequal bins, this graph fails to show the shape or skew of the distribution. 3.4.3 Expectation We call a variable or process with a numerical outcome a random variable, and we usually represent this random variable with a capital letter such as X, Y , or Z. The amount of money a single student will spend on her statistics books is a random variable, and we represent it by X. RANDOM VARIABLE A random process or variable with a numerical outcome. The possible outcomes of X are labeled with a corresponding lower case letter x and subscripts. For example, we write x1 = $0, x2 = $137, and x3 = $170, which occur with probabilities 0.20, 0.55, and 0.25. The distribution of X is summarized in Figure 3.18 and Figure 3.23. i xi P (xi) 1 $0 0.20 2 $137 0.55 3 $170 0.25 Total – 1.00 Figure 3.23: The probability distribution for the random variable X, representing the bookstore’s revenue from a single student. We use P (xi) to represent the probability of xi. We computed the average outcome of X as $117.85 in Example 3.65. We call this average the expected value of X, denoted by E(X). The expected value of a random variable is computed by adding each outcome weighted by its probability: E(X) = 0 · P (0) + 137 · P (137) + 170 · P (170) = 0 · 0.20 + 137 · 0.55 + 170 · 0.25 = 117.85 EXPECTED VALUE OF A DISCRETE RANDOM VARIABLE If X takes outcomes x1, x2, ..., xn with probabilities P (x1), P (x2), ..., P (xn), the mean, or expected value, of X is the sum of each outcome multiplied by its corresponding probability: µX = E(X) = x1 · P (x1) + x2 · P (x2) + · · · + xn · P (xn) = n i=1 xi · P (xi) 0−2525−5050−100100+0.000.050.100.150.200.25US household incomes ($1000s)probability 3.4. RANDOM VARIABLES 181 The expected value for a random variable represents the average outcome. For example, E(X) = 117.85 represents the average amount the bookstore expects to make from a single student, which we could also write as µ = 117.85. While the bookstore will make more than this on some students and less than this on other students, the average of many randomly selected students will be near $117.85. It is also possible to compute the expected value of a continuous random variable. However, it requires a little calculus and we save it for a later class.52 In physics, the expectation holds the same meaning as the center of gravity. The distribution can be represented by a series of weights at each outcome, and the mean represents the balancing point. This is represented in Figures 3.18 and 3.24. The idea of a center of gravity also expands to continuous probability distributions. Figure 3.25 shows a continuous probability distribution balanced atop a wedge placed at the mean. Figure 3.24: A weight system representing the probability distribution for X.
The string holds the distribution at the mean to keep the system balanced. Figure 3.25: A continuous distribution can also be balanced at its mean. 52µX = xf (x)dx where f (x) represents a function for the density curve. 0137170117.85m 182 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.4.4 Variability in random variables Suppose you ran the university bookstore. Besides how much revenue you expect to generate, you might also want to know the volatility (variability) in your revenue. The variance and standard deviation can be used to describe the variability of a random variable. Section 2.2.2 introduced a method for ﬁnding the variance and standard deviation for a data set. We ﬁrst computed deviations from the mean (xi − µ), squared those deviations, and took an average to get the variance. In the case of a random variable, we again compute squared deviations. However, we take their sum weighted by their corresponding probabilities, just like we did for the expectation. This weighted sum of squared deviations equals the variance, and we calculate the standard deviation by taking the square root of the variance, just as we did in Section 2.2.2. VARIANCE AND STANDARD DEVIATION OF A DISCRETE RANDOM VARIABLE If X takes outcomes x1, x2, ..., xn with probabilities P (x1), P (x2), ..., P (xn) and expected value µX = E(X), then to ﬁnd the standard deviation of X, we ﬁrst ﬁnd the variance and then take its square root. V ar(X) = σ2 x = (x1 − µX)2 · P (x1) + (x2 − µX)2 · P (x2) + · · · + (xn − µX)2 · P (xn) n = (xi − µX)2 · P (xi) SD(X) = σX = i=1 n i=1 (xi − µX)2 · P (xi) Just as it is possible to compute the mean of a continuous random variable using calculus, we can also use calculus to compute the variance.53 However, this topic is beyond the scope of the AP exam. EXAMPLE 3.67 Compute the expected value, variance, and standard deviation of X, the revenue of a single statistics student for the bookstore. It is useful to construct a table that holds computations for each outcome separately, then add up the results. i xi P (xi) xi · P (xi) 1 $0 0.20 0 2 $137 0.55 75.35 3 $170 0.25 42.50 Total 117.85 Thus, the expected value is µX = 117.85, which we computed earlier. The variance can be constructed using a similar table: i xi P (xi) xi − µX (xi − µX)2 (xi − µX)2 · P (xi) 1 $0 0.20 -117.85 13888.62 2777.7 2 $137 0.55 19.15 366.72 201.7 3 $170 0.25 52.15 2719.62 679.9 Total 3659.3 The variance of X is σ2 X = 3659.3, which means the standard deviation is σX = √ 3659.3 = $60.49. 53σ2 x = (x − µX )2f (x)dx where f (x) represents a function for the density curve. 3.4. RANDOM VARIABLES 183 GUIDED PRACTICE 3.68 The bookstore also oﬀers a chemistry textbook for $159 and a book supplement for $41. From past experience, they know about 25% of chemistry students just buy the textbook while 60% buy both the textbook and supplement.54 (a) What proportion of students don’t buy either book? Assume no students buy the supplement without the textbook. (b) Let Y represent the revenue from a single student. Write out the probability distribution of Y , i.e. a table for each outcome and its associated probability. (c) Compute the expected revenue from a single chemistry student. (d) Find the standard deviation to describe the variability associated with the revenue from a single student. 3.4.5 Linear transformations of a random variable An online store is selling a limited edition t-shirt. The maximum a person is allowed to buy is 3. Let X be a random variable that represents how many of the t-shirts a t-shirt buyer orders. The probability distribution of X is given in the following table. xi P (xi) 1 0.6 2 0.3 3 0.1 Using the methods of the previous section we can ﬁnd that the mean µX = 1.5 and the standard deviation σX = 0.67. Suppose that the cost of each t-shirt is $30 and that there is ﬂat rate $5 shipping fee. The amount of money a t-shirt buyer pays, then, is 30X + 5, where X is the number of t-shirts ordered. To calculate the mean and standard deviation for the amount of money a t-shirt buyers pays, we could deﬁne a new variable Y as follows: Y = 30X + 5 GUIDED PRACTICE 3.69 Verify that the distribution of Y is given by the table below.55 yi P (yi) $35 0.6 $65 0.3 $95 0.1 54(a) 100% - 25% - 60% = 15% of students do not buy any books for the class. Part (b) is represented by the ﬁrst two lines in the table below. The expectation for part (c) is given as the total on the line yi · P (yi). The result of part (d) is the square-root of the variance listed on in the total on the last line: σY = V ar(Y ) = 4800 = 69.28. √ i (scenario) yi P (yi) yi · P (yi) yi − µY (yi − µY)2 (yi − µY)2 · P (yi) 5530 × 1 + 5 = 35; 30 × 2 + 5 = 65; 30 × 3 + 5 = 95 1 (noBook) 0.00 0.15 0.00 -159.75 25520.06 3828.0 2 (textbook) 159.00 0.25 39.75 -0.75 0.56 0.1 3 (both) 200.00 0.60 120.00 40.25 1620.06 972.0 Total E(Y ) = 159.75 V ar(Y ) ≈ 4800 184 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Using this new table, we can compute the mean and standard deviation of the cost for tshirt orders. However, because Y is a linear transformation of X, we can use the properties from Section 2.2.7. Recall that multiplying every X by 30 multiplies both the mean and standard deviation by 30. Adding 5 only adds 5 to the mean, not the standard deviation. Therefore, µ30X+5 = E(30X + 5) = 30 × E(X) + 5 = 30 × 1.5 + 5 = 50.00 σ30X+5 = SD(30X + 5) = 30 × SD(X) = 30 × 0.67 = 20.10 Among t-shirt buyers, they spend an average of $50.00, with a standard deviation of $20.10. LINEAR TRANSFORMATIONS OF A RANDOM VARIABLE If X is a random variable, then a linear transformation is given by aX + b, where a and b are some ﬁxed numbers. E(aX + b) = a × E(X) + b SD(aX + b) = \|a\| × SD(X) 3.4.6 Linear combinations of random variables So far, we have thought of each variable as being a complete story in and of itself. Sometimes it is more appropriate to use a combination of variables. For instance, the amount of time a person spends commuting to work each week can be broken down into several daily commutes. Similarly, the total gain or loss in a stock portfolio is the sum of the gains and losses in its components. EXAMPLE 3.70 John travels to work ﬁve days a week. We will use X1 to represent his travel time on Monday, X2 to represent his travel time on Tuesday, and so on. Write an equation using X1, ..., X5 that represents his travel time for the week, denoted by W . His total weekly travel time is the sum of the ﬁve daily values: W = X1 + X2 + X3 + X4 + X5 Breaking the weekly travel time W into pieces provides a framework for understanding each source of randomness and is useful for modeling W . EXAMPLE 3.71 It takes John an average of 18 minutes each day to commute to work. What would you expect his average commute time to be for the week? We were told that the average (i.e. expected value) of the commute time is 18 minutes per day: E(Xi) = 18. To get the expected time for the sum of the ﬁve days, we can add up the expected time for each individual day: E(W ) = E(X1 + X2 + X3 + X4 + X5) = E(X1) + E(X2) + E(X3) + E(X4) + E(X5) = 18 + 18 + 18 + 18 + 18 = 90 minutes The expectation of the total time is equal to the sum of the expected individual times. More generally, the expectation of a sum of random variables is always the sum of the expectation for each random variable. 3.4. RANDOM VARIABLES 185 GUIDED PRACTICE 3.72 Elena is selling a TV at a cash auction and also intends to buy a toaster oven in the auction. If X represents the proﬁt for selling the TV and Y represents the cost of the toaster oven, write an equation that represents the net change in Elena’s cash.56 GUIDED PRACTICE 3.73 Based on past auctions, Elena ﬁgures she should expect to make about $175 on the TV and pay about $23 for the toaster oven. In total, how much should she expect to make or spend?57 GUIDED PRACTICE 3.74 Would you be surprised if John’s weekly commute wasn’t exactly 90 minutes or if Elena didn’t make exactly $152? Explain.58 Two important concepts concerning combinations of random variables have so far been introduced. First, a ﬁnal value can sometimes be described as the sum of its parts in an equation. Second, intuition suggests that putting the individual average values into this equation gives the average value we would expect in total. This second point needs clariﬁcation – it is guaranteed to be true in what are called linear combinations of random variables. A linear combination of two random variables X and Y is a fancy phrase to describe a combination aX + bY where a and b are some ﬁxed and known numbers. For John’s commute time, there were ﬁve random variables – one for each work day – and each random variable could be written as having a ﬁxed coeﬃcient of 1: 1X1 + 1X2 + 1X3 + 1X4 + 1X5 For Elena’s net gain or loss, the X random variable had a coeﬃcient of +1 and the Y random variable had a coeﬃcient of -1. When considering the average of a linear combination of random variables, it is safe to plug in the mean of each random variable and then compute the ﬁnal result. For a few examples of nonlinear combinations of random variables – cases where we cannot simply plug in the means – see the footnote.59 LINEAR COMBINATIONS OF RANDOM VARIABLES AND THE AVERAGE RESULT If X and Y are random variables, then a linear combination of the random variables is given by aX + bY , where a and b are some ﬁxed numbers. To compute the average value of a linear combination of random variables, plug in the average of each individual random variable and compute the result: E(aX + bY ) = a × E(X) + b × E(Y ) Recall that the expected value is the same as the mean, i.e. E(X) = µX. 56She will make X dollars on the TV but spend Y dollars on the toaster oven: X − Y . 57E(X − Y ) = E(X) − E(Y ) = 175 − 23 = $152. She should expect to make about $152. 58No, since there is probably some variability. For example, the traﬃc will vary from one day to next, and auction prices will vary depending on the quality of the mer
chandise and the interest of the attendees. 59If X and Y are random variables, consider the following combinations: X 1+Y , X × Y , X/Y . In such cases, plugging in the average value for each random variable and computing the result will not generally lead to an accurate average value for the end result. 186 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS EXAMPLE 3.75 Leonard has invested $6000 in Google Inc. (stock ticker: GOOG) and $2000 in Exxon Mobil Corp. (XOM). If X represents the change in Google’s stock next month and Y represents the change in Exxon Mobil stock next month, write an equation that describes how much money will be made or lost in Leonard’s stocks for the month. For simplicity, we will suppose X and Y are not in percents but are in decimal form (e.g. if Google’s stock increases 1%, then X = 0.01; or if it loses 1%, then X = −0.01). Then we can write an equation for Leonard’s gain as $6000 × X + $2000 × Y If we plug in the change in the stock value for X and Y , this equation gives the change in value of Leonard’s stock portfolio for the month. A positive value represents a gain, and a negative value represents a loss. GUIDED PRACTICE 3.76 Suppose Google and Exxon Mobil stocks have recently been rising 2.1% and 0.4% per month, respectively. Compute the expected change in Leonard’s stock portfolio for next month.60 GUIDED PRACTICE 3.77 You should have found that Leonard expects a positive gain in Guided Practice 3.76. However, would you be surprised if he actually had a loss this month?61 3.4.7 Variability in linear combinations of random variables Quantifying the average outcome from a linear combination of random variables is helpful, but it is also important to have some sense of the uncertainty associated with the total outcome of that combination of random variables. The expected net gain or loss of Leonard’s stock portfolio was considered in Guided Practice 3.76. However, there was no quantitative discussion of the volatility of this portfolio. For instance, while the average monthly gain might be about $134 according to the data, that gain is not guaranteed. Figure 3.26 shows the monthly changes in a portfolio like Leonard’s during the 36 months from 2009 to 2011. The gains and losses vary widely, and quantifying these ﬂuctuations is important when investing in stocks. Figure 3.26: The change in a portfolio like Leonard’s for the 36 months from 2009 to 2011, where $6000 is in Google’s stock and $2000 is in Exxon Mobil’s. Just as we have done in many previous cases, we use the variance and standard deviation to describe the uncertainty associated with Leonard’s monthly returns. To do so, the standard deviations and variances of each stock’s monthly return will be useful, and these are shown in Figure 3.27. The stocks’ returns are nearly independent. 60E($6000 × X + $2000 × Y ) = $6000 × 0.021 + $2000 × 0.004 = $134. 61No. While stocks tend to rise over time, they are often volatile in the short term. Monthly returns (2009−2011)−1000−50005001000 3.4. RANDOM VARIABLES 187 GOOG XOM Mean (¯x) 0.0210 0.0038 Standard deviation (s) Variance (s2) 0.0072 0.0849 0.0027 0.0520 Figure 3.27: The mean, standard deviation, and variance of the GOOG and XOM stocks. These statistics were estimated from historical stock data, so notation used for sample statistics has been used. We want to describe the uncertainty of Leonard’s monthly returns by ﬁnding the standard deviation of the return on his combined portfolio. First, we note that the variance of a sum has a nice property: the variance of a sum is the sum of the variances. That is, if X and Y are independent random variables: V ar(X + Y ) = V ar(X) + V ar(Y ) Because the standard deviation is the square root of the variance, we can rewrite this equation using standard deviations: (SDX+Y )2 = (SDX )2 + (SDY )2 This equation might remind you of a theorem from geometry: c2 = a2 + b2. The equation for the standard deviation of the sum of two independent random variables looks analogous to the Pythagorean Theorem. Just as the Pythagorean Theorem only holds for right triangles, this equation only holds when X and Y are independent.62 STANDARD DEVIATION OF THE SUM AND DIFFERENCE OF RANDOM VARIABLES If X and Y are independent random variables: SDX+Y = SDX−Y = (SDX )2 + (SDY )2 Because SDY = SD−Y , the standard deviation of the diﬀerence of two variables equals the standard deviation of the sum of two variables. This property holds for more than two variables as well. For example, if X, Y, and Z are independent random variables: SDX+Y +Z = SDX−Y −Z = (SDX )2 + (SDY )2 + (SDZ)2 If we need the standard deviation of a linear combination of independent variables, such as aX + bY , we can consider aX and bY as two new variables. Recall that multiplying all of the values of variable by a positive constant multiplies the standard deviation by that constant. Thus, SDaX = a × SDX and SDbY = b × SDY . It follows that: SDaX+bY = (a × SDX )2 + (b × SDY )2 This equation can be used to compute the standard deviation of Leonard’s monthly return. Recall that Leonard has $6,000 in Google stock and $2,000 in Exxon Mobil’s stock. From Figure 3.27, the standard deviation of Google stock is 0.0849 and the standard deviation of Exxon Mobile stock is 0.0520. SD6000X+2000Y = (6000 × SDX )2 + (2000 × SDY )2 = (6000 × 0.0849)2 + (2000 × .0520)2 = 270,304 = 520 The standard deviation of the total is $520. While an average monthly return of $134 on an $8000 investment is nothing to scoﬀ at, the monthly returns are so volatile that Leonard should not expect this income to be very stable. 62Another word for independent is orthogonal, meaning right angle! When X and Y are dependent, the equation for SDX+Y becomes analogous to the law of cosines. 188 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS STANDARD DEVIATION OF LINEAR COMBINATIONS OF RANDOM VARIABLES To ﬁnd the standard deviation of a linear combination of random variables, we ﬁrst consider aX and bY separately. We ﬁnd the standard deviation of each, and then we apply the equation for the standard deviation of the sum of two variables: SDaX+bY = (a × SDX )2 + (b × SDY )2 This equation is valid as long as the random variables X and Y are independent of each other. EXAMPLE 3.78 Suppose John’s daily commute has a standard deviation of 4 minutes. What is the uncertainty in his total commute time for the week? The expression for John’s commute time is X1 + X2 + X3 + X4 + X5 Each coeﬃcient is 1, so the standard deviation of the total weekly commute time is SD = (1 × 4)2 + (1 × 4)2 + (1 × 4)2 + (1 × 4)2 + (1 × 4)2 = 5 × (4)2 = 8.94 The standard deviation for John’s weekly work commute time is about 9 minutes. GUIDED PRACTICE 3.79 The computation in Example 3.78 relied on an important assumption: the commute time for each day is independent of the time on other days of that week. Do you think this is valid? Explain.63 GUIDED PRACTICE 3.80 Consider Elena’s two auctions from Guided Practice 3.72 on page 185. Suppose these auctions are approximately independent and the variability in auction prices associated with the TV and toaster oven can be described using standard deviations of $25 and $8. Compute the standard deviation of Elena’s net gain.64 Consider again Guided Practice 3.80. The negative coeﬃcient for Y in the linear combination was eliminated when we squared the coeﬃcients. This generally holds true: negatives in a linear combination will have no impact on the variability computed for a linear combination, but they do impact the expected value computations. 63One concern is whether traﬃc patterns tend to have a weekly cycle (e.g. Fridays may be worse than other days). If that is the case, and John drives, then the assumption is probably not reasonable. However, if John walks to work, then his commute is probably not aﬀected by any weekly traﬃc cycle. 64The equation for Elena can be written as: (1) × X + (−1) × Y . To ﬁnd the SD of this new variable we do: SD(1)×X+(−1)×Y = (1 × SDX )2 + (−1 × SDY )2 = (1 × 25)2 + (−1 × 8)2 = 26.25 The SD is about $26.25. 3.4. RANDOM VARIABLES 189 3.4.8 Normal approximation for sums of random variables We have seen that many distributions are approximately normal. The sum and the diﬀerence of normally distributed variables is also normal. While we cannot prove this here, the usefulness of it is seen in the following example. EXAMPLE 3.81 Three friends are playing a cooperative video game in which they have to complete a puzzle as fast as possible. Assume that the individual times of the 3 friends are independent of each other. The individual times of the friends in similar puzzles are approximately normally distributed with the following means and standard deviations. Friend 1 Friend 2 Friend 3 Mean 5.6 5.8 6.1 SD 0.11 0.13 0.12 To advance to the next level of the game, the friends’ total time must not exceed 17.1 minutes. What is the probability that they will advance to the next level? Because each friend’s time is approximately normally distributed, the sum of their times is also approximately normally distributed. We will do a normal approximation, but ﬁrst we need to ﬁnd the mean and standard deviation of the sum. We learned how to do this in Section 3.4. Let the three friends be labeled X, Y , Z. We want P (X + Y + Z < 17.1). The mean and standard deviation of the sum of X, Y , and Z is given by: µsum = E(X + Y + Z) = E(X) + E(Y ) + E(Z) = 5.6 + 5.8 + 6.1 = 17.5 Now we can ﬁnd the Z-score. σsum = (SDX )2 + (SDY )2 + (SDZ)2 = (0.11)2 + (0.13)2 + (0.12)2 = 0.208 Z = = xsum − µsum σsum 17.1 − 17.5 0.208 = −1.92 Finally, we want the probability that the sum is less than 17.5, so we shade the area to the left of Z = −1.92. Using technology, we get There is a 2.7% chance that the friends will advance to the next level. P (Z < −1.92) = 0.027 GUIDED PRACTICE 3.82 What is the probability that Friend 2 will complete the puzzle with a faster time than Friend 1? Hint: ﬁnd P (Y < X), or P (Y − X <
0).65 65First ﬁnd the mean and standard deviation of Y − X. The mean of Y − X is µY −X = 5.8 − 5.6 = 0.2. The 0.170 = −1.18 and P (Z < −1.18) = .119. standard deviation is SDY −X = (0.13)2 + (0.11)2 = 0.170. Then Z = 0−0.2 There is an 11.9% chance that Friend 2 will complete the puzzle with a faster time than Friend 1. 190 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Section summary • A discrete probability distribution can be summarized in a table that consists of all possible outcomes of a random variable and the probabilities of those outcomes. The outcomes must be disjoint, and the sum of the probabilities must equal 1. • A probability distribution can be represented with a histogram and, like the distributions of data that we saw in Chapter 2, can be summarized by its center, spread, and shape. • When given a probability distribution table, we can calculate the mean (expected value) and standard deviation of a random variable using the following formulas. E(X) = µX = xi · P (xi) = x1 · P (x1) + x2 · P (x2) + · · · + xn · P (xn) V ar(X) = σ2 x = (xi − µX)2 · P (xi) SD(X) = σX = (xi − µX)2 · P (xi) = (x1 − µX)2 · P (x1) + (x2 − µX)2 · P (x2) + · · · + (xn − µX)2 · P (xn) We can think of P (xi) as the weight, and each term is weighted its appropriate amount. • The mean of a probability distribution does not need to be a value in the distribution. It represents the average of many, many repetitions of a random process. The standard deviation represents the typical variation of the outcomes from the mean, when the random process is repeated over and over. • Linear transformations. Adding a constant to every value in a probability distribution adds that value to the mean, but it does not aﬀect the standard deviation. When multiplying every value by a constant, this multiplies the mean by the constant and it multiplies the standard deviation by the absolute value of the constant. • Combining random variables. Let X and Y be random variables and let a and b be constants. – The expected value of the sum is the sum of the expected values. E(X + Y ) = E(X) + E(Y ) E(aX + bY ) = a × E(X) + b × E(Y ) – When X and Y are independent: The standard deviation of a sum or a diﬀerence is the square root of the sum of each standard deviation squared. SD(X + Y ) = (SD(X))2 + (SD(Y ))2 SD(X − Y ) = (SD(X))2 + (SD(Y ))2 SD(aX + bY ) = (a × SD(X))2 + (b × SD(Y ))2 The SD properties require that X and Y be independent. The expected value properties hold true whether or not X and Y are independent. • Because the sum or diﬀerence of two normally distributed variables is itself a normally distributed variable, the normal approximation is also used in the following type of problem. Find the probability that a sum X + Y or a diﬀerence X − Y is greater/less than some value. 1. Verify that the distribution of X and the distribution of Y are approximately normal. 2. Find the mean of the sum or diﬀerence. Recall: the mean of a sum is the sum of the means. The mean of a diﬀerence is the diﬀerence of the means. Find the SD of the sum or diﬀerence using: SD(X + Y ) = SD(X − Y ) = (SD(X))2 + (SD(Y ))2. 3. Calculate the Z-score. Use the calculated mean and SD to standardize the given sum or diﬀerence. 4. Find the appropriate area under the normal curve. 3.4. RANDOM VARIABLES 191 Exercises 3.31 Patreon contributions. The distribution of monthly contributions made to a particular Patreon creator is given as follows: Monthly Contribution Proportion $3 0.50 $5 0.30 $10 0.15 $25 0.05 (a) Compute the average monthly contribution made to this creator. (b) Compute the standard deviation of the monthly contributions made to this creator. 3.32 Experiment launch impact. An online shopping website develops new features and iterates on its product recommendations on a regular basis. The team also uses an experiment when launching each new feature to assess the impact of the change on the total sales. Below is a summary table of the impacts that the team observes: Number of Sales From Launch Proportion of Launches -200 0.1 0 0.5 +200 +500 0.3 0.1 (a) Compute the average impact for the experiments. (b) Compute the standard deviation of the impact from these experiments. 3.33 Hearts win. In a new card game, you start with a well-shuﬄed full deck and draw 3 cards without replacement. If you draw 3 hearts, you win $50. If you draw 3 black cards, you win $25. For any other draws, you win nothing. (a) Create a probability model for the amount you win at this game, and ﬁnd the expected winnings. Also compute the standard deviation of this distribution. (b) If the game costs $5 to play, what would be the expected value and standard deviation of the net proﬁt (or loss)? (Hint: proﬁt = winnings − cost; X − 5) (c) If the game costs $5 to play, should you play this game? Explain. 3.34 Ace of clubs wins. Consider the following card game with a well-shuﬄed deck of cards. If you draw a red card, you win nothing. If you get a spade, you win $5. For any club, you win $10 plus an extra $20 for the ace of clubs. (a) Create a probability model for the amount you win at this game. Also, ﬁnd the expected winnings for a single game and the standard deviation of the winnings. (b) What is the maximum amount you would be willing to pay to play this game? Explain your reasoning. 3.35 Portfolio return. A portfolio’s value increases by 18% during a ﬁnancial boom and by 9% during normal times. It decreases by 12% during a recession. What is the expected return on this portfolio if each scenario is equally likely? 3.36 Baggage fees. An airline charges the following baggage fees: $25 for the ﬁrst bag and $35 for the second. Suppose 54% of passengers have no checked luggage, 34% have one piece of checked luggage and 12% have two pieces. We suppose a negligible portion of people check more than two bags. (a) Build a probability model, compute the average revenue per passenger, and compute the corresponding standard deviation. (b) About how much revenue should the airline expect for a ﬂight of 120 passengers? With what standard deviation? Note any assumptions you make and if you think they are justiﬁed. 3.37 American roulette. The game of American roulette involves spinning a wheel with 38 slots: 18 red, 18 black, and 2 green. A ball is spun onto the wheel and will eventually land in a slot, where each slot has an equal chance of capturing the ball. Gamblers can place bets on red or black. If the ball lands on their color, they double their money. If it lands on another color, they lose their money. Suppose you bet $1 on red. What’s the expected value and standard deviation of your winnings? 192 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.38 European roulette. The game of European roulette involves spinning a wheel with 37 slots: 18 red, 18 black, and 1 green. A ball is spun onto the wheel and will eventually land in a slot, where each slot has an equal chance of capturing the ball. Gamblers can place bets on red or black. If the ball lands on their color, they double their money. If it lands on another color, they lose their money. (a) Suppose you play roulette and bet $3 on a single round. What is the expected value and standard deviation of your total winnings? (b) Suppose you bet $1 in three diﬀerent rounds. What is the expected value and standard deviation of your total winnings? (c) How do your answers to parts (a) and (b) compare? What does this say about the riskiness of the two games? 3.39 Lemonade at The Cafe. Drink pitchers at The Cafe are intended to hold about 64 ounces of lemonade and glasses hold about 12 ounces. However, when the pitchers are ﬁlled by a server, they do not always ﬁll it with exactly 64 ounces. There is some variability. Similarly, when they pour out some of the lemonade, they do not pour exactly 12 ounces. The amount of lemonade in a pitcher is normally distributed with mean 64 ounces and standard deviation 1.732 ounces. The amount of lemonade in a glass is normally distributed with mean 12 ounces and standard deviation 1 ounce. (a) How much lemonade would you expect to be left in a pitcher after pouring one glass of lemonade? (b) What is the standard deviation of the amount left in a pitcher after pouring one glass of lemonade? (c) What is the probability that more than 50 ounces of lemonade is left in a pitcher after pouring one glass of lemonade? 3.40 Spray paint, Part I. Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribution with a mean of 25 square feet and a standard deviation of 3 square feet. Suppose also that you buy three cans of spray paint. (a) How much area would you expect to cover with these three cans of spray paint? (b) What is the standard deviation of the area you expect to cover with these three cans of spray paint? (c) The area you wanted to cover is 80 square feet. What is the probability that you will be able to cover this entire area with these three cans of spray paint? In Exercises 2.27 and 2.29 we saw two distributions for GRE scores: 3.41 GRE scores, Part III. N (µ = 151, σ = 7) for the verbal part of the exam and N (µ = 153, σ = 7.67) for the quantitative part. Suppose performance on these two sections is independent. Use this information to compute each of the following: (a) The probability of a combined (verbal + quantitative) score above 320. (b) The score of a student who scored better than 90% of the test takers overall. 3.42 Betting on dinner, Part I . Suppose a restaurant is running a promotion where prices of menu items are random following some underlying distribution. If you’re lucky, you can get a basket of fries for $3, or if you’re not so lucky you might end up having to pay $10 for the same menu item. The price of basket of fries is drawn from a normal distribution with mean $6 and standard deviation of $2. The price of a fountain drink is drawn from a normal distribution with mean $3 and standard deviation of $1. What i
s the probability that you pay more than $10 for a dinner consisting of a basket of fries and a fountain drink? 3.5. GEOMETRIC DISTRIBUTION 193 3.5 Geometric distribution How many times should we expect to roll a die until we get a 1? How many people should we expect to see at a hospital until we get someone with blood type O+? These questions can be answered using the geometric distribution. Learning objectives 1. Determine if a scenario is geometric. 2. Calculate the probabilities of the possible values of a geometric random variable. 3. Find and interpret the mean (expected value) and standard deviation of a geometric distribu- tion. 4. Understand the shape of the geometric distribution. 3.5.1 Bernoulli distribution Many health insurance plans in the United States have a deductible, where the insured individual is responsible for costs up to the deductible, and then the costs above the deductible are shared between the individual and insurance company for the remainder of the year. Suppose a health insurance company found that 70% of the people they insure stay below their deductible in any given year. Each of these people can be thought of as a trial. We label a person a success if her healthcare costs do not exceed the deductible. We label a person a failure if she does exceed her deductible in the year. Because 70% of the individuals will not exceed their deductible, we denote the probability of a success as p = 0.7. The probability of a failure is sometimes denoted with q = 1 − p, which would be 0.3 in for the insurance example. When an individual trial only has two possible outcomes, often labeled as success or failure, it is called a Bernoulli random variable. We chose to label a person who does not exceed her deductible as a “success” and all others as “failures”. However, we could just as easily have reversed these labels. The mathematical framework we will build does not depend on which outcome is labeled a success and which a failure, as long as we are consistent. 194 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Bernoulli random variables are often denoted as 1 for a success and 0 for a failure. In addition to being convenient in entering data, it is also mathematically handy. Suppose we observe ten trials Then the sample proportion, ˆp, is the sample mean of these observations: ˆp = # of successes # of trials = 10 = 0.6 This mathematical inquiry of Bernoulli random variables can be extended even further. Because 0 and 1 are numerical outcomes, we can deﬁne the mean and standard deviation of a Bernoulli random variable.66 BERNOULLI RANDOM VARIABLE If X is a random variable that takes value 1 with probability of success p and 0 with probability 1 − p, then X is a Bernoulli random variable with mean and standard deviation µ = p σ = p(1 − p) In general, it is useful to think about a Bernoulli random variable as a random process with only two outcomes: a success or failure. Then we build our mathematical framework using the numerical labels 1 and 0 for successes and failures, respectively. 66See Exercises 3.47 and 3.48 3.5. GEOMETRIC DISTRIBUTION 195 3.5.2 Geometric distribution The geometric distribution is used to describe how many trials it takes to observe a success. Let’s ﬁrst look at an example. EXAMPLE 3.83 Suppose we are working at the insurance company and need to ﬁnd a case where the person did not exceed her deductible as a case study. If the probability a person will not exceed her deductible is 0.7 and we are drawing people at random, what are the chances that the ﬁrst person will not have exceeded her deductible, i.e. be a success? The second person? The third? What about the probability that we pull x − 1 cases before we ﬁnd the ﬁrst success, i.e. the ﬁrst success is the xth person? (If the ﬁrst success is the ﬁfth person, then we say x = 5.) The probability of stopping after the ﬁrst person is just the chance the ﬁrst person will not exceed her deductible: 0.7. The probability the second person is the ﬁrst to exceed her deductible is P (second person is the ﬁrst to exceed deductible) = P (the ﬁrst won’t, the second will) = (0.3)(0.7) = 0.21 Likewise, the probability it will be the third case is (0.3)(0.3)(0.7) = 0.063. If the ﬁrst success is on the xth person, then there are x − 1 failures and ﬁnally 1 success, which corresponds to the probability (0.3)x−1(0.7). This is the same as (1 − 0.7)x−1(0.7). Example 3.83 illustrates what the geometric distribution, which describes the waiting time until a success for independent and identically distributed (iid) Bernoulli random variables. In this case, the independence aspect just means the individuals in the example don’t aﬀect each other, and identical means they each have the same probability of success. The geometric distribution from Example 3.83 is shown in Figure 3.28. In general, the proba- bilities for a geometric distribution decrease exponentially fast. Figure 3.28: The geometric distribution when the probability of success is p = 0.7. While this text will not derive the formulas for the mean (expected) number of trials needed to ﬁnd the ﬁrst success or the standard deviation of this distribution, we present general formulas for each. ProbabilityNumber of Trials Until a Success for p = 0.7123456780.00.20.40.6 196 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS GEOMETRIC DISTRIBUTION Let X have a geometric distribution with one parameter p, where p is the probability of a success in one trial. Then the probability of ﬁnding the ﬁrst success in the xth trial is given by P (X = x) = (1 − p)x−1p where x = 1, 2, 3, . . . The mean (i.e. expected value) and standard deviation of this wait time are given by µX = 1 p σX = √ 1 − p p It is no accident that we use the symbol µ for both the mean and expected value. The mean and the expected value are one and the same. It takes, on average, 1/p trials to get a success under the geometric distribution. This mathematical result is consistent with what we would expect intuitively. If the probability of a success is high (e.g. 0.8), then we don’t usually wait very long for a success: 1/0.8 = 1.25 trials on average. If the probability of a success is low (e.g. 0.1), then we would expect to view many trials before we see a success: 1/0.1 = 10 trials. GUIDED PRACTICE 3.84 The probability that a particular case would not exceed their deductible is said to be 0.7. If we were to examine cases until we found one that where the person did not exceed her deductible, how many cases should we expect to check?67 EXAMPLE 3.85 What is the chance that we would ﬁnd the ﬁrst success within the ﬁrst 3 cases? This is the chance the ﬁrst (X = 1), second (X = 2), or third (X = 3) case is the ﬁrst success, which are three disjoint outcomes. Because the individuals in the sample are randomly sampled from a large population, they are independent. We compute the probability of each case and add the separate results: P (X = 1, 2, or 3) = P (X = 1) + P (X = 2) + P (X = 3) = (0.3)1−1(0.7) + (0.3)2−1(0.7) + (0.3)3−1(0.7) = 0.973 There is a probability of 0.973 that we would ﬁnd a successful case within 3 cases. GUIDED PRACTICE 3.86 Determine a more clever way to solve Example 3.85. Show that you get the same result.68 67We would expect to see about 1/0.7 ≈ 1.43 individuals to ﬁnd the ﬁrst success. 68First ﬁnd the probability of the complement: P (no success in ﬁrst 3 trials) = 0.33 = 0.027. Next, compute one minus this probability: 1 − P (no success in 3 trials) = 1 − 0.027 = 0.973. 3.5. GEOMETRIC DISTRIBUTION 197 EXAMPLE 3.87 Suppose a car insurer has determined that 88% of its drivers will not exceed their deductible in a given year. If someone at the company were to randomly draw driver ﬁles until they found one that had not exceeded their deductible, what is the expected number of drivers the insurance employee must check? What is the standard deviation of the number of driver ﬁles that must be drawn? In this example, a success is again when someone will not exceed the insurance deductible, which has probability p = 0.88. The expected number of people to be checked is 1/p = 1/0.88 = 1.14 and the standard deviation is √ √ 1−p p = 1−0.88 0.88 = 0.39. GUIDED PRACTICE 3.88 Using the results from Example 3.87, where µX = 1.14 and σX = 0.39, would it be appropriate to use the empirical rule to ﬁnd what proportion of experiments would end in 1.14 ± 0.39 trials?69 The independence assumption is crucial to the geometric distribution’s accurate description of a scenario. Mathematically, we can see that to construct the probability of the success on the xth trial, we had to use the General Multiplication Rule for independent processes. It is no simple task to generalize the geometric model for dependent trials. 3.5.3 Technology: geometric probablities Get started quickly with this Desmos Geometric Calculator (available at openintro.org/ahss/desmos). 69No. The geometric distribution is always right skewed and can never be well-approximated by a normal model. 198 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Section summary • It is useful to model yes/no, success/failure with the values 1 and 0, respectively. We call the prob- ability of success p and the probability of failure 1 − p. • When the trials are independent and the value of p is constant, the probability of ﬁnding the ﬁrst success on the xth trial is given by (1 − p)x−1p. We can see the reasoning behind this formula as follows: for the ﬁrst success to happen on the xth trial, it has to not happen the ﬁrst x − 1 trials (with probability 1 − p), and then happen on the xth trial (with probability p). • When we consider the entire distribution of possible values for the how long until the ﬁrst success, we get a discrete probability distribution known as the geometric distribution. The geometric distribution describes the waiting time until the ﬁrst success, when the trials are independent and the probability of success, p, is constant. If X has a geometric di
stribution with parameter p, then P (X = x) = (1 − p)x−1p, where x = 1, 2, 3 . . . . • The geometric distribution is always right skewed and, in fact, has no maximum value. The probabil- ities, though, decrease exponentially fast. • Even though the geometric distribution has an inﬁnite number of values, it has a well-deﬁned mean: 10 , then on average . If the probability of success is 1 1−p µX = 1 p and standard deviation: σX = p it takes 10 trials until we see the ﬁrst success. √ • Note that when the trials are not independent, we can modify the geometric formula to ﬁnd the probability that the ﬁrst success happens on the xth trial. Instead of simply raising (1 − p) to the x − 1, multiply the appropriate conditional probabilities. 3.5. GEOMETRIC DISTRIBUTION 199 Exercises 3.43 Is it Bernoulli? Determine if each trial can be considered an independent Bernoulli trial for the following situations. (a) Cards dealt in a hand of poker. (b) Outcome of each roll of a die. 3.44 With and without replacement. In the following situations assume that half of the speciﬁed population is male and the other half is female. (a) Suppose you’re sampling from a room with 10 people. What is the probability of sampling two females in a row when sampling with replacement? What is the probability when sampling without replacement? (b) Now suppose you’re sampling from a stadium with 10,000 people. What is the probability of sampling two females in a row when sampling with replacement? What is the probability when sampling without replacement? (c) We often treat individuals who are sampled from a large population as independent. Using your ﬁndings from parts (a) and (b), explain whether or not this assumption is reasonable. 3.45 Eye color, Part I. A husband and wife both have brown eyes but carry genes that make it possible for their children to have brown eyes (probability 0.75), blue eyes (0.125), or green eyes (0.125). (a) What is the probability the ﬁrst blue-eyed child they have is their third child? Assume that the eye colors of the children are independent of each other. (b) On average, how many children would such a pair of parents have before having a blue-eyed child? What is the standard deviation of the number of children they would expect to have until the ﬁrst blue-eyed child? 3.46 Defective rate. A machine that produces a special type of transistor (a component of computers) has a 2% defective rate. The production is considered a random process where each transistor is independent of the others. (a) What is the probability that the 10th transistor produced is the ﬁrst with a defect? (b) What is the probability that the machine produces no defective transistors in a batch of 100? (c) On average, how many transistors would you expect to be produced until the ﬁrst with a defect? What is the standard deviation? (d) Another machine that also produces transistors has a 5% defective rate where each transistor is produced independent of the others. On average how many transistors would you expect to be produced with this machine before the ﬁrst with a defect? What is the standard deviation? (e) Based on your answers to parts (c) and (d), how does increasing the probability of an event aﬀect the mean and standard deviation of the wait time until success? 3.47 Bernoulli, the mean. Use the probability rules from Section 3.4 to derive the mean of a Bernoulli random variable, i.e. a random variable X that takes value 1 with probability p and value 0 with probability 1 − p. That is, compute the expected value of a generic Bernoulli random variable. 3.48 Bernoulli, the standard deviation. Use the probability rules from Section 3.4 to derive the standard deviation of a Bernoulli random variable, i.e. a random variable X that takes value 1 with probability p and value 0 with probability 1 − p. That is, compute the square root of the variance of a generic Bernoulli random variable. 200 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.6 Binomial distribution What is the probability of exactly 50 heads in 100 coin tosses? Or the probability of randomly sampling 12 people and having more than 9 of them identify as male? If the probability of a defective part is 1%, how many defective items would we expect in a random shipment of 200 of those parts? We can model these scenarios and answer these questions using the binomial distribution. Learning objectives 1. Calculate the number of possible scenarios for obtaining x successes in n trials. 2. Determine whether a scenario is binomial or not. 3. Calculate the probabilities of the possible values of a binomial random variable using the binomial formula. 4. Recognize that the binomial formula uses the special Addition Rule for mutually exclusive events. 5. Find probabilities of the form “at least” or “at most” by applying the binomial formula multiple times. 6. Calculate and interpret the mean (expected value) and standard deviation of the number of successes in n binomial trials. 7. Determine whether a binomial distribution can be modeled as approximately normal. If so, use normal approximation to estimate cumulative binomial probabilities. 3.6.1 Introducing the binomial formula Let’s again imagine ourselves back at the insurance agency where 70% of individuals do not exceed their deductible. Each people is thought of as a trial. We label a trial a success if the individual healthcare costs do not exceed the deductible. We label a trial a failure if the individual healthcare costs do exceed the deductible. Because 70% of the individuals will not exceed their deductible, we denote the probability of a success as p = 0.7. EXAMPLE 3.89 Suppose the insurance agency is considering a random sample of four individuals they insure. What is the chance exactly one of them will exceed the deductible and the other three will not? Let’s call the four people Ariana (A), Brittany (B), Carlton (C), and Damian (D) for convenience. Let’s consider a scenario where one person exceeds the deductible: P (A = exceed, B = not, C = not, D = not) = P (A = exceed) P (B = not) P (C = not) P (D = not) = (0.3)(0.7)(0.7)(0.7) = (0.7)3(0.3)1 = 0.103 But there are three other scenarios: Brittany, Carlton, or Damian could have been the one to exceed the deductible. In each of these cases, the probability is again (0.7)3(0.3)1. These four scenarios exhaust all the possible ways that exactly one of these four people could have exceeded the deductible, so the total probability is 4 × (0.7)3(0.3)1 = 0.412. GUIDED PRACTICE 3.90 Verify that the scenario where Brittany is the only one to exceed the deductible has probability (0.7)3(0.3)1. 70 70 P (A = not, B = exceed, C = not, D = not) = (0.7)(0.3)(0.7)(0.7) = (0.7)3(0.3)1. 3.6. BINOMIAL DISTRIBUTION 201 The binomial distribution describes the probability of having exactly x successes in n independent trials with probability of a success p (in Example 3.89, n = 4, x = 3, p = 0.7). We would like to determine the probabilities associated with the binomial distribution more generally, i.e. we want a formula where we can use n, x, and p to obtain the probability. To do this, we reexamine each part of Example 3.89. There were four individuals who could have been the one to exceed the deductible, and each of these four scenarios had the same probability. Thus, we could identify the ﬁnal probability as [# of scenarios] × P (single scenario) The ﬁrst component of this equation is the number of ways to arrange the x = 3 successes among the n = 4 trials. The second component is the probability of any of the four (equally probable) scenarios. Consider P (single scenario) under the general case of x successes and n − x failures in the n trials. In any such scenario, we apply the Multiplication Rule for independent events: px(1 − p)n−x This is our general formula for P (single scenario). Secondly, we introduce the binomial coeﬃcient, which gives the number of ways to choose x successes in n trials, i.e. arrange x successes and n − x failures: n x = n! x!(n − x)! The quantity n expression. x is read n choose x.71 The exclamation point notation (e.g. n!) denotes a factorial 0! = 1 1! = 1 2! = 2 × 1 = 2 3! = 4 × 3 × 2 × 1 = 24 ... n! = n × (n − 1) × ... × 3 × 2 × 1 Using the formula, we can compute the number of ways to choose x = 3 successes in n = 4 trials: 4 3 = 4! 3!(4 − 3)! = 4! 3!13 × 2 × 1)(1) = 4 This result is exactly what we found by carefully thinking of each possible scenario in Example 3.89. Substituting n choose x for the number of scenarios and px(1 − p)n−x for the single scenario probability yields the binomial formula. BINOMIAL FORMULA Suppose the probability of a single trial being a success is p. Then the probability of observing exactly x successes in n independent trials is given by P (X = x) = n x px(1 − p)n−x = n! x!(n − x)! px(1 − p)n−x 71Other notations for n choose x includes nCx, Cx n, and C(n, x). 202 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.6.2 When and how to apply the formula IS IT BINOMIAL? FOUR CONDITIONS TO CHECK. (1) The trials are independent. (2) The number of trials, n, is ﬁxed. (3) Each trial outcome can be classiﬁed as a success or failure. (4) The probability of a success, p, is the same for each trial. EXAMPLE 3.91 What is the probability that 3 of 8 randomly selected individuals will have exceeded the insurance deductible, i.e. that 5 of 8 will not exceed the deductible? Recall that 70% of individuals will not exceed the deductible. We would like to apply the binomial model, so we check the conditions. The number of trials is ﬁxed (n = 8) (condition 2) and each trial outcome can be classiﬁed as a success or failure (condition 3). Because the sample is random, the trials are independent (condition 1) and the probability of a success is the same for each trial (condition 4). In the outcome of interest, there are x = 5 successes in n = 8 trials (recall that a success is an individual who does not exceed the deductible, and the probability of
a success is p = 0.7. So the probability that 5 of 8 will not exceed the deductible and 3 will exceed the deductible is given by (0.7)5(1 − 0.7)8−5 = 8 5 8! 5!(5 − 3)! (0.7)5(1 − 0.7)8−5 = 8! 5!3! (0.7)5(0.3)3 Dealing with the factorial part: 8! 5!35 × 4 × 3 × 2 × 1)(3 × 2 × 1 = 56 Using (0.7)5(0.3)3 ≈ 0.00454, the ﬁnal probability is about 56 × 0.00454 ≈ 0.254. If you must calculate the binomial coeﬃcient by hand, it’s often useful to cancel out as many terms as possible in the top and bottom. See Section 3.6.3 for how to evaluate the binomial coeﬃcient and the binomial formula using a calculator. COMPUTING BINOMIAL PROBABILITIES The ﬁrst step in using the binomial model is to check that the model is appropriate. The second step is to identify n, p, and x. Finally, apply the binomial formula to determine the probability and interpret the results. 3.6. BINOMIAL DISTRIBUTION 203 EXAMPLE 3.92 Approximately 35% of a population has blood type O+. Suppose four people show up at a hospital and we want to ﬁnd the probability that exactly one of them has blood type O+. Can we use the binomial formula? To check if the binomial model is appropriate, we must verify the conditions. 1. We will suppose that these 4 people comprise a random sample. This seems reasonable, since one person with a particular blood type showing up at a hospital seems unlikely to aﬀect the chance that other people with that blood type would show up at the hospital. This sample is without replacement, though, not with replacement, so the observations are not entirely independent. However, when the sample size is very small compared to the population size, we can treat the observations as if they were with replacement, since the composition of the population changes very little with each additional person sampled. Since we have a random sample of a very small percent of the population, we will consider the independence condition met. 2. We have a ﬁxed number of trials (n = 4). 3. Each outcome is a success or failure (blood type O+ or not blood type O+). 4. The probability of a success is the same for each trial since the individuals are like a random sample (p = 0.35 if we say a “success” is someone having blood type O+). SAMPLING WITHOUT REPLACEMENT When randomly sampling without replacement, if the sample size is small relative to the population size (rule of thumb: sample size less than 1/10 of the population size), we will consider the observations to be independent. EXAMPLE 3.93 Given that 35% of a population has blood type O+, what is the probabilty that in a random sample of 4 people: (a) none of them have blood type O+? (b) one will have blood type O+? (c) no more than one will have blood type O+? (a) P (X = 0) = 4 (0.35)0(0.65)4 = 1 × 1 × 0.654 = 0.654 = 0.179 0 Note that we could have answered this question without the binomial formula, using methods from the previous section. (b) P (X = 1) = 4 (c) This can be computed as the sum of parts (a) and (b): P (X = 0) + P (X = 1) = 0.179 + 0.384 = 0.563. (0.35)1(0.65)3 = 0.384. 1 That is, there is about a 56.3% chance that no more than one of them will have blood type O+. GUIDED PRACTICE 3.94 What is the probability that at least 3 of 4 people in a random sample will have blood type O+ if 35% of the population has blood type O+?72 GUIDED PRACTICE 3.95 The probability that a random smoker will develop a severe lung condition in her lifetime is about 0.3. If you have 4 friends who smoke and you want to ﬁnd the probability that 1 of them will develop a severe lung condition in her lifetime, can you apply the binomial formula?73 72P (at least 3 of 4 have blood type O+) = P (X = 3) + P (X = 4) = 4 3 0.126 (0.35)3(0.65)1 + (0.35)4 = 0.111 + 0.015 = 73While conditions (2) and (3) are met, most likely the friends know each other, so the independence assumption (1) is probably not satisﬁed. For example, acquaintances may have similar smoking habits, or those friends might make a pact to quit together. Condition (4) is also not satisﬁed since this is not a random sample of people. 204 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS EXAMPLE 3.96 There are 13 marbles in a bag. 4 are blue and 9 are red. Randomly draw 5 marbles without replacement. Find the probability you get exactly 3 blue marbles. Because we are drawing without replacement the probability of success p is not the same for each trial. Also, the sample size is large compared to the population size (much greater than 1/10 of the population size), so we cannot treat these observations as independent and we cannot use the binomial formula. However, we can use the same logic to arrive at the following answer. P (X = 3) = (# of combinations with 3 blue) × P (3 blue and 2 red in a speciﬁc order) 5 3 × P (BBBRR) 5 3 4 13 × 3 12 × 2 11 × 9 10 × 8 9 = = = 0.1119 GUIDED PRACTICE 3.97 Draw 4 cards without replacement from a deck of 52 cards. What is the probability that you get at least two hearts?74 Lastly, we consider the binomial coeﬃcient,, n choose x, under some special scenarios. GUIDED PRACTICE 3.98 Why is it true that n 0 = 1 and n = 1 for any number n?75 n GUIDED PRACTICE 3.99 How many ways can you arrange one success and n − 1 failures in n trials? How many ways can you arrange n − 1 successes and one failure in n trials?76 74P (at least 2 hearts in 4 draws from a deck) = 1 − [P (X = 0) + P (X = 1)] = 1 − [( 39 ( 13 52 )( 39 75Frame these expressions into words. How many diﬀerent ways are there to arrange 0 successes and n failures in 49 )] = 1 − [.0.3038 + 0.4388] = 0.2574. 50 )( 36 52 )( 38 51 )( 37 50 )( 37 51 )( 38 49 ) + 4 1 n trials? (1 way.) How many diﬀerent ways are there to arrange n successes and 0 failures in n trials? (1 way.) 76One success and n − 1 failures: there are exactly n unique places we can put the success, so there are n ways to arrange one success and n − 1 failures. A similar argument is used for the second question. Mathematically, we show these results by verifying the following two equations: n 1 = n, n n − 1 = n 3.6. BINOMIAL DISTRIBUTION 205 3.6.3 Technology: binomial probabilities Get started quickly with this Desmos Binomial Calculator (available at openintro.org/ahss/desmos). Calculator instructions n n TI-83/84: COMPUTING THE BINOMIAL COEFFICIENT n x x x Use MATH, PRB, nCr to evaluate n choose r. Here r and x are diﬀerent letters for the same quantity. 1. Type the value of n. 2. Select MATH. 3. Right arrow to PRB. 4. Choose 3:nCr. 5. Type the value of x. 6. Hit ENTER. Example: 5 nCr 3 means 5 choose 3. n n CASIO FX-9750GII: COMPUTING THE BINOMIAL COEFFICIENT n x x x 1. Navigate to the RUN-MAT section (hit MENU, then hit 1). 2. Enter a value for n. 3. Go to CATALOG (hit buttons SHIFT and then 7). 4. Type C (hit the ln button), then navigate down to the bolded C and hit EXE. 5. Enter the value of x. Example of what it should look like: 7C3. 6. Hit EXE. 206 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS P (X = x) = n P (X = x) = n TI-84: COMPUTING THE BINOMIAL FORMULA, P (X = x) = n px(1 − p)n−x px(1 − p)n−x px(1 − p)n−x x x x Use 2ND VARS, binompdf to evaluate the probability of exactly x occurrences out of n independent trials of an event with probability p. 1. Select 2ND VARS (i.e. DISTR) 2. Choose A:binompdf (use the down arrow to scroll down). 3. Let trials be n. 4. Let p be p 5. Let x value be x. 6. Select Paste and hit ENTER. TI-83: Do step 1, choose 0:binompdf, then enter n, p, and x separated by commas: binompdf(n, p, x). Then hit ENTER. P (X ≤ x) = n TI-84: COMPUTING P (X ≤ x) = n P (X ≤ x) = n p0(1 − p)n−0 + ... + n p0(1 − p)n−0 + ... + n p0(1 − p)n−0 + ... + n Use 2ND VARS, binomcdf to evaluate the cumulative probability of at most x occurrences out of n independent trials of an event with probability p. px(1 − p)n−x px(1 − p)n−x px(1 − p)n−x x x x 0 0 0 1. Select 2ND VARS (i.e. DISTR) 2. Choose B:binomcdf (use the down arrow). 3. Let trials be n. 4. Let p be p 5. Let x value be x. 6. Select Paste and hit ENTER. TI-83: Do steps 1-2, then enter the values for n, p, and x separated by commas as follows: binomcdf(n, p, x). Then hit ENTER. CASIO FX-9750GII: BINOMIAL CALCULATIONS 1. Navigate to STAT (MENU, then hit 2). 2. Select DIST (F5), and then BINM (F5). 3. Choose whether to calculate the binomial distribution for a speciﬁc number of successes, P (X = k), or for a range P (X ≤ k) of values (0 successes, 1 success, ..., x successes). • For a speciﬁc number of successes, choose Bpd (F1). • To consider the range 0, 1, ..., x successes, choose Bcd(F1). 4. If needed, set Data to Variable (Var option, which is F2). 5. Enter the value for x (x), Numtrial (n), and p (probability of a success). 6. Hit EXE. GUIDED PRACTICE 3.100 Find the number of ways of arranging 3 blue marbles and 2 red marbles.77 GUIDED PRACTICE 3.101 There are 13 marbles in a bag. 4 are blue and 9 are red. Randomly draw 5 marbles with replacement. Find the probability you get exactly 3 blue marbles.78 GUIDED PRACTICE 3.102 There are 13 marbles in a bag. 4 are blue and 9 are red. Randomly draw 5 marbles with replacement. Find the probability you get at most 3 blue marbles (i.e. less than or equal to 3 blue marbles).79 77Here n = 5 and x = 3. Doing 5 nCr 3 gives the number of combinations as 10. 78Here, n = 5, p = 4/13, and x = 3, so set trials = 5, p = 4/13 and x value = 3. The probability is 0.1396. 79Similarly, set trials = 5, p = 4/13 and x value = 3. The cumulative probability is 0.9662. 3.6. BINOMIAL DISTRIBUTION 207 3.6.4 An example of a binomial distribution In Guided Practice 3.93, we asked various probability questions regarding the number of people out of 4 with blood type O+. We veriﬁed that the scenario was binomial and that each problem could be solved using the binomial formula. Instead of looking at it piecewise, we could describe the entire distribution of possible values and their corresponding probabilities. Since there are 4 people, there are several possible outcomes for the number who might have blood type
O+: 0, 1, 2, 3, 4. We can make a distribution table with these outcomes. Recall that the probability of a randomly sampled person being blood type O+ is about 0.35. The binomial distribution is used to describe the number of successes in a ﬁxed number of trials. This is diﬀerent from the geometric distribution, which described the number of trials we must wait before we observe a success. 1 xi P (xi0.35)0(0.65)4 = 0.179 (0.35)1(0.65)3 = 0.384 (0.35)2(0.65)2 = 0.311 (0.35)3(0.65)1 = 0.111 (0.35)4(0.65)0 = 0.015 2 3 4 Figure 3.29: Probability distribution for the number with blood type O+ in a random sample of 4 people. This is a binomial distribution. Correcting for rounding error, the probabilities add up to 1, as they must for any probability distribution. 3.6.5 The mean and standard deviation of a binomial distribution Since this is a probability distribution we could ﬁnd its mean and standard deviation using the formulas from Chapter 3. Those formulas require a lot of calculations, so it is fortunate there’s an easier way to compute the mean and standard deviation for a binomial random variable. MEAN AND STANDARD DEVIATION OF THE BINOMIAL DISTRIBUTION For a binomial distribution with parameters n and p, where n is the number of trials and p is the probability of a success, the mean and standard deviation of the number of observed successes are µX = np σX = np(1 − p) EXAMPLE 3.103 If the probability that a person has blood type O+ is 0.35 and you have 40 randomly selected people, about how many would you expect to have blood type O+? What is the standard deviation of the number of people who would have blood type O+ among the 40 people? We are asked to determine the expected number (the mean) and the standard deviation, both of which can be directly computed from the formulas above. µX = np = 40(0.35) = 14 σX = np(1 − p) = 40(0.35)(0.65) = 3.0 The exact distribution is shown in Figure 3.30. 012340.00.10.20.30.4ProbabilityNumber With Blood Type O+ in a Random Sample of Size 4 208 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Figure 3.30: Distribution for the number of people with blood type O+ in a random sample of size 40, where p = 0.35. The distribution is binomial and is centered on 14 with a standard deviation of 3. 3.6.6 Normal approximation to the binomial distribution The binomial formula is cumbersome when the sample size (n) is large, particularly when we consider a range of observations. EXAMPLE 3.104 Find the probability that fewer than 12 out of 40 randomly selected people would have blood type O+, where probability of blood type O+ is 0.35. This is equivalent to asking, what is the probability of observing X = 0, 1, 2, ..., or 11 with blood type O+ in a sample of size 40 when p = 0.35? We previously veriﬁed that this scenario is binomial. We can compute each of the 12 probabilities using the binomial formula and add them together to ﬁnd the answer: P (X = 0 or X = 1 or · · · or X = 11) = P (X = 0) + P (X = 1) + · · · + P (X = 11) (0.35)0(0.65)40 + 40 1 (0.35)1(0.65)39 + · · · + 40 11 (0.35)11(0.65)29 40 0 = = 0.21 If the true proportion with blood type O+ in the population is p = 0.35, then the probability of observing fewer than 12 in a sample of n = 40 is 0.21. The computations in Example 3.6.6 are tedious and long. In general, we should avoid such work if an alternative method exists that is faster, easier, and still accurate. Recall that calculating probabilities of a range of values is much easier in the normal model. In some cases we may use the normal distribution to estimate binomial probabilities. While a normal approximation for the distribution in Figure 3.29 when the sample size was n = 4 would not be appropriate, it might not be too bad for the distribution in Figure 3.30 where n = 40. We might wonder, when is it reasonable to use the normal model to approximate a binomial distribution? GUIDED PRACTICE 3.105 Here we consider the binomial model when the probability of a success is p = 0.10. Figure 3.31 shows four hollow histograms for simulated samples from the binomial distribution using four diﬀerent sample sizes: n = 10, 30, 100, 300. What happens to the shape of the distributions as the sample size increases? How does the binomial distribution change as n gets larger?80 The shape of the binomial distribution depends upon both n and p. Here we introduce a rule of thumb for when normal approximation of a binomial distribution is reasonable. We will use this rule of thumb in many applications going forward. 80The distribution is transformed from a blocky and skewed distribution into one that rather resembles the normal distribution in the last hollow histogram. Number With Blood Type O+ in a Random Sample of Size 40Probability8111417200.000.050.10 3.6. BINOMIAL DISTRIBUTION 209 Figure 3.31: Hollow histograms of samples from the binomial model when p = 0.10. The sample sizes for the four plots are n = 10, 30, 100, and 300, respectively. NORMAL APPROXIMATION OF THE BINOMIAL DISTRIBUTION The binomial distribution with probability of success p is nearly normal when the sample size n is suﬃciently large that np ≥ 10 and n(1 − p) ≥ 10. The approximate normal distribution has parameters corresponding to the mean and standard deviation of the binomial distribution: µ = np σ = np(1 − p) The normal approximation may be used when computing the range of many possible successes. For instance, we may apply the normal distribution to the setting described in Figure 3.30. EXAMPLE 3.106 Use the normal approximation to estimate the probability of observing fewer than 12 with blood type O+ in a random sample of 40, if the true proportion with blood type O+ in the population is p = 0.35. First we verify that np and n(1 − p) are at least 10 so that we can apply the normal approximation to the binomial model: np = 40(0.35) = 14 ≥ 10 n(1 − p) = 40(0.65) = 26 ≥ 10 With these conditions checked, we may use the normal distribution to approximate the binomial distribution with the following mean and standard deviation: µ = np = 40(0.35) = 14 σ = np(1 − p) = 40(0.35)(0.65) = 3.0 We want to ﬁnd the probability of observing fewer than 12 with blood type O+ using this model. We note that 12 is less than 1 standard deviation below the mean: Next, we compute the Z-score as Z = 12−14 3 = −0.67 to ﬁnd the shaded area in the picture: P (Z < −0.67) = 0.25. This probability of 0.25 using the normal approximation is reasonably close to the true probability of 0.21 computed using the binomial distribution. n = 100246n = 300246810n = 10005101520n = 3001020304050581114172023 210 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS EXAMPLE 3.107 Use the normal approximation to estimate the probability of observing fewer than 120 people with blood type O+ in a random sample of 400, if the true proportion with blood type O+ in the population is p = 0.35. We have previously veriﬁed that the binomial model is reasonable for this context. Now we will verify that both np and n(1 − p) are at least 10 so we can apply the normal approximation to the binomial model: np = 400(0.35) = 140 ≥ 10 n(1 − p) = 400(0.35) = 260 ≥ 10 With these conditions checked, we may use the normal approximation in place of the binomial distribution with the following mean and standard deviation: µ = np = 400(0.35) = 140 σ = np(1 − p) = 400(0.35)(0.65) = 9.5 We want to ﬁnd the probability of observing fewer than 120 with blood type O+ using this model. We note that 120 is just over 2 standard deviations below the mean: Next, we compute the Z-score as Z = 120−140 9.5 = −2.1 to ﬁnd the shaded area in the picture: P (Z < −2.1) = 0.0179. This probability of 0.0179 using the normal approximation is very close to the true probability of 0.0196 from the binomial distribution. GUIDED PRACTICE 3.108 Use normal approximation, if applicable, to estimate the probability of getting greater than 15 sixes in 100 rolls of a fair die.81 3.6.7 Normal approximation breaks down on small intervals (special topic) THE NORMAL APPROXIMATION MAY FAIL ON SMALL INTERVALS The normal approximation to the binomial distribution tends to perform poorly when estimating the probability of a small range of counts, even when the conditions are met. We consider again our example where 35% of people are blood type O+. Suppose we want to ﬁnd the probability that between 129 and 131 people, inclusive, have blood type O+ in a random sample of 400 people. We want to compute the probability of observing 129, 130, or 131 people with blood type O+ when p = 0.20 and n = 400. With such a large sample, we might be tempted to apply the normal approximation and use the range 129 to 131. However, we would ﬁnd that the binomial solution and the normal approximation notably diﬀer: Binomial: 0.0732 Normal: 0.0483 81np = 100(1/6) = 16.7 ≥ 10 and n(1 − p) = 100(5/6) = 83.3 ≥ 10 µ = np = 100(1/6) = 16.7; σ = np(1 − p) = 100(1/6)(5/6) = 3.7 Z = 15−16.7 3.7 = −0.46. P (Z > −0.46) = 0.677 111.5121130.5140149.5159168.5 3.6. BINOMIAL DISTRIBUTION 211 We can identify the cause of this discrepancy using Figure 3.32, which shows the areas representing the binomial probability (outlined) and normal approximation (shaded). Notice that the width of the area under the normal distribution is 0.5 units too slim on both sides of the interval. The binomial distribution is a discrete distribution, and the each bar is centered over an integer value. Looking closely at Figure 3.32, we can see that the bar corresponding to 129 begins at 128.5 and ends at 129.5, the bar corresponding to 131 begins at 130.5 and ends at 131.5, etc. Figure 3.32: A normal curve with the area between 129 and 131 shaded. The outlined area from 128.5 to 131.5 represents the exact binomial probability. IMPROVING ACCURACY OF THE NORMAL APPROXIMATION TO THE BINOMIAL DISTRIBUTION The normal approximation to the binomial distribution for intervals of values is usually improved if cutoﬀ values for the lower end of a shaded region are reduced by 0.5 and the
cutoﬀ value for the upper end are increased by 0.5. This correction is called the continuity correction and accounts for the fact that the binomial distribution is discrete. EXAMPLE 3.109 Use the method described to ﬁnd a more accurate estimate for the probability of observing 129, 130, or 131 people with blood type O+ in 400 randomly selected people when p = 0.35. Instead of standardizing 129 and 131, we will standardize 128.5 and 131.5: Zlef t = Zright = 128.5 − 140 9.5 131.5 − 140 9.5 = −1.263 = −0.895 P (−1.263 < Z < −0.895) = 0.0772 The probability 0.0772 is much closer to the true value of 0.0732 than the previous estimate of 0.0483 we calculated using normal approximation without the continuity correction. It is always possible to apply the continuity correction when ﬁnding a normal approximation to the binomial distribution. However, when n is very large or when the interval is wide, the beneﬁt of the modiﬁcation is limited since the added area becomes negligible compared to the overall area being calculated. 120130140150160 212 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Section summary • n x , the binomial coeﬃcient, describes the number of combinations for arranging x successes among n trials. n = x n! x!(n−x)! , where n! = 1 × 2 × 3 × ...n, and 0!=0. • The binomial formula can be used to ﬁnd the probability that something happens exactly x times in n trials. Suppose the probability of a single trial being a success is p. Then the probability of observing exactly x successes in n independent trials is given by n x px(1 − p)n−x = n! x!(n − x)! px(1 − p)n−x • To apply the binomial formula, the events must be independent from trial to trial. Additionally, n, the number of trials must be ﬁxed in advance, and p, the probability of the event occurring in a given trial, must be the same for each trial. • To use the binomial formula, ﬁrst conﬁrm that the binomial conditions are met. Next, identify the number of trials n, the number of times the event is to be a “success” x, and the probability that a single trial is a success p. Finally, plug these three numbers into the formula to get the probability of exactly x successes in n trials. • To ﬁnd a probability involving at least or at most, ﬁrst determine if the scenario is binomial. If so, apply the binomial formula as many times as needed and add up the results, e.g. P (at least 3 Heads in 5 tosses of a fair coin) = P (exactly 3 Heads) + P (exactly 4 Heads) + P (exactly 5 Heads), where each probability can be found using the binomial formula. • The distribution of the number of successes in n independent trials gives rise to a binomial distri- bution. If X has a binomial distribution with parameters n and p, then px(1 − p)n−x, where x = 0, 1, 2, 3 . . . , n. P (X = x) = n x • To write out a binomial probability distribution table, list all possible values for x, the number of successes, then use the binomial formula to ﬁnd the probability of each of those values. • If X follows a binomial distribution with parameters n and p, then: – The mean is given by µX = np. – The standard deviation is given by σX = np(1 − p). – When np ≥ 10 and n(1 − p) ≥ 10, the binomial distribution is approximately normal. (spread ) (center ) (shape) 3.6. BINOMIAL DISTRIBUTION 213 Exercises 3.49 Exploring combinations. A coin is tossed 5 times. How many sequences / combinations of Heads/Tails are there that have: (a) Exactly 1 Tail? (b) Exactly 4 Tails? (c) Exactly 3 Tails? (d) At least 3 Tails? 3.50 Political afﬁliation. Suppose that in a large population, 51% identify as Democrat. A researcher takes a random sample of 3 people. (a) Use the binomial model to calculate the probability that two of them identify as Democrat. (b) Write out all possible orderings of 3 people, 2 of whom identify as Democrat. Use these scenarios to calculate the same probability from part (a) but using the Addition Rule for disjoint events. Conﬁrm that your answers from parts (a) and (b) match. (c) If we wanted to calculate the probability that a random sample of 8 people will have 3 that identify as Democrat, brieﬂy describe why the approach from part (b) would be more tedious than the approach from part (a). 3.51 Underage drinking, Part I. Data collected by the Substance Abuse and Mental Health Services Administration (SAMSHA) suggests that 69.7% of 18-20 year olds consumed alcoholic beverages in any given year.82 (a) Suppose a random sample of ten 18-20 year olds is taken. Is the use of the binomial distribution appropriate for calculating the probability that exactly six consumed alcoholic beverages? Explain. (b) Calculate the probability that exactly 6 out of 10 randomly sampled 18- 20 year olds consumed an alcoholic drink. (c) What is the probability that exactly four out of ten 18-20 year olds have not consumed an alcoholic beverage? (d) What is the probability that at most 2 out of 5 randomly sampled 18-20 year olds have consumed alcoholic beverages? (e) What is the probability that at least 1 out of 5 randomly sampled 18-20 year olds have consumed alcoholic beverages? 3.52 Chickenpox, Part I. Boston Children’s Hospital estimates that 90% of Americans have had chickenpox by the time they reach adulthood.83 (a) Suppose we take a random sample of 100 American adults. Is the use of the binomial distribution appropriate for calculating the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood? Explain. (b) Calculate the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood. (c) What is the probability that exactly 3 out of a new sample of 100 American adults have not had chickenpox in their childhood? (d) What is the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox? (e) What is the probability that at most 3 out of 10 randomly sampled American adults have not had chickenpox? 82SAMHSA, Oﬃce of Applied Studies, National Survey on Drug Use and Health, 2007 and 2008. 83Boston Children’s Hospital, Chickenpox summary page, referenced April 29, 2021. 214 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.53 Game of dreidel. A dreidel is a four-sided spinning top with the Hebrew letters nun, gimel, hei, and shin, one on each side. Each side is equally likely to come up in a single spin of the dreidel. Suppose you spin a dreidel three times. Calculate the probability of getting (a) at least one nun? (b) exactly 2 nuns? (c) exactly 1 hei? (d) at most 2 gimel s? Photo by Staccabees, cropped (http://ﬂic.kr/p/7gLZTf) CC BY 2.0 license 3.54 Sickle cell anemia. Sickle cell anemia is a genetic blood disorder where red blood cells lose their ﬂexibility and assume an abnormal, rigid, “sickle” shape, which results in a risk of various complications. If both parents are carriers of the disease, then a child has a 25% chance of having the disease, 50% chance of being a carrier, and 25% chance of neither having the disease nor being a carrier. If two parents who are carriers of the disease have 3 children, what is the probability that (a) two will have the disease? (b) none will have the disease? (c) at least one will neither have the disease nor be a carrier? (d) the ﬁrst child with the disease will the be 3rd child? We learned in Exercise 3.51 that about 70% of 18-20 year olds 3.55 Underage drinking, Part II. consumed alcoholic beverages in any given year. We now consider a random sample of ﬁfty 18-20 year olds. (a) How many people would you expect to have consumed alcoholic beverages? And with what standard deviation? (b) Would you be surprised if there were 45 or more people who have consumed alcoholic beverages? (c) What is the probability that 45 or more people in this sample have consumed alcoholic beverages? How does this probability relate to your answer to part (b)? 3.56 Chickenpox, Part II. We learned in Exercise 3.52 that about 90% of American adults had chickenpox before adulthood. We now consider a random sample of 120 American adults. (a) How many people in this sample would you expect to have had chickenpox in their childhood? And with what standard deviation? (b) Would you be surprised if there were 105 people who have had chickenpox in their childhood? (c) What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood? How does this probability relate to your answer to part (b)? 3.6. BINOMIAL DISTRIBUTION 215 Chapter highlights This chapter focused on understanding likelihood and chance variation, ﬁrst by solving individual probability questions and then by investigating probability distributions. The main probability techniques covered in this chapter are as follows: • The General Multiplication Rule for and probabilities (intersection), along with the special case when events are independent. • The General Addition Rule for or probabilities (union), along with the special case when events are mutually exclusive. • The Conditional Probability Rule. • Tree diagrams and Bayes’ Theorem to solve more complex conditional problems. • Simulations and the use of random digits to estimate probabilities. Fundamental to all of these problems is understanding when events are independent and when they are mutually exclusive. Two events are independent when the outcome of one does not aﬀect the outcome of the other, i.e. P (A\|B) = P (A). Two events are mutually exclusive when they cannot both happen together, i.e. P (A and B) = 0. Moving from solving individual probability questions to studying probability distributions helps us better understand chance processes and quantify expected chance variation. • For a discrete probability distribution, the sum of the probabilities must equal 1. • As with any distribution, one can calculate the mean and standard deviation of a probability distribution. In the context of a probability distribution, the mean and standard deviation describe the average and the typical deviation from t
he average, respectively, after many, many repetitions of the chance process. • A probability distribution can be summarized by its center (mean, median), spread (SD, IQR), and shape (right skewed, left skewed, approximately symmetric). • Adding a constant to every value in a probability distribution adds that value to the mean, but it does not aﬀect the standard deviation. When multiplying every value by a constant, this multiplies the mean by the constant and it multiplies the standard deviation by the absolute value of the constant. • The mean of the sum of two random variables equals the sum of the means. However, this is not true for standard deviations. Instead, when ﬁnding the standard deviation of a sum or diﬀerence of random variables, take the square root of the sum of each of the standard deviations squared. • The geometric distribution provides a model for the number of trials until the ﬁrst success, when the trials are independent. • The binomial distribution provides a model for the number of successes in n independent trials. • The geometric distribution is always right skewed. However, when the success-failure rule is met (at least 10 success and 10 failures), the binomial distribution can be modeled using a normal distribution with mean = np and standard deviation np(1 − p). The study of probability is useful for measuring uncertainty and assessing risk. In addition, probability serves as the foundation for inference, providing a framework for evaluating when an outcome falls outside of the range of what would be expected by chance alone. 216 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Chapter exercises 3.57 Grade distributions. Each row in the table below is a proposed grade distribution for a class. Identify each as a valid or invalid probability distribution, and explain your reasoning. A 0.3 0 0.3 0.3 0.2 0 (a) (b) (c) (d) (e) (f) Grades C 0.3 1 0.3 0.2 0.2 1.1 B 0.3 0 0.3 0.5 0.4 -0.1 D 0.2 0 0 0.1 0.1 0 F 0.1 0 0 -0.1 0.1 0 3.58 Health coverage, frequencies. The Behavioral Risk Factor Surveillance System (BRFSS) is an annual telephone survey designed to identify risk factors in the adult population and report emerging health trends. The following table summarizes two variables for the respondents: health status and health coverage, which describes whether each respondent had health insurance.84 Health Coverage No Yes Total Health Status Excellent Very good Good 854 4,821 5,675 459 4,198 4,657 727 6,245 6,972 Fair Poor 99 385 578 1,634 677 2,019 Total 2,524 17,476 20,000 (a) If we draw one individual at random, what is the probability that the respondent has excellent health and doesn’t have health coverage? (b) If we draw one individual at random, what is the probability that the respondent has excellent health or doesn’t have health coverage? 3.59 HIV in Eswatini. Eswatini (formerly named in English as Swaziland) has the highest HIV prevalence in the world: 25.9% of this country’s population is infected with HIV.85 The ELISA test is one of the ﬁrst and most accurate tests for HIV. For those who carry HIV, the ELISA test is 99.7% accurate. For those who do not carry HIV, the test is 92.6% accurate. If an individual from Eswatini has tested positive, what is the probability that he carries HIV? 3.60 Twins. About 30% of human twins are identical, and the rest are fraternal. Identical twins are necessarily the same sex – half are males and the other half are females. One-quarter of fraternal twins are both male, one-quarter both female, and one-half are mixes: one male, one female. You have just become a parent of twins and are told they are both girls. Given this information, what is the probability that they are identical? 3.61 Cost of breakfast. Sally gets a cup of coﬀee and a muﬃn every day for breakfast from one of the many coﬀee shops in her neighborhood. She picks a coﬀee shop each morning at random and independently of previous days. The average price of a cup of coﬀee is $1.40 with a standard deviation of 30¢ ($0.30), the average price of a muﬃn is $2.50 with a standard deviation of 15¢, and the two prices are independent of each other. (a) What is the mean and standard deviation of the amount she spends on breakfast daily? (b) What is the mean and standard deviation of the amount she spends on breakfast weekly (7 days)? 84Oﬃce of Surveillance, Epidemiology, and Laboratory Services Behavioral Risk Factor Surveillance System, BRFSS 2010 Survey Data. 85Source: CIA Factbook, Country Comparison: HIV/AIDS - Adult Prevalence Rate. 3.6. BINOMIAL DISTRIBUTION 217 3.62 Scooping ice cream. Ice cream usually comes in 1.5 quart boxes (48 ﬂuid ounces), and ice cream scoops hold about 2 ounces. However, there is some variability in the amount of ice cream in a box as well as the amount of ice cream scooped out. We represent the amount of ice cream in the box as X and the amount scooped out as Y . Suppose these random variables have the following means, standard deviations, and variances: mean 48 2 SD variance 1 0.25 1 0.0625 X Y (a) An entire box of ice cream, plus 3 scoops from a second box is served at a party. How much ice cream do you expect to have been served at this party? What is the standard deviation of the amount of ice cream served? (b) How much ice cream would you expect to be left in the box after scooping out one scoop of ice cream? That is, ﬁnd the expected value of X − Y . What is the standard deviation of the amount left in the box? (c) Using the context of this exercise, explain why we add variances when we subtract one random variable from another. 3.63 College smokers. At a university, 13% of students smoke. (a) Calculate the expected number of smokers in a random sample of 100 students from this university. (b) The university gym opens at 9 am on Saturday mornings. One Saturday morning at 8:55 am there are 27 students outside the gym waiting for it to open. Should you use the same approach from part (a) to calculate the expected number of smokers among these 27 students? 3.64 Speeding on the I-5, Part II. Exercise 2.53 states that the distribution of speeds of cars traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. The speed limit on this stretch of the I-5 is 70 miles/hour. (a) A highway patrol oﬃcer is hidden on the side of the freeway. What is the probability that 5 cars pass and none are speeding? Assume that the speeds of the cars are independent of each other. (b) On average, how many cars would the highway patrol oﬃcer expect to watch until the ﬁrst car that is speeding? What is the standard deviation of the number of cars he would expect to watch? 3.65 Roulette winnings. In the game of roulette, a wheel is spun and you place bets on where it will stop. One popular bet is that it will stop on a red slot; such a bet has an 18/38 chance of winning. If it stops on red, you double the money you bet. If not, you lose the money you bet. Suppose you play 3 times, each time with a $1 bet. Let Y represent the total amount won or lost. Write a probability model for Y. 3.66 Multiple choice quiz. In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz at all, and decides to randomly guess the answers. What is the probability that (a) the ﬁrst question she gets right is the 3rd question? (b) she gets exactly 3 or exactly 4 questions right? (c) she gets the majority of the questions right? 218 Chapter 4 Sampling distributions 4.1 Sampling distribution for a sample proportion 4.2 Sampling distribution for a sample mean 4.3 Sampling distribution for a difference of proportions or means 219 At the end of the previous chapter, we explored the Geometric Distribution and the Binomial Distribution. In this chapter, we consider what is called the sampling distribution, which is a term for describing the distribution of a statistic. For example, if we take a survey of a random sample of a population and compute the proportion of “yes” responses to a particular question, then it’s helpful to consider the distribution of all possible sample proportions. We explore sampling distributions for two common statistics: the sample proportion and the sample mean. We then consider the distributions of a diﬀerence of sample proportions and a diﬀerence of sample means. In each case, the goal is the same – to describe the center, spread, and shape of the sampling distribution and determine when the normal approximation to the sampling distribution is reasonable. For videos, slides, and other resources, please visit www.openintro.org/ahss 220 CHAPTER 4. SAMPLING DISTRIBUTIONS 4.1 Sampling distribution for a sample proportion Often, instead of the number of successes in n trials, we are interested in the proportion of successes in n trials. We can use the sampling distribution for a sample proportion to answer questions such as the following: • Given a fair coin, what is the probability that in 200 tosses you would get greater than 52% Tails just by random variation? • In a particular state, 48% support a controversial measure. When estimating the percent through polling, what is the probability that a random sample of size 200 will mistakenly estimate the percent support to be greater than 50%? Learning objectives 1. Understand the concept of a sampling distribution. 2. Describe the center, spread, and shape of the sampling distribution for a sample proportion. 3. Recognize the relationship between the distribution of a sample proportion and the corre- sponding binomial distribution. 4. Explain the Central Limit Theorem and what it says about the shape of the sampling distri- bution for a sample proportion. 5. Verify appropriate conditions and, if met, carry out normal approximation for a sample pro- portion or sample count. 4.1.1 The mean and standard deviation of ˆpˆpˆp To answer the two questions posed at the beginning of
this section, we investigate the distribution of the sample proportion ˆp. In the previous section, we saw that the number of people with blood type O+ in a random sample of size 40 follows a binomial distribution with n = 40 and p = 0.35 that is centered on 14 and has standard deviation 3.0. What does the distribution of the proportion of people with blood type O+ in a sample of size 40 look like? To convert from a count to a proportion, we divide the count (i.e. number of yeses) by the sample size, n = 40. For example, 8 becomes 8/40 = 0.20 as a proportion and 11 becomes 11/40 = 0.275. We can ﬁnd the general formula for the mean (expected value) and standard deviation of a sample proportion ˆp using our tools that we’ve learned so far. To get the sample mean for ˆp, we divide the binomial mean µbinomial = np by n: µ ˆp = µbinomial n = np n = p As one might expect, the sample proportion ˆp is centered on the true proportion p. Likewise, the standard deviation of ˆp is equal to the standard deviation of the binomial distribution divided by n: σ ˆp = σbinomial n = np(1 − p) n = p(1 − p) n 4.1. SAMPLING DISTRIBUTION FOR A SAMPLE PROPORTION 221 MEAN AND STANDARD DEVIATION OF A SAMPLE PROPORTION The mean and standard deviation of the sample proportion describe the center and spread of the distribution of all possible sample proportions ˆp from a random sample of size n with true population proportion p. µ ˆp = p σ ˆp = p(1 − p) n The standard deviation of a sample proportion, σ ˆp, tells us about the “typical” deviation in the sample proportions from the true population proportion. In analyses, we think of the standard deviation of a sample proportion as describing the uncertainty associated with the estimate ˆp. That is, σ ˆp can be thought of as a way to quantify the typical error in our sample estimate ˆp of the true proportion p. Understanding the variability of statistics such as ˆp is a central component in the study of statistics. In our blood type O+ example, we have n = 40 and p = 0.35. If we look at the distribution of all possible values of a sample proportion for random samples of size 40 from this population, it is = 0.075. We see in Figure 4.1 centered on µ ˆp = 0.35 and has standard deviation σ ˆp = that the distribution of proportion of people in a sample of size 40 with blood type O+ is equivalent to the distribution of number of people in a sample with blood type O+ out of 40, but with a change of scale. Instead of counts along the horizontal axis, we have proportions. 0.35(1−0.35) 40 (a) (b) Figure 4.1: Two distributions where p = 0.35 and n = 40: the binomial distribution for the number with blood type O+ and the sampling distribution for the proportion with blood type O+. EXAMPLE 4.1 If the proportion of people in the county with blood type O+ is really 35%, ﬁnd and interpret the mean and standard deviation of the sample proportion for a random sample of size 400. The mean of the distribution of the sample proportion is the population proportion: 0.35. That is, the distribution of all possible values for the sample proportion is centered on 0.35. In other words, if we took many, many samples and calculated ˆp, these values would average out to 0.35. The standard deviation of ˆp is described by the standard deviation for the proportion: σ ˆp = p(1 − p) n = 0.35(0.65) 400 = 0.024 The sample proportion will typically be about 0.024 or 2.4% away from the true proportion of p = 0.35. We’ll become more rigorous about quantifying how close ˆp will tend to be to p in Chapter 5. Number With Blood Type O+ in a Random Sample of Size 40Probability8111417200.000.050.10Proportion With Blood Type O+ in a Random Sample of Size 40Probability0.200.2750.350.4250.500.000.050.10 222 CHAPTER 4. SAMPLING DISTRIBUTIONS 4.1.2 Central Limit Theorem The distribution in Figure 4.1 looks an awful lot like a normal distribution. That is no anomaly; it is the result of a general principle called the Central Limit Theorem. CENTRAL LIMIT THEOREM AND THE SUCCESS-FAILURE CONDITION When observations are independent and the sample size is suﬃciently large, the sample proportion ˆp will tend to follow a normal distribution with the following mean and standard deviation: µ ˆp = p σ ˆp = p(1 − p) n In order for the Central Limit Theorem to hold, the sample size is typically considered suﬃciently large when np ≥ 10 and n(1 − p) ≥ 10, which is called the success-failure condition. The Central Limit Theorem is incredibly important, and it provides a foundation for much of statistics. As we begin applying the Central Limit Theorem, be mindful of the two technical conditions: the observations must be independent, and the sample size must be suﬃciently large such that np ≥ 10 and n(1 − p) ≥ 10. HOW TO VERIFY SAMPLE OBSERVATIONS ARE INDEPENDENT If the observations are from a random process such as tossing a coin, then they are independent. If the observations are from a random sample with replacement, then they are independent. If the observations are from a simple random sample (without replacement), we can treat them as independent if the sample size is less than 10% of the population size. If a sample is from a seemingly random process, e.g. an occasional error on an assembly line, checking independence is more diﬃcult. In this case, use your best judgement. When the sample exceeds 10% of the population size, the methods we discuss tend to overes- timate the sampling error slightly versus what we would get using more advanced methods.1 An interesting question to answer is, what happens when np < 10 or n(1 − p) < 10? We can simulate drawing samples of diﬀerent sizes where, say, the true proportion is p = 0.25. Here’s a sample of size 10: no, no, yes, yes, no, no, no, no, no, no In this sample, we observe a sample proportion of yeses of ˆp = 2 10 = 0.2. We can simulate many such proportions to understand the sampling distribution for ˆp when n = 10 and p = 0.25, which we’ve plotted in Figure 4.2 alongside a normal distribution with the same mean and variability. These distributions have a number of imprrtant diﬀerences. Normal: N (0.25, 0.14) n = 10, p = 0.25 Unimodal? Yes Yes Smooth? Yes No Symmetric? Yes No Notice that the success-failure condition was not satisﬁed when n = 10 and p = 0.25: np = 10 × 0.25 = 2.5 n(1 − p) = 10 × 0.75 = 7.5 This single sampling distribution does not show that the success-failure condition is the perfect guideline, but we have found that the guideline did correctly identify that a normal distribution might not be appropriate. 1For example, we could use what’s called the ﬁnite population correction factor: if the sample is of size n N −1 to obtain a and the population size is N , then we can multiple the typical standard deviation formula by smaller, more precise estimate of the actual standard deviation. When n < 0.1 × N , this correction factor is relatively close to 1. N −n 4.1. SAMPLING DISTRIBUTION FOR A SAMPLE PROPORTION 223 Figure 4.2: Left: simulations of ˆp when the sample size is n = 10 and the population proportion is p = 0.25. Right: a normal distribution with the same mean (0.25) and standard deviation (0.137). We can complete several additional simulations, shown in Figures 4.3 and 4.4, and we can see some trends: 1. When either np or n(1 − p) is small, the distribution is more discrete, i.e. not continuous. 2. When np or n(1 − p) is smaller than 10, the skew in the distribution is more noteworthy. 3. The larger both np and n(1 − p), the more normal the distribution. This may be a little harder to see for the larger sample size in these plots as the variability also becomes much smaller. 4. When np and n(1 − p) are both very large, the distribution’s discreteness is hardly evident, and the distribution looks much more like a normal distribution. So far we’ve only focused on the skew and discreteness of the distributions. We haven’t considered how the mean and standard error of the distributions change. Take a moment to look back at the graphs, and pay attention to three things: 1. The centers of the distribution are always at the population proportion, p, that was used to generate the simulation. Because the sampling distribution for ˆp is always centered at the population parameter p, it means the sample proportion ˆp is unbiased when the data are independent and drawn from such a population. 2. For a particular population proportion p, the variability in the sampling distribution decreases as the sample size n becomes larger. This will likely align with your intuition: an estimate based on a larger sample size will tend to be more accurate. 3. For a particular sample size, the variability will be largest when p = 0.5. The diﬀerences may be a little subtle, so take a close look. This reﬂects the role of the proportion p in the standard error formula: SE = . The standard error is largest when p = 0.5. p(1−p) n At no point will the distribution of ˆp look perfectly normal, since ˆp will always be take discrete values (x/n). It is always a matter of degree, and we will use the standard success-failure condition with minimums of 10 for np and n(1 − p) as our guideline within this book. THREE IMPORTANT FACTS ABOUT THE DISTRIBUTION OF A SAMPLE PROPORTION ˆpˆpˆp When the observations can be considered independent, such as from a random sample of less than 10% of the population, the distribution of the sample proportion can be described as follows. 1. CENTER: The mean of a sample proportion is p. 2. SPREAD: The SD of a sample proportion is p(1−p) n . 3. SHAPE: When np ≥ 10 and n(1 − p) ≥ 10, the sample proportion closely follows a normal distribution. Using these facts, we can now answer the question posed at the beginning of this section. Sample Proportions0.00.10.20.30.40.50.60.70.8010002000Frequency−0.20.00.20.40.6 224 CHAPTER 4. SAMPLING DISTRIBUTIONS Figure 4.3: Sampling distributions for several scenarios of p and n. Rows: p = 0.10, p = 0.20, p = 0.50, p = 0.80, and p = 0.90. Columns: n = 10 a
nd n = 25. n = 10n = 25p = 0.10.00.20.40.60.81.00.00.20.40.60.81.0p = 0.20.00.20.40.60.81.00.00.20.40.60.81.0p = 0.50.00.20.40.60.81.00.00.20.40.60.81.0p = 0.80.00.20.40.60.81.00.00.20.40.60.81.0p = 0.90.00.20.40.60.81.00.00.20.40.60.81.0 4.1. SAMPLING DISTRIBUTION FOR A SAMPLE PROPORTION 225 Figure 4.4: Sampling distributions for several scenarios of p and n. Rows: p = 0.10, p = 0.20, p = 0.50, p = 0.80, and p = 0.90. Columns: n = 50, n = 100, and n = 250. n = 50n = 100n = 2500.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.0 226 CHAPTER 4. SAMPLING DISTRIBUTIONS 4.1.3 Normal approximation for the sampling distribution for ˆpˆpˆp EXAMPLE 4.2 Find the probability that less than 30% of a random sample of 400 people will be blood type O+ if the population proportion is 35%. In the previous section we veriﬁed that the observations can be treated as independent and that np and n(1 − p) are at least 10. The mean of the sample proportion is 0.35 and the standard deviation for the sample proportion is given by approximation to estimate the probability: 0.35(1−0.35) 400 = 0.024. We can ﬁnd a Z-score and use normal Z = ˆp − µ ˆp σ ˆp = 0.30 − 0.35 0.024 = −2.1 P (Z < −2.1) = 0.0179 We leave it to the reader to construct a ﬁgure for this example. EXAMPLE 4.3 The probability 0.0179 is the same probability we calculated when we found the probability of getting fewer than 120 with blood type O+ out of 400! Why is this? Notice that 120/400 = 0.30. Using the binomial distribution to ﬁnd the probability of fewer than 120 with blood type O+ in the sample is equivalent to using the distribution of ˆp to ﬁnd the probability of a sample proportion less than 0.30. GUIDED PRACTICE 4.4 Given a population that is 50% male, what is the probability that a random sample of size 200 would have greater than 55% males? Remember to verify that conditions for normal approximation are met.2 2First, verify the conditions: There is a random sample, and the sample size is much smaller than the population size, so observations can be considered independent. Also, np = 200(0.5) = 100 ≥ 10 and n(1 − p) = 200(0.5) = 100 ≥ 10, so the normal approximation is reasonable. Next we ﬁnd the mean and standard deviation of ˆp: µ ˆp = p = 0.50 σ ˆp = p(1 − p) n = 0.5(0.5) 200 = 0.0354 Then we ﬁnd a Z-score and ﬁnd the upper tail of the normal distribution: Z = ˆp − µ ˆp σ ˆp = 0.55 − 0.5 0.0354 = 1.412 → P (Z > 1.412) = 0.07 The probability of getting a sample proportion greater than 55% is about 0.07. 4.1. SAMPLING DISTRIBUTION FOR A SAMPLE PROPORTION 227 Section summary • A Z-score represents the number of standard deviations a value in a data set is above or below the mean. To calculate a Z-score use: Z = x−mean SD . • The standard deviation of ˆp describes the typical error or distance of the sample proportion from the population proportion. It also tells us how much the sample proportion is likely to vary from one random sample to another. • The sampling distribution for the sample proportion ˆp for a random sample of size n is identical to the binomial distribution with parameters n and p, but with a change of scale, i.e. diﬀerent mean and diﬀerent SD, but same shape. • The same success-failure condition for the binomial distribution holds for a sample propor- tion ˆp. • Three important facts about the sampling distribution for the sample proportion ˆp, where the observations can be considered independent: – The mean of a sample proportion is denoted by µ ˆp, and it is equal to p. (center ) – The SD of a sample proportion is denoted by σ ˆp, and it is equal to p(1−p) n . (spread ) – When np ≥ 10 and n(1 − p) ≥ 10, the distribution of the sample proportion will be approximately normal. (shape) • We use these properties when solving the following type of normal approximation problems involving a sample proportion. Find the probability of getting more / less than % yeses in a sample of size n. 1. Identify n and p. Verify than observations can be treated as independent and that np ≥ 10 and n(1 − p) ≥ 10, which implies that normal approximation is reasonable. 2. Calculate the Z-score. Use µ ˆp = p and σ ˆp = tion. p(1−p) n to standardize the sample propor- 3. Find the appropriate area under the normal curve. 228 CHAPTER 4. SAMPLING DISTRIBUTIONS Exercises 4.1 Distribution of ˆpˆpˆp. Suppose the true population proportion were p = 0.95. The ﬁgure below shows what the distribution of a sample proportion looks like when the sample size is n = 20, n = 100, and n = 500. (a) What does each point (observation) in each of the samples represent? (b) Describe the distribution of the sample proportion, ˆp. How does the distribution of the sample proportion change as n becomes larger? 4.2 Distribution of ˆpˆpˆp. Suppose the true population proportion were p = 0.5. The ﬁgure below shows what the distribution of a sample proportion looks like when the sample size is n = 20, n = 100, and n = 500. What does each point (observation) in each of the samples represent? Describe how the distribution of the sample proportion, ˆp, changes as n becomes larger. 0.700.750.800.850.900.951.000.700.750.800.850.900.951.000.700.750.800.850.900.951.000.20.40.60.80.20.40.60.80.20.40.60.8 4.1. SAMPLING DISTRIBUTION FOR A SAMPLE PROPORTION 229 4.3 Vegetarian college students. Suppose that 8% of college students are vegetarians. Determine if the following statements are true or false, and explain your reasoning. (a) The distribution of the sample proportions of vegetarians in random samples of size 60 is approximately normal since n ≥ 30. (b) The distribution of the sample proportions of vegetarian college students in random samples of size 50 is right skewed. (c) A random sample of 125 college students where 12% are vegetarians would be considered unusual. (d) A random sample of 250 college students where 12% are vegetarians would be considered unusual. (e) The standard error would be reduced by one-half if we increased the sample size from 125 to 250. 4.4 Young Americans, Part I. About 77% of young adults think they can achieve the American dream. Determine if the following statements are true or false, and explain your reasoning.3 (a) The distribution of sample proportions of young Americans who think they can achieve the American dream in random samples of size 20 is left skewed. (b) The distribution of sample proportions of young Americans who think they can achieve the American dream in random samples of size 40 is approximately normal since n ≥ 30. (c) A random sample of 60 young Americans where 85% think they can achieve the American dream would be considered unusual. (d) A random sample of 120 young Americans where 85% think they can achieve the American dream would be considered unusual. 4.5 Distribution of ˆpˆpˆp. simple random sample of size n = 50. Suppose the true population proportion were p = 0.5 and a researcher takes a (a) Find and interpret the standard deviation of the sample proportion ˆp. (b) Calculate the probability that the sample proportion will be larger than 0.55 for a random sample of size 50. 4.6 Distribution of ˆpˆpˆp. Suppose the true population proportion were p = 0.6 and a researcher takes a simple random sample of size n = 50. (a) Find and interpret the standard deviation of the sample proportion ˆp. (b) Calculate the probability that the sample proportion will be larger than 0.65 for a random sample of size 50. It is believed that nearsightedness aﬀects about 8% of all children. We 4.7 Nearsighted children. are interested in ﬁnding the probability that fewer than 12 out of 200 randomly sampled children will be nearsighted. (a) Estimate this probability using the normal approximation to the binomial distribution. (b) Estimate this probability using the distribution of the sample proportion. (c) How do your answers from parts (a) and (b) compare? 4.8 Social network use. The Pew Research Center estimates that as of January 2014, 89% of 18-29 year olds in the United States use social networking sites.4 Calculate the probability that at least 95% of 500 randomly sampled 18-29 year olds use social networking sites. 4.9 CLT for proportions, Part 1. Deﬁne the term “sampling distribution” of the sample proportion, and describe how the shape, center, and spread of the sampling distribution change as the sample size increases when p = 0.1. 4.10 CLT for proportions, Part 2. Deﬁne the term “sampling distribution” of the sample proportion, and describe how the shape, center, and spread of the sampling distribution change as the sample size increases when p = 0.8. 3A. Vaughn. “Poll ﬁnds young adults optimistic, but not about money”. In: Los Angeles Times (2011). 4Pew Research Center, Washington, D.C. Social Networking Fact Sheet, accessed on May 9, 2015. 230 CHAPTER 4. SAMPLING DISTRIBUTIONS 4.2 Sampling distribution for a sample mean If bags of chips are produced with an average weight of 15 oz and a standard deviation of 0.1 oz, what is the probability that the average weight of 30 bags will be within 0.1 oz of the mean? The answer is not 68%! To answer this question we must visualize and understand what is called the sampling distribution for a sample mean. Learning objectives 1. Describe the center, spread, and shape of the sampling distribution for a sample mean. 2. Distinguish between the standard deviation of a population and the standard deviation of a sampling distribution. 3. Explain the content and importance of the Central Limit Theorem. 4. Identify and explain the conditions for using normal approximation involving a sample mean. 5. Check the appropriate conditions and, when met, carry out normal approximation involving a sample mean or sample sum. 4.2.1 The mean and standard deviation of ¯x¯x¯x In this section we consider a data set called ru
n17, which represents all 19,961 runners who ﬁnished the 2017 Cherry Blossom 10 mile run in Washington, DC. Part of this data set is shown in Figure 4.5, and the variables are described in Figure 4.6. ID 1 2 ... 16923 16924 time 92.25 106.35 ... 122.87 93.30 age 38.00 33.00 ... 37.00 27.00 gender state M MD M DC ... ... VA F DC F Figure 4.5: Four observations from the run17 data set. variable time age gender state description Ten mile run time, in minutes Age, in years Gender (M for male, F for female) Home state (or country if not from the US) Figure 4.6: Variables and their descriptions for the run17 data set. 4.2. SAMPLING DISTRIBUTION FOR A SAMPLE MEAN 231 These data are special because they include the results for the entire population of runners who ﬁnished the 2017 Cherry Blossom Run. We took a simple random sample of this population, which is represented in Figure 4.7. A histogram summarizing the time variable in the run17samp data set is shown in Figure 4.8. ID 1983 8192 ... 1287 time 88.31 100.67 ... 89.49 age 59 32 ... 26 gender state M MD VA M ... ... M DC Figure 4.7: Three observations for the run17samp data set, which represents a simple random sample of 100 runners from the 2017 Cherry Blossom Run. Figure 4.8: Histogram of time for a single sample of size 100. The average of the sample is in the mid-90s and the standard deviation of the sample s ≈ 17 minutes. From the random sample represented in run17samp, we guessed the average time it takes to run 10 miles is 95.61 minutes. Suppose we take another random sample of 100 individuals and take its mean: 95.30 minutes. Suppose we took another (93.43 minutes) and another (94.16 minutes), and so on. If we do this many many times – which we can do only because we have the entire population data set – we can build up a sampling distribution for the sample mean when the sample size is 100, shown in Figure 4.9. SAMPLING DISTRIBUTION The sampling distribution represents the distribution of the point estimates based on samples of a ﬁxed size from a certain population. It is useful to think of a point estimate as being drawn from such a distribution. Understanding the concept of a sampling distribution is central to understanding statistical inference. The sampling distribution shown in Figure 4.9 is unimodal and approximately symmetric. It is also centered exactly at the true population mean: µ = 94.52. Intuitively, this makes sense. The sample mean should be an unbiased estimator of the population mean. Because we are considering the distribution of the sample mean, we will use µ¯x = 94.52 to describe the true mean of this distribution. Time (Minutes)Frequency608010012014005101520 232 CHAPTER 4. SAMPLING DISTRIBUTIONS Figure 4.9: A histogram of 1000 sample means for run time, where the samples are of size n = 100. This histogram approximates the true sampling distribution for the sample mean, with mean µ¯x and standard deviation σ¯x. We can see that the sample mean has some variability around the population mean, which can be quantiﬁed using the standard deviation of this distribution of sample means. The standard deviation of the sample mean tells us how far the typical estimate is away from the actual population mean, 94.52 minutes. It also describes the typical error of a single estimate, and is denoted by the symbol σ¯x. STANDARD DEVIATION OF AN ESTIMATE The standard deviation associated with an estimate describes the typical error or uncertainty associated with the estimate. EXAMPLE 4.5 Looking at Figures 4.8 and 4.9, we see that the standard deviation of the sample mean with n = 100 is much smaller than the standard deviation of a single sample. Interpret this statement and explain why it is true. The variation from one sample mean to another sample mean is much smaller than the variation from one individual to another individual. This makes sense because when we average over 100 values, the large and small values tend to cancel each other out. While many individuals have a time under 90 minutes, it would be unlikely for the average of 100 runners to be less than 90 minutes. Sample meanFrequency708090100110120050100The distribution of sample means,shown here, is much narrower thanthe distribution of raw observations. 4.2. SAMPLING DISTRIBUTION FOR A SAMPLE MEAN 233 GUIDED PRACTICE 4.6 (a) Would you rather use a small sample or a large sample when estimating a parameter? Why? (b) Using your reasoning from (a), would you expect a point estimate based on a small sample to have smaller or larger standard deviation than a point estimate based on a larger sample?5 When considering how to calculate the standard deviation of a sample mean, there is one problem: there is no obvious way to estimate this from a single sample. However, statistical theory provides a helpful tool to address this issue. In the sample of 100 runners, the standard deviation of the sample mean is equal to one-tenth of the population standard deviation: 15.93/10 = 1.59. In other words, the standard deviation of the sample mean based on 100 observations is equal to SD¯x = σ¯x = σx√ n = 15.93 √ 100 = 1.59 where σx is the standard deviation of the individual observations. This is no coincidence. We can show mathematically that this equation is correct when the observations are independent using the probability tools of Section 3.4. COMPUTING SD FOR THE SAMPLE MEAN Given n independent observations from a population with standard deviation σ, the standard deviation of the sample mean is equal to SD¯x = σ¯x = σ √ n (4.7) A reliable method to ensure sample observations are independent is to conduct a simple random sample consisting of less than 10% of the population. GUIDED PRACTICE 4.8 The average of the runners’ ages is 35.05 years with a standard deviation of σ = 8.97. A simple random sample of 100 runners is taken. (a) What is the standard deviation of the sample mean? (b) Would you be surprised to get a sample of size 100 with an average of 36 years?6 GUIDED PRACTICE 4.9 (a) Would you be more trusting of a sample that has 100 observations or 400 observations? (b) We want to show mathematically that our estimate tends to be better when the sample size is larger. If the standard deviation of the individual observations is 10, what is our estimate of the standard deviation of the mean when the sample size is 100? What about when it is 400? (c) Explain how your answer to (b) mathematically justiﬁes your intuition in part (a).7 5(a) Consider two random samples: one of size 10 and one of size 1000. Individual observations in the small sample are highly inﬂuential on the estimate while in larger samples these individual observations would more often average each other out. The larger sample would tend to provide a more accurate estimate. (b) If we think an estimate is better, we probably mean it typically has less error. Based on (a), our intuition suggests that a larger sample size corresponds to a smaller standard deviation. 6(a) Use Equation (4.7) with the population standard deviation to compute the standard deviation of the sample mean: SD¯y = 8.97/ 100 = 0.90 years. (b) It would not be surprising. 36 years is about 1 standard deviation from the true mean of 35.05. Based on the 68, 95 rule, we would get a sample mean at least this far away from the true mean approximately 100% − 68% = 32% of the time. √ √ 7(a) Extra observations are usually helpful in understanding the population, so a point estimate with 400 observations seems more trustworthy. (b) The standard deviation of the mean when the sample size is 100 is given by 400 = 0.5. The larger sample has a smaller standard deviation of SD100 = 10/ the mean. (c) The standard deviation of the mean of the sample with 400 observations is lower than that of the sample with 100 observations. The standard deviation of ¯x describes the typical error, and since it is lower for the larger sample, this mathematically shows the estimate from the larger sample tends to be better – though it does not guarantee that every large sample will provide a better estimate than a particular small sample. 100 = 1. For 400: SD400 = 10/ √ 234 CHAPTER 4. SAMPLING DISTRIBUTIONS 4.2.2 The Central Limit Theorem revisited In Figure 4.9, the sampling distribution for the sample mean looks approximately normally distributed. Will the sampling distribution for a mean always be nearly normal? To address this question, we will investigate three cases to see roughly when the approximation is reasonable. We consider three data sets: one from a uniform distribution, one from an exponential distribution, and the other from a normal distribution. These distributions are shown in the top panels of Figure 4.10. The uniform distribution is symmetric, and the exponential distribution may be considered as having moderate skew since its right tail is relatively short (few outliers). The left panel in the n = 2 row represents the sampling distribution for ¯x if it is the sample mean of two observations from the uniform distribution shown. The dashed line represents the closest approximation of the normal distribution. Similarly, the center and right panels of the n = 2 row represent the respective distributions of ¯x for data from exponential and log-normal distributions. GUIDED PRACTICE 4.10 Examine the distributions in each row of Figure 4.10. What do you notice about the sampling distribution for the mean as the sample size, n, becomes larger?8 EXAMPLE 4.11 In general, would normal approximation for a sample mean be appropriate when the sample size is at least 30? Yes, the sampling distributions when n = 30 all look very much like the normal distribution. However, the more non-normal a population distribution, the larger a sample size is necessary for the sampling distribution to look nearly normal. DETERMINING IF THE SAMPLE MEAN IS NORMALLY DISTRIBUTED If the population is normal, the sampling distribution for ¯x will be normal for any sample size. The less normal the population,
the larger n needs to be for the sampling distribution for ¯x to be nearly normal. However, a good rule of thumb is that for almost all populations, the sampling distribution for ¯x will be approximately normal if n ≥ 30. This brings us to the Central Limit Theorem, the most fundamental theorem in Statistics. CENTRAL LIMIT THEOREM When taking a random sample of independent observations from a population with a ﬁxed mean and standard deviation, the distribution of ¯x approaches the normal distribution as n increases. 8The normal approximation becomes better as larger samples are used. However, in the case when the population is normally distributed, the normal distribution of the sample mean is normal for all sample sizes. 4.2. SAMPLING DISTRIBUTION FOR A SAMPLE MEAN 235 Figure 4.10: Sampling distributions for the mean at diﬀerent sample sizes and for three diﬀerent distributions. The dashed red lines show normal distributions. 0:1populationdistributionsuniform−20246exponential−20246normal−202460:1n = 2−202402020:1n = 50120120120:1n = 12012012012n = 300.51.01.50.51.01.50.51.01.5 236 CHAPTER 4. SAMPLING DISTRIBUTIONS EXAMPLE 4.12 Sometimes we do not know what the population distribution looks like. We have to infer it based on the distribution of a single sample. Figure 4.11 shows a histogram of 20 observations. These represent winnings and losses from 20 consecutive days of a professional poker player. Based on this sample data, can the normal approximation be applied to the distribution of the sample mean? We should consider each of the required conditions. (1) These are referred to as time series data, because the data arrived in a particular sequence. If the player wins on one day, it may inﬂuence how she plays the next. To make the assumption of independence we should perform careful checks on such data. (2) The sample size is 20, which is smaller than 30. (3) There are two outliers in the data, both quite extreme, which suggests the population may not be normal and instead may be very strongly skewed or have distant outliers. Outliers can play an important role and aﬀect the distribution of the sample mean and the estimate of the standard deviation of the sample mean. Since we should be skeptical of the independence of observations and the extreme upper outliers pose a challenge, we should not use the normal model for the sample mean of these 20 observations. If we can obtain a much larger sample, then the concerns about skew and outliers would no longer apply. Figure 4.11: Sample distribution of poker winnings. These data include two very clear outliers. These are problematic when considering the normality of the sample mean. For example, outliers are often an indicator of very strong skew. EXAMINE DATA STRUCTURE WHEN CONSIDERING INDEPENDENCE Some data sets are collected in such a way that they have a natural underlying structure between observations, e.g. when observations occur consecutively. Be especially cautious about independence assumptions regarding such data sets. WATCH OUT FOR STRONG SKEW AND OUTLIERS Strong skew in the population is often identiﬁed by the presence of clear outliers in the data. If a data set has prominent outliers, then a larger sample size will be needed for the sampling distribution for ¯x to be normal. There are no simple guidelines for what sample size is big enough for each situation. However, we can use the rule of thumb that, in general, an n of at least 30 is suﬃcient for most cases. Poker Winnings and Losses (US$)Frequency−200−1000100200300400500024681012 4.2. SAMPLING DISTRIBUTION FOR A SAMPLE MEAN 237 4.2.3 Normal approximation for the sampling distribution for ¯x¯x¯x At the beginning of this chapter, we used normal approximation for populations or for data that had an approximately normal distribution. When appropriate conditions are met, we can also use the normal approximation to estimate probabilities about a sample average. We must remember to verify that the conditions are met and use the mean µ¯x and standard deviation σ¯x for the sampling distribution for the sample average. THREE IMPORTANT FACTS ABOUT THE DISTRIBUTION OF A SAMPLE MEAN ¯x¯x¯x When the observations can be considered independent, such as from a random sample of less than 10% of the population, the distribution of the sample mean can be described as follows. 1. The mean of a sample mean is denoted by µ¯x, and it is equal to µ. 2. The SD of a sample mean is denoted by σ¯x, and it is equal to σ√ n . 3. When the population is normal or when n ≥ 30, the sample mean closely follows a normal distribution. EXAMPLE 4.13 In the 2017 Cherry Blossom 10 mile run, the average time for all of the runners is 94.52 minutes with a standard deviation of 8.97 minutes. The distribution of run times is approximately normal. Find the probabiliy that a randomly selected runner completes the run in less than 90 minutes. Because the distribution of run times is approximately normal, we can use normal approximation. Z = x − µ σ = 90 − 94.52 8.97 P (Z < −0.504) = 0.3072 = −0.504 There is a 30.72% probability that a randomly selected runner will complete the run in less than 90 minutes. EXAMPLE 4.14 Find the probabiliy that the average of 20 runners is less than 90 minutes. Here, n = 20 < 30, but the distribution of the population, that is, the distribution of run times is stated to be approximately normal. Because of this, the sampling distribution will be normal for any sample size. = σ¯x = = 2.01 σ √ n ¯x − µ¯x σ¯x 8.97 √ 20 90 − 94.52 2.01 P (Z < −2.25) = 0.0123 Z = = = −2.25 There is a 1.23% probability that the average run time of 20 randomly selected runners will be less than 90 minutes. 238 CHAPTER 4. SAMPLING DISTRIBUTIONS EXAMPLE 4.15 The average of all the runners’ ages is 35.05 years with a standard deviation of σ = 8.97. The distribution of age is somewhat skewed. What is the probability that a randomly selected runner is older than 37 years? Because the distribution of age is skewed and is not normal, we cannot use normal approximation for this problem. In order to answer this question, we would need to look at all of the data. GUIDED PRACTICE 4.16 What is the probability that the average of 50 randomly selected runners is greater than 37 years?9 REMEMBER TO DIVIDE BY When ﬁnding the probability that an average or mean is greater or less than a particular value, n to calculate the correct SD. remember to divide the standard deviation of the population by √ √ √ √ n n n 9Because n = 50 ≥ 30, the sampling distribution for the mean is approximately normal, so we can use normal approximation for this problem. The mean is given as 35.05 years. σ¯x = σ √ n = 8.97 √ 50 = 1.27 z = ¯x − µ¯x σ¯x = 37 − 35.05 1.27 = 1.535 P (Z > 1.535) = 0.062 There is a 6.2% chance that the average age of 50 runners will be greater than 37. 4.2. SAMPLING DISTRIBUTION FOR A SAMPLE MEAN 239 Section summary • The symbol ¯x denotes the sample average. ¯x for any particular sample is a number. However, ¯x can vary from sample to sample. The distribution of all possible values of ¯x for repeated samples of a ﬁxed size from a certain population is called the sampling distribution for ¯x. • The standard deviation of ¯x describes the typical error or distance of the sample mean from the population mean. It also tells us how much the sample mean is likely to vary from one random sample to another. • The standard deviation of ¯x will be smaller than the standard deviation of the population by a factor of n. The larger the sample, the better the estimate tends to be. √ • Consider taking a simple random sample from a population with a ﬁxed mean and standard deviation. The Central Limit Theorem ensures that regardless of the shape of the original population, as the sample size increases, the distribution of the sample average ¯x becomes more normal. • Three important facts about the sampling distribution for the sample average ¯x where the observations can be treated as independent: – The mean of a sample mean is denoted by µ¯x, and it is equal to µ. (center ) – The SD of a sample mean is denoted by σ¯x, and it is equal to σ√ n . (spread ) – When the population is normal or when n ≥ 30, the sample mean closely follows a normal distribution. (shape) • These facts are used when solving the following two types of normal approximation prob- lems involving a sample mean or a sample sum. A: Find the probability that a sample average will be greater/less than a certain value. 1. Verify that the observations can be treated as independent and that either the pop- ulation is approximately normal or n ≥ 30. 2. Calculate the Z-score. Use µ¯x = µ and σ¯x = σ√ 3. Find the appropriate area under the normal curve. n to standardize the sample average. B: Find the probability that a sample sum/total will be greater/less than a certain value. 1. Convert the sample sum into a sample average, using ¯x = sum n . 2. Do steps 1-3 from Part A above. 240 CHAPTER 4. SAMPLING DISTRIBUTIONS Exercises 4.11 Ages of pennies, Part I . The histogram below shows the distribution of ages of pennies at a bank. (a) Describe the distribution. (b) Sampling distributions for means from simple random samples of 5, 30, and 100 pennies is shown in the histograms below. Describe the shapes of these distributions and comment on whether they look like what you would expect to see based on the Central Limit Theorem. 4.12 Ages of pennies, Part II. The mean age of the pennies from Exercise 4.11 is 10.44 years with a standard deviation of 9.2 years. Using the Central Limit Theorem, calculate the means and standard deviations of the distribution of the mean from random samples of size 5, 30, and 100. Comment on whether the sampling distributions shown in Exercise 4.11 agree with the values you compute. A housing survey was conducted to determine the price of a typical home in 4.13 Housing prices. Topanga, CA. The mean price of a house was roughly $1.3 million with a standard deviation of $300,000. T
here were no houses listed below $600,000 but a few houses above $3 million. (a) Is the distribution of housing prices in Topanga symmetric, right skewed, or left skewed? Hint: Sketch the distribution. (b) Would you expect most houses in Topanga to cost more or less than $1.3 million? (c) Can we estimate the probability that a randomly chosen house in Topanga costs more than $1.4 million using the normal distribution? (d) What is the probability that the mean of 60 randomly chosen houses in Topanga is more than $1.4 million? (e) How would doubling the sample size aﬀect the standard deviation of the mean? 4.14 Stats ﬁnal scores. Each year about 1500 students take the introductory statistics course at a large university. This year scores on the ﬁnal exam are distributed with a median of 74 points, a mean of 70 points, and a standard deviation of 10 points. There are no students who scored above 100 (the maximum score attainable on the ﬁnal) but a few students scored below 20 points. (a) Is the distribution of scores on this ﬁnal exam symmetric, right skewed, or left skewed? (b) Would you expect most students to have scored above or below 70 points? (c) Can we calculate the probability that a randomly chosen student scored above 75 using the normal distribution? (d) What is the probability that the average score for a random sample of 40 students is above 75? (e) How would cutting the sample size in half aﬀect the standard deviation of the mean? Penny ages01020304050x n = 5051015202530x n = 30681012141618x n = 100891011121314 4.2. SAMPLING DISTRIBUTION FOR A SAMPLE MEAN 241 4.15 Identify distributions, Part I. Four plots are presented below. The plot at the top is a distribution for a population. The mean is 10 and the standard deviation is 3. Also shown below is a distribution of (1) a single random sample of 100 values from this population, (2) a distribution of 100 sample means from random samples with size 5, and (3) a distribution of 100 sample means from random samples with size 25. Determine which plot (A, B, or C) is which and explain your reasoning. 4.16 Identify distributions, Part II. Four plots are presented below. The plot at the top is a distribution for a population. The mean is 60 and the standard deviation is 18. Also shown below is a distribution of (1) a single random sample of 500 values from this population, (2) a distribution of 500 sample means from random samples of each size 18, and (3) a distribution of 500 sample means from random samples of each size 81. Determine which plot (A, B, or C) is which and explain your reasoning. 4.17 Weights of pennies. The distribution of weights of United States pennies is approximately normal with a mean of 2.5 grams and a standard deviation of 0.03 grams. (a) What is the probability that a randomly chosen penny weighs less than 2.4 grams? (b) Describe the sampling distribution for the mean weight of 10 randomly chosen pennies. (c) What is the probability that the mean weight of 10 pennies is less than 2.4 grams? (d) Sketch the two distributions (population and sampling) on the same scale. (e) Could you estimate the probabilities from (a) and (c) if the weights of pennies had a skewed distribution? 05101520Populationm = 10s = 3Plot A67891011121301020Plot B5101501020Plot C0102091011020406080100Populationm = 60s = 18Plot A54565860626466050100Plot B020406080100050100Plot C5055606570050100 242 CHAPTER 4. SAMPLING DISTRIBUTIONS 4.18 CFLBs. A manufacturer of compact ﬂuorescent light bulbs advertises that the distribution of the lifespans of these light bulbs is nearly normal with a mean of 9,000 hours and a standard deviation of 1,000 hours. (a) What is the probability that a randomly chosen light bulb lasts more than 10,500 hours? (b) Describe the distribution of the mean lifespan of 15 light bulbs. (c) What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours? (d) Sketch the two distributions (population and sampling) on the same scale. (e) Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution? 4.19 Songs on an iPod. Suppose an iPod has 3,000 songs. The histogram below shows the distribution of the lengths of these songs. We also know that, for this iPod, the mean length is 3.45 minutes and the standard deviation is 1.63 minutes. (a) Calculate the probability that a randomly selected song lasts more than 5 minutes. (b) You are about to go for an hour run and you make a random playlist of 15 songs. What is the probability that your playlist lasts for the entire duration of your run? Hint: If you want the playlist to last 60 minutes, what should be the minimum average length of a song? (c) You are about to take a trip to visit your parents and the drive is 6 hours. You make a random playlist of 100 songs. What is the probability that your playlist lasts the entire drive? 4.20 Spray paint, Part II. Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribution with a mean of 25 square feet and a standard deviation of 3 square feet. (a) What is the probability that the area covered by a can of spray paint is more than 27 square feet? (b) Suppose you want to spray paint an area of 540 square feet using 20 cans of spray paint. On average, how many square feet must each can be able to cover to spray paint all 540 square feet? (c) What is the probability that you can cover a 540 square feet area using 20 cans of spray paint? (d) If the area covered by a can of spray paint had a slightly skewed distribution, could you still calculate the probabilities in parts (a) and (c) using the normal distribution? 4.21 Wireless routers. John is shopping for wireless routers and is overwhelmed by the number of available options. In order to get a feel for the average price, he takes a random sample of 75 routers and ﬁnds that the average price for this sample is $75 and the standard deviation is $25. (a) Based on this information, how much variability should he expect to see in the mean prices of repeated samples, each containing 75 randomly selected wireless routers? (b) A consumer website claims that the average price of routers is $80. Is a true average of $80 consistent with John’s sample? 4.22 Betting on dinner, Part II. Exercise 3.42 introduces a promotion at a restaurant where prices of menu items are determined randomly following some underlying distribution. We are told that the price of basket of fries is drawn from a normal distribution with mean 6 and standard deviation of 2. You want to get 5 baskets of fries but you only have $28 in your pocket. What is the probability that you would have enough money to pay for all ﬁve baskets of fries? Length of song02468100200400600800 4.3. SAMPLING DISTRIBUTION FOR A DIFFERENCE OF PROPORTIONS OR MEANS 243 4.3 Sampling distribution for a difference In this section, we consider the case where instead of having one random sample from a population of interest, we have two independent random samples and we are interested in how far the sample values might be from one another. We describe the sampling distribution for the diﬀerence of sample proportions and the diﬀerence of sample means, and we ﬁnd the probability that the diﬀerence would be greater than or less than a certain amount. Learning objectives 1. Describe the center, spread, and shape of the sampling distribution for a diﬀerence of sample proportions. 2. Describe the center, spread, and shape of the sampling distribution for a diﬀerence of sample means. 3. Verify appropriate conditions and, if met, carry out the normal approximation using a diﬀer- ence of sample proportions or a diﬀerence of sample means. 4.3.1 The mean and SD for a difference of two random variables (review) In Section 3.4, we saw how to ﬁnd the mean and standard deviation of a diﬀerence of two random variables X and Y . We found that: E(X − Y ) = E(X) − E(Y ) That is, if we have the mean of each random variable, µX and µY , then the mean of the diﬀerence of the random variables is simply the diﬀerences of the individual means. This should seem very straightforward. The situation is a little more complex when looking at the variability of the diﬀerence of X and Y . Here we’re going to require a condition be met, speciﬁcally that X and Y are independent random variables. When that independence condition is met, then the following formula for the standard deviation of the diﬀerence, X − Y , holds: SD(X − Y ) = (SD(X))2 + (SD(Y ))2 These two formulas from probability play an important role in applications where we look at the diﬀerence of two sample proportions (ˆp1 − ˆp2) or the diﬀerence of sample means (¯x1 − ¯x2). 244 CHAPTER 4. SAMPLING DISTRIBUTIONS 4.3.2 Difference of sample proportions In Section 4.1 we considered a county where the proportion of people with blood type O+ is known to be 35%, and we described the distribution of the sample statistic ˆp from a random sample of size n from this population. Let us now consider two counties. In County 1, it is known that 35% of people have blood type O+. In County 2, it is known that 30% of people have blood type O+. If we take a random sample of size 50 from County 1 and a random sample of size 50 from County 2, what is the probability that we will get a higher proportion with blood type O+ in the County 2 sample? We know that the expected proportion with blood type O+ is lower in County 2. However, due to random variability, there is still some chance that the sample from County 2 will have a higher proportion with blood type O+. First, we need to ﬁnd the mean (expected value) and the standard deviation for the diﬀerence of sample proportions ˆp1 − ˆp2. For this, we use the two formulas from Section 4.3.1 in the context of the diﬀerence of sample proportions by substituting X = ˆp1 and Y = ˆp2: E(ˆp1 − ˆp2) = E(ˆp1) − E(ˆp2) = p1 − p2 SD(ˆp1 − ˆp2) = (SD(ˆp1))2 + (SD(ˆp2))2
= p1(1 − p1) n1 + p2(1 − p2) n2 . That is, given two independent random samples, the distribution of all possible values of ˆp1 − ˆp2 is centered on the true diﬀerence p1 − p2 and the typical distance or error of ˆp1 − ˆp2 from p1 − p2 is given by . Using µ for mean and σ for SD, we summarize this as follows. p1(1−p1) n1 + p2(1−p2) n2 MEAN AND STANDARD DEVIATION OF A DIFFERENCE OF SAMPLE PROPORTIONS The mean and standard deviation of the diﬀerence of sample proportions describe the center and spread of the distribution of all possible diﬀerences ˆp1 − ˆp2. Given population proportions p1 and p2 and random samples of size n1 and n2 where the independence condition is satisﬁed, we have the following. µ ˆp1− ˆp2 = p1 − p2 σ ˆp1− ˆp2 = p1(1 − p1) n1 + p2(1 − p2) n2 As we saw previously, the independence condition is satisﬁed if the data is collected from an experiment with two randomly assigned treatments or collected from 2 independent random samples, where each sample size is less than 10% of the population size if done without replacement. Having described the center and spread of the distribution of the diﬀerence of sample proportions, we need to understand the shape of the distribution. When looking at the sum or diﬀerence of random variables, if each variable is nearly normal, then the sum and diﬀerence are also nearly normal. This property will be very useful, as it says that when each sample proportion has a nearly normal distribution, then the diﬀerence of sample proportions will also be nearly normal, and we can use normal approximation to estimate probabilities that the diﬀerence will be greater or less than some value. 4.3. SAMPLING DISTRIBUTION FOR A DIFFERENCE OF PROPORTIONS OR MEANS 245 EXAMPLE 4.17 Let’s return to the blood type example. In County 1, it is known that 35% of people have blood type O+. In County 2, it is known that 30% of people have blood type O+. If we take a random sample of size 50 from County 1 and a random sample of size 50 from County 2, what is the probability that we will get a higher proportion with blood type O+ in the County 2 sample? We want to ﬁnd P (ˆp1 < ˆp2). This is equivalent to P (ˆp1 − ˆp2 < 0). Let us ﬁrst determine whether the independence condition is satisﬁed and whether normal approximation will be appropriate. We have two random samples, and the samples are independent of each other as they are from distinct populations. It is reasonable to assume that each sample size is less than 10% of the county population size. We now check the success-failure condition for each group. n1p1 = 50 × 0.35 = 17.5 ≥ 10 n2p2 = 50 × 0.30 = 15.0 ≥ 10 n1(1 − p1) = 50 × (1 − 0.35) = 32.5 ≥ 10 n2(1 − p2) = 50 × (1 − 0.30) = 35.0 ≥ 10 The independence condition is satisﬁed and the success-failure condition is met for both groups, so the distribution of ˆp1 − ˆp2 can be said to be nearly normal. We now ﬁnd the mean and the standard deviation of ˆp1 − ˆp2. µ ˆp1− ˆp2 = p1 − p2 = 0.35 − 0.30 = 0.05 σ ˆp1− ˆp2 = p1(1 − p1) n1 + p2(1 − p2) n2 = 0.35(1 − 0.35) 50 + 0.30(1 − 0.30) 50 = 0.0935 Recall, we want to ﬁnd P (ˆp1 − ˆp2 < 0). We use the calculated mean and standard deviation of ˆp1 − ˆp2 to ﬁnd a Z-score for the value of interest, which is 0.05 0.0935 = −0.535 Using technology to ﬁnd the area to the left of -0.535 under the standard normal curve, we obtain P (ˆp1 − ˆp2) ≈ P (Z < −0.535) = 0.296. Even though County 2 has a lower proportion of people with blood type O+, with these small sample sizes, there is still a 29.6% chance that the sample from County 2 ends up with a higher proportion with blood type O+ than the sample from County 1. −0.140.050.24x = 0 246 CHAPTER 4. SAMPLING DISTRIBUTIONS 4.3.3 Difference of sample means In Section 4.2 we started with all of the data from the 2017 Cherry Blossom Run, and we considered what the sampling distribution for a mean would look like for random samples of size n. The population mean for all the runners is 94.52 minutes and the population standard deviation is 8.97 minutes. Imagine taking two independent random samples of size 50 from this population. We know that the sampling distribution for the mean would have the same center and spread in each sample, but what about the sampling distribution for the diﬀerence of the sample means? What is the likelihood that the sample means from these two independent random samples would diﬀer by more than 3 minutes? To answer this, we consider two scenarios. First, consider the possibility that the mean of sample 1 is more than 3 minutes greater than the mean of sample 2. That is, we want to ﬁnd P(¯x1 − ¯x2 > 3). Second, it is also possible that the mean of sample 2 is more than 3 minutes greater than the mean of sample 1, so we also need to ﬁnd P(¯x2 − ¯x1 > 3), which we could also write as P(¯x1 − ¯x2 < −3). We now see that we are interested in the upper and lower tail of the distribution of ¯x1 − ¯x2. As we will soon see, the distribution of ¯x1 − ¯x2 follows a normal distribution when certain conditions are met, and in those cases we can ﬁnd one tail and then double it thanks to the symmetry of the normal distribution. Let’s ﬁnd the mean (expected value) and the standard deviation for the diﬀerence of sample means, ¯x1 − ¯x2. We again use the formulas discussed in Section 4.3.1 for the diﬀerence in two independent random normal variables, X − Y , but this time we substitute in X = ¯x1 and Y = ¯x2: E(¯x1 − ¯x2) = E(¯x1) − E(¯x2) = µ1 − µ2 SD(¯x1 − ¯x2) = (SD(¯x1))2 + (SD(¯x2))2 = σ2 1 n1 + σ2 2 n2 . That is, given two independent random samples, the distribution of all possible values of ¯x1 − ¯x2 is centered on the true diﬀerence µ1 − µ2 and the typical distance or error of ¯x1 − ¯x2 from µ1 − µ2 is given by . σ2 1 n1 + σ2 2 n2 MEAN AND STANDARD DEVIATION OF A DIFFERENCE OF SAMPLE MEANS The mean and standard deviation of the diﬀerence of sample means describe the center and spread of the distribution of all possible diﬀerences ¯x1 − ¯x2. Given population means µ1 and µ2, individual population standard deviations σ1 and σ2, and two random samples of size n1 and n2, then if the independence condition is satisﬁed, we have the following: µ¯x1−¯x2 = µ1 − µ2 σ¯x1−¯x2 = σ2 1 n1 + σ2 2 n2 Recall that in Section 4.3.2, we also noted that the sum or diﬀerence of random variables will also nearly normal, if each variable is itself nearly normal and the two random variables are also independent of each other. We will frequently use this principle to determine when the diﬀerence of sample means can be modeled using a normal distribution. 4.3. SAMPLING DISTRIBUTION FOR A DIFFERENCE OF PROPORTIONS OR MEANS 247 EXAMPLE 4.18 Let’s return to the Cherry Blossom Run application. We have that the population mean for all the runners in the 2017 Cherry Blossom Run is 94.52 minutes and the population standard deviation is 8.97 minutes. If we take two independent random samples of 50 runners, what is the probability that the sample means from these two samples will diﬀer by more than 3 minutes? We want to ﬁnd P (¯x1 − ¯x2 > 3) + P (¯x1 − ¯x2 < −3). Let us ﬁrst determine whether the independence condition is satisﬁed and whether normal approximation will be appropriate. We have two random samples, and the samples are distinct and independent of each other. The sample sizes are each less than 10% of total population of runners. Also, because the sample sizes n1 and n2 are both 50, and 50 ≥ 30, we don’t need the distribution of all run times to be nearly normal. With these conditions met, we have shown that the distribution of ¯x1 − ¯x2 is nearly normal. Next, we ﬁnd the mean and the standard deviation of ¯x1 − ¯x2: µ¯x1−¯x2 = µ1 − µ2 = 94.52 − 94.52 = 0 σ¯x1−¯x2 = σ2 1 n1 + σ2 2 n2 = 8.972 50 + 8.972 50 = 1.79 To ﬁnd P (¯x1 − ¯x2 > 3), we use the calculated mean and standard deviation of ¯x1 − ¯x2 to ﬁrst ﬁnd a Z-score for the diﬀerence of interest, which is 3.79 = 1.676 Using technology, to ﬁnd the area to the right of 1.676 under the standard normal curve, we have: P (¯x1 − ¯x2 > 3) ≈ P (Z > 1.676) = 0.047. Because the normal distribution of interest is centered at zero, the P (¯x1 − ¯x2 < −3) tail area will have the same size. So to get the total of the two tail areas, we double 0.047. Even though the samples are from the same population of runners, there is still a 2 × 0.047 = 0.094 probability, or a 9.4% chance, that the sample means will diﬀer by more than 3 minutes. −5.37−3.58−1.7901.793.585.37x = −3x = 3 248 CHAPTER 4. SAMPLING DISTRIBUTIONS Section summary • When two random variables each follow a nearly normal distribution, the distribution of their diﬀerence also follows a nearly normal distribution. • Both ˆp1 − ˆp2 and ¯x1 − ¯x2 are statistics that can take on diﬀerent values from one random sample to the next. As such, they have sampling distributions that can be described by their center, spread, and shape. • Three important facts about the sampling distribution for the diﬀerence of sample proportions ˆp1 − ˆp2 where the observations can be treated as independent: – The mean of the diﬀerence of sample proportions, denoted by µ ˆp1− ˆp2, is equal to p1 − p2. (center) – The SD of the diﬀerence of sample proportions, denoted by σ ˆp1− ˆp2, is equal to p1(1−p1) n1 + p2(1−p2) n2 . (spread) – When both groups meet the success-failure condition, the diﬀerence of sample proportions can be modeled using a normal distribution. (shape) • Three important facts about the sampling distribution for the diﬀerence of sample means ¯x1 − ¯x2 where the observations can be treated as independent: – The mean of the diﬀerence of sample means, denoted by µ¯x1−¯x2 , is equal to µ1 − µ2. (center) – The SD of the diﬀerence of sample means, denoted by σ¯x1−¯x2 , is equal to σ2 1 n1 + σ2 2 n2 . (spread) – When both populations are nearly normal or when n1 ≥ 30 and n2 ≥ 30, the diﬀerence of sample means can be modeled using a normal distribution. (shape) • When the diﬀerence of sample proportions ˆp1 − ˆp2 or the diﬀerence of samp
le means ¯x1 − ¯x2 follow a nearly normal distribution, we can ﬁnd the probability that the diﬀerence is greater than or less than a certain amount by ﬁnding a Z-score and using the normal approximation. 4.3. SAMPLING DISTRIBUTION FOR A DIFFERENCE OF PROPORTIONS OR MEANS 249 Exercises 4.23 Difference of proportions, Part 1. The fraction of workers who are considered “supercommuters”, because they commute more than 90 minutes to get to work, varies by state. Suppose the following were the exact values for Nebraska and New York: State Nebraska New York Proportion Supercommuters 0.01 0.06 Now suppose that we plan a study to survey 1000 people from each state, and we will compute the sample proportions ˆpN E for Nebraska and ˆpN Y for New York. (a) What is the associated mean and standard deviation of ˆpN E? (b) What is the associated mean and standard deviation of ˆpN Y ? (c) Calculate and interpret the mean and standard deviation associated with the diﬀerence in sample pro- portions for the two groups, ˆpN Y − ˆpN E. (d) How are the standard deviations from parts (a), (b), and (c) related? 4.24 Difference of proportions, Part 2. The fraction of workers who are considered “supercommuters”, because they commute more than 90 minutes to get to work, varies by state. Suppose the following were the exact values for Nebraska and New York: State Nebraska New York Proportion Supercommuters 0.01 0.06 Now suppose that we plan a study to survey 1000 people from each state, and we will compute the sample proportions ˆpN E for Nebraska and ˆpN Y for New York. (a) What distribution is associated with the diﬀerence ˆpN Y − ˆpN E? Justify your answer. (b) Determine the probability that ˆpN Y − ˆpN E will be larger than 0.055. (c) Determine the probability that ˆpN Y − ˆpN E will be smaller than 0.4. 4.25 Difference of means, Part 1. Suppose we will collect two random samples from the following distributions: Sample 1 Sample 2 Mean 15 20 Standard Deviation 20 10 Sample Size 50 30 In each of the parts below, consider the sample means ¯x1 and ¯x2 that we might observe from these two samples. (a) What is the associated mean and standard deviation of ¯x1? (b) What is the associated mean and standard deviation of ¯x2? (c) Calculate and interpret the mean and standard deviation associated with the diﬀerence in sample means for the two groups, ¯x2 − ¯x1. (d) How are the standard deviations from parts (a), (b), and (c) related? 4.26 Difference of means, Part 2. Suppose we will collect two random samples from the following distributions: Sample 1 Sample 2 Mean 15 20 Standard Deviation 20 10 Sample Size 50 30 In each of the parts below, consider the sample means ¯x1 and ¯x2 that we might observe from these two samples. (a) What distribution is associated with the diﬀerence ¯x2 − ¯x1? Justify your answer. (b) Determine the probability that ¯x2 − ¯x1 will be larger than 7. (c) Determine the probability that ¯x2 − ¯x1 will be smaller than 3. (d) Determine the probability that ¯x2 − ¯x1 will be smaller than 0. 250 CHAPTER 4. SAMPLING DISTRIBUTIONS Chapter highlights This chapter began by introducing the idea of a sampling distribution. As with any distribution, we can summarize a sampling distribution with regard to its center, spread, and shape. A common thread that ran through this chapter is the application of normal approximation (introduced in Section 2.3) to diﬀerent sampling distributions. The key steps are included for each of the normal approximation scenarios below. To verify that observations can be considered independent, verify that you have one of the following: a random process, a random sample with replacement, or a random sample without replacement of less than 10% of the population. To satisfy the independence condition when working with two groups, we require 2 independent random samples with replacement, 2 independent samples without replacement of less than 10% of their populations, or an experiment with 2 randomly assigned treatments. For completion and comparison purposes, we include cases introduced in earlier chapters as well in the overview below. 1. Normal approximation for numerical data: (introduced in Section 2.3) • Verify that observations can be treated as independent and that population is approxi- mately normal. • Use a normal model with mean µ and SD σ. 2. Normal approximation for a sample proportion (with categorical data): • Verify that observations can be treated as independent and that np ≥ 10 and n(1−p) ≥ 10. • Use a normal model with mean µ ˆp = p and SD σ ˆp = p(1−p) n . 3. Normal approximation for a sample mean (with numerical data): • Verify that observations can be treated as independent and that population is approxi- mately normal or that n ≥ 30. • Use a normal model with mean µ¯x = µ and SD σ¯x = σ√ n . 4. Normal approximation for a diﬀerence of sample proportions: • Verify that observations can be treated as independent and that n1p1 ≥ 10, n1(1 − p1) ≥ 10, n2(1 − p2) ≥ 10, and n2(1 − p2) ≥ 10. • Use a normal model with mean µ ˆp1− ˆp2 = p1 − p2 and SD σ ˆp1− ˆp2 = p1(1−p1) n1 + p2(1−p2) n2 . 5. Normal approximation for a diﬀerence of sample means: • Verify that observations can be treated as independent and that both populations are nearly normal or both n1 and n2 are ≥ 30. • Use a normal model with mean: µ¯x1−¯x2 = µ1 − µ2 and SD: σ¯x1−¯x2 = σ2 1 n1 + σ2 2 n2 . Cases 1, 3 and 5 are for numerical variables, while cases 2 and 4 are for categorical yes/no variables. In the case of proportions and counts, we never look to see if the population is normal. That would not make sense because a yes/no variable cannot have a normal distribution. The Central Limit Theorem is the mathematical rule that ensures that when the sample size is suﬃciently large, the sample mean/sum and sample proportion/count will be approximately normal. 4.3. SAMPLING DISTRIBUTION FOR A DIFFERENCE OF PROPORTIONS OR MEANS 251 Chapter exercises 4.27 University admissions. Suppose a university announced that it admitted 2,500 students for the following year’s freshman class. However, the university has dorm room spots for only 1,786 freshman students. If there is a 70% chance that an admitted student will decide to accept the oﬀer and attend this university, what is the approximate probability that the university will not have enough dormitory room spots for the freshman class? 4.28 SAT scores. SAT scores (out of 1600) are distributed normally with a mean of 1100 and a standard deviation of 200. Suppose a school council awards a certiﬁcate of excellence to all students who score at least 1350 on the SAT, and suppose we pick one of the recognized students at random. What is the probability this student’s score will be at least 1500? (The material covered in Section 3.2 would be useful for this question.) 4.29 Overweight baggage. Suppose weights of the checked baggage of airline passengers follow a nearly normal distribution with mean 45 pounds and standard deviation 3.2 pounds. Most airlines charge a fee for baggage that weigh in excess of 50 pounds. Determine what percent of airline passengers incur this fee. 4.30 Survey response rate. Pew Research reported that the typical response rate to their surveys is only 9%. If for a particular survey 15,000 households are contacted, what is the probability that at least 1,500 will agree to respond?10 Suppose weights of the checked baggage of airline passengers 4.31 Overweight baggage, Part II. follow a nearly normal distribution with mean 45 pounds and standard deviation 3.2 pounds. What is the probability that the total weight of 10 bags is greater than 460 lbs? 4.32 Chocolate chip cookies. Students are asked to count the number of chocolate chips in 22 cookies for a class activity. The packaging for these cookies claims that there are an average of 20 chocolate chips per cookie with a standard deviation of 4.37 chocolate chips. (a) Based on this information, about how much variability should they expect to see in the mean number of chocolate chips in random samples of 22 chocolate chip cookies? (b) What is the probability that a random sample of 22 cookies will have an average less than 14.77 chocolate chips if the company’s claim on the packaging is true? Assume that the distribution of chocolate chips in these cookies is approximately normal. (c) Assume the students got 14.77 as the average in their sample of 22 cookies. Do you have conﬁdence or not in the company’s claim that the true average is 20? Explain your reasoning. 4.33 Young Hispanics in the US. The 2019 Current Population Survey (CPS) estimates that 36.0% of the people of Hispanic origin in the Unites States are under 21 years old.11 Calculate the probability that at least 35 people among a random sample of 100 Hispanic people living in the United States are under 21 years old. 4.34 Poverty in the US. The 2019 Current Population Survey (CPS) estimates that 19.2% of Mississippians live in poverty, which makes Mississippi the state with the highest poverty rate in the United States.12 We are interested in ﬁnding out the probability that at least 200 people among a random sample of 1,000 Mississippians live in poverty. (a) Estimate this probability using the normal approximation to the binomial distribution. (b) Estimate this probability using the distribution of the sample proportion. (c) How do your answers from parts (a) and (b) compare? 10Pew Research Center, Assessing the Representativeness of Public Opinion Surveys, May 15, 2012. 11United States Census Bureau. https://www.census.gov/data/tables/2019/demo/hispanic-origin/2019-cps.html. The Hispanic Population in the United States: 2019. Web. Census 12United Bureau. States https://www.census.gov/data/tables/time-series/demo/income- poverty/historical-poverty-people.html. Historical Poverty Tables: People and Families - 1959 to 2020. Web. 252 Chapter 5 Foundations for inference 5.1 Estimating unknown parameters 5.2 Conﬁdence intervals 5.3 Introducing hypothesis testing 253
Statistical inference is primarily concerned with understanding and quantifying the uncertainty of parameter estimates. While the equations and details change depending on the setting, the foundations for inference are the same throughout all of statistics. We start with a familiar topic: the idea of using a sample proportion to estimate a population proportion. Next, we create what’s called a conﬁdence interval, which is a range of values where the true population value is likely to lie. Finally, we introduce a hypothesis testing framework, which allows us use data to formally evaluate claims about the population, such as whether a survey provides strong evidence that a candidate has the support of a majority of the voting population. For videos, slides, and other resources, please visit www.openintro.org/ahss 254 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.1 Estimating unknown parameters Companies such as the Gallup and Pew Research frequently conduct polls as a way to understand the state of public opinion or knowledge on many topics, including politics, scientiﬁc understanding, brand recognition, and more. How well do these polls estimate the opinion or knowledge of the broader population? Why is a larger sample generally preferable to a smaller sample? And what role does the concept of a sampling distribution, introduced in the previous chapter, play in answering these questions? Learning objectives 1. Explain the diﬀerence between probability and inference and identify when to use which one. 2. Understand the purpose and use of a point estimate. 3. Understand how to measure the variability/error in a point estimate. 4. Recognize the relationship between the standard error of a point estimate and the standard deviation of a sample statistic. 5. Understand how changing the sample size aﬀects the variability/error in a point estimate. 5.1.1 Point estimates With this chapter, we move from the world of probability to the world of inference. Whereas probability involves using a known population value (parameter) to make a prediction about the likelihood of a particular sample value (statistic), inference involves using a calculated sample value (statistic) to estimate or better understand an unknown population value (parameter). For both of these, the concept of the sampling distribution is fundamental. Suppose a poll suggested the US President’s approval rating is 45%. We would consider 45% to be a point estimate of the approval rating we might see if we collected responses from the entire population. This entire-population response proportion is generally referred to as the parameter of interest, and when the parameter is a proportion, we denote it with the letter p. We typically estimate the parameter by collecting information from a sample of the population; we compute the observed proportion in the sample and we denote this sample proportion as ˆp. Unless we collect responses from every individual in the sample, p remains unknown, and we use ˆp as our point estimate for p. 5.1. ESTIMATING UNKNOWN PARAMETERS 255 The diﬀerence we observe from the poll versus the parameter is called the error in the estimate. Generally, the error consists of two aspects: sampling error and bias. Bias describes a systematic tendency to over- or under-estimate the true population value. For instance, if we took a political poll but our sample didn’t include a roughly representative distribution of the political parties, the sample would likely skew in a particular direction and be biased. Taking a truly random sample helps avoid bias. However, as we saw in Chapter 1, even with a random sample, various types of response bias can still be present. For example, if we were taking a student poll asking about support for a new college stadium, we’d probably get a biased estimate of the stadium’s level of student support by wording the question as, Do you support your school by supporting funding for the new stadium? We try to minimize bias through thoughtful data collection procedures, but bias can creep into our estimates without us even be aware. Sampling error is uncertainty in a point estimate that happens naturally from one sample to the next. Much of statistics, including much of this book, is focused on understanding and quantifying sampling error. Remember though, that sampling error does not account for the possible eﬀects of leading questions or other types of response bias. When we measure sampling error, we are measuring the expected variability in a point estimate that arises from randomly sampling only a subset of the population. EXAMPLE 5.1 In Chapter 2, we found the summary statistics for the number of characters in a set of 50 email data. These values are summarized below. ¯x median sx 11,160 6,890 13,130 Estimate the population mean based on the sample. The best estimate for the population mean is the sample mean. That is, ¯x = 11, 160 is our best estimate for µ. GUIDED PRACTICE 5.2 Using the email data, what quantity should we use as a point estimate for the population standard deviation σ?1 1Again, intuitively we would use the sample standard deviation s = 13, 130 as our best estimate for σ. 256 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.1.2 Understanding the variability of a point estimate Suppose the proportion of American adults who support the expansion of solar energy is p = 0.88, which is our parameter of interest.2 If we were to take a poll of 1000 American adults on this topic, the estimate would not be perfect, but how close might we expect the sample proportion in the poll would be to 88%? We want to understand, how does the sample proportion ˆp behave when the true population proportion is 0.88.3 Let’s ﬁnd out! We can simulate responses we would get from a simple random sample of 1000 American adults, which is only possible because we know the actual support expanding solar energy to be 0.88. Here’s how we might go about constructing such a simulation: 1. There were about 250 million American adults in 2018. On 250 million pieces of paper, write “support” on 88% of them and “not” on the other 12%. 2. Mix up the pieces of paper and pull out 1000 pieces to represent our sample of 1000 American adults. 3. Compute the fraction of the sample that say “support”. Any volunteers to conduct this simulation? Probably not. While this physical simulation is totally impractical, we can simulate it thousands, even millions, of times using computer code. We’ve written a short computer simulation and run it 10,000 times. The results are show in Figure 5.1 in case you are curious what the computer code looks like. In this simulation, the sample gave a point estimate of ˆp1 = 0.894. We know the population proportion for the simulation was p = 0.88, so we know the estimate had an error of 0.894 − 0.88 = +0.014. and 12% are "not". # 1. Create a set of 250 million entries, where 88% of them are "support" # pop size <- 250000000 possible entries <- c(rep("support", 0.88 * pop size), rep("not", 0.12 * pop size)) # 2. Sample 1000 entries without replacement. sampled entries <- sample(possible entries, size = 1000) # 3. Compute p-hat: count the number that are "support", then divide by # sum(sampled entries == "support") / 1000 the sample size. Figure 5.1: For those curious, this is code for a single ˆp simulation using the statistical software called R. Each line that starts with # is a code comment, which is used to describe in regular language what the code is doing. We’ve provided software labs in R at openintro.org/stat/labs for anyone interested in learning more. One simulation isn’t enough to get a great sense of the distribution of estimates we might expect in the simulation, so we should run more simulations. In a second simulation, we get ˆp2 = 0.885, which has an error of +0.005. In another, ˆp3 = 0.878 for an error of -0.002. And in another, an estimate of ˆp4 = 0.859 with an error of -0.021. With the help of a computer, we’ve run the simulation 10,000 times and created a histogram of the results from all 10,000 simulations in Figure 5.2. This distribution of sample proportions is called a sampling distribution. We can characterize this sampling distribution as follows: 2We haven’t actually conducted a census to measure this value perfectly. However, a very large sample has suggested the actual level of support is about 88%. 3Note: 88% written as a proportion would be 0.88. It is common to switch between proportion and percent. 5.1. ESTIMATING UNKNOWN PARAMETERS 257 Center. The center of the distribution is µ ˆp = 0.880, which is the same as the parameter. Notice that the simulation mimicked a simple random sample of the population, which is a straightforward sampling strategy that helps avoid sampling bias. Spread. The standard deviation of the distribution is σ ˆp = 0.010. Shape. The distribution is symmetric and bell-shaped, and it resembles a normal distribution. These ﬁndings are encouraging! When the population proportion is p = 0.88 and the sample size is n = 1000, the sample proportion ˆp tends to give a pretty good estimate of the population proportion. We also have the interesting observation that the histogram resembles a normal distribution. Figure 5.2: A histogram of 10,000 sample proportions sampled from a population where the population proportion is 0.88 and the sample size is n = 1000. SAMPLING DISTRIBUTIONS ARE NEVER OBSERVED, BUT WE KEEP THEM IN MIND In real-world applications, we never actually observe the sampling distribution, yet it is useful to always think of a point estimate as coming from such a hypothetical distribution. Understanding the sampling distribution will help us characterize and make sense of the point estimates that we do observe. EXAMPLE 5.3 If we used a much smaller sample size of n = 50, would you guess that the standard error for ˆp would be larger or smaller than when we used n = 1000? Intuitively, it seems like more data is better than less data, and generally that is correct!
The typical error when p = 0.88 and n = 50 would be larger than the error we would expect when n = 1000. Example 5.3 highlights an important property we will see again and again: a bigger sample tends to provide a more precise point estimate than a smaller sample. Remember though, that this is only true for random samples. Additionally, a bigger sample cannot correct for response bias or other types of bias that may be present. Sample Proportions0.840.860.880.900.920250500750Frequency 258 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.1.3 Introducing the standard error Point estimates only approximate the population parameter. How can we quantify the expected variability in a point estimate ˆp? The discussion in Section 4.1 tells us how. The variability in the distribution of ˆp is given by its standard deviation. If we know the population proportion, we can calculate the standard deviation of the point estimate ˆp. In our simulation we knew p was 0.88. Thus we can calculate the standard deviation as SD ˆp = p(1 − p) n = (.088)(1 − 0.88) n = 0.01 If we now look at the sampling distribution, we see that the typical distance sample proportions are from the true value of 0.88 is about 0.01. EXAMPLE 5.4 Consider a random sample of size 80 from a county population. We ﬁnd that 15% of the sample support a controversial new ballot measure at the state level. How far is our estimate likely to be from the true percent that support the measure? We would like to calculate the standard deviation of ˆp, but we run into a serious problem: p is unknown. In fact, when doing inference, p must be unknown, otherwise it is illogical to try to estimate it. We cannot calculate the SD, but we can estimate it using, you might have guessed, the sample proportion ˆp. This estimate of the standard deviation is known as the standard error, or SESESE for short. SE ˆp = ˆp(1 − ˆp) n EXAMPLE 5.5 Calculate and interpret the SE of ˆp for the previous example. SE ˆp = ˆp(1 − ˆp) n = 0.15(1 − 0.15) 80 = 0.04 The typical or expected error in our estimate is 4%. EXAMPLE 5.6 If we quadruple the sample size from 80 to 320, what will happen to the SE? SE ˆp = ˆp(1 − ˆp) n = 0.15(1 − 0.15) 320 = 0.02 The larger the sample size, the smaller our standard error. This is consistent with intuition: the more data we have, the more reliable an estimate will tend to be. However, quadrupling the sample size does not reduce the error by a factor of 4. Because of the square root, the eﬀect is to reduce the error by a factor 4, or 2. √ 5.1. ESTIMATING UNKNOWN PARAMETERS 259 5.1.4 Basic properties of point estimates We achieved three goals in this section. First, we determined that point estimates from a sample may be used to estimate population parameters. We also determined that these point estimates are not exact: they vary from one sample to another. Lastly, we quantiﬁed the uncertainty of the sample proportion using what we call the standard error. Remember that the standard error only measures sampling error. It does not account for bias that results from leading questions or other types of response bias. When our sampling method produces estimates in an unbiased way, the sampling distribution will be centered on the true value and we call the method accurate. When the sampling method produces estimates that have low variability, the sampling distribution will have a low standard error, and we call the method precise. EXAMPLE 5.7 Using Figure 5.3, which of the distributions were produced by methods that are biased? That are accurate? Order the distributions from most precise to least precise (that is, from lowest variability to highest variability). Distributions (b) and (d) are centered on the parameter (the true value), so those methods are accurate. The methods that produced distributions (a) and (c) are biased, because those distributions are not centered on the parameter. From most precise to least precise, we have (a), (b), (c), (d). Figure 5.3: Four sampling distributions shown, with parameter identiﬁed. Explore these distributions through a Desmos activity at openintro.org/ahss/desmos. EXAMPLE 5.8 Why do we want a point estimate to be both precise and accurate? If the point estimate is precise, but highly biased, then we will consistently get a bad estimate. On the other hand, if the point estimate is unbiased but not at all precise, then by random chance, we may get an estimate far from the true value. Remember, when taking a sample, we generally get only one chance. sampling distribution that tell us how much conﬁdence we can have in the estimate. It is the properties of the The strategy of using a sample statistic to estimate a parameter is quite common, and it’s a strategy that we can apply to other statistics besides a proportion. For instance, if we want to estimate the average salary for graduates from a particular college, we could survey a random sample of recent graduates; in that example, we’d be using a sample mean ¯x to estimate the population mean µ for all graduates. As another example, if we want to estimate the diﬀerence in product prices for two websites, we might take a random sample of products available on both sites, check the prices on each, and use then compute the average diﬀerence; this strategy certainly wouldn’t give us a perfect measurement of the actual diﬀerence, but it would give us a point estimate. While this chapter emphases a single proportion context, we’ll encounter many diﬀerent contexts throughout this book where these methods will be applied. The principles and general ideas are the same, even if the details change a little. 260 CHAPTER 5. FOUNDATIONS FOR INFERENCE Section summary • In this section we laid the groundwork for our study of inference. Inference involves using known sample values to estimate or better understand unknown population values. • A sample statistic can serve as a point estimate for an unknown parameter. For example, the sample mean is a point estimate for an unknown population mean, and the sample proportion is a point estimate for an unknown population proportion. • It is helpful to imagine a point estimate as being drawn from a particular sampling distribution. • The standard error (SE) of a point estimate tells us the typical error or uncertainty associated with the point estimate. It is also an estimate of the spread of the sampling distribution. • A point estimate is unbiased (accurate) if the sampling distribution (i.e., the distribution of all possible outcomes of the point estimate from repeated samples from the same population) is centered on the true population parameter. • A point estimate has lower variability (more precise) when the standard deviation of the sampling distribution is smaller. • In a random sample, increasing the sample size n will make the standard error smaller. This is consistent with the intuition that larger samples tend to be more reliable, all other things being equal. • In general, we want a point estimate to be unbiased and to have low variability. Remember: the terms unbiased (accurate) and low variability (precise) are properties of generic point estimates, which are variables that have a sampling distribution. These terms do not apply to individual values of a point estimate, which are numbers. 5.1. ESTIMATING UNKNOWN PARAMETERS 261 Exercises 5.1 Identify the parameter, Part I. For each of the following situations, state whether the parameter of interest is a mean or a proportion. It may be helpful to examine whether individual responses are numerical or categorical. (a) In a survey, one hundred college students are asked how many hours per week they spend on the Internet. (b) In a survey, one hundred college students are asked: “What percentage of the time you spend on the Internet is part of your course work?” (c) In a survey, one hundred college students are asked whether or not they cited information from Wikipedia in their papers. (d) In a survey, one hundred college students are asked what percentage of their total weekly spending is on alcoholic beverages. (e) In a sample of one hundred recent college graduates, it is found that 85 percent expect to get a job within one year of their graduation date. 5.2 Identify the parameter, Part II. For each of the following situations, state whether the parameter of interest is a mean or a proportion. (a) A poll shows that 64% of Americans personally worry a great deal about federal spending and the budget deﬁcit. (b) A survey reports that local TV news has shown a 17% increase in revenue within a two year period while newspaper revenues decreased by 6.4% during this time period. (c) In a survey, high school and college students are asked whether or not they use geolocation services on their smart phones. (d) In a survey, smart phone users are asked whether or not they use a web-based taxi service. (e) In a survey, smart phone users are asked how many times they used a web-based taxi service over the last year. 5.3 Quality control. As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She ﬁnds that 27 of the chips are defective. (a) What population is under consideration in the data set? (b) What parameter is being estimated? (c) What is the point estimate for the parameter? (d) What is the name of the statistic that we use to measure the uncertainty of the point estimate? (e) Compute the value from part (d) for this context. (f) The historical rate of defects is 10%. Should the engineer be surprised by the observed rate of defects during the current week? (g) Suppose the true population value was found to be 10%. If we use this proportion to recompute the value in part (e) using p = 0.1 instead of ˆp, does the resulting value change much? 262 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.4 Unexpected expense. In a random sample 765 adults in the United States, 322 say they coul
d not cover a $400 unexpected expense without borrowing money or going into debt. (a) What population is under consideration in the data set? (b) What parameter is being estimated? (c) What is the point estimate for the parameter? (d) What is the name of the statistic that we can use to measure the uncertainty of the point estimate? (e) Compute the value from part (d) for this context. (f) A cable news pundit thinks the value is actually 50%. Should she be surprised by the data? (g) Suppose the true population value was found to be 40%. If we use this proportion to recompute the value in part (e) using p = 0.4 instead of ˆp, does the resulting value change much? 5.5 Repeated water samples. A nonproﬁt wants to understand the fraction of households that have elevated levels of lead in their drinking water. They expect at least 5% of homes will have elevated levels of lead, but not more than about 30%. They randomly sample 800 homes and work with the owners to retrieve water samples, and they compute the fraction of these homes with elevated lead levels. They repeat this 1,000 times and build a distribution of sample proportions. (a) What is this distribution called? (b) Would you expect the shape of this distribution to be symmetric, right skewed, or left skewed? Explain your reasoning. (c) If the proportions are distributed around 8%, what is the variability of the distribution? (d) What is the formal name of the value you computed in (c)? (e) Suppose the researchers’ budget is reduced, and they are only able to collect 250 observations per sample, but they can still collect 1,000 samples. They build a new distribution of sample proportions. How will the variability of this new distribution compare to the variability of the distribution when each sample contained 800 observations? 5.6 Repeated student samples. Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class project, students randomly sample 40 students and check if those students made the list. They repeat this 1,000 times and build a distribution of sample proportions. (a) What is this distribution called? (b) Would you expect the shape of this distribution to be symmetric, right skewed, or left skewed? Explain your reasoning. (c) Calculate the variability of this distribution. (d) What is the formal name of the value you computed in (c)? (e) Suppose the students decide to sample again, this time collecting 90 students per sample, and they again collect 1,000 samples. They build a new distribution of sample proportions. How will the variability of this new distribution compare to the variability of the distribution when each sample contained 40 observations? 5.2. CONFIDENCE INTERVALS 263 5.2 Conﬁdence intervals The site ﬁvethirtyeight.com regularly forecasts support for each candidate in Congressional races, i.e. races in the US House of Representatives and the US Senate. In addition to point estimates, they report conﬁdence intervals.4 What are conﬁdence intervals, and how do we interpret them? Learning objectives 1. Explain the purpose and use of conﬁdence intervals. 2. Construct 95% conﬁdence intervals assuming the point estimate follows a normal distribution. 3. Calculate the critical value for a C% conﬁdence interval when the point estimate follows a normal distribution. 4. Describe how sample size and conﬁdence level aﬀect the width of a conﬁdence interval. 5. Interpret a conﬁdence interval and the conﬁdence level in context. 6. Draw conclusions with a speciﬁed conﬁdence level about the values of unknown parameters. 7. Calculate and interpret the margin of error for a C% conﬁdence interval. Distinguish between margin of error and standard error. 5.2.1 Capturing the population parameter A point estimate provides a single plausible value for a parameter. However, a point estimate isn’t perfect and will have some standard error associated with it. When estimating a parameter, it is better practice to provide a plausible range of values instead of supplying just the point estimate. A plausible range of values for the population parameter is called a conﬁdence interval. Using only a point estimate is like ﬁshing in a murky lake with a spear, and using a conﬁdence interval is like ﬁshing with a net. We can throw a spear where we saw a ﬁsh, but we will probably miss. On the other hand, if we toss a net in that area, we have a good chance of catching the ﬁsh. If we report a point estimate, we probably will not hit the exact population parameter. On the other hand, if we report a range of plausible values – a conﬁdence interval – we have a good shot at capturing the parameter. 4See: https://projects.fivethirtyeight.com/2018-midterm-election-forecast/senate 264 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.2.2 Constructing a 95% conﬁdence interval A point estimate is our best guess for the value of the parameter, so it makes sense to build the conﬁdence interval around that value. The standard error is a measure of the uncertainty associated with the point estimate. EXAMPLE 5.9 How many standard errors should we extend above and below the point estimate if we want to be 95% conﬁdent of capturing the true value? First, we observe that the area under the standard normal curve between -1.96 and 1.96 is 95%. When conditions for a normal model are met, the point estimate we observe will be within 1.96 standard deviations of the true value about 95% of the time. Thus, if we want to be 95% conﬁdent of capturing the true value, we should go 1.96 standard errors on either side of the point estimate. CONSTRUCTING A 95% CONFIDENCE INTERVAL USING A NORMAL MODEL When the sampling distribution for a point estimate can reasonably be modeled as normal, a 95% conﬁdence interval for the unknown parameter can be constructed as: point estimate ± 1.96 × SE of estimate (5.10) We can be 95% conﬁdent that this interval captures the true value. In the next chapters we will determine when we can apply a normal model to a point estimate. For now, we will assume that a normal model is reasonable. EXAMPLE 5.11 The point estimate of the support for a ballot measure from Section 5.1 was 15% from a sample size of 80. The standard error for this point estimate was calculated to be SE = 0.04. Assuming that conditions for a normal model are met, construct and interpret a 95% conﬁdence interval. point estimate ± 1.96 × SE of estimate 0.15 ± 1.96 × 0.04 (0.0716, 0.2284) We are 95% conﬁdent that the true percent of support in this population for the ballot measure is between 7.16% and 22.84%. EXAMPLE 5.12 Suppose the level of support at the state level for the ballot measure has been precisely estimated at 0.20 (20%). Based on the conﬁdence interval above, is there evidence that a smaller proportion support the ballot measure in the speciﬁc county we’re looking at relative to the state as a whole? While the point estimate of 0.15 is lower than 0.20, it appears this deviation may be due to random chance. Because the conﬁdence interval includes the value 0.20, the value of 0.20 is a reasonable value for the proportion of the county population that support the ballot measure. Therefore, based on this conﬁdence interval, we do not have evidence that a smaller proportion support the ballot measure in this county than in the state as a whole. 5.2. CONFIDENCE INTERVALS 265 We can be 95% conﬁdent that a 95% conﬁdence interval captures the true population parameter. However, conﬁdence intervals are imperfect. About 1-in-20 (5%) properly constructed 95% conﬁdence intervals will fail to capture the parameter of interest. Figure 5.4 shows 25 conﬁdence intervals for a proportion that were constructed from simulations where the true proportion was p = 0.3. However, 1 of these 25 conﬁdence intervals happened not to include the true value. Figure 5.4: Twenty-ﬁve samples of size n = 300 were simulated when p = 0.30. For each sample, a conﬁdence interval was created to try to capture the true proportion p. However, 1 of these 25 intervals did not capture p = 0.30. GUIDED PRACTICE 5.13 In Figure 5.4, one interval does not capture the true proportion, p = 0.3. Does this imply that there was a problem with the simulations?5 5No. Just as some observations occur more than 1.96 standard deviations from the mean, some point estimates will be more than 1.96 standard errors from the parameter. A conﬁdence interval only provides a plausible range of values for a parameter. While we might say other values are implausible based on the data, this does not mean they are impossible. 0.25p = 0.300.35llllllllllllllllllllllllll 266 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.2.3 Changing the conﬁdence level Suppose we want to construct a conﬁdence interval with a conﬁdence level somewhat greater than 95%: perhaps we would like a conﬁdence level of 99%. EXAMPLE 5.14 Other things being equal, would a 99% conﬁdence interval be wider or narrower than a 95% conﬁdence interval? Using a previous analogy: if we want to be more conﬁdent that we will catch a ﬁsh, we should use a wider net, not a smaller one. To be 99% conﬁdence of capturing the true value, we must use a wider interval. On the other hand, if we want an interval with lower conﬁdence, such as 90%, we would use a narrower interval. The 95% conﬁdence interval structure provides guidance in how to make intervals with new conﬁdence levels. Below is a general 95% conﬁdence interval for a point estimate that comes from a nearly normal distribution: point estimate ± 1.96 × SE of estimate (5.15) There are three components to this interval: the point estimate, “1.96”, and the standard error. The choice of 1.96 × SE was based on capturing 95% of the distribution since the estimate is within 1.96 standard deviations of the true value about 95% of the time. The choice of 1.96 corresponds to a 95% conﬁdence level. GUIDED PRACTICE 5.16 If X is a normally distributed random variable, how often will X be within 2.58 standard deviations of the mean?6 Figure
5.5: The area between -z and z increases as \|z\| becomes larger. If the conﬁdence level is 99%, we choose z such that 99% of the normal curve is between -z and z, which corresponds to 0.5% in the lower tail and 0.5% in the upper tail: z = 2.58. 6This is equivalent to asking how often the Z-score will be larger than -2.58 but less than 2.58. (For a picture, see Figure 5.5.) There is ≈ 0.99 probability that the unobserved random variable X will be within 2.58 standard deviations of the mean. standard deviations from the mean−3−2−1012395%, extends −1.96 to 1.9699%, extends −2.58 to 2.58 5.2. CONFIDENCE INTERVALS 267 To create a 99% conﬁdence interval, change 1.96 in the 95% conﬁdence interval formula to be 2.58. Guided Practice 5.16 highlights that 99% of the time a normal random variable will be within 2.58 standard deviations of its mean. Thus, the formula for a 99% conﬁdence interval is point estimate ± 2.58 × SE of estimate (5.17) Figure 5.5 provides a picture of how to identify z based on a conﬁdence level. The number of standard errors we go above and below the point estimate is called the critical value. When the critical value is determined based on a normal model, we call the critical value z. CONFIDENCE INTERVAL FOR ANY CONFIDENCE LEVEL If the point estimate follows a normal model with standard error SE, then a conﬁdence interval for the population parameter is point estimate ± z × SE of estimate where z depends on the conﬁdence level selected. Finding the value of z that corresponds to a particular conﬁdence level is most easily accomplished by using a new table, called the t-table. For now, what is noteworthy about this table is that the bottom row corresponds to conﬁdence levels. The numbers inside the table are the critical values, but which row should we use? Later in this book, we will see that a t-curve with inﬁnite degrees of freedom corresponds to the normal curve. For this reason, when ﬁnding z, we use the t-table at row ∞. df one tail 1 2 3 ... 1000 0.100 3.078 1.886 1.638 ... 1.282 ∞ 1.282 80% 0.050 6.314 2.920 2.353 ... 1.646 1.645 90% 0.025 12.71 4.303 3.182 ... 1.962 1.960 95% 0.010 31.82 6.965 4.541 ... 2.330 2.326 98% 0.005 63.66 9.925 5.841 2.581 2.576 99% Conﬁdence level C Figure 5.6: An abbreviated look at the t-table. The columns correspond to conﬁdence levels. Row ∞ corresponds to the normal curve. FINDING zzz FOR A PARTICULAR CONFIDENCE LEVEL We select z so that the area between -z and z in the normal model corresponds to the conﬁdence level. Use a calculator or use the t-table at row ∞ to ﬁnd the critical value z. GUIDED PRACTICE 5.18 Find the appropriate z value for an 80% conﬁdence interval.7 The normal approximation is crucial to the precision of these conﬁdence intervals. The next two chapters provide detailed discussions about when a normal model can safely be applied to a variety of situations. When a normal model is not a good ﬁt, we will use alternate distributions that better characterize the sampling distribution. 7Using row ∞ on the t-table, we see that the value that corresponds to an 80% conﬁdence level is 1.282. Therefore, we should use 1.282 as the z value. 268 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.2.4 Margin of error The conﬁdence intervals we have encountered thus far have taken the form point estimate ± z∗ × SE of estimate Conﬁdence intervals are also often reported as point estimate ± margin of error For example, instead of reporting an interval as 0.09 ± 1.645 × 0.028 or (0.044, 0.136), it could be reported as 0.09 ± 0.046. The margin of error equals the critical value times the standard error of the estimate. The researcher chooses a conﬁdence level, and then calculates the margin of error that corresponds to that conﬁdence level. The margin of error for a 95% conﬁdence interval tells us that we are 95% conﬁdent that the sample statistic is within that margin of error of the true value. Numerically, the margin of error corresponds to the distance between the point estimate and the lower or upper bound of a conﬁdence interval, and thus is half of the total width of the interval. MARGIN OF ERROR When the point estimate follows a normal distribution, margin of error = z × SE of estimate. EXAMPLE 5.19 All other things being equal, will the margin of error be bigger for a 68% conﬁdence interval or a 95% conﬁdence interval? A 95% conﬁdence interval is wider than a 68% conﬁdence interval and has a larger z value, so the 95% conﬁdence interval will have a larger margin of error. In order to be more conﬁdent of capturing the true value, we must create a wider interval. GUIDED PRACTICE 5.20 What is the margin of error for the conﬁdence interval: (0.035, 0.145)?8 8The margin of error is half of the total width of the interval. The margin of error for this interval is 0.145−0.035 = 2 0.055. 5.2. CONFIDENCE INTERVALS 269 5.2.5 Interpreting conﬁdence intervals A careful eye might have observed the somewhat awkward language used to describe conﬁdence intervals. Correct interpretation: We are C% conﬁdent that the population parameter is between and . Incorrect language might try to describe the conﬁdence interval as capturing the population parameter with a certain probability.9 Applying the language of probability to a ﬁxed interval or to a ﬁxed parameter is one of the most common errors. As we saw in Figure 5.4, the 95% conﬁdence interval method has a 95% probability of producing an interval that will capture the population parameter. A correct interpretation of the conﬁdence level is that such intervals will capture the population parameter that percent of the time (assuming conditions are met and the probability model is true). However, each individual interval either does or does not capture the population parameter. A correct interpretation of an individual conﬁdence interval cannot involve the vocabulary of probability. Another especially important consideration of conﬁdence intervals is that they only try to capture the population parameter. Our intervals say nothing about the conﬁdence of capturing individual observations, a proportion of the observations, or point estimates. Conﬁdence intervals only attempt to capture population parameters. 5.2.6 Conﬁdence interval procedures: a ﬁve step process Use a conﬁdence interval to estimate a parameter with a particular conﬁdence level, C. (AP EXAM TIP) WHEN CARRYING OUT A CONFIDENCE INTERVAL PROCEDURE, FOLLOW THESE FIVE STEPS: • Identify: Identify the parameter and the conﬁdence level. • Choose: Choose the appropriate interval procedure and identify it by name. • Check: Check that the conditions for the interval procedure are met. • Calculate: Calculate the conﬁdence interval and record it in interval form. CI: point estimate ± critical value × SE of estimate • Conclude: Interpret the interval and, if applicable, draw a conclusion based on whether the interval is entirely above, is entirely below, or contains the value of interest. 9To see that this interpretation is incorrect, imagine taking two random samples and constructing two 95% conﬁdence intervals for an unknown proportion. If these intervals are disjoint, can we say that there is a 95%+95%=190% chance that the ﬁrst or the second interval captures the true value? 270 CHAPTER 5. FOUNDATIONS FOR INFERENCE Section summary • A point estimate is not perfect; there is almost always some error in the estimate. It is often useful to supply a plausible range of values for the parameter, which we call a conﬁdence interval. • A conﬁdence interval is centered on the point estimate and extends a certain number of standard errors on either side of the estimate, depending upon how conﬁdent one wants to be. For a ﬁxed sample size, to be more conﬁdent of capturing the true value requires a wider interval. • When the sampling distribution for a point estimate can reasonably be modeled as normal, such as with a sample proportion, then the following are true: – A 95% conﬁdence interval is given by: point estimate ± 1.96 × SE of estimate. We can be 95% conﬁdent this interval captures the true value. – A C% conﬁdence interval is given by: point estimate ± z × SE of estimate. We can be C% conﬁdent this interval captures the true value. • For a C% conﬁdence interval described above, we select z such that the area between -z and z under the standard normal curve is C%. Use a t-table at row ∞ to ﬁnd the critical value z. 10 • After interpreting the interval, we can usually draw a conclusion, with C% conﬁdence, about whether a given value X is a reasonable value for the population parameter. When drawing a conclusion based on a conﬁdence interval, there are three possibilities. – We have evidence that the true [parameter]: ...is greater than X, because the entire interval is above X. ...is less than X, because the entire interval is below X. – We do not have evidence that the true [parameter] is not X, because X is in the interval. Interpreting conﬁdence intervals and conﬁdence levels • 68% and 95% are examples of conﬁdence levels. The conﬁdence level tells us the capture rate with repeated sampling. For example, a correct interpretation of a 95% conﬁdence level is that if many samples of the same size were taken from the population, about 95% of the resulting conﬁdence intervals would capture the true population parameter (assuming the conditions are met and the probability model is true). Note that this is a relative frequency interpretation. • We cannot use the language of probability to interpret an individual conﬁdence interval, once it has been calculated. The conﬁdence level tells us what percent of the intervals will capture the population parameter, not the probability that a calculated interval captures the population parameter. Each calculated interval either does or does not capture the population parameter. Margin of error • Conﬁdence intervals are often reported as: point estimate ± margin of error. The margin of error (M E) = critical value × SE of estimate, and it tells us,
with a particular conﬁdence, how much we expect our point estimate to deviate from the true population value due to chance. • The margin of error depends on the conﬁdence level ; the standard error does not. Other things being constant, a higher conﬁdence level leads to a larger margin of error. • For a ﬁxed conﬁdence level, increasing the sample size decreases the margin of error. This assumes a random sample. • The margin of error formula only applies if a sample is random. Moreover, the margin of error measures only sampling error ; it does not account for additional error introduced by response bias and non-response bias. Even with a perfectly random sample, the actual error in a poll is likely higher than the reported margin of error.11 10We explain the relationship between z and t in Chapter 7 11nytimes.com/2016/10/06/upshot/when-you-hear-the-margin-of-error-is-plus-or-minus-3-percent-think-7- instead.html 5.2. CONFIDENCE INTERVALS 271 Exercises 5.7 Chronic illness, Part I. In 2013, the Pew Research Foundation reported that “45% of U.S. adults report that they live with one or more chronic conditions”.12 However, this value was based on a sample, so it may not be a perfect estimate for the population parameter of interest on its own. The study reported a standard error of about 1.2%, and a normal model may reasonably be used in this setting. Create a 95% conﬁdence interval for the proportion of U.S. adults who live with one or more chronic conditions. Also interpret the conﬁdence interval in the context of the study. 5.8 Twitter users and news, Part I. A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter.13. The standard error for this estimate was 2.4%, and a normal distribution may be used to model the sample proportion. Construct a 99% conﬁdence interval for the fraction of U.S. adult Twitter users who get some news on Twitter, and interpret the conﬁdence interval in context. 5.9 Waiting at an ER, Part I. A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a simple random sample of 64 patients and determines the time (in minutes) between when they checked in to the ER until they were ﬁrst seen by a doctor. A 95% conﬁdence interval based on this sample is (128 minutes, 147 minutes), which is based on the normal model for the mean. Determine whether the following statements are true or false, and explain your reasoning. (a) We are 95% conﬁdent that the average waiting time of these 64 emergency room patients is between 128 and 147 minutes. (b) We are 95% conﬁdent that the average waiting time of all patients at this hospital’s emergency room is between 128 and 147 minutes. (c) 95% of random samples have a sample mean between 128 and 147 minutes. (d) A 99% conﬁdence interval would be narrower than the 95% conﬁdence interval since we need to be more sure of our estimate. (e) The margin of error is 9.5 and the sample mean is 137.5. (f) In order to decrease the margin of error of a 95% conﬁdence interval to half of what it is now, we would need to double the sample size. 5.10 Mental health. The General Social Survey asked the question: “For how many days during the past 30 days was your mental health, which includes stress, depression, and problems with emotions, not good?” Based on responses from 1,151 US residents, the survey reported a 95% conﬁdence interval of 3.40 to 4.24 days in 2010. (a) Interpret this interval in context of the data. (b) What does “95% conﬁdent” mean? Explain in the context of the application. (c) Suppose the researchers think a 99% conﬁdence level would be more appropriate for this interval. Will this new interval be smaller or wider than the 95% conﬁdence interval? (d) If a new survey were to be done with 500 Americans, do you think the standard error of the estimate be larger, smaller, or about the same. 12Pew Research Center, Washington, D.C. The Diagnosis Diﬀerence, November 26, 2013. 13Pew Research Center, Washington, D.C. Twitter News Consumers: Young, Mobile and Educated, November 4, 2013. 272 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.11 Cyberbullying rates. Teens were surveyed about cyberbullying, and 54% to 64% reported experiencing cyberbullying (95% conﬁdence interval).14 Answer the following questions based on this interval. (a) A newspaper claims that a majority of teens have experienced cyberbullying. Is this claim supported by the conﬁdence interval? Explain your reasoning. (b) A researcher conjectured that 70% of teens have experienced cyberbullying. Is this claim supported by the conﬁdence interval? Explain your reasoning. (c) Without actually calculating the interval, determine if the claim of the researcher from part (b) would be supported based on a 90% conﬁdence interval? 5.12 Waiting at an ER, Part II. Exercise 5.9 provides a 95% conﬁdence interval for the mean waiting time at an emergency room (ER) of (128 minutes, 147 minutes). Answer the following questions based on this interval. (a) A local newspaper claims that the average waiting time at this ER exceeds 3 hours. Is this claim supported by the conﬁdence interval? Explain your reasoning. (b) The Dean of Medicine at this hospital claims the average wait time is 2.2 hours. Is this claim supported by the conﬁdence interval? Explain your reasoning. (c) Without actually calculating the interval, determine if the claim of the Dean from part (b) would be supported based on a 99% conﬁdence interval? 14Pew Research Center, A Majority of Teens Have Experienced Some Form of Cyberbullying. September 27, 2018. 5.3. INTRODUCING HYPOTHESIS TESTING 273 5.3 Introducing hypothesis testing In an experiment, one treatment reduces cholesterol by 10% while another treatment reduces it by 17%. Is this strong enough evidence that the second treatment is more eﬀective? In this section, we will set up a framework for answering questions such as this and will look at the diﬀerent types of decision errors that researcher can make when drawing conclusions based on data. Learning objectives 1. Explain the logic of hypothesis testing, including setting up hypotheses and drawing a conclu- sion based on the set signiﬁcance level and the calculated p-value. 2. Set up the null and alternative hypothesis in words and in terms of population parameters. 3. Interpret a p-value in context and recognize how the calculation of the p-value depends upon the direction of the alternative hypothesis. 4. Deﬁne and interpret the concept statistically signiﬁcant. 5. Interpret Type I, Type II Error, and power in the context of hypothesis testing. 6. Distinguish between statistically signiﬁcant and practically signiﬁcant, and recognize the role that sample size plays here. 7. Understand the two general conditions for when the conﬁdence interval and hypothesis testing procedures apply and explain why these conditions are necessary. 5.3.1 Case study: medical consultant People providing an organ for donation sometimes seek the help of a special medical consultant. These consultants assist the patient in all aspects of the surgery, with the goal of reducing the possibility of complications during the medical procedure and recovery. Patients might choose a consultant based in part on the historical complication rate of the consultant’s clients. One consultant tried to attract patients by noting the overall complication rate for liver donor surgeries in the US is about 10%, but her clients have had only 9 complications in the 142 liver donor surgeries she has facilitated. She claims this is strong evidence that her work meaningfully contributes to reducing complications (and therefore she should be hired!). EXAMPLE 5.21 We will let p represent the true complication rate for liver donors working with this consultant. Calculate the best estimate for p using the data. Label the point estimate as ˆp. The sample proportion for the complication rate is 9 complications divided by the 142 surgeries the consultant has worked on: ˆp = 9/142 = 0.063. EXAMPLE 5.22 Is it possible to prove that the consultant’s work reduces complications? No. The claim implies that there is a causal connection, but the data are observational. For example, maybe patients who can aﬀord a medical consultant can aﬀord better medical care, which can also lead to a lower complication rate. 274 CHAPTER 5. FOUNDATIONS FOR INFERENCE EXAMPLE 5.23 While it is not possible to assess the causal claim, it is still possible to ask whether the low complication rate of ˆp = 0.063 provides evidence that the consultant’s true complication rate is diﬀerent than the US complication rate. Why might we be tempted to immediately conclude that the consultant’s true complication rate is diﬀerent than the US complication rate? Can we draw this conclusion? Her sample complication rate is ˆp = 0.063, which is 0.037 lower than the US complication rate of 10%. However, we cannot yet be sure if the observed diﬀerence represents a real diﬀerence or is just the result of random variation. We wouldn’t expect the sample proportion to be exactly 0.10, even if the truth was that her real complication rate was 0.10. 5.3.2 Setting up the null and alternative hypothesis We can set up two competing hypotheses about the consultant’s true complication rate. The ﬁrst is call the null hypothesis and represents either a skeptical perspective or a perspective of no diﬀerence. The second is called the alternative hypothesis (or alternative hypothesis) and represents a new perspective such as the possibility that there has been a change or that there is a treatment eﬀect in an experiment. NULL AND ALTERNATIVE HYPOTHESES The null hypothesis is abbreviated H0. It represents a skeptical perspective and is often a claim of no change or no diﬀerence. The alternative hypothesis is abbreviated HA. It is the claim researchers hope to prove or ﬁnd evidence for, and it often asserts that there has been a change or
an eﬀect. Our job as data scientists is to play the skeptic: before we buy into the alternative hypothesis, we need to see strong supporting evidence. EXAMPLE 5.24 Identify the null and alternative claim regarding the consultant’s complication rate. H0: The true complication rate for the consultant’s clients is the same as the US complication rate of 10%. HA: The true complication rate for the consultant’s clients is diﬀerent than 10%. Often it is convenient to write the null and alternative hypothesis in mathematical or numerical terms. To do so, we must ﬁrst identify the quantity of interest. This quantity of interest is known as the parameter for a hypothesis test. PARAMETERS AND POINT ESTIMATES A parameter for a hypothesis test is the “true” value of the population of interest. When the parameter is a proportion, we call it p. A point estimate is calculated from a sample. When the point estimate is a proportion, we call it ˆp. The observed or sample proportion of 0.063 is a point estimate for the true proportion. The parameter in this problem is the true proportion of complications for this consultant’s clients. The parameter is unknown, but the null hypothesis is that it equals the overall proportion of complications: p = 0.10. This hypothesized value is called the null value. 5.3. INTRODUCING HYPOTHESIS TESTING 275 NULL VALUE OF A HYPOTHESIS TEST The null value is the value hypothesized for the parameter in H0, and it is sometimes represented with a subscript 0, e.g. p0 (just like H0). In the medical consultant case study, the parameter is p and the null value is p0 = 0.10. We can write the null and alternative hypothesis as numerical statements as follows. • H0: p = 0.10 (The complication rate for the consultant’s clients is equal to the US complication rate of 10%.) • HA: p = 0.10 (The complication rate for the consultant’s clients is not equal to the US complication rate of 10%.) HYPOTHESIS TESTING These hypotheses are part of what is called a hypothesis test. A hypothesis test is a statistical technique used to evaluate competing claims using data. Often times, the null hypothesis takes a stance of no diﬀerence or no eﬀect. If the null hypothesis and the data notably disagree, then we will reject the null hypothesis in favor of the alternative hypothesis. Don’t worry if you aren’t a master of hypothesis testing at the end of this section. We’ll discuss these ideas and details many times in this chapter and the two chapters that follow. The null claim is always framed as an equality: it tells us what quantity we should use for the parameter when carrying out calculations for the hypothesis test. There are three choices for the alternative hypothesis, depending upon whether the researcher is trying to prove that the value of the parameter is greater than, less than, or not equal to the null value. ALWAYS WRITE THE NULL HYPOTHESIS AS AN EQUALITY We will ﬁnd it most useful if we always list the null hypothesis as an equality (e.g. p = 7) while the alternative always uses an inequality (e.g. p = 0.7, p > 0.7, or p < 0.7). GUIDED PRACTICE 5.25 According to the 2010 US Census, 7.6% of residents in the state of Alaska were under 5 years old. A researcher plans to take a random sample of residents from Alaska to test whether or not this is still the case. Write out the hypotheses that the researcher should test in both plain and statistical language.15 When the alternative claim uses a =, we call the test a two-sided test, because either extreme provides evidence against H0. When the alternative claim uses a < or a >, we call it a one-sided test. ONE-SIDED AND TWO-SIDED TESTS If the researchers are only interested in showing an increase or a decrease, but not both, use a one-sided test. If the researchers would be interested in any diﬀerence from the null value – an increase or decrease – then the test should be two-sided. 15H0: p = 0.076; The proportion of residents under 5 years old in Alaska is unchanged from 2010. HA: p = 0.076; The proportion of residents under 5 years old in Alaska has changed from 2010. Note that it could have increased or decreased. 276 CHAPTER 5. FOUNDATIONS FOR INFERENCE EXAMPLE 5.26 For the example of the consultant’s complication rate, we knew that her sample complication rate was 0.063, which was lower than the US complication rate of 0.10. Why did we conduct a two-sided hypothesis test for this setting? The setting was framed in the context of the consultant being helpful, but what if the consultant actually performed worse than the US complication rate? Would we care? More than ever! Since we care about a ﬁnding in either direction, we should run a two-sided test. ONE-SIDED HYPOTHESES ARE ALLOWED ONLY BEFORE SEEING DATA After observing data, it is tempting to turn a two-sided test into a one-sided test. Avoid this temptation. Hypotheses must be set up before observing the data. If they are not, the test must be two-sided. 5.3.3 Evaluating the hypotheses with a p-value EXAMPLE 5.27 There were 142 patients in the consultant’s sample. If the null claim is true, how many would we expect to have had a complication? If the null claim is true, we would expect about 10% of the patients, or about 14.2 to have a complication. The consultant’s complication rate for her 142 clients was 0.063 (0.063 × 142 ≈ 9). What is the probability that a sample would produce a number of complications this far from the expected value of 14.2, if her true complication rate were 0.10, that is, if H0 were true? The probability, which is estimated in Section 5.7 on page 276, is about 0.1754. We call this quantity the p-value. Figure 5.7: The shaded area represents the p-value. We observed ˆp = 0.063, so any observations smaller than this are at least as extreme relative to the null value, p0 = 0.1, and so the lower tail is shaded. However, since this is a two-sided test, values above 0.137 are also at least as extreme as 0.063 (relative to 0.1), and so they also contribute to the p-value. The tail areas together total of about 0.1754 when calculated using a simulation technique in Section 5.3.4. Sample Proportions Under the Null Hypothesis0.020.050.070.10.130.150.18 5.3. INTRODUCING HYPOTHESIS TESTING 277 Figure 5.8: When the alternative hypothesis takes the form p < null value, the p-value is represented by the lower tail. When it takes the form p > null value, the p-value is represented by the upper tail. When using p = null value, then the p-value is represented by both tails. FINDING AND INTERPRETING THE P-VALUE We ﬁnd and interpret the p-value according to the nature of the alternative hypothesis. HA: parameter > null value. The p-value corresponds to the area in the upper tail and is probability of getting a test statistic larger than the observed test statistic if the null hypothesis is true and the probability model is accurate. HA: parameter < null value. The p-value corresponds to the area in the lower tail and is the probability of observing a test statistic smaller than the observed test statistic if the null hypothesis is true and the probability model is accurate. HA: parameter = null value. The p-value corresponds to the area in both tails and is the probability of observing a test statistic larger in magnitude than the observed test statistic if the null hypothesis is true and the probability model is accurate. More generally, we can say that the p-value is the probability of getting a test statistic as extreme or more extreme than the observed test statistic in the direction of HA if the null hypothesis is true and the probability model is accurate. When working with proportions, we can also say that the p-value is the probability of getting a sample proportion as far from or farther from the null proportion in the direction of HA if the null hypothesis is true and the normal model holds. When the p-value is small, i.e. less than a previously set threshold, we say the results are statistically signiﬁcant. This means the data provide such strong evidence against H0 that we reject the null hypothesis in favor of the alternative hypothesis. The threshold is called the signiﬁcance level and is represented by α (the Greek letter alpha). The signiﬁcance level is typically set to α = 0.05, but can vary depending on the ﬁeld or the application. STATISTICAL SIGNIFICANCE If the p-value is less than the signiﬁcance level α (usually 0.05), we say that the result is statistically signiﬁcant. We reject H0, and we have strong evidence favoring HA. If the p-value is greater than the signiﬁcance level α, we say that the result is not statistically signiﬁcant. We do not reject H0, and we do not have strong evidence for HA. Recall that the null claim is the claim of no diﬀerence. If we reject H0, we are asserting that there is a real diﬀerence. If we do not reject H0, we are saying that the null claim is reasonable, but we are not saying that the null claim has been proven. GUIDED PRACTICE 5.28 Because the p-value is 0.1754, which is larger than the signiﬁcance level 0.05, we do not reject the null hypothesis. Explain what this means in the context of the problem using plain language.16 16The data do not provide evidence that the consultant’s complication rate is signiﬁcantly lower or higher than the US complication rate of 10%. HA: p<null valueHA: p>null valueHA: p„null value 278 CHAPTER 5. FOUNDATIONS FOR INFERENCE EXAMPLE 5.29 In the previous exercise, we did not reject H0. This means that we did not disprove the null claim. Is this equivalent to proving the null claim is true? No. We did not prove that the consultant’s complication rate is exactly equal to 10%. Recall that the test of hypothesis starts by assuming the null claim is true. That is, the test proceeds as an argument by contradiction. If the null claim is true, there is a 0.1754 chance of seeing sample data as divergent from 10% as we saw in our sample. Because 0.1754 is large, it is within the realm of chance error, and we cannot say the null
hypothesis is unreasonable.17 DOUBLE NEGATIVES CAN SOMETIMES BE USED IN STATISTICS In many statistical explanations, we use double negatives. For instance, we might say that the null hypothesis is not implausible or we failed to reject the null hypothesis. Double negatives are used to communicate that while we are not rejecting a position, we are also not saying that we know it to be true. EXAMPLE 5.30 Does the conclusion in Guided Practice 5.28 ensure that there is no real association between the surgical consultant’s work and the risk of complications? Explain. No. It is possible that the consultant’s work is associated with a lower or higher risk of complications. If this was the case, the sample may have been too small to reliable detect this eﬀect. EXAMPLE 5.31 An experiment was conducted where study participants were randomly divided into two groups. Both were given the opportunity to purchase a DVD, but one half was reminded that the money, if not spent on the DVD, could be used for other purchases in the future, while the other half was not. The half that was reminded that the money could be used on other purchases was 20% less likely to continue with a DVD purchase. We determined that such a large diﬀerence would only occur about 1-in-150 times if the reminder actually had no inﬂuence on student decision-making. What is the p-value in this study? Was the result statistically signiﬁcant? The p-value was 0.006 (about 1/150). Since the p-value is less than 0.05, the data provide statistically signiﬁcant evidence that US college students were actually inﬂuenced by the reminder. WHAT’S SO SPECIAL ABOUT 0.05? We often use a threshold of 0.05 to determine whether a result is statistically signiﬁcant. But why 0.05? Maybe we should use a bigger number, or maybe a smaller number. If you’re a little puzzled, that probably means you’re reading with a critical eye – good job! We’ve made a video to help clarify why 0.05 : www.openintro.org/why05 Sometimes it’s a good idea to deviate from the standard. We’ll discuss when to choose a threshold diﬀerent than 0.05 in Section 5.3.7. Statistical inference is the practice of making decisions and conclusions from data in the context of uncertainty. Just as a conﬁdence interval may occasionally fail to capture the true value of the parameter, a test of hypothesis may occasionally lead us to an incorrect conclusion. While a given data set may not always lead us to a correct conclusion, statistical inference gives us tools to control and evaluate how often these errors occur. 17The p-value is a conditional probability. It is P(getting data at least as divergent from the null value as we observed \| H0 is true). It is NOT P( H0 is true \| we got data this divergent from the null value). 5.3. INTRODUCING HYPOTHESIS TESTING 279 5.3.4 Calculating the p-value by simulation (special topic) When conditions for the applying a normal model are met, we use a normal model to ﬁnd the p-value of a test of hypothesis. In the complication rate example, the distribution is not normal. It is, however, binomial, because we are interested in how many out of 142 patients will have complications. We could calculate the p-value of this test using binomial probabilities. A more general approach, though, for calculating p-values when a normal model does not apply is to use what is known as simulation. While performing this procedure is outside of the scope of the course, we provide an example here in order to better understand the concept of a p-value. We simulate 142 new patients to see what result might happen if the complication rate really is 0.10. To do this, we could use a deck of cards. Take one red card, nine black cards, and mix them up. If the cards are well-shuﬄed, drawing the top card is one way of simulating the chance if the card is red, we say the patient had a a patient has a complication if the true rate is 0.10: If we repeat this complication, and if it is black then we say they did not have a complication. process 142 times and compute the proportion of simulated patients with complications, ˆpsim, then this simulated proportion is exactly a draw from the null distribution. There were 12 simulated cases with a complication and 130 simulated cases without a compli- cation: ˆpsim = 12/142 = 0.085. One simulation isn’t enough to get a sense of the null distribution, so we repeated the simulation 10,000 times using a computer. Figure 5.9 shows the null distribution from these 10,000 simulations. The simulated proportions that are less than or equal to ˆp = 0.063 are shaded. There were 0.0877 simulated sample proportions with ˆpsim ≤ 0.063, which represents a fraction 0.0877 of our simulations: left tail = Number of observed simulations with ˆpsim ≤ 0.063 10000 = 877 10000 = 0.0877 However, this is not our p-value! Remember that we are conducting a two-sided test, so we should double the one-tail area to get the p-value:18 p-value = 2 × left tail = 2 × 0.0877 = 0.1754 Figure 5.9: The null distribution for ˆp, created from 10,000 simulated studies. The left tail contains 8.77% of the simulations. For a two-sided test, we double the tail area to get the p-value. This doubling accounts for the observations we might have observed in the upper tail, which are also at least as extreme (relative to 0.10) as what we observed, ˆp = 0.063. 18This doubling approach is preferred even when the distribution isn’t symmetric, as in this case. p^sim 0.000.050.100.150.2002004006008001000Observations over hereare just as extreme,so they should also counttowards the p−valueNumber of Simulations 280 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.3.5 Hypothesis testing: a ﬁve step process Use a hypothesis test to test H0 versus HA at a particular signﬁcance level, α. (AP EXAM TIP) WHEN CARRYING OUT A HYPOTHESIS TEST PROCEDURE, FOLLOW THESE FIVE STEPS: • Identify: Identify the hypotheses and the signiﬁcance level. • Choose: Choose the appropriate test procedure and identify it by name. • Check: Check that the conditions for the test procedure are met. • Calculate: Calculate the test statistic and the p-value. test statistic = point estimate − null value SE of estimate • Conclude: Compare the p-value to the signiﬁcance level to determine whether to reject H0 or not reject H0. Draw a conclusion in the context of HA. 5.3.6 Decision errors The hypothesis testing framework is a very general tool, and we often use it without a second thought. If a person makes a somewhat unbelievable claim, we are initially skeptical. However, if there is suﬃcient evidence that supports the claim, we set aside our skepticism. The hallmarks of hypothesis testing are also found in the US court system. EXAMPLE 5.32 A US court considers two possible claims about a defendant: she is either innocent or guilty. If we set these claims up in a hypothesis framework, which would be the null hypothesis and which the alternative? The jury considers whether the evidence is so convincing (strong) that there is evidence beyond a reasonable doubt of the person’s guilt. That is, the starting assumption (null hypothesis) is that the person is innocent until evidence is presented that convinces the jury that the person is guilty (alternative hypothesis). In statistics, our evidence comes in the form of data, and we use the signiﬁcance level to decide what is beyond a reasonable doubt. Jurors examine the evidence to see whether it convincingly shows a defendant is guilty. Notice that a jury ﬁnds a defendant either guilty or not guilty. They either reject the null claim or they do not reject the null claim. They never prove the null claim, that is, they never ﬁnd the defendant innocent. If a jury ﬁnds a defendant not guilty, this does not necessarily mean the jury is conﬁdent in the person’s innocence. They are simply not convinced of the alternative that the person is guilty. This is also the case with hypothesis testing: even if we fail to reject the null hypothesis, we typically do not accept the null hypothesis as truth. Failing to ﬁnd strong evidence for the alternative hypothesis is not equivalent to providing evidence that the null hypothesis is true. Hypothesis tests are not ﬂawless. Just think of the court system: innocent people are sometimes wrongly convicted and the guilty sometimes walk free. Similarly, data can point to the wrong conclusion. However, what distinguishes statistical hypothesis tests from a court system is that our framework allows us to quantify and control how often the data lead us to the incorrect conclusion. There are two competing hypotheses: the null and the alternative. In a hypothesis test, we make a statement about which one might be true, but we might choose incorrectly. There are four possible scenarios in a hypothesis test, which are summarized in Figure 5.10. 5.3. INTRODUCING HYPOTHESIS TESTING 281 Test conclusion Truth H0 true HA true do not reject H0 correct conclusion reject H0 in favor of HA Type I Error Type II Error correct conclusion Figure 5.10: Four diﬀerent scenarios for hypothesis tests. TYPE I AND TYPE II ERRORS A Type I Error is rejecting H0 when H0 is actually true. When we reject the null hypothesis, it is possible that we make a Type I Error. A Type II Error is failing to reject H0 when HA is actually true. When we dod not reject the null hypothesis, it is possible that we make a Type II Error. EXAMPLE 5.33 In a US court, the defendant is either innocent (H0) or guilty (HA). What does a Type I Error represent in this context? What does a Type II Error represent? Figure 5.10 may be useful. If the court makes a Type I Error, this means the defendant is innocent (H0 true) but wrongly convicted. A Type II Error means the court failed to reject H0 (i.e. failed to convict the person) when they were in fact guilty (HA true). EXAMPLE 5.34 How could we reduce the Type I Error rate in US courts? What inﬂuence would this have on the Type II Error rate? To lower the Type I Error rate, we might raise our standard fo
r conviction from “beyond a reasonable doubt” to “beyond a conceivable doubt” so fewer people would be wrongly convicted. However, this would also make it more diﬃcult to convict the people who are actually guilty, so we would make more Type II Errors. GUIDED PRACTICE 5.35 How could we reduce the Type II Error rate in US courts? What inﬂuence would this have on the Type I Error rate?19 GUIDED PRACTICE 5.36 A group of women bring a class action lawsuit that claims discrimination in promotion rates. What would a Type I Error represent in this context?20 These examples provide an important lesson: if we reduce how often we make one type of error, we generally make more of the other type. 19To lower the Type II Error rate, we want to convict more guilty people. We could lower the standards for conviction from “beyond a reasonable doubt” to “beyond a little doubt”. Lowering the bar for guilt will also result in more wrongful convictions, raising the Type I Error rate. 20We must ﬁrst identify which is the null hypothesis and which is the alternative. The alternative hypothesis is the one that bears the burden of proof, so the null hypothesis is that there was no discrimination and the alternative hypothesis is that there was discrimination. Making a Type I Error in this context would mean that in fact there was no discrimination, even though we concluded that women were discriminated against. Notice that this does not necessarily mean something was wrong with the data or that we made a computational mistake. Sometimes data simply point us to the wrong conclusion, which is why scientiﬁc studies are often repeated to check initial ﬁndings. 282 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.3.7 Choosing a signiﬁcance level If H0 is true, what is the probability that we will incorrectly reject it? In hypothesis testing, we perform calculations under the premise that H0 is true, and we reject H0 if the p-value is smaller than the signiﬁcance level α. That is, α is the probability of making a Type I Error. The choice of what to make α is not arbitrary. It depends on the gravity of the consequences of a Type I Error. RELATIONSHIP BETWEEN TYPE I AND TYPE II ERRORS The probability of a Type I Error is called α and corresponds to the signiﬁcance level of a test. The probability of a Type II Error is called β. As we make α smaller, β typically gets larger, and vice versa. EXAMPLE 5.37 If making a Type I Error is especially dangerous or especially costly, should we choose a smaller signiﬁcance level or a higher signiﬁcance level? Under this scenario, we want to be very cautious about rejecting the null hypothesis, so we demand very strong evidence before we are willing to reject the null hypothesis. Therefore, we want a smaller signiﬁcance level, maybe α = 0.01. EXAMPLE 5.38 If making a Type II Error is especially dangerous or especially costly, should we choose a smaller signiﬁcance level or a higher signiﬁcance level? We should choose a higher signiﬁcance level (e.g. 0.10). Here we want to be cautious about failing to reject H0 when the null is actually false. SIGNIFICANCE LEVELS SHOULD REFLECT CONSEQUENCES OF ERRORS The signiﬁcance level selected for a test should reﬂect the real-world consequences associated with making a Type I or Type II Error. If a Type I Error is very dangerous, make α smaller. 5.3.8 Statistical power of a hypothesis test When the alternative hypothesis is true, the probability of not making a Type II Error is called power. It is common for researchers to perform a power analysis to ensure their study collects enough data to detect the eﬀects they anticipate ﬁnding. As you might imagine, if the eﬀect they care about is small or subtle, then if the eﬀect is real, the researchers will need to collect a large sample size in order to have a good chance of detecting the eﬀect. However, if they are interested in large eﬀect, they need not collect as much data. The Type II Error rate β and the magnitude of the error for a point estimate are controlled by the sample size. As the sample size n goes up, the Type II Error rate goes down, and power goes up. Real diﬀerences from the null value, even large ones, may be diﬃcult to detect with small samples. However, if we take a very large sample, we might ﬁnd a statistically signiﬁcant diﬀerence but the size of the diﬀerence might be so small that it is of no practical value. 5.3. INTRODUCING HYPOTHESIS TESTING 283 5.3.9 Statistical signiﬁcance versus practical signiﬁcance When the sample size becomes larger, point estimates become more precise and any real diﬀerences in the mean and null value become easier to detect and recognize. Even a very small diﬀerence would likely be detected if we took a large enough sample. Sometimes researchers will take such large samples that even the slightest diﬀerence is detected. While we still say that diﬀerence is statistically signiﬁcant, it might not be practically signiﬁcant. Statistically signiﬁcant diﬀerences are sometimes so minor that they are not practically relevant. This is especially important to research: if we conduct a study, we want to focus on ﬁnding a meaningful result. We don’t want to spend lots of money ﬁnding results that hold no practical value. The role of a data scientist in conducting a study often includes planning the size of the study. The data scientist might ﬁrst consult experts or scientiﬁc literature to learn what would be the smallest meaningful diﬀerence from the null value. She also would obtain some reasonable estimate for the standard deviation. With these important pieces of information, she would choose a suﬃciently large sample size so that the power for the meaningful diﬀerence is perhaps 80% or 90%. While larger sample sizes may still be used, she might advise against using them in some cases, especially in sensitive areas of research. 5.3.10 Statistical signiﬁcance versus a real difference When a result is statistically signiﬁcant at the α = 0.05 level, we have evidence that the result is real. However, when there is no diﬀerence or eﬀect, we can expect that 5% of the time the test conclusion will lead to a Type I Error and incorrectly reject the null hypothesis. Therefore we must beware of what is called p-hacking, in which researchers may test many, many hypotheses and then publish the ones that come out statistically signiﬁcant. As we noted, we can expect 5% of the results to be signiﬁcant when the null hypothesis is true and there really is no diﬀerence or eﬀect.21 5.3.11 When to retreat We must point out that statistical tools rely on conditions. When the conditions are not met, these tools are unreliable and drawing conclusions from them is treacherous. The conditions for these tools typically come in two forms. • The individual observations must be independent. A random sample from less than 10% of the population ensures the observations are independent. In experiments, we generally require that subjects are randomized into groups. If independence fails, then advanced techniques must be used, and in some such cases, inference may not be possible. • Other conditions focus on sample size and skew. For example, in Chapter 4 we looked at the success-failure condition and sample size condition for when ˆp and ¯x will follow a nearly normal distribution. Veriﬁcation of conditions for statistical tools is always necessary. When conditions are not satisﬁed for a given statistical technique, it is necessary to investigate new methods that are appropriate for the data. Finally, we caution that there may be no inference tools helpful when considering data that include unknown biases, such as convenience samples. For this reason, there are books, courses, and researchers devoted to the techniques of sampling and experimental design. See Sections 1.3-1.5 for basic principles of data collection. 21 The problem is even greater than p-hacking. In what has been called the “reproducibility crisis”, researchers have failed to reproduce a large proportion of results that were found signiﬁcant and were published in scientiﬁc journals. This problem highlights the importance of research that reproduces earlier work rather than taking the word of a single study. Also keep in mind that the probability that a diﬀerence will be found to be signiﬁcant given that there is no real diﬀerence is not the same as the probability that a diﬀerence is not real, given that it was found signiﬁcant. Depending upon the veracity of the hypotheses tested, the latter can be upwards of 80%, leading some to assert that “most published research is false”. https://www.economist.com/briefing/2013/10/18/trouble-at-the-lab 284 CHAPTER 5. FOUNDATIONS FOR INFERENCE Section summary • A hypothesis test is a statistical technique used to evaluate competing claims based on data. • The competing claims are called hypotheses and are often about population parameters (e.g. µ and p); they are never about sample statistics. – The null hypothesis is abbreviated H0. It represents a skeptical perspective or a per- spective of no diﬀerence or no change. – The alternative hypothesis is abbreviated HA. It represents a new perspective or a perspective of a real diﬀerence or change. Because the alternative hypothesis is the stronger claim, it bears the burden of proof. • The logic of a hypothesis test: In a hypothesis test, we begin by assuming that the null hypothesis is true. Then, we calculate how unlikely it would be to get a sample value as extreme as we actually got in our sample, assuming that the null value is correct. If this likelihood is too small, it casts doubt on the null hypothesis and provides evidence for the alternative hypothesis. • We set a signiﬁcance level, denoted α, which represents the threshold below which we will reject the null hypothesis. The most common signiﬁcance level is α = 0.05. If we require more evidence to reject the null hypothesis, we use a smaller α. • After verifying that the relevant conditions are met, we can calc
ulate the test statistic. The test statistic tells us how many standard errors the point estimate (sample value) is from the null value (i.e. the value hypothesized for the parameter in the null hypothesis). When investigating a single mean or proportion or a diﬀerence of means or proportions, the test statistic is calculated as: point estimate − null value . SE of estimate • After the test statistic, we calculate the p-value. We ﬁnd and interpret the p-value according to the nature of the alternative hypothesis. The three possibilities are: HA: parameter > null value. The p-value corresponds to the area in the upper tail. HA: parameter < null value. The p-value corresponds to the area in the lower tail. HA: parameter = null value. The p-value corresponds to the area in both tails. The p-value is the probability of getting a test statistic as extreme or more extreme than the observed test statistic in the direction of HA if the null hypothesis is true and the probability model is accurate. • The conclusion or decision of a hypothesis test is based on whether the p-value is smaller or larger than the preset signiﬁcance level α. – When the p-value < α, we say the results are statistically signiﬁcant at the α level and we have evidence of a real diﬀerence or change. The observed diﬀerence is beyond what would have been expected from chance variation alone. This leads us to reject H0 and gives us evidence for HA. – When the p-value > α, we say the results are not statistically signiﬁcant at the α level and we do not have evidence of a real diﬀerence or change. The observed diﬀerence was within the realm of expected chance variation. This leads us to not reject H0 and does not give us evidence for HA. • Decision errors. In a hypothesis test, there are two types of decision errors that could be made. These are called Type I and Type II Errors. – A Type I Error is rejecting H0, when H0 is actually true. We commit a Type I Error if we call a result signiﬁcant when there is no real diﬀerence or eﬀect. P(Type I Error) = α. – A Type II Error is not rejecting H0, when HA is actually true. We commit a Type II Error if we call a result not signiﬁcant when there is a real diﬀerence or eﬀect. P(Type II Error) = β. 5.3. INTRODUCING HYPOTHESIS TESTING 285 – The probability of a Type I Error (α) and a Type II Error (β) are inversely related. Decreasing α makes β larger; increasing α makes β smaller. – Once a decision is made, only one of the two types of errors is possible. If the test rejects H0, for example, only a Type I Error is possible. • The power of a test. – When a particular HA is true, the probability of not making a Type II Error is called power. Power = 1 − β. – The power of a test is the probability of detecting an eﬀect of a particular size when it is present. – Increasing the signiﬁcance level decreases the probability of a Type II Error and increases power. α ↑, β ↓, power ↑. – For a ﬁxed α, increasing the sample size n makes it easier to detect an eﬀect and therefore decreases the probability of a Type II Error and increases power. n ↑, β ↓, power ↑. • A small percent of the time (α), a signiﬁcant result will not be a real result. If many tests are run, a small percent of them will produce signiﬁcant results due to chance alone.22 • With a very large sample, a signiﬁcant result may point to a result that is real but not practically signiﬁcant. That is, the diﬀerence detected may be so small as to be unimportant or meaningless. • The inference procedures in this book all require two broad conditions to be met. The ﬁrst is that some type of random sampling or random assignment must be involved. The second condition focuses on sample size and skew to determine whether the point estimate follows the intended distribution. 22Similarly, if many conﬁdence intervals are constructed, a small percent (100 - C%) of them will fail to capture a true value due to chance alone. A value outside the conﬁdence interval is not an impossible value. 286 CHAPTER 5. FOUNDATIONS FOR INFERENCE Exercises 5.13 Identify hypotheses, Part I. Write the null and alternative hypotheses in words and then symbols for each of the following situations. (a) A tutoring company would like to understand if most students tend to improve their grades (or not) after they use their services. They sample 200 of the students who used their service in the past year and ask them if their grades have improved or declined from the previous year. (b) Employers at a ﬁrm are worried about the eﬀect of March Madness, a basketball championship held each spring in the US, on employee productivity. They estimate that on a regular business day employees spend on average 15 minutes of company time checking personal email, making personal phone calls, etc. They also collect data on how much company time employees spend on such non-business activities during March Madness. They want to determine if these data provide convincing evidence that employee productivity changed during March Madness. 5.14 Identify hypotheses, Part II. Write the null and alternative hypotheses in words and using symbols for each of the following situations. (a) Since 2008, chain restaurants in California have been required to display calorie counts of each menu item. Prior to menus displaying calorie counts, the average calorie intake of diners at a restaurant was 1100 calories. After calorie counts started to be displayed on menus, a nutritionist collected data on the number of calories consumed at this restaurant from a random sample of diners. Do these data provide convincing evidence of a diﬀerence in the average calorie intake of a diners at this restaurant? (b) The state of Wisconsin would like to understand the fraction of its adult residents that consumed alcohol in the last year, speciﬁcally if the rate is diﬀerent from the national rate of 70%. To help them answer this question, they conduct a random sample of 852 residents and ask them about their alcohol consumption. 5.15 Online communication. A study suggests that 60% of college student spend 10 or more hours per week communicating with others online. You believe that this is incorrect and decide to collect your own sample for a hypothesis test. You randomly sample 160 students from your dorm and ﬁnd that 70% spent 10 or more hours a week communicating with others online. A friend of yours, who oﬀers to help you with the hypothesis test, comes up with the following set of hypotheses. Indicate any errors you see. H0 : ˆp < 0.6 HA : ˆp > 0.7 5.16 Married at 25. A study suggests that the 25% of 25 year olds have gotten married. You believe that this is incorrect and decide to collect your own sample for a hypothesis test. From a random sample of 25 year olds in census data with size 776, you ﬁnd that 24% of them are married. A friend of yours oﬀers to help you with setting up the hypothesis test and comes up with the following hypotheses. Indicate any errors you see. H0 : ˆp = 0.24 HA : ˆp = 0.24 5.3. INTRODUCING HYPOTHESIS TESTING 287 5.17 Unemployment and relationship problems. A USA Today/Gallup poll asked a group of unemployed and underemployed Americans if they have had major problems in their relationships with their spouse or another close family member as a result of not having a job (if unemployed) or not having a full-time job (if underemployed). 27% of the 1,145 unemployed respondents and 25% of the 675 underemployed respondents said they had major problems in relationships as a result of their employment status. (a) What are the hypotheses for evaluating if the proportions of unemployed and underemployed people who had relationship problems were diﬀerent? (b) The p-value for this hypothesis test is approximately 0.35. Explain what this means in context of the hypothesis test and the data. 5.18 Which is higher? In each part below, there is a value of interest and two scenarios (I and II). For each part, report if the value of interest is larger under scenario I, scenario II, or whether the value is equal under the scenarios. (a) The standard error of ˆp when (I) n = 125 or (II) n = 500. (b) The margin of error of a conﬁdence interval when the conﬁdence level is (I) 90% or (II) 80%. (c) The p-value for a Z-statistic of 2.5 calculated based on a (I) sample with n = 500 or based on a (II) sample with n = 1000. (d) The probability of making a Type 2 Error when the alternative hypothesis is true and the signiﬁcance level is (I) 0.05 or (II) 0.10. 5.19 Testing for Fibromyalgia. A patient named Diana was diagnosed with Fibromyalgia, a long-term syndrome of body pain, and was prescribed anti-depressants. Being the skeptic that she is, Diana didn’t initially believe that anti-depressants would help her symptoms. However after a couple months of being on the medication she decides that the anti-depressants are working, because she feels like her symptoms are in fact getting better. (a) Write the hypotheses in words for Diana’s skeptical position when she started taking the anti-depressants. (b) What is a Type 1 Error in this context? (c) What is a Type 2 Error in this context? 288 CHAPTER 5. FOUNDATIONS FOR INFERENCE Chapter highlights Statistical inference is the practice of making decisions from data in the context of uncertainty. In this chapter, we introduced two frameworks for inference: conﬁdence intervals and hypothesis tests. • Conﬁdence intervals are used for estimating unknown population parameters by providing an interval of reasonable values for the unknown parameter with a certain level of conﬁdence. • Hypothesis tests are used to assess how reasonable a particular value is for an unknown pop- ulation parameter by providing degrees of evidence against that value. • The results of conﬁdence intervals and hypothesis tests are, generally speaking, consistent.23 That is: – Values that fall inside a 95% conﬁdence interval (implying they are reasonable) will not be rejected by a test at the 5% signiﬁcance level (implying they ar
e reasonable), and vice-versa. – Values that fall outside a 95% conﬁdence interval (implying they are not reasonable) will be rejected by a test at the 5% signiﬁcance level (implying they are not reasonable), and vice-versa. – When the conﬁdence level and the signiﬁcance level add up to 100%, the conclusions of the two procedures are consistent. • Many values fall inside of a conﬁdence interval and will not be rejected by a hypothesis test. “Not rejecting H0” is NOT equivalent to accepting H0. When we “do not reject H0”, we are asserting that the null value is reasonable, not that the parameter is exactly equal to the null value. • For a 95% conﬁdence interval, 95% is not the probability that the true value lies inside the conﬁdence interval (it either does or it doesn’t). Likewise, for a hypothesis test, α is not the probability that H0 is true (it either is or it isn’t). In both frameworks, the probability is about what would happen in a random sample, not about what is true of the population. • The conﬁdence interval procedures and hypothesis tests described in this book should not be applied unless particular conditions (described in more detail in the following chapters) are met. If these procedures are applied when the conditions are not met, the results may be unreliable and misleading. While a given data set may not always lead us to a correct conclusion, statistical inference gives us tools to control and evaluate how often errors occur. 23In the context of proportions there will be a small range of cases where this is not true. This is because when working with proportions, the SE used for conﬁdence intervals and the SE used for tests are slightly diﬀerent, as we will see in the next chapter. 5.3. INTRODUCING HYPOTHESIS TESTING 289 Chapter exercises 5.20 Twitter users and news, Part II. A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter, and the standard error for this estimate was 2.4%. Identify each of the following statements as true or false. Provide an explanation to justify each of your answers. (a) The data provide statistically signiﬁcant evidence that more than half of U.S. adult Twitter users get some news through Twitter. Use a signiﬁcance level of α = 0.01. (b) Since the standard error is 2.4%, we can conclude that 97.6% of all U.S. adult Twitter users were included in the study. (c) If we want to reduce the standard error of the estimate, we should collect less data. (d) If we construct a 90% conﬁdence interval for the percentage of U.S. adults Twitter users who get some news through Twitter, this conﬁdence interval will be wider than a corresponding 99% conﬁdence interval. 5.21 Chronic illness, Part II. In 2013, the Pew Research Foundation reported that “45% of U.S. adults report that they live with one or more chronic conditions”, and the standard error for this estimate is 1.2%. Identify each of the following statements as true or false. Provide an explanation to justify each of your answers. (a) We can say with certainty that the conﬁdence interval from Exercise 5.7 contains the true percentage of U.S. adults who suﬀer from a chronic illness. (b) If we repeated this study 1,000 times and constructed a 95% conﬁdence interval for each study, then approximately 950 of those conﬁdence intervals would contain the true fraction of U.S. adults who suﬀer from chronic illnesses. (c) The poll provides statistically signiﬁcant evidence (at the α = 0.05 level) that the percentage of U.S. adults who suﬀer from chronic illnesses is below 50%. (d) Since the standard error is 1.2%, only 1.2% of people in the study communicated uncertainty about their answer. 5.22 Relaxing after work. The General Social Survey asked the question: “After an average work day, about how many hours do you have to relax or pursue activities that you enjoy?” to a random sample of 1,155 Americans.24 A 95% conﬁdence interval for the mean number of hours spent relaxing or pursuing activities they enjoy was (1.38, 1.92). (a) Interpret this interval in context of the data. (b) Suppose another set of researchers reported a conﬁdence interval with a larger margin of error based on the same sample of 1,155 Americans. How does their conﬁdence level compare to the conﬁdence level of the interval stated above? (c) Suppose next year a new survey asking the same question is conducted, and this time the sample size is 2,500. Assuming that the population characteristics, with respect to how much time people spend relaxing after work, have not changed much within a year. How will the margin of error of the 95% conﬁdence interval constructed based on data from the new survey compare to the margin of error of the interval stated above? 24National Opinion Research Center, General Social Survey, 2018. 290 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.23 Testing for food safety. A food safety inspector is called upon to investigate a restaurant with a few customer reports of poor sanitation practices. The food safety inspector uses a hypothesis testing framework to evaluate whether regulations are not being met. If he decides the restaurant is in gross violation, its license to serve food will be revoked. (a) Write the hypotheses in words. (b) What is a Type 1 Error in this context? (c) What is a Type 2 Error in this context? (d) Which error is more problematic for the restaurant owner? Why? (e) Which error is more problematic for the diners? Why? (f) As a diner, would you prefer that the food safety inspector requires strong evidence or very strong evidence of health concerns before revoking a restaurant’s license? Explain your reasoning. 5.24 True or false. Determine if the following statements are true or false, and explain your reasoning. If false, state how it could be corrected. (a) If a given value (for example, the null hypothesized value of a parameter) is within a 95% conﬁdence interval, it will also be within a 99% conﬁdence interval. (b) Decreasing the signiﬁcance level (α) will increase the probability of making a Type 1 Error. (c) Suppose the null hypothesis is p = 0.5 and we fail to reject H0. Under this scenario, the true population proportion is 0.5. (d) With large sample sizes, even small diﬀerences between the null value and the observed point estimate, a diﬀerence often called the eﬀect size, will be identiﬁed as statistically signiﬁcant. 5.25 Practical vs. statistical signiﬁcance. Determine whether the following statement is true or false, and explain your reasoning: “With large sample sizes, even small diﬀerences between the null value and the observed point estimate can be statistically signiﬁcant.” 5.26 Same observation, different sample size. Suppose you conduct a hypothesis test based on a sample where the sample size is n = 50, and arrive at a p-value of 0.08. You then refer back to your notes and discover that you made a careless mistake, the sample size should have been n = 500. Will your p-value increase, decrease, or stay the same? Explain. 291 Chapter 6 Inference for categorical data 6.1 Inference for a single proportion 6.2 Inference for the difference of two proportions 6.3 Testing for goodness of ﬁt using chi-square 6.4 Homogeneity and independence in two-way tables 292 In this chapter, we apply the methods and ideas from Chapter 5 in several contexts for categorical data. We’ll start by revisiting what we learned for a single proportion, where a normal distribution can be used to model the uncertainty in the sample proportion. Next, we apply these same ideas to analyze the diﬀerence of two proportions using a normal model. Later in the chapter we will encounter contingency tables, and we will use a diﬀerent distribution, though the core ideas of hypothesis testing remain the same. For videos, slides, and other resources, please visit www.openintro.org/ahss 6.1. INFERENCE FOR A SINGLE PROPORTION 293 6.1 Inference for a single proportion In this section, we will apply the inferential procedures introduced in Chapter 5 to the context of a single proportion, and we will explore how to do sample size calculations for data collection purposes. We will answer questions such as the following: • Do greater than half of adults in the U.S. oppose nuclear energy? • What percent of adults in the U.S. approve of the way the Supreme Court is handling its job? • What is the standard error is associated with this estimate? • How do we construct a conﬁdence interval for this value? • What sample size is required to estimate this within a 3% margin of error using a 95% conﬁ- dence level? Learning objectives 1. State and verify whether or not the conditions for inference on a proportion using a normal distribution are met. 2. Recognize that the success-failure condition and the standard error calculation are diﬀerent for the test and for the conﬁdence interval and explain why this is the case. 3. Carry out a complete hypothesis test and conﬁdence interval procedure for a single proportion. 4. Find the minimum sample size needed to estimate a proportion with C% conﬁdence and a margin of error no greater than a certain value. 5. Recognize that margin of error calculations only measure sampling error, and that other types of errors may be present. 294 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA 6.1.1 Distribution of a sample proportion (review) The distribution of a sample proportion, such as the distribution of all possible values for the proportion of people who share a particular opinion in a poll, was introduced in Section 4.1. When the sampling distribution for a sample proportion, ˆp, is approximately normal, we can use conﬁdence intervals and hypothesis tests based on a normal distribution. We call these Z-intervals and Z-tests for short. Here, we review the conditions necessary for a sample proportion to be modeled using a normal distribution. CONDITIONS FOR THE SAMPLING DISTRIBUTION FOR ˆPˆPˆP BEING NEARLY NORMAL The sampling distribution for a sample proportion, ˆp, based on a r
andom sample of size n from a population with a true proportion p, is nearly normal when 1. the sample observations are independent and 2. np ≥ 10 and n(1 − p) ≥ 10. This is called the success-failure condition. If these conditions are met, then the sampling distribution for ˆp is nearly normal with mean µ ˆp = p and standard deviation σ ˆp = p(1−p) . n 6.1.2 Checking conditions for inference using a normal distribution We can use a normal model for inference for a proportion when the sampling distribution for the sample proportion is nearly normal. We check that this assumption is reasonable by assessing the independence assumption and verifying that the success-failure condition is met. Independence. Observations can be considered independent when the data are collected from a random process, such as tossing a coin, or from a random sample. Without a random sample or process, the standard error formula would not apply, and it is unclear to what population the inference would apply. When sampling without replacement from a ﬁnite population, the observations can be considered independent when sampling less than 10% of the population.1 Success-failure condition. We saw in Section 4.1 that, when that the observations are independent, the sampling distribution for a sample proportion will be nearly normal if the successfailure condition is met, i.e. when the expected number of successes and failures are both at least 10. 1When sampling without replacement and sampling greater than 10% of the population, a modiﬁed standard error formula should be used. 6.1. INFERENCE FOR A SINGLE PROPORTION 295 6.1.3 Conﬁdence intervals for a proportion The Gallup organization began measuring the public’s view of the Supreme Court’s job performance in 2000, and has measured it every year since then with the question: “Do you approve or disapprove of the way the Supreme Court is handling its job?”. In 2018, the Gallup poll randomly sampled 1,033 adults in the U.S. and found that 53% of them approved. We know that 53% is just a point estimate. What range of values are reasonable estimates for the percent of the population that approved of the job the Supreme Court is doing? We can use the conﬁdence interval procedure introduced in the previous chapter to answer this question, but ﬁrst we must clearly identify the parameter we’re trying to estimate and be sure that a Z-interval will be appropriate. The following examples walk through the various steps for carrying out a conﬁdence interval procedure using the Gallup poll data. EXAMPLE 6.1 Identify the population of interest and the parameter of interest for the Gallup poll about the U.S. Supreme Court. Gallup sampled from U.S. adults, therefore the population of interest, and the population to which we can make an inference, is U.S. adults. We know the percent of the sample that said they approve of the job the Supreme Court is doing. However, we do not know what percent of the population would approve. The parameter of interest, which is unknown, is the percent of all U.S. adults that approve of the job the Supreme Court is doing. This is the quantity that we seek to estimate with the conﬁdence interval. EXAMPLE 6.2 Can the sample proportion ˆp be modeled using a normal distribution? In order to construct a Z-interval, the sample statistic must be able to be modeled using a normal distribution. Gallup took a random sample of adults in the U.S. The sample is random and the sample size is much less than 10% of the population size, so the ﬁrst condition (the independence condition) is satisﬁed. We must also test the second condition (the success-failure condition) to ensure that the sample size is large enough for the central limit theorem to apply. The successfailure condition is met when np and n(1 − p) are at least 10. Since p is always unknown when constructing a conﬁdence interval for p, we use the sample proportion ˆp to check this condition. Here we have: nˆp = 1033(0.53) = 547 (“successes”) n(1 − ˆp) = 1033(1 − 0.53) = 486 (“failures”) The second condition is satisﬁed since 547 and 486 are both at least 10. With the two conditions satisﬁed, we can model the sample proportion ˆp using a normal model and we can construct a Z-interval. 296 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA EXAMPLE 6.3 Calculate the point estimate and the SE of the estimate. The point estimate for the unknown parameter p (the proportion of all U.S. adults) who approve of the job the Supreme Court is doing) is the sample proportion. The point estimate here is ˆp = 0.53. Because the point estimate is the sample proportion, the SE of the estimate is the SE of ˆp. In Section 4.1, we learned that the formula for the standard deviation of ˆp is σ ˆp = p(1 − p) n The proportion p is unknown, so we use the sample proportion ˆp to ﬁnd the SE of ˆp. SE = ˆp(1 − ˆp) n Here ˆp = 0.53 and n = 1, 033, so the SE of the sample proportion is: SE = 0.53(1 − 0.53) 1033 = 0.016 EXAMPLE 6.4 Construct a 90% conﬁdence interval for p, the proportion of all U.S. adults that approve of the job the Supreme Court is doing. Recall that the general form of a conﬁdence interval is: point estimate ± critical value × SE of estimate We have already found the point estimate and the SE of the estimate. Because we previously veriﬁed that ˆp can be modeled using a normal distribution, the critical value is a z. The z value can be found in the t-table on page 514, using the bottom row (∞), where the column corresponds to the conﬁdence level. Here the conﬁdence level is 90%, so z=1.65. We can now construct the 90% conﬁdence interval as follows. point estimate ± z × SE of estimate 0.53 ± 1.65 × 0.016 = (0.504, 0.556) We are 90% conﬁdent that the true proportion of U.S. adults who approve of the job the Supreme Court is doing is between 0.504 and .556. EXAMPLE 6.5 Based on the interval, is there evidence that more than half of U.S. adults approve of the job the Supreme Court is doing? The 90% conﬁdence interval (0.504, 0.556) provides an interval of reasonable values for the parameter. The value 0.50 is not in the interval, therefore can be considered unreasonable. Because the entire interval is above 0.50, we do have evidence, at the 90% conﬁdence level, that more than half of U.S. adults (at the time of this poll) approve of the job the Supreme Court is doing. 6.1. INFERENCE FOR A SINGLE PROPORTION 297 EXAMPLE 6.6 Do we have evidence at the 95% conﬁdence level that more than half of U.S. adults approve of the job the Supreme Court is doing? First, we observe that a 95% conﬁdence interval will be wider than a 90% conﬁdence interval. For a 95% Z-interval, z = 1.96. The 95% conﬁdence interval is: 0.53 ± 1.96 × 0.016 = (0.499, 0.561) Now, we see that 0.50 is just barely inside the interval, making it within the range of reasonable values. Therefore, we do not have evidence, at the 95% conﬁdence level, that more than half of U.S. adults (at the time of this poll) approve of the job the Supreme Court is doing. Notice that we come to a diﬀerent conclusion based on diﬀerent conﬁdence levels, which may feel a little jarring. However, this will happen with real data, and it highlights why it is important to be explicit in identifying the conﬁdence level being used. Having worked through this example, we now summarize the steps for constructing a conﬁdence interval for a proportion using the ﬁve step framework introduce in Chapter 5. CONSTRUCTING A CONFIDENCE INTERVAL FOR A PROPORTION To carry out a complete conﬁdence interval procedure to estimate a single proportion p, Identify: Identify the parameter and the conﬁdence level, C%. The parameter will be a population proportion, e.g. the proportion of all U.S. adults that approve of the job the Supreme Court is doing. Choose: Choose the correct interval procedure and identify it by name. To estimate a single proportion we use a 1-proportion Z-interval. Check: Check conditions for the sampling distribution for ˆp to be nearly normal. 1. Independence: Data come from a random sample or random process. When sampling without replacement, check that sample size is less than 10% of the population size. 2. Success-failure: nˆp ≥ 10 and n(1 − ˆp) ≥ 10. Calculate: Calculate the conﬁdence interval and record it in interval form. point estimate ± z × SE of estimate point estimate: the sample proportion ˆp ˆp(1− ˆp) SE of estimate: n z: use a t-table at row ∞ and conﬁdence level C% ( , ) Conclude: Interpret the interval and, if applicable, draw a conclusion in context. We are C% conﬁdent that the true proportion of [...] is between . If applicable, draw a conclusion based on whether the interval is entirely above, is entirely below, or contains the value of interest. and 298 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA EXAMPLE 6.7 A February 2018 Marist Poll reports: “Many Americans (68%) think there is intelligent life on other planets.” The results were based on a random sample of 1,033 adults in the U.S. Does this poll provide evidence at the 95% conﬁdence level that greater than half of all U.S. adults think there is intelligent life on other planets? Carry out a conﬁdence interval procedure to answer this question. Use the ﬁve step framework to organize your work. Identify: First we identify the parameter of interest. Here the parameter is the true proportion of U.S. adults that think there is intelligent life on other planets. We will estimate this at the 95% conﬁdence level. Choose: Because the parameter to be estimated is a single proportion, we will use a 1-proportion Z-interval. Check: We must check that a Z-interval is appropriate. The problem states that the data come from a random sample, and since the population is adults in the U.S., the population size is much more than 10 times larger than the sample size. Next we must check the success-failure condition. Here, we have that 1033(.68) ≥ 10 and 1033(1 − 0.68) ≥ 10. The nearly normal sampling distribution conditions are met, so we can proceed with a 1-proporti
on Z-interval. Calculate: We will calculate the interval: point estimate ± z × SE of estimate The point estimate is the sample proportion: ˆp = 0.68 The SE of the sample proportion is: ˆp(1− ˆp) n 0.68(1−0.68) 1033 = = 0.015. z is found using the t-table at row ∞ and conﬁdence level C%. For a 95% conﬁdence level, z = 1.96. The 95% conﬁdence interval is given by: 0.68 ± 1.96 × 0.68(1 − 0.68) 1033 0.68 ± 1.96 × 0.015 = (0.651, 0.709) Conclude: We are 95% conﬁdent that the true proportion of U.S. adults that think there is intelligent life on other planets is between 0.651 and 0.709. Because the entire interval is above 0.5 we have evidence that greater than half of all U.S. adults think there is intelligent life on other planets. GUIDED PRACTICE 6.8 True or False: There is a 95% probability that between 65.1% and 70.9% of U.S. adults think that there is intelligent life on other planets.2 2False. The true percent of U.S. adults that think there is intelligent life on other planets either falls in that interval or it doesn’t. A correct interpretation of the conﬁdence level would be that if we were to repeat this process over and over, about 95% of the 95% conﬁdence intervals constructed would contain the true value. 6.1. INFERENCE FOR A SINGLE PROPORTION 299 6.1.4 Technology: the 1-proportion Z-interval A calculator can be helpful for evaluating the ﬁnal interval in the Calculate step. However, it should not be used as a substitute for understanding. TI-83/84: 1-PROPORTION Z-INTERVAL Use STAT, TESTS, 1-PropZInt. 1. Choose STAT. 2. Right arrow to TESTS. 3. Down arrow and choose A:1-PropZInt. 4. Let x be the number of yeses (must be an integer). 5. Let n be the sample size. 6. Let C-Level be the desired conﬁdence level. 7. Choose Calculate and hit ENTER, which returns the conﬁdence interval the sample proportion the sample size ( , ) ^p n CASIO FX-9750GII: 1-PROPORTION Z-INTERVAL 1. Navigate to STAT (MENU button, then hit the 2 button or select STAT). 2. Choose the INTR option (F4 button). 3. Choose the Z option (F1 button). 4. Choose the 1-P option (F3 button). 5. Specify the interval details: • Conﬁdence level of interest for C-Level. • Enter the number of successes, x. • Enter the sample size, n. 6. Hit the EXE button, which returns Left, Right ^p n ends of the conﬁdence interval sample proportion sample size GUIDED PRACTICE 6.9 Using a calculator, evaluate the conﬁdence interval from Example 6.7. Recall that we wanted to ﬁnd a 95% conﬁdence interval for the proportion of U.S. adults who think there is intelligent life on other planets. The sample percent was 68% and the sample size was 1,033.3 3Navigate to the 1-proportion Z-interval on the calculator. To ﬁnd x, the number of yes responses in the sample, we multiply the sample proportion by the sample size. Here 0.68 × 1033 = 702.44. We must round this to an integer, so we use x = 702. Also, n =1033 and C-Level = 0.95. The calculator output of (0.651, 0.708) matches our previously computed interval of (0.651, 0.709) with minor rounding diﬀerence. 300 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA 6.1.5 Choosing a sample size when estimating a proportion Planning a sample size before collecting data is important. If we collect too little data, the standard error of our point estimate may be so large that the estimate is not very useful. On the other hand, collecting data in some contexts is time-consuming and expensive, so we don’t want to waste resources on collecting more data than we need. When considering the sample size, we want to put an upper bound on the margin of error. Recall that the margin of error is measured as the distance between the point estimate and the lower or upper bound of a conﬁdence interval. MARGIN OF ERROR The margin of error of a conﬁdence interval is given by: critical value × SE of estimate The margin of error for a C% conﬁdence interval tells us that we can be C% conﬁdent that our point estimate is within that margin of error of the true value. EXAMPLE 6.10 Suppose we are conducting a university survey to determine whether students support a $200 per year increase in fees to pay for a new football stadium. Find the smallest sample size n so that the margin of error of the point estimate ˆp will be no larger than 0.04 when using a 95% conﬁdence level. Because we are working with proportions, the critical value is a z value. We want the margin of error to be less than or equal to 0.04, so we have: z × p(1 − p) n ≤ 0.04 There are two unknowns in the inequality: p and n. If we have an estimate of p, perhaps from a similar survey, we could use that value. If we have no such estimate, we must use some other value for p. It turns out that the margin of error is largest when p is 0.5, so we typically use this worst case estimate of p = 0.5 if no other estimate is available. 1.96 × 1.962 × 1.962 × 0.5(1 − 0.5) n 0.5(1 − 0.5) n 0.5(1 − 0.5) 0.042 ≤ 0.04 ≤ 0.042 ≤ n 600.25 ≤ n n = 601 The sample size must be an integer and we round up because n must be greater than or equal to 600.25. We need at least 601 participants to ensure the sample proportion is within 0.04 of the true proportion with 95% conﬁdence. No estimate of the true proportion is required in sample size computations for a proportion. However, if we have a reliable estimate of the proportion, we should use it in place of the worst case estimate of 0.5. 6.1. INFERENCE FOR A SINGLE PROPORTION 301 EXAMPLE 6.11 A recent estimate of Congress’ approval rating was 17%. If another poll were taken, what minimum sample size does this estimate suggest should be used to have a margin of error no greater than 0.04 with 95% conﬁdence? We complete the same computations as before, except now we use 0.17 instead of 0.5 for p: 1.96 × 0.17(1 − 0.17) n ≤ 0.04 n ≥ 338.8 n = 339 If the true proportion is 0.17, then 339 is the minimum sample size that will ensure a margin of error no greater than 0.04 with 95% conﬁdence. IDENTIFY A SAMPLE SIZE FOR A PARTICULAR MARGIN OF ERROR When estimating a single proportion, we ﬁnd the minimum sample size n needed to achieve a margin of error no greater than m with a speciﬁed conﬁdence level as follows: z × p(1 − p) n ≤ m where z depends on the conﬁdence level. If no reliable estimate of p exists, use p = 0.5. GUIDED PRACTICE 6.12 All other things being equal, what would we have to do to the sample size in order to halve the margin of error (decrease it by a factor of 2)?4 GUIDED PRACTICE 6.13 A manager is about to oversee the mass production of a new tire model in her factory, and she would like to estimate the proportion of these tires that will be rejected through quality control. The quality control team has previously found that about 6.2% of tires fail inspection. (a) How many tires should the manager examine to estimate the failure rate of the new tire model to within 2% with a 90% conﬁdence level?5 (b) What if the estimate of p is 1.7% rather than 6.2%?6 4To decrease the error, we would need to increase the sample size. We note that n is in the denominator of the SE formula, so we would have to quadruple the sample size in order to decrease the SE by a factor of 2. The margin of error as well as the width of the conﬁdence interval is proportional to 1√ n 5The z corresponding to a 90% conﬁdence level is 1.645. Since we have an estimate for p of 6.2%, we use it. So . 0.062(1−0.062) n we have: 1.645 × ≤ 0.02. Rearranging for n gives: n ≥ 393.4, so she should use n = 394. 6Substituting 0.017 for p gives an n of 114. We can note that in this case n × p = 114 × 0.017 = 1.9 < 10. Since the success-failure condition is not met, the use of z = 1.645 based on a normal model is not appropriate. We would need additional methods than what we’ve covered so far to get a good estimate for the minimum sample size in this scenario. √ 302 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA 6.1.6 Hypothesis testing for a proportion While a conﬁdence interval provides a range of reasonable values for an unknown parameter, a hypothesis test evaluates a speciﬁc claim. In a hypothesis test, we set up competing hypotheses and ﬁnd degrees of evidence against the null hypothesis. EXAMPLE 6.14 Deborah Toohey is running for Congress, and her campaign manager claims she has more than 50% support from the district’s electorate. A newspaper collects a random sample of 500 likely voters in the district and estimates Toohey’s support to be 52%. (a) Identify the null and the alternative hypothesis. What value should we use as the null value, p0? (b) Can we model ˆp using a normal model? Check the conditions. (a) The alternative hypothesis, the one that bears the burden of proof, argues that Toohey has more than 50% support. Therefore, HA will be one-sided and the null value will be p0 = 0.5. So we have H0: p = 0.5 and HA: p > 0.5. Note that the hypotheses are about a population parameter. The hypotheses are never about the sample. (b) First, we observe that the problem states that a random sample was chosen. We assume that the size of the electorate in Toohey’s district is more than 10 times the size of the sample. Next, we check the success-failure condition. Because we assume that p = p0 for the calculations of the hypothesis test, we use the hypothesized value p0 rather than the sample value ˆp when verifying the success-failure condition. np0 ≥ 10 → 500(0.5) ≥ 10 n(1 − p0) ≥ 10 → 500(1 − 0.5) ≥ 10 The conditions for a normal model are met. In Chapter 5, we saw that the general form of the test statistic for a hypothesis test takes the following form: test statistic = point estimate − null value SE of estimate When the conditions for a normal model are met: • We use Z as the test statistic and call the test a Z-test. • The point estimate is the sample proportion ˆp (just like for a conﬁdence interval). • Since we compute the test statistic assuming the null hypothesis (that p = p0) is true, we compute the standard error of the sample proportion using the null value p0. SE = p0(1 − p0) n CONFIDENCE INTE
RVALS VERSUS HYPOTHESIS TESTS FOR A SINGLE PROPORTION 1-proportion Z-interval Check: nˆp ≥ 10 and n(1 − ˆp) ≥ 10 SE = 1-proportion Z-test Check: np0 ≥ 10 and n(1 − p0) ≥ 10 SE = ˆp(1 − ˆp) n p0(1 − p0) n 6.1. INFERENCE FOR A SINGLE PROPORTION 303 EXAMPLE 6.15 (Continues previous example). Deborah Toohey’s campaign manager claimed she has more than 50% support from the district’s electorate. A newspaper poll ﬁnds that 52% of 500 likely voters who were sampled support Toohey. Does this provide convincing evidence for the claim by Toohey’s manager at the 5% signiﬁcance level? We will use a one-sided test with the following hypotheses: H0: p = 0.5. Toohey’s support is 50%. HA: p > 0.5. Toohey’s manager is correct, and her support is higher than 50%. We will use a signiﬁcance level of α = 0.05 for the test. We can compute the standard error as SE = p0(1 − p0) n = 0.5(1 − 0.5 ) 500 = 0.022 The test statistic can be computed as: Z = point estimate − null value SE of estimate = 0.52 − 0.50 0.022 = 0.89 Because the alternative hypothesis uses a greater than sign (>), this is an upper-tail test. We ﬁnd the area under the standard normal curve to the right of Z = 0.89. A ﬁgure featuring the p-value is shown in Figure 6.1 as the shaded region. Figure 6.1: Sampling distribution for the sample proportion if the null hypothesis is true for Example 6.15. The p-value for the test is shaded. Using a table or a calculator, we ﬁnd the p-value is 0.19. This p-value of 0.19 is greater than α = 0.05, so we do not reject H0. That is, we do not have suﬃcient evidence to support Toohey’s campaign manager’s claims that she has more than 50% support within the district. EXAMPLE 6.16 Based on the result above, do we have evidence that Toohey’s support equals 50%? No. In a hypothesis test we look for degrees of evidence against the null hypothesis. We cannot ever prove the null hypothesis directly. The value 0.5 is reasonable, but many other values are reasonable as well. There are many values that would not get rejected by this test. We now summarize the steps for carrying out a hypothesis test for a proportion using the ﬁve step framework introduced in the previous chapter. 00.89 304 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA HYPOTHESIS TESTING FOR A PROPORTION To carry out a complete hypothesis test to test the claim that a single proportion p is equal to a null value p0, Identify: Identify the hypotheses and the signiﬁcance level, α. H0: p = p0 HA: p = p0; HA: p > p0; or HA: p < p0 Choose: Choose the correct test procedure and identify it by name. To test hypotheses about a single proportion we use a 1-proportion Z-test. Check: Check conditions for the sampling distribution for ˆp to be nearly normal, assuming H0: p = p0 is true. 1. Independence: Data come from a random sample or random process. When sampling without replacement, check that sample size is less than 10% of the population size. 2. Success-failure: np0 ≥ 10 and n(1 − p0) ≥ 10 Calculate: Calculate the Z-statistic and p-value. Z = point estimate − null value SE of estimate point estimate: the sample proportion ˆp SE of estimate: null value: p0 p0(1−p0) n p-value = (based on the Z-statistic and the direction of HA) Conclude: Compare the p-value to α, and draw a conclusion in context. If the p-value is < α, reject H0; there is suﬃcient evidence that [HA in context]. If the p-value is > α, do not reject H0; there is not suﬃcient evidence that [HA in context]. 6.1. INFERENCE FOR A SINGLE PROPORTION 305 EXAMPLE 6.17 A Gallup poll conducted in March of 2016 found that 54% of respondents oppose nuclear energy. This was the ﬁrst time since Gallup ﬁrst asked the question in 1994 that a majority of respondents said they oppose nuclear energy. The survey was based on telephone interviews from a random sample of 1,019 adults in the United States. Does this poll provide evidence that greater than half of U.S. adults oppose nuclear energy? Carry out an appropriate test at the 0.10 signiﬁcance level. Use the ﬁve step framework to organize your work. Identify: We will test the following hypotheses at the α = 10% signiﬁcance level. H0: p = 0.5 HA: p > 0.5 Greater than half of all U.S. adults oppose nuclear energy. Note: p > 0.5 is what we want to ﬁnd evidence for; this bears the burden of proof, so this corresponds to HA. Choose: Because the hypotheses are about a single proportion, we choose the 1-proportion Z-test. Check: We must check the independence and success-failure conditions to show that the sample proportion can be modeled using a normal distribution. The problem states that the data come from a random sample. Again, the population is adults in the U.S., so the sample size is much smaller than 10% of the population size. Also, 1019(0.5) ≥ 10 and 1019(1 − 0.5) ≥ 10. (Remember to use the hypothesized proportion, not the sample proportion, when checking the conditions for this test.) Calculate: We will calculate the Z-statistic and the p-value. Z = point estimate − null value SE of estimate The point estimate is the sample proportion: ˆp = 0.54. The value hypothesized for the parameter in H0 is the null value: p0 = 0.5 The SE of the sample proportion, assuming H0 is true, is: p0(1−p0) n 0.5(1−0.5) 1019 = Z = 0.54 − 0.5 0.5(1−0.5) 1019 = 2.5 Because HA uses a greater than sign (>), meaning that it is an upper-tail test, the p-value is the area to the right of Z = 2.5 under the standard normal curve. This area can be found using a normal table or a calculator. The area or p-value = 0.006. Conclude: The p-value of 0.006 is < 0.10, so we reject H0; there is suﬃcient evidence that greater than half of U.S. adults oppose nuclear energy (as of March 2016). GUIDED PRACTICE 6.18 In context, interpret the p-value of 0.006 from the previous example.7 7Assuming the normal model is accurate and assuming the null hypothesis is true, i.e. that the true proportion of U.S. adults that oppose nuclear energy really is 0.5, there is a 0.006 probability of getting a test statistic as large or larger than 2.5 (H0 uses a > sign, so the p-value is the area in the right tail). Note: We start by assuming H0 is true, that p really equals 0.5. Then, assuming this, we estimate the probability of getting a sample proportion of 0.54 or larger by ﬁnding the area under the standard normal curve to the right of 2.5. This probability is very small, which casts doubt on the null hypothesis and leads us to reject it. 306 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA 6.1.7 Technology: the 1-proportion Z-test A calculator can be useful for evaluating the test statistic and computing the p-value. TI-83/84: 1-PROPORTION Z-TEST Use STAT, TESTS, 1-PropZTest. 1. Choose STAT. 2. Right arrow to TESTS. 3. Down arrow and choose 5:1-PropZTest. 4. Let p0 be the null or hypothesized value of p. 5. Let x be the number of yeses (must be an integer). 6. Let n be the sample size. 7. Choose =, <, or > to correspond to HA. 8. Choose Calculate and hit ENTER, which returns z Z-statistic p-value p the sample proportion ^p the sample size n CASIO FX-9750GII: 1-PROPORTION Z-TEST The steps closely match those of the 1-proportion conﬁdence interval. 1. Navigate to STAT (MENU button, then hit the 2 button or select STAT). 2. Choose the TEST option (F3 button). 3. Choose the Z option (F1 button). 4. Choose the 1-P option (F3 button). 5. Specify the test details: • Specify the sidedness of the test using the F1, F2, and F3 keys. • Enter the null value, p0. • Enter the number of successes, x. • Enter the sample size, n. 6. Hit the EXE button, which returns z Z-statistic p-value p the sample proportion ^p the sample size n GUIDED PRACTICE 6.19 Using a calculator, ﬁnd the test statistic and p-value for the earlier Example 6.17. Recall that we were looking for evidence that more than half of U.S. adults oppose nuclear energy. The sample percent was 54%, and the sample size was 1019.8 8Navigate to the 1-proportion Z-test on the calculator. Let p0 = 0.5. To ﬁnd x, do 0.54 × 1019 = 550.26. This needs to be an integer, so round to the closest integer. Here x = 550. Also, n = 1019. We are looking for evidence that greater than half oppose, so choose > p0. When we do Calculate, we get the test statistic: Z = 2.64 and the p-value: p = 0.006. 6.1. INFERENCE FOR A SINGLE PROPORTION 307 Section summary Most of the conﬁdence interval procedures and hypothesis tests of this book involve: a point estimate, the standard error for the point estimate, and an assumption about the shape of the sampling distribution of the point estimate. In this section, we explore inference when the parameter of interest is a proportion. • We use the sample proportion ˆp as the point estimate for the unknown population proportion p. The sampling distribution for ˆp is approximately normal when the success-failure condition is met and the observations are independent. When the sampling distribution for ˆp is normal, the standardized test statistic also follows a normal distribution. • When verifying the success-failure condition and calculating the SE, – use the sample proportion ˆp for the conﬁdence interval, but – use the null/hypothesized proportion p0 for the hypothesis test. • When there is one sample and the parameter of interest is a single proportion: – Estimate p at the C% conﬁdence level using a 1-proportion Z-interval. – Test H0: p = p0 at the α signiﬁcance level using a 1-proportion Z-test. • The one proportion Z-test and Z-interval require the sampling distribution for ˆp to be nearly normal. For this reason we must check that the following conditions are met. 1. Independence: The data should come from a random sample or random process. When sampling without replacement, check that the sample size is less than 10% of the population size. 2. Success-failure for Interval: nˆp ≥ 10 and n(1 − ˆp) ≥ 10. Success-failure for Test, assuming H0: p = p0 is true: np0 ≥ 10 and n(1 − p0) ≥ 10. • When the conditions are met, we calculate the conﬁdence interval and the test statistic as follo
ws. Conﬁdence interval: point estimate ± z × SE of estimate Test statistic: Z = point estimate − null value SE of estimate Here the point estimate is the sample proportion ˆp. The SE of estimate is the SE of the sample proportion. – For an Interval, use SE = ˆp(1− ˆp) n . – For a Test with H0: p = p0, use SE = p0(1−p0) n . • The margin of error (M E) for a one-sample conﬁdence interval for a proportion is z ˆp(1− ˆp) n , which is proportional to 1√ n . • To ﬁnd the minimum sample size needed to estimate a proportion with a given conﬁdence level and a given margin of error, m, set up an inequality of the form: z ˆp(1 − ˆp) n < m z depends on the desired conﬁdence level. Unless a particular proportion is given in the problem, use ˆp = 0.5. We solve for the sample size n. The ﬁnal answer should be an integer, since n refers to a number of people or things. 308 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA Exercises 6.1 Orange tabbies. Suppose that 90% of orange tabby cats are male. Determine if the following statements are true or false, and explain your reasoning. (a) The distribution of sample proportions of random samples of size 30 is left skewed. (b) Using a sample size that is 4 times as large will reduce the standard error of the sample proportion by one-half. (c) The distribution of sample proportions of random samples of size 140 is approximately normal. (d) The distribution of sample proportions of random samples of size 280 is approximately normal. 6.2 Young Americans, Part II. About 25% of young Americans have delayed starting a family due to the continued economic slump. Determine if the following statements are true or false, and explain your reasoning.9 (a) The distribution of sample proportions of young Americans who have delayed starting a family due to the continued economic slump in random samples of size 12 is right skewed. (b) In order for the distribution of sample proportions of young Americans who have delayed starting a family due to the continued economic slump to be approximately normal, we need random samples where the sample size is at least 40. (c) A random sample of 50 young Americans where 20% have delayed starting a family due to the continued economic slump would be considered unusual. (d) A random sample of 150 young Americans where 20% have delayed starting a family due to the continued economic slump would be considered unusual. (e) Tripling the sample size will reduce the standard error of the sample proportion by one-third. 6.3 Gender equality. The General Social Survey asked a random sample of 1,390 Americans the following question: “On the whole, do you think it should or should not be the government’s responsibility to promote equality between men and women?” 82% of the respondents said it “should be”. At a 95% conﬁdence level, this sample has 2% margin of error. Based on this information, determine if the following statements are true or false, and explain your reasoning.10 (a) We are 95% conﬁdent that between 80% and 84% of Americans in this sample think it’s the government’s responsibility to promote equality between men and women. (b) We are 95% conﬁdent that between 80% and 84% of all Americans think it’s the government’s respon- sibility to promote equality between men and women. (c) If we considered many random samples of 1,390 Americans, and we calculated 95% conﬁdence intervals for each, 95% of these intervals would include the true population proportion of Americans who think it’s the government’s responsibility to promote equality between men and women. (d) In order to decrease the margin of error to 1%, we would need to quadruple (multiply by 4) the sample size. (e) Based on this conﬁdence interval, there is suﬃcient evidence to conclude that a majority of Americans think it’s the government’s responsibility to promote equality between men and women. 6.4 Elderly drivers. The Marist Poll published a report stating that 66% of adults nationally think licensed drivers should be required to retake their road test once they reach 65 years of age. It was also reported that interviews were conducted on a random sample of 1,018 American adults, and that the margin of error was 3% using a 95% conﬁdence level.11 (a) Verify the margin of error reported by The Marist Poll. (b) Based on a 95% conﬁdence interval, does the poll provide convincing evidence that more than two thirds of the population think that licensed drivers should be required to retake their road test once they turn 65? 9Demos.org. “The State of Young America: The Poll”. In: (2011). 10National Opinion Research Center, General Social Survey, 2018. 11Marist Poll, Road Rules: Re-Testing Drivers at Age 65?, March 4, 2011. 6.1. INFERENCE FOR A SINGLE PROPORTION 309 6.5 Fireworks on July 4th. A local news outlet reported that 56% of 600 randomly sampled Kansas residents planned to set oﬀ ﬁreworks on July 4th. Determine the margin of error for the 56% point estimate using a 95% conﬁdence level.12 6.6 Life rating in Greece. Greece has faced a severe economic crisis since the end of 2009. A Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that 25% of them said they would rate their lives poorly enough to be considered “suﬀering”.13 (a) Describe the population parameter of interest. What is the value of the point estimate of this parameter? (b) Check if the conditions required for constructing a conﬁdence interval based on these data are met. (c) Construct a 95% conﬁdence interval for the proportion of Greeks who are “suﬀering”. (d) Without doing any calculations, describe what would happen to the conﬁdence interval if we decided to use a higher conﬁdence level. (e) Without doing any calculations, describe what would happen to the conﬁdence interval if we used a larger sample. 6.7 Study abroad. A survey on 1,509 high school seniors who took the SAT and who completed an optional web survey shows that 55% of high school seniors are fairly certain that they will participate in a study abroad program in college.14 (a) Is this sample a representative sample from the population of all high school seniors in the US? Explain your reasoning. (b) Let’s suppose the conditions for inference are met. Even if your answer to part (a) indicated that this approach would not be reliable, this analysis may still be interesting to carry out (though not report). Construct a 90% conﬁdence interval for the proportion of high school seniors (of those who took the SAT) who are fairly certain they will participate in a study abroad program in college, and interpret this interval in context. (c) What does “90% conﬁdence” mean? (d) Based on this interval, would it be appropriate to claim that the majority of high school seniors are fairly certain that they will participate in a study abroad program in college? 6.8 Legalization of marijuana, Part I. The General Social Survey asked a random sample of 1,578 US residents: “Do you think the use of marijuana should be made legal, or not?” 61% of the respondents said it should be made legal.15 (a) Is 61% a sample statistic or a population parameter? Explain. (b) Construct a 95% conﬁdence interval for the proportion of US residents who think marijuana should be made legal, and interpret it in the context of the data. (c) A critic points out that this 95% conﬁdence interval is only accurate if the statistic follows a normal distribution, or if the normal model is a good approximation. Is this true for these data? Explain. (d) A news piece on this survey’s ﬁndings states, “Majority of Americans think marijuana should be legal- ized.” Based on your conﬁdence interval, is this news piece’s statement justiﬁed? 6.9 National Health Plan, Part I. A Kaiser Family Foundation poll for a random sample of US adults in 2019 found that 79% of Democrats, 55% of Independents, and 24% of Republicans supported a generic “National Health Plan”. There were 347 Democrats, 298 Republicans, and 617 Independents surveyed.16 (a) A political pundit on TV claims that a majority of Independents support a National Health Plan. Do these data provide strong evidence to support this type of statement? (b) Would you expect a conﬁdence interval for the proportion of Independents who oppose the public option plan to include 0.5? Explain. 12Survey USA, News Poll #19333, data collected on June 27, 2012. 13Gallup World, More Than One in 10 “Suﬀering” Worldwide, data collected throughout 2011. 14studentPOLL, College-Bound Students’ Interests in Study Abroad and Other International Learning Activities, January 2008. 15National Opinion Research Center, General Social Survey, 2018. 16Kaiser Family Foundation, The Public On Next Steps For The ACA And Proposals To Expand Coverage, data collected between Jan 9-14, 2019. 310 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA 6.10 Is college worth it? Part I. Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not aﬀord school.17 (a) A newspaper article states that only a minority of the Americans who decide not to go to college do so because they cannot aﬀord it and uses the point estimate from this survey as evidence. Conduct a hypothesis test to determine if these data provide strong evidence supporting this statement. (b) Would you expect a conﬁdence interval for the proportion of American adults who decide not to go to college because they cannot aﬀord it to include 0.5? Explain. Some people claim that they can tell the diﬀerence between a diet soda and a regular 6.11 Taste test. soda in the ﬁrst sip. A researcher wanting to test this claim randomly sampled 80 such people. He then ﬁlled 80 plain white cups with soda, half diet and half regular through random assignment, and asked each person to take one sip from their cup and identify the soda as diet or regular. 53 participants correctly identiﬁed the soda. (a) Do these data provide strong
evidence that these people are able to detect the diﬀerence between diet and regular soda, in other words, are the results signiﬁcantly better than just random guessing? Carry out an appropriate test and include all steps of the Identify, Choose, Check, Calculate, Conclude framework. (b) Interpret the p-value in this context. Is college worth it? Part II. Exercise 6.10 presents a poll where 48% of 331 randomly selected 6.12 Americans reported that they decided not to go to college because they cannot aﬀord it. (a) Calculate a 90% conﬁdence interval to estimate the proportion of Americans who decide to not go to college because they cannot aﬀord it. Include all steps of the Identify, Choose, Check, Calculate, Conclude framework. (b) Suppose we wanted the margin of error for the 90% conﬁdence level to be about 1.5%. How large of a survey would you recommend? 6.13 National Health Plan, Part II. Exercise 6.9 presents the results of a poll evaluating support for a generic “National Health Plan” in the US in 2019, reporting that 55% of Independents are supportive. If we want to estimate the percent of Independents who are supportive this year to within 1% with 90% conﬁdence, what would be an appropriate sample size? 6.14 Legalize Marijuana, Part II. As discussed in Exercise 6.8, the General Social Survey reported a sample where about 61% of US residents thought marijuana should be made legal. If we wanted to limit the margin of error of a 95% conﬁdence interval to 2%, about how many Americans would we need to survey? 17Pew Research Center Publications, Is College Worth It?, data collected between March 15-29, 2011. 6.2. INFERENCE FOR THE DIFFERENCE OF TWO PROPORTIONS 311 6.2 Inference for the difference of two proportions We often wish to compare two groups to each other. In this section, we will answer the following questions: • How much more eﬀective is a blood thinner than a placebo for those who undergo CPR for a heart attack? • How diﬀerent is the approval of the 2010 healthcare law under two diﬀerent question phrasings? • Does the use of ﬁsh oils reduce heart attacks better than a placebo? Learning objectives 1. State and verify whether or not the conditions for inference on the diﬀerence of two proportions using a normal distribution are met. 2. Recognize that the standard error calculation is diﬀerent for the test and for the interval, and explain why that is the case. 3. Know how to calculate the pooled proportion and when to use it. 4. Carry out a complete conﬁdence interval procedure for the diﬀerence of two proportions. 5. Carry out a complete hypothesis test for the diﬀerence of two proportions. 6.2.1 Sampling distribution for the difference of two proportions (review) In this section we want to compare proportions from two independent groups. When comparing two proportions, the quantity that we generally want to estimate is the diﬀerence p1 − p2, which tells us how far apart the two population proportions are. Before we perform inference for the two proportion case, we must review the sampling distribution for ˆp1 − ˆp2, which will represent our point estimate. We know from Section 4.3 that when the independence condition is satisﬁed, the sampling distribution for ˆp1 − ˆp2 is centered on p1 − p2 and the standard deviation is given by: σ ˆp1− ˆp2 = p1(1 − p1) n1 + p2(1 − p2) n2 . When the individual population proportions are unknown, we estimate the standard deviation of ˆp1 − ˆp2 using the Standard Error, abbreviated SE. The SE of ˆp1 − ˆp2 is found by substituting in our best estimates of p1 and p2 using the sample values: SE ˆp1− ˆp2 = ˆp1(1 − ˆp1) n1 + ˆp2(1 − ˆp2) n2 . The diﬀerence of two sample proportions ˆp1 − ˆp2 follows a nearly normal distribution when two conditions are met. First, the sampling distribution for each sample proportion must be nearly normal. Second, the observations must be independent, both within and between groups. We cover these conditions in greater detail next. 312 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA 6.2.2 Checking conditions for inference using a normal distribution When comparing two proportions, we carry out inference on p1 − p2. To model the test statistic with a normal distribution, we need the sampling distribution for ˆp1 − ˆp2 to be nearly normal, and this assumption is reasonable when two conditions are met: Independence. The observations across the two samples are independent. This condition is generally satisﬁed by checking whether the data are collected from two independent random samples or from an experiment with two randomly assigned treatments. Randomly assigning subjects to treatments is equivalent to randomly assigning treatments to subjects. When sampling without replacement, the observations can be considered independent when the sample sizes is less than 10% of the population size for both samples. Success-failure condition. In the two-sample case, the number of successes and failures should be at least 10 for both groups, so there are four inequalities to check. 6.2.3 Conﬁdence interval for the difference of two proportions We consider an experiment for patients who underwent CPR for a heart attack and were subsequently admitted to a hospital. These patients were randomly divided into a treatment group where they received a blood thinner or the control group where they did not receive a blood thinner. The outcome variable of interest was whether the patients survived for at least 24 hours. The results are shown in Figure 6.2. Treatment Control Total Survived Died 14 11 25 26 39 65 Total 40 50 90 Figure 6.2: Results for the CPR study. Patients in the treatment group were given a blood thinner, and patients in the control group were not. Here, the parameter of interest is a diﬀerence of population proportions, speciﬁcally, the diﬀerence in the proportion of similar patients that would survive for at least 24 hours if in the treatment group versus if in the control group. Let: p1 : proportion that would survive in treatment group, and p2 : proportion that would survive in control group Then the parameter of interest is p1 − p2. In order to use a Z-interval to estimate this diﬀerence, we must see if the point estimate, ˆp1 − ˆp2, follows a normal distribution. Because the patients were randomly assigned to one of the two groups and one heart attack patient is unlikely to inﬂuence the next that was in the study, the observations are considered independent, both within the samples and between the samples (since there is no sampling, there is no need to check the 10% condition). Next, the success-failure condition should be veriﬁed for each group. We use the sample proportions along with the sample sizes to check the condition. n1 ˆp1 ≥ 10 40 × 14 40 ≥ 10 n1(1 − ˆp1) ≥ 10 14 40 ) ≥ 10 40 × (1 − n2 ˆp2 ≥ 10 50 × 11 50 ≥ 10 n2(1 − ˆp2) ≥ 10 11 50 ) ≥ 10 50 × (1 − Because all conditions are met, the normal model can be used for the point estimate of the diﬀerence in survival rate. The point estimate is: ˆp1 − ˆp2 = 14 40 − 11 50 = 0.35 − 0.22 = 0.13 6.2. INFERENCE FOR THE DIFFERENCE OF TWO PROPORTIONS 313 We compute the standard error for the diﬀerence of sample proportions in the same way that we compute the standard deviation for the diﬀerence of sample proportions – the only diﬀerence is that we use the sample proportions in place of the population proportions: SE = ˆp1(1 − ˆp1) n1 + ˆp2(1 − ˆp2) n2 = 0.35(1 − 0.35) 40 + 0.22(1 − 0.22) 50 = 0.095 Let us estimate the true diﬀerence in survival rate with 90% conﬁdence. For a 90% conﬁdence level, we use z = 1.645. The 90% conﬁdence interval is calculated as: point estimate ± z × SE of estimate 0.13 ± 1.65 × 0.095 (−0.027, 0.287) We are 90% conﬁdent that the true diﬀerence in the survival rate (treatment − control) lies between -0.027 and 0.287. That is, we are 90% conﬁdent that the treatment of blood thinners changes survival rate for patients like those in the study by -2.7% to +28.7% percentage points. Because this interval contains both negative and positive values, we do not have enough information to say with conﬁdence whether blood thinners harm or help heart attack patients who have been admitted after they have undergone CPR. CONSTRUCTING A CONFIDENCE INTERVAL FOR THE DIFFERENCE OF TWO PROPORTIONS To carry out a complete conﬁdence interval procedure to estimate the diﬀerence of two proportions p1 − p2, Identify: Identify the parameter and the conﬁdence level, C%. The parameter will be a diﬀerence of proportions, e.g. the true diﬀerence in the proportion of 17 and 18 year olds with a summer job (proportion of 18 year olds − proportion of 17 year olds). Choose: Identify the correct interval procedure and identify it by name. To estimate a diﬀerence of proportions we choose the 2-proportion Z-interval. Check: Check conditions for the sampling distribution for ˆp1 − ˆp2 to be nearly normal. 1. Independence: Data come from 2 independent random samples or from a randomized experiment with two treatments. When sampling without replacement, check that the sample size is less than 10% of the population size for both samples. 2. Success-failure: n1 ˆp1 ≥ 10, n1(1 − ˆp1) ≥ 10, n2 ˆp2 ≥ 10, and n2(1 − ˆp2) ≥ 10 Calculate: Calculate the conﬁdence interval and record it in interval form. point estimate ± z × SE of estimate point estimate: the diﬀerence of sample proportions ˆp1 − ˆp2 SE of estimate: ˆp1(1− ˆp1) n1 + ˆp2(1− ˆp2) n2 z: use a t-table at row ∞ and conﬁdence level C% ( , ) Conclude: Interpret the interval and, if applicable, draw a conclusion in context. We are C% conﬁdent that the true diﬀerence in the proportion of [...] is between and . If applicable, draw a conclusion based on whether the interval is entirely above, is entirely below, or contains the value 0. 314 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA EXAMPLE 6.20 A remote control car company is considering a new manufacturer for wheel gears. The new manufacturer would be more expensive but their higher quality gears are more reliable,
resulting in happier customers and fewer warranty claims. However, management must be convinced that the more expensive gears are worth the conversion before they approve the switch. The quality control engineer collects a sample of gears from each supplier, examining 1000 gears from each company, and ﬁnds that 879 gears pass inspection from the current supplier and 958 pass inspection from the prospective supplier. Using these data, construct a 95% conﬁdence interval for the diﬀerence in the proportion from each supplier that would pass inspection. Use the ﬁve step framework described above to organize your work. Identify: First we identify the parameter of interest. Here the parameter we wish to estimate is the true diﬀerence in the proportion of gears from each supplier that would pass inspection, p1 − p2. We will take the diﬀerence as: current − prospective, so p1 is the true proportion that would pass from the current supplier and p2 is the true proportion that would pass from the prospective supplier. We will estimate the diﬀerence using a 95% conﬁdence level. Choose: Because the parameter to be estimated is a diﬀerence of proportions, we will use a 2- proportion Z-interval. Check: The samples are independent, but not necessarily random, so to proceed we must assume the gears are all independent. For this sample we will suppose this assumption is reasonable, but the engineer would be more knowledgeable as to whether this assumption is appropriate. We will also assume that the 1000 gears represents less than 10% of the total gears from each supplier. Next, we verify the minimum sample size conditions: 1000 × 879 1000 ≥ 10 1000 × 121 1000 ≥ 10 1000 × 958 1000 ≥ 10 1000 × 42 1000 ≥ 10 The success-failure condition is met for both samples. Calculate: We will calculate the interval: point estimate ± z × SE of estimate The point estimate is the diﬀerence of sample proportions: ˆp1 − ˆp2 = 0.879 − 0.958 = −0.079. The SE of the diﬀerence of sample proportions is: ˆp1(1− ˆp1) 0.879(1−0.879) 1000 = + ˆp2(1− ˆp2) n2 + 0.958(1−0.958) 1000 n1 = 0.0121 So the 95% conﬁdence interval is given by: 0.879 − 0.958 ± 1.96 × 0.879(1 − 0.879) 1000 + 0.958(1 − 0.958) 1000 −0.079 ± 1.96 × 0.0121 (−0.103, −0.055) Conclude: We are 95% conﬁdent that the true diﬀerence (current − prospective) in the proportion that would pass inspection is between -0.103 and -0.055, meaning that we are 95% conﬁdent that the prospective supplier would have between a 5.5% and 10.3% greater rate of passing inspection. Because the entire interval is below zero, the data provide suﬃcient evidence that the prospective gears pass inspection more often than the current gears. The remote control car company should go with the new manufacturer. 6.2. INFERENCE FOR THE DIFFERENCE OF TWO PROPORTIONS 315 6.2.4 Technology: the 2-proportion Z-interval As with the 1-proportion Z-interval, a calculator can be helpful for evaluating the ﬁnal interval. TI-83/84: 2-PROPORTION Z-INTERVAL Use STAT, TESTS, 2-PropZInt. 1. Choose STAT. 2. Right arrow to TESTS. 3. Down arrow and choose B:2-PropZInt. 4. Let x1 be the number of yeses (must be an integer) in sample 1 and let n1 be the size of sample 1. 5. Let x2 be the number of yeses (must be an integer) in sample 2 and let n2 be the size of sample 2. 6. Let C-Level be the desired conﬁdence level. 7. Choose Calculate and hit ENTER, which returns: the conﬁdence interval sample 1 proportion sample 2 proportion ( , ) ^p1 ^p2 n1 n2 size of sample 1 size of sample 2 CASIO FX-9750GII: 2-PROPORTION Z-INTERVAL 1. Navigate to STAT (MENU button, then hit the 2 button or select STAT). 2. Choose the INTR option (F4 button). 3. Choose the Z option (F1 button). 4. Choose the 2-P option (F4 button). 5. Specify the interval details: • Conﬁdence level of interest for C-Level. • Enter the number of successes for each group, x1 and x2. • Enter the sample size for each group, n1 and n2. 6. Hit the EXE button, which returns Left, Right ^p1, ^p2 n1, n2 the ends of the conﬁdence interval the sample proportions sample sizes GUIDED PRACTICE 6.21 From Example 6.20, we have that a quality control engineer collects a sample of gears, examining 1000 gears from each company and ﬁnds that 879 gears pass inspection from the current supplier and 958 pass inspection from the prospective supplier. Use a calculator to ﬁnd a 95% conﬁdence interval for the diﬀerence (current − prospective) in the proportion that would pass inspection.18 18Navigate to the 2-proportion Z-interval on the calculator. Let x1 = 879, n1 = 1000, x2 = 958, and n2 = 1000. C-Level is .95. This should lead to an interval of (-0.1027, -0.0553), which matches what we found previously. 316 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA H0: p1 = p2 H0: p1 = p2 6.2.5 Hypothesis testing when H0: p1 = p2 Here we use a new example to examine a special estimate of the standard error when the null hypothesis is that two population proportions equal each other, i.e. H0: p1 = p2. We investigate whether the way a question is phrased can inﬂuence a person’s response. Pew Research Center conducted a survey with the following question: As you may know, by 2014 nearly all Americans will be required to have health insurance. [People who do not buy insurance will pay a penalty] while [People who cannot aﬀord it will receive ﬁnancial help from the government]. Do you approve or disapprove of this policy? For each randomly sampled respondent, the statements in brackets were randomized: either they were kept in the original order given above, or they were reversed. Results are presented in Figure 6.3 sample size Approve law (%) 47 771 Disapprove law (%) 49 732 34 63 Other 3 3 “People who do not buy insurance will pay a penalty” is given ﬁrst (original order) “People who cannot aﬀord it will receive ﬁnancial help from the government” is given ﬁrst (reversed order) Figure 6.3: Results for a Pew Research Center poll where the ordering of two statements in a question regarding healthcare were randomized. GUIDED PRACTICE 6.22 Is this study an experiment or an observational study?19 The approval percents of 47% and 34% seem far apart. However, could this diﬀerence be due to random chance? We will answer this question using a hypothesis test. To simplify things, let p1: the proportion of respondents that would approve of policy with the original statement ordering, and p2: the proportion of respondents that would approve of policy with the reversed statement ordering. EXAMPLE 6.23 Set up hypotheses to test whether the two statement orders produce the same response. The null claim is that the question order does not matter, that is, that the two proportions should be equal. The alternate claim, the one that bears the burden of proof, is that the question ordering does matter. H0: p1 = p2 HA: p1 = p2 19There is a random sample involved, but there are also two treatments. Half of the respondents are given the original statement order and the other half, randomly, are given the reversed statement order. This is an experiment because there are randomly assigned treatments. 6.2. INFERENCE FOR THE DIFFERENCE OF TWO PROPORTIONS 317 Now, we can note that: p1 = p2 is equivalent to p1 − p2 = 0, and p1 = p2 is equivalent to p1 − p2 = 0. We can now see that the hypotheses are really about a diﬀerence of proportions: p1 − p2. In the last section, we used a 2-proportion Z-interval to estimate the parameter p1 − p2; here, we will use a 2-proportion Z-test to test the null hypothesis that p1 − p2 = 0, i.e. that p1 = p2. Recall that the test statistic Z has the form: Z = point estimate − null value SE of estimate The parameter of interest is p1 − p2, so the point estimate will be the observed diﬀerence of sample proportions: ˆp1 − ˆp2 = 0.47 − 0.34 = 0.13. The null value depends on the null hypothesis. The null hypothesis is that the approval rate would be the same for both statement orderings, i.e. that the diﬀerence is 0, therefore, the null value is 0. In this section we consider only the case where H0: p1 = p2, so the null value for the diﬀerence will always be 0. The SD of a diﬀerence of sample proportions has the form: SD = p1(1 − p1) n1 + p2(1 − p2) n2 However, in a hypothesis test, the distribution of the point estimate is always examined assuming the null hypothesis is true, i.e. in this case, p1 = p2. Both the success-failure check and the standard error formula should reﬂect this equality in the null hypothesis. We will use pc to represent the common proportion that support healthcare law regardless of statement order: SD = pc(1 − pc) n1 + pc(1 − pc) n2 = pc(1 − pc) 1 n1 + 1 n2 We don’t know the true proportion pc, but we can obtain a good estimate of it, ˆpc, by pooling the results of both samples. We ﬁnd the total number of “yeses” or “successes” and divide that by the total number of cases. This is equivalent to taking a weighted average of ˆp1 and ˆp2. We call ˆpc the pooled sample proportion, and we use it to check the success-failure condition and to compute the standard error when the null hypothesis is that p1 = p2. Here: ˆpc = 771(0.47) + 732(0.34) 771 + 732 = 0.407 POOLED SAMPLE PROPORTION When the null hypothesis is p1 = p2, it is useful to ﬁnd the pooled sample proportion: ˆpc = number of “successes” number of cases = x1 + x2 n1 + n2 = n1 ˆp1 + n2 ˆp2 n1 + n2 Here x1 represents the number of successes in sample 1. If x1 is not given, it can be computed as n1 × ˆp1. Similarly, x2 represents the number of successes in sample 2 and can be computed as n2 × ˆp2. 318 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA H0: p1 = p2 USE THE POOLED SAMPLE PROPORTION WHEN H0: p1 = p2 H0: p1 = p2 When the null hypothesis states that the proportions are equal, we use the pooled sample proportion (ˆpc) to check the success-failure condition and to estimate the standard error: SE = ˆpc(1 − ˆpc) 1 n1 + 1 n2 EXAMPLE 6.24 Verify that conditions for using the normal are met and ﬁnd the SE of estimate for this hypothesis test. Recall that the pooled
proportion ˆpc = 0.407, n1 = 771, and n2 = 732. The data do come from a randomized experiment, where the treatments are the two diﬀerent orderings of the question regarding healthcare (because this is an experiment, the 10% condition does not need to be checked). Also, the success-failure condition (minimums of 10) easily holds for each group. 771 × 0.407 ≥ 10 771 × (1 − 0.407) ≥ 10 732 × 0.407 ≥ 10 732 × (1 − 0.407) ≥ 10xf Here, we compute the SE for the diﬀerence of sample proportions as: SE = ˆpc(1 − ˆpc) 1 n1 + 1 n2 = 0.407(1 − 0.407) 1 771 + 1 732 = 0.025 EXAMPLE 6.25 Complete the hypothesis test using a signiﬁcance level of 0.01. We have already set up the hypotheses and veriﬁed that the diﬀerence of proportions can be modeled using a normal distribution. We can now calculate the test statistic and p-value. Z = point estimate − null value SE of estimate = (0.47 − 0.34) − 0 0.025 = 5.2 This is a two-tailed test as HA is that p1 = p2. We can ﬁnd the area in one tail and double it. Here, the p-value ≈ 0. Because the p-value is smaller than α = 0.01, we reject the null hypothesis and conclude that the order of the statements aﬀects how likely a respondent is to support the 2010 healthcare law. 6.2. INFERENCE FOR THE DIFFERENCE OF TWO PROPORTIONS 319 HYPOTHESIS TESTING FOR THE DIFFERENCE OF TWO PROPORTIONS To carry out a complete hypothesis test to test the claim that two proportions p1 and p2 are equal to each other, Identify: Identify the hypotheses and the signiﬁcance level, α. H0: p1 = p2 HA: p1 = p2; HA: p1 > p2; or HA: p1 < p2 Choose: Choose the correct test procedure and identify it by name. To test hypotheses about a diﬀerence of proportions we use a 2-proportion Z-test. Check: Check conditions for the sampling distribution for ˆp1− ˆp2 to be nearly normal, assuming H0: p1 = p2 is true. 1. Independence: Data come from 2 independent random samples or from a randomized experiment with two treatments. When sampling without replacement, check that the sample size is less than 10% of the population size for both samples. 2. Success-failure: n1 ˆpc ≥ 10, n1(1 − ˆpc) ≥ 10, n2 ˆpc ≥ 10, and n2(1 − ˆpc) ≥ 10 Calculate: Calculate the Z-statistic and p-value. Z = point estimate − null value SE of estimate point estimate: the diﬀerence of sample proportions ˆp1 − ˆp2 1 SE of estimate: ˆpc(1 − ˆpc) n1 null value: 0 + 1 n2 , where ˆpc is the pooled proportion p-value = (based on the Z-statistic and the direction of HA) Conclude: Compare the p-value to α, and draw a conclusion in context. If the p-value is < α, reject H0; there is suﬃcient evidence that [HA in context]. If the p-value is > α, do not reject H0; there is not suﬃcient evidence that [HA in context]. 320 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA EXAMPLE 6.26 A 5-year experiment was conducted to evaluate the eﬀectiveness of ﬁsh oils on reducing heart attacks, where each subject was randomized into one of two treatment groups. We’ll consider heart attack outcomes in these patients: ﬁsh oil placebo heart attack 145 200 no event Total 12933 12938 12788 12738 Carry out a complete hypothesis test at the 10% signiﬁcance level to test whether the use of ﬁsh oils is eﬀective in reducing heart attacks. Identify: Deﬁne p1 and p2 as follows: p1: the true proportion that would suﬀer a heart attack if given ﬁsh oil p2: the true proportion that would suﬀer a heart attack if given placebo We will test the following hypotheses at the α = 0.10 signiﬁcance level. H0: p1 = p2 HA: p1 < p2 Fish oil and placebo are equally eﬀective. Fish oil is eﬀective in reducing heart attacks. Choose: Because we are testing whether two proportions equal each other, we choose the 2- proportion Z-test for a diﬀerence of proportions. Check: We must verify that the diﬀerence of sample proportions can be modeled using a normal distribution. First we note that there is a randomized experiment with two treatments: ﬁsh oil and placebo. Second, we calculate the pooled proportion as follows: ˆpc = x1 + x2 n1 + n2 = 145 + 200 12933 + 12938 = 0.0133 We can now verify: 12933(0.0133) ≥ 10, 12933(1 − 0.0133) ≥ 10, 12938(0.0133) ≥ 10, and 12938(1 − 0.0133) ≥ 10, so both conditions are met. Calculate: We will calculate the Z-statistic and the p-value. Z = point estimate − null value SE of estimate The point estimate is the diﬀerence of sample proportions: ˆp1− ˆp2 = 0.0112−0.0155 = −0.0043. The value hypothesized for the parameter in H0 is the null value: null value = 0. The pooled proportion, calculated above, is: ˆpc = 0.0133. The SE of the diﬀerence of sample proportions, assuming H0 is true, is: ˆpc(1 − ˆpc) = 0.0133(1 − 0.0133) 12938 = 0.00142. 12933 + 1 1 1 n1 + 1 n2 Z = −0.0043 − 0 0.00142 = −3.0 Because HA uses a less than, meaning that it is a lower-tail test, the p-value is the area to the left of Z = −3.0 under the standard normal curve. This area can be found using a normal table or a calculator. The area or p-value = 0.0013. Conclude: The p-value of 0.0013 is < 0.10, so we reject H0; there is suﬃcient evidence that ﬁsh oil is eﬀective in reducing heart attacks. 6.2. INFERENCE FOR THE DIFFERENCE OF TWO PROPORTIONS 321 6.2.6 Technology: the 2-proportion Z-test TI-83/84: 2-PROPORTION Z-TEST Use STAT, TESTS, 2-PropZTest. 1. Choose STAT. 2. Right arrow to TESTS. 3. Down arrow and choose 6:2-PropZTest. 4. Let x1 be the number of yeses (must be an integer) in sample 1 and let n1 be the size of sample 1. 5. Let x2 be the number of yeses (must be an integer) in sample 2 and let n2 be the size of sample 2. 6. Choose =, <, or > to correspond to HA. 7. Choose Calculate and hit ENTER, which returns: z ^p1 ^p2 Z-statistic sample 1 proportion sample 2 proportion p ^p p-value pooled sample proportion CASIO FX-9750GII: 2-PROPORTION Z-TEST 1. Navigate to STAT (MENU button, then hit the 2 button or select STAT). 2. Choose the TEST option (F3 button). 3. Choose the Z option (F1 button). 4. Choose the 2-P option (F4 button). 5. Specify the test details: • Specify the sidedness of the test using the F1, F2, and F3 keys. • Enter the number of successes for each group, x1 and x2. • Enter the sample size for each group, n1 and n2. 6. Hit the EXE button, which returns z Z-statistic p-value p ^p1, ^p2 ^p n1, n2 sample proportions pooled proportion sample sizes GUIDED PRACTICE 6.27 Use a calculator to ﬁnd the test statistic, p-value, and pooled proportion for a test with: HA: p for ﬁsh oil < p for placebo.20 ﬁsh oil placebo heart attack 145 200 no event Total 12933 12938 12788 12738 20Correctly going through the calculator steps should lead to a solution with the test statistic z = -2.977 and the p-value p = 0.00145. These two values match our calculated values from the previous example to within rounding error. The pooled proportion is given as ^p = 0.0133. Note: values for x1 and x2 were given in the table. If, instead, proportions are given, ﬁnd x1 and x2 by multiplying the proportions by the sample sizes and rounding the result to an integer. 322 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA Section summary In the previous section, we looked at inference for a single proportion. In this section, we compared two groups to each other with respect to a proportion or a percent. • We are interested in whether the true proportion of yeses is the same or diﬀerent between two distinct groups. Call these proportions p1 and p2. The diﬀerence, p1 − p2 tells us whether p1 is greater than, less than, or equal to p2. • When comparing two proportions to each other, the parameter of interest is the diﬀerence of proportions, p1 − p2, and we use the diﬀerence of sample proportions, ˆp1 − ˆp2, as the point estimate. • The sampling distribution for ˆp1 − ˆp2 is nearly normal when the success-failure condition is met for both groups and when the observations are independent between and within groups. When the sampling distribution for ˆp1 − ˆp2 is nearly normal, the standardized test statistic also follows a normal distribution. • When the null hypothesis is that the two populations proportions are equal to each other, use the pooled sample proportion ˆpc = x1+x2 , i.e. the combined number of yeses over the n1+n2 combined sample sizes, when verifying the success-failure condition and when ﬁnding the SE. For the conﬁdence interval, do not use the pooled sample proportion; use the separate values of ˆp1 and ˆp2. • When there are two samples or treatments and the parameter of interest is a diﬀerence of proportions, e.g. the true diﬀerence in proportion of 17 and 18 year olds with a summer job (proportion of 18 year olds − proportion of 17 year olds): – Estimate p1 − p2 at the C% conﬁdence level using a 2-proportion Z-interval. – Test H0: p1 − p2 = 0 at the α signiﬁcance level using a 2-proportion Z-test. • The two proportion Z-interval and Z-test require the sampling distribution for ˆp1 − ˆp2 to be nearly normal. For this reason we must check that the following conditions are met. 1. Independence: Data come from 2 independent random samples or from a randomized experiment with 2 treatments. When sampling without replacement, check that the sample size is less than 10% of the population size for both samples. 2. Success-failure for CI: n1 ˆp1 ≥ 10, n1(1 − ˆp1) ≥ 10, n2 ˆp2 ≥ 10, and n2(1 − ˆp2) ≥ 10. Success-failure for Test: n1 ˆpc ≥ 10, n1(1 − ˆpc) ≥ 10, n2 ˆpc ≥ 10, and n2(1 − ˆpc) ≥ 10. • When the conditions are met, we calculate the conﬁdence interval and the test statistic using the same structure as in the previous section. Conﬁdence interval: point estimate ± z × SE of estimate Test statistic: Z = point estimate − null value SE of estimate Here the point estimate is the diﬀerence of sample proportions ˆp1 − ˆp2. The SE of estimate is the SE of a diﬀerence of sample proportions. – For a CI, use: SE = – For a Test, use: SE = ˆpc(1 − ˆpc) . + ˆp2(1− ˆp2) n2 1 n1 + 1 n2 . ˆp1(1− ˆp1) n1 6.2. INFERENCE FOR THE DIFFERENCE OF TWO PROPORTIONS 323 Exercises 6.15 Social experiment, Part I. A “social experiment” conducted by a TV program questione
d what people do when they see a very obviously bruised woman getting picked on by her boyfriend. On two diﬀerent occasions at the same restaurant, the same couple was depicted. In one scenario the woman was dressed “provocatively” and in the other scenario the woman was dressed “conservatively”. The table below shows how many restaurant diners were present under each scenario, and whether or not they intervened. Scenario Provocative Conservative Total Intervene Yes No Total 5 15 20 15 10 25 20 25 45 Explain why the sampling distribution for the diﬀerence between the proportions of interventions under provocative and conservative scenarios does not follow an approximately normal distribution. 6.16 Heart transplant success. The Stanford University Heart Transplant Study was conducted to determine whether an experimental heart transplant program increased lifespan. Each patient entering the program was oﬃcially designated a heart transplant candidate, meaning that he was gravely ill and might beneﬁt from a new heart. Patients were randomly assigned into treatment and control groups. Patients in the treatment group received a transplant, and those in the control group did not. The table below displays how many patients survived and died in each group.21 survived died control 4 30 treatment 24 45 Suppose we are interested in estimating the diﬀerence in survival rate between the control and treatment groups using a conﬁdence interval. Explain why we cannot construct such an interval using the normal approximation. What might go wrong if we constructed the conﬁdence interval despite this problem? 6.17 Gender and color preference. A study asked 1,924 male and 3,666 female undergraduate college students their favorite color. A 95% conﬁdence interval for the diﬀerence between the proportions of males and females whose favorite color is black (pmale − pf emale) was calculated to be (0.02, 0.06). Based on this information, determine if the following statements about undergraduate college students are true or false, and explain your reasoning for each statement you identify as false.22 (a) We are 95% conﬁdent that the true proportion of males whose favorite color is black is 2% lower to 6% higher than the true proportion of females whose favorite color is black. (b) We are 95% conﬁdent that the true proportion of males whose favorite color is black is 2% to 6% higher than the true proportion of females whose favorite color is black. (c) 95% of random samples will produce 95% conﬁdence intervals that include the true diﬀerence between the population proportions of males and females whose favorite color is black. (d) We can conclude that there is a signiﬁcant diﬀerence between the proportions of males and females whose favorite color is black and that the diﬀerence between the two sample proportions is too large to plausibly be due to chance. (e) The 95% conﬁdence interval for (pf emale − pmale) cannot be calculated with only the information given in this exercise. 21B. Turnbull et al. “Survivorship of Heart Transplant Data”. In: Journal of the American Statistical Association 69 (1974), pp. 74–80. 22L Ellis and C Ficek. “Color preferences according to gender and sexual orientation”. In: Personality and Individual Diﬀerences 31.8 (2001), pp. 1375–1379. 324 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA 6.18 The Daily Show. A Pew Research foundation poll indicates that among a random sample of 1,099 college graduates, 33% watch The Daily Show. Meanwhile, 22% of the 1,110 people with a high school degree but no college degree in the poll watch The Daily Show. A 95% conﬁdence interval for (pcollege grad − pHS or less), where p is the proportion of those who watch The Daily Show, is (0.07, 0.15). Based on this information, determine if the following statements are true or false, and explain your reasoning if you identify the statement as false.23 (a) At the 5% signiﬁcance level, the data provide convincing evidence of a diﬀerence between the proportions of college graduates and those with a high school degree or less who watch The Daily Show. (b) We are 95% conﬁdent that 7% less to 15% more college graduates watch The Daily Show than those with a high school degree or less. (c) 95% of random samples of 1,099 college graduates and 1,110 people with a high school degree or less will yield diﬀerences in sample proportions between 7% and 15%. (d) A 90% conﬁdence interval for (pcollege grad − pHS or less) would be wider. (e) A 95% conﬁdence interval for (pHS or less − pcollege grad) is (-0.15,-0.07). 6.19 National Health Plan, Part III. Exercise 6.9 presents the results of a poll evaluating support for a generically branded “National Health Plan” in the United States. 79% of 347 Democrats and 55% of 617 Independents support a National Health Plan. (a) Calculate a 95% conﬁdence interval for the diﬀerence between the proportion of Democrats and Independents who support a National Health Plan (pD − pI ), and interpret it in this context. We have already checked conditions for you. (b) True or false: If we had picked a random Democrat and a random Independent at the time of this poll, it is more likely that the Democrat would support the National Health Plan than the Independent. 6.20 Sleep deprivation, CA vs. OR, Part I. According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insuﬃcient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Calculate a 95% conﬁdence interval to estimate the diﬀerence between the proportions of Californians and Oregonians who are sleep deprived. Include all steps of the Identify, Choose, Check, Calculate, Conclude framework.24 6.21 Sleep deprived transportation workers. The National Sleep Foundation conducted a survey on the sleep habits of randomly sampled transportation workers and randomly sampled non-transportation workers that serve as a “control” for comparison. The results of the survey are shown below.25 Transportation Professionals Train Truck Less than 6 hours of sleep 6 to 8 hours of sleep More than 8 hours Total Control 35 193 64 292 Pilots Drivers Operators 19 132 51 202 35 117 51 203 29 119 32 180 Bus/Taxi/Limo Drivers 21 131 58 210 Conduct a hypothesis test to evaluate if these data provide evidence of a diﬀerence between the proportion of truck drivers and non-transportation workers (the “control” group) who get less than 6 hours of sleep per day (i.e. are considered sleep deprived). Include all steps of the Identify, Choose, Check, Calculate, Conclude framework. 23The Pew Research Center, Americans Spending More Time Following the News, data collected June 8-28, 2010. 24CDC, Perceived Insuﬃcient Rest or Sleep Among Adults — United States, 2008. 25National Sleep Foundation, 2012 Sleep in America Poll: Transportation Workers’ Sleep, 2012. 6.2. INFERENCE FOR THE DIFFERENCE OF TWO PROPORTIONS 325 6.22 Sleep deprivation, CA vs. OR, Part II. Exercise 6.20 provides data on sleep deprivation rates of Californians and Oregonians. The proportion of California residents who reported insuﬃcient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. (a) Conduct a hypothesis test to determine if these data provide strong evidence that the rate of sleep deprivation is diﬀerent for the two states. (Reminder: Check conditions) (b) It is possible the conclusion of the test in part (a) is incorrect. If this is the case, what type of error was made? 6.23 Prenatal vitamins and Autism. Researchers studying the link between prenatal vitamin use and autism surveyed the mothers of a random sample of children aged 24 - 60 months with autism and conducted another separate random sample for children with typical development. The table below shows the number of mothers in each group who did and did not use prenatal vitamins during the three months before pregnancy (periconceptional period).26 Autism Periconceptional No vitamin prenatal vitamin Vitamin 111 143 254 (a) State appropriate hypotheses to test for independence of use of prenatal vitamins during the three Autism Typical development Total 181 70 302 159 483 229 Total months before pregnancy and autism. (b) Complete the hypothesis test and state an appropriate conclusion. (Reminder: Verify any necessary conditions for the test.) (c) A New York Times article reporting on this study was titled “Prenatal Vitamins May Ward Oﬀ Autism”. Do you ﬁnd the title of this article to be appropriate? Explain your answer. Additionally, propose an alternative title.27 6.24 An apple a day keeps the doctor away. A physical education teacher at a high school wanting to increase awareness on issues of nutrition and health asked her students at the beginning of the semester whether they believed the expression “an apple a day keeps the doctor away”, and 40% of the students responded yes. Throughout the semester she started each class with a brief discussion of a study highlighting positive eﬀects of eating more fruits and vegetables. She conducted the same apple-a-day survey at the end of the semester, and this time 60% of the students responded yes. Can she used a two-proportion method from this section for this analysis? Explain your reasoning. 26R.J. Schmidt et al. “Prenatal vitamins, one-carbon metabolism gene variants, and risk for autism”. In: Epidemi- ology 22.4 (2011), p. 476. 27R.C. Rabin. “Patterns: Prenatal Vitamins May Ward Oﬀ Autism”. In: New York Times (2011). 326 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA 6.3 Testing for goodness of ﬁt using chi-square In this section, we develop a method for assessing a null model when the data take on more than two categories, such as yes/no/maybe
instead of simply yes/no. This allows us to answer questions such as the following: • Are juries representative of the population in terms of race/ethnicity, or is there a bias in jury selection? • Is the color distribution of actual M&M’s consistent with what was reported on the Mars website? • Do people choose rock, paper, scissors with the same likelihood, or is one choice favored over another? Learning objectives 1. Calculate the expected counts and degrees of freedom for a one-way table. 2. Calculate and interpret the test statistic χ2. 3. State and verify whether or not the conditions for the chi-square goodness of ﬁt are met. 4. Carry out a complete hypothesis test to evaluate if the distribution of a categorical variable follows a hypothesized distribution. 5. Understand how the degrees of freedom aﬀect the shape of the chi-square curve. 6.3.1 Creating a test statistic for one-way tables Data is collected from a random sample of 275 jurors in a small county. Jurors identiﬁed their racial/ethnic group, as shown in Figure 6.4, and we would like to determine if these jurors are representative of the population with respect to race/ethnicity. If the jury is representative of the population, then the proportions in the sample should roughly reﬂect the population of eligible jurors, i.e. registered voters. Race/Ethnicity Representation in juries Registered voters White Black Hispanic Other 205 0.72 26 0.07 25 0.12 19 0.09 Total 275 1.00 Figure 6.4: Representation by race in a city’s juries and population. While the proportions in the juries do not precisely represent the population proportions, it is unclear whether these data provide convincing evidence that the sample is not representative. If the jurors really were randomly sampled from the registered voters, we might expect small diﬀerences due to chance. However, unusually large diﬀerences may provide convincing evidence that the juries were not representative. 6.3. TESTING FOR GOODNESS OF FIT USING CHI-SQUARE 327 EXAMPLE 6.28 Of the people in the city, 275 served on a jury. If the individuals are randomly selected to serve on a jury, about how many of the 275 people would we expect to be White? How many would we expect to be Black? About 72% of the population is White, so we would expect about 72% of the jurors to be White: 0.72 × 275 = 198. Similarly, we would expect about 7% of the jurors to be Black, which would correspond to about 0.07 × 275 = 19.25 Black jurors. GUIDED PRACTICE 6.29 Twelve percent of the population is Hispanic and 9% represent other racial/ethnic groups. How many of the 275 jurors would we expect to be Hispanic or from another racial/ethnic group? Answers can be found in Figure 6.5. Race/Ethnicity Observed data Expected counts White Black Hispanic Other 205 198 26 19.25 25 33 19 24.75 Total 275 275 Figure 6.5: Actual and expected make-up of the jurors. The sample proportion represented from each race/ethnicity among the 275 jurors was not a precise match for any ethnic group. While some sampling variation is expected, we would expect the sample proportions to be fairly similar to the population proportions if there is no bias on juries. We need to test whether the diﬀerences are strong enough to provide convincing evidence that the jurors are not a random sample. These ideas can be organized into hypotheses: H0: The jurors are a random sample, i.e. there is no racial/ethnic bias in who serves on a jury, and the observed counts reﬂect natural sampling ﬂuctuation. HA: The jurors are not randomly sampled, i.e. there is racial/ethnic bias in juror selection. To evaluate these hypotheses, we quantify how diﬀerent the observed counts are from the expected counts. Strong evidence for the alternative hypothesis would come in the form of unusually large deviations in the groups from what would be expected based on sampling variation alone. 6.3.2 The chi-square test statistic In previous hypothesis tests, we constructed a test statistic of the following form: Z = point estimate − null value SE of point estimate This construction was based on (1) identifying the diﬀerence between a point estimate and an expected value if the null hypothesis was true, and (2) standardizing that diﬀerence using the standard error of the point estimate. These two ideas will help in the construction of an appropriate test statistic for count data. In this example we have four categories: White, Black, Hispanic, and other. Because we have four values rather than just one or two, we need a new tool to analyze the data. Our strategy will be to ﬁnd a test statistic that measures the overall deviation between the observed and the expected counts. We ﬁrst ﬁnd the diﬀerence between the observed and expected counts for the four groups: observed - expected 205 − 198 26 − 19.25 White Black Hispanic 25 − 33 Other 19 − 24.75 328 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA Next, we square the diﬀerences: (observed - expected)2 White (205 − 198)2 Black (26 − 19.25)2 Hispanic (25 − 33)2 Other (19 − 24.75)2 We must standardize each term. To know whether the squared diﬀerence is large, we compare it to what was expected. If the expected count was 5, a squared diﬀerence of 25 is very large. However, if the expected count was 1,000, a squared diﬀerence of 25 is very small. We will divide each of the squared diﬀerences by the corresponding expected count. (observed - expected)2 expected (205 − 198)2 198 (26 − 19.25)2 19.25 (25 − 33)2 33 (19 − 24.75)2 24.75 White Black Hispanic Other Finally, to arrive at the overall measure of deviation between the observed counts and the expected counts, we add up the terms. χ2 = (observed - expected)2 expected = (205 − 198)2 198 + (26 − 19.25)2 19.25 + (25 − 33)2 33 + (19 − 24.75)2 24.75 We can write an equation for χ2 using the observed counts and expected counts: χ2 = (observed count1 − expected count1)2 expected count1 + · · · + (observed count4 − expected count4)2 expected count4 The ﬁnal number χ2 summarizes how strongly the observed counts tend to deviate from the null counts. In Section 6.3.4, we will see that if the null hypothesis is true, then χ2 follows a new distribution called a chi-square distribution. Using this distribution, we will be able to obtain a p-value to evaluate whether there appears to be racial/ethnic bias in the juries for the city we are considering. 6.3.3 The chi-square distribution and ﬁnding areas The chi-square distribution is sometimes used to characterize data sets and statistics that are always positive and typically right skewed. Recall a normal distribution had two parameters – mean and standard deviation – that could be used to describe its exact characteristics. The chisquare distribution has just one parameter called degrees of freedom (df ), which inﬂuences the shape, center, and spread of the distribution. GUIDED PRACTICE 6.30 Figure 6.6 shows three chi-square distributions. (a) How does the center of the distribution change when the degrees of freedom is larger? (b) What about the variability (spread)? (c) How does the shape change?28 Figure 6.6 and Guided Practice 6.30 demonstrate three general properties of chi-square distributions as the degrees of freedom increases: the distribution becomes more symmetric, the center moves to the right, and the variability inﬂates. 28(a) The center becomes larger. If we look carefully, we can see that the center of each distribution is equal to the distribution’s degrees of freedom. (b) The variability increases as the degrees of freedom increases. (c) The distribution is very strongly right skewed for df = 2, and then the distributions become more symmetric for the larger degrees of freedom df = 4 and df = 9. In fact, as the degrees of freedom increase, the χ2 distribution approaches a normal distribution. 6.3. TESTING FOR GOODNESS OF FIT USING CHI-SQUARE 329 Figure 6.6: Three chi-square distributions with varying degrees of freedom. Our principal interest in the chi-square distribution is the calculation of p-values, which (as we have seen before) is related to ﬁnding the relevant area in the tail of a distribution. To do so, a new table is needed: the chi-square table, partially shown in Figure C.5. A more complete table is presented in Appendix C.4 on page 518. This table is very similar to the t-table from Sections 7.1 and 7.3: we identify a range for the area, and we examine a particular row for distributions with diﬀerent degrees of freedom. One important diﬀerence from the t-table is that the chi-square table only provides upper tail values. Upper tail 1 df 2 3 4 5 6 7 0.3 1.07 2.41 3.66 4.88 6.06 7.23 8.38 0.2 1.64 3.22 4.64 5.99 7.29 8.56 9.80 0.1 2.71 4.61 6.25 7.78 9.24 10.64 12.02 0.05 3.84 5.99 7.81 9.49 11.07 12.59 14.07 0.02 5.41 7.82 9.84 11.67 13.39 15.03 16.62 0.01 6.63 9.21 11.34 13.28 15.09 16.81 18.48 0.005 7.88 10.60 12.84 14.86 16.75 18.55 20.28 0.001 10.83 13.82 16.27 18.47 20.52 22.46 24.32 Figure 6.7: A section of the chi-square table. A complete table is in Appendix C.4. EXAMPLE 6.31 Figure 6.8(a) shows a chi-square distribution with 3 degrees of freedom and an upper shaded tail starting at 6.25. Use Figure C.5 to estimate the shaded area. This distribution has three degrees of freedom, so only the row with 3 degrees of freedom (df) is relevant. This row has been italicized in the table. Next, we see that the value – 6.25 – falls in the column with upper tail area 0.1. That is, the shaded upper tail of Figure 6.8(a) has area 0.1. EXAMPLE 6.32 We rarely observe the exact value in the table. For instance, Figure 6.8(b) shows the upper tail of a chi-square distribution with 2 degrees of freedom. The lower bound for this upper tail is at 4.3, which does not fall in Figure C.5. Find the approximate tail area. The cutoﬀ 4.3 falls between the second and third columns in the 2 degrees of freedom row. Because these columns correspond to tail areas of 0.2 and 0.1, we can be certain that the area shaded in Figure 6.8(b) is between 0.1 and 0.2. Using
a calculator or statistical software allows us to get more precise areas under the chi-square curve than we can get from the table alone. 0510152025Degrees of Freedom249 330 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA (a) (c) (e) (b) (d) (f) Figure 6.8: (a) Chi-square distribution with 3 degrees of freedom, area above 6.25 shaded. (b) 2 degrees of freedom, area above 4.3 shaded. (c) 5 degrees of freedom, area above 5.1 shaded. (d) 7 degrees of freedom, area above 11.7 shaded. (e) 4 degrees of freedom, area above 10 shaded. (f ) 3 degrees of freedom, area above 9.21 shaded. 051015202505101520250510152025051015202505101520250510152025 6.3. TESTING FOR GOODNESS OF FIT USING CHI-SQUARE 331 TI-84: FINDING AN UPPER TAIL AREA UNDER THE CHI-SQUARE CURVE Use the χ2cdf command to ﬁnd areas under the chi-square curve. 1. Hit 2ND VARS (i.e. DISTR). 2. Choose 8:χ2cdf. 3. Enter the lower bound, which is generally the chi-square value. 4. Enter the upper bound. Use a large number, such as 1000. 5. Enter the degrees of freedom. 6. Choose Paste and hit ENTER. TI-83: Do steps 1-2, then type the lower bound, upper bound, and degrees of freedom separated by commas. e.g. χ2cdf(5, 1000, 3), and hit ENTER. CASIO FX-9750GII: FINDING AN UPPER TAIL AREA UNDER THE CHI-SQ. CURVE 1. Navigate to STAT (MENU button, then hit the 2 button or select STAT). 2. Choose the DIST option (F5 button). 3. Choose the CHI option (F3 button). 4. Choose the Ccd option (F2 button). 5. If necessary, select the Var option (F2 button). 6. Enter the Lower bound (generally the chi-square value). 7. Enter the Upper bound (use a large number, such as 1000). 8. Enter the degrees of freedom, df. 9. Hit the EXE button. GUIDED PRACTICE 6.33 Figure 6.8(c) shows an upper tail for a chi-square distribution with 5 degrees of freedom and a cutoﬀ of 5.1. Find the tail area using a calculator.29 GUIDED PRACTICE 6.34 Figure 6.8(d) shows a cutoﬀ of 11.7 on a chi-square distribution with 7 degrees of freedom. Find the area of the upper tail.30 GUIDED PRACTICE 6.35 Figure 6.8(e) shows a cutoﬀ of 10 on a chi-square distribution with 4 degrees of freedom. Find the area of the upper tail.31 GUIDED PRACTICE 6.36 Figure 6.8(f) shows a cutoﬀ of 9.21 with a chi-square distribution with 3 df. Find the area of the upper tail.32 29Use a lower bound of 5.1, an upper bound of 1000, and df = 5. The upper tail area is 0.4038. 30The area is 0.1109. 31The area is 0.0404. 32The area is 0.0266. 332 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA 6.3.4 Finding a p-value for a chi-square distribution In Section 6.3.2, we identiﬁed a new test statistic (χ2) within the context of assessing whether there was evidence of racial/ethnic bias in how jurors were sampled. The null hypothesis represented the claim that jurors were randomly sampled and there was no racial/ethnic bias. The alternative hypothesis was that there was racial/ethnic bias in how the jurors were sampled. We determined that a large χ2 value would suggest strong evidence favoring the alternative hypothesis: that there was racial/ethnic bias. However, we could not quantify what the chance was of observing such a large test statistic (χ2 = 5.89) if the null hypothesis actually was true. This is where the chi-square distribution becomes useful. If the null hypothesis was true and there was no racial/ethnic bias, then χ2 would follow a chi-square distribution, with three degrees of freedom in this case. Under certain conditions, the statistic χ2 follows a chi-square distribution with k − 1 degrees of freedom, where k is the number of bins or categories of the variable. EXAMPLE 6.37 How many categories were there in the juror example? How many degrees of freedom should be associated with the chi-square distribution used for χ2? In the jurors example, there were k = 4 categories: White, Black, Hispanic, and other. According to the rule above, the test statistic χ2 should then follow a chi-square distribution with k − 1 = 3 degrees of freedom if H0 is true. Just like we checked sample size conditions to use the normal model in earlier sections, we must also check a sample size condition to safely model χ2 with a chi-square distribution. Each expected count must be at least 5. In the juror example, the expected counts were 198, 19.25, 33, and 24.75, all easily above 5, so we can model the χ2 test statistic, using a chi-square distribution. EXAMPLE 6.38 If the null hypothesis is true, the test statistic χ2 = 5.89 would be closely associated with a chisquare distribution with three degrees of freedom. Using this distribution and test statistic, identify the p-value and state whether or not there is evidence of racial/ethnic bias in the juror selection. The chi-square distribution and p-value are shown in Figure 6.9. Because larger chi-square values correspond to stronger evidence against the null hypothesis, we shade the upper tail to represent the p-value. Using a calculator, we look at the chi-square curve with 3 degrees of freedom and ﬁnd the area to the right of χ2 = 5.89. This area, which corresponds to the p-value, is equal to 0.117. This p-value is larger than the default signiﬁcance level of 0.05, so we do not reject the null hypothesis. In other words, the data do not provide convincing evidence of racial/ethnic bias in the juror selection. Figure 6.9: The p-value for the juror hypothesis test is shaded in the chi-square distribution with df = 3. The test that we just carried out regarding jury selection is known as the χ2χ2χ2 goodness of ﬁt test. It is called “goodness of ﬁt” because we test whether or not the proposed or expected distribution is a good ﬁt for the observed data. 051015 6.3. TESTING FOR GOODNESS OF FIT USING CHI-SQUARE 333 CHI-SQUARE GOODNESS OF FIT TEST FOR ONE-WAY TABLE Suppose we are to evaluate whether there is convincing evidence that a set of observed counts O1, O2, ..., Ok in k categories are unusually diﬀerent from what might be expected under a null hypothesis. Calculate the expected counts that are based on the null hypothesis E1, E2, ..., Ek. If each expected count is at least 5 and the null hypothesis is true, then the test statistic below follows a chi-square distribution with k − 1 degrees of freedom: χ2 = (O1 − E1)2 E1 + (O2 − E2)2 E2 + · · · + (Ok − Ek)2 Ek The p-value for this test statistic is found by looking at the upper tail of this chi-square distribution. We consider the upper tail because larger values of χ2 would provide greater evidence against the null hypothesis. CONDITIONS FOR THE CHI-SQUARE GOODNESS OF FIT TEST The chi-square goodness of ﬁt test requires the test statistic to be well modeled by a chi-square distribution. This will be valid when the observations are independent and the expected counts are large. If these conditions are not met, the chi-square goodness of ﬁt test should not be used. Independence. The observations can be considered independent if the data come from a random process. If randomly sampling without replacement from a ﬁnite population, the observations can be considered independent when sampling less than 10% of the population. Large expected counts. In order for the χ2-statistic to follow the chi-square distribution, each particular bin or category must have at least 5 expected cases under the assumption that the null hypothesis is true. 334 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA 6.3.5 Evaluating goodness of ﬁt for a distribution GOODNESS OF FIT TEST FOR A ONE-WAY TABLE When there is one sample and we are comparing the distribution of a categorical variable to a speciﬁed or population distribution, e.g. using sample values to determine if a machine is producing M&M’s with the speciﬁed distribution of color, Identify: Identify the hypotheses and the signiﬁcance level, α. H0: The distribution of [...] matches the speciﬁed or population distribution. HA: The distribution of [...] doesn’t match the speciﬁed or population distribution. Choose: Choose the correct test procedure and identify it by name. Here we use a χ2χ2χ2 goodness of ﬁt test. Check: Check that the test statistic follows a chi-square distribution. 1. Independence: Data come from a random sample or random process. If sampling without replacement, check that the sample size is less than 10% of the population size. 2. Expected counts: All expected counts are ≥ 5. Calculate: Calculate the χ2-statistic, df , and p-value. test statistic: χ2 = (observed − expected)2 df = # of categories − 1 p-value = (area to the right of χ2-statistic with the appropriate df ) expected Conclude: Compare the p-value to α, and draw a conclusion in context. If the p-value is < α, reject H0; there is suﬃcient evidence that [HA in context]. If the p-value is > α, do not reject H0; there is not suﬃcient evidence that [HA in context]. Have you ever wondered about the color distribution of M&M’s®? If so, then you will be glad to know that Rick Wicklin, a statistician working at the statistical software company SAS, wondered about this too. But he did more than wonder; he decided to collect data to test whether the distribution of M&M colors was consistent with the stated distribution published on the Mars website in 2008. Starting at end of 2016, over the course of several weeks, he collected a sample of 712 candies, or about 1.5 pounds. We will investigate his results in the next example. You can read about his adventure in the Quartz article linked in the Data Appendix, which starts on page 503. 6.3. TESTING FOR GOODNESS OF FIT USING CHI-SQUARE 335 EXAMPLE 6.39 The stated color distribution of M&M’s on the Mars website in 2008 is shown in the table below, along with the observed percentages from Rick Wicklin’s sample of size 712. (See the paragraph before this example for more background.) website percentages (2008): observed percentages: Brown Blue 24% 13% 18.7% 18.7% 19.5% 14.5% 15.1% 13.5% Orange Green Yellow 16% Red 13% 14% 20% Is there evidence at the 5% signiﬁcance level that the distribution of M&M’s in 2016 were diﬀerent from
the stated distribution on the website in 2008? Use the ﬁve step framework to organize your work. Identify: We will test the following hypotheses at the α = 0.05 signiﬁcance level. H0: The distribution of M&M colors is the same as the stated distribution in 2008. HA: The distribution of M&M colors is diﬀerent than the stated distribution in 2008. Choose: Because we have one variable (color), broken up into multiple categories, we choose the chi-square goodness of ﬁt test. Check: We must verify that the test statistic follows a chi-square distribution. Note that there is only one sample here. The website percentages are considered ﬁxed – they are not the result of a sample and do not have sampling variability associated with them. To carry out the chi-square goodness of ﬁt test, we will have to assume that Wicklin’s sample can be considered a random sample of M&M’s. We note that the total population size of M&M’s is much larger than 10 times the sample size of 712. Next, we need to ﬁnd the expected counts. Here, n = 712. If H0 is true, then we would expect 24% of the M&M’s to be Blue, 20% to be Orange, etc. So the expected counts can be found as: expected counts: Blue 0.24(712) = 170.9 Orange 0.20(712) = 142.4 Green 0.16(712) = 113.9 Yellow 0.14(712) = 99.6 Red 0.13(712) = 92.6 Brown 0.13(712) = 92.6 Calculate: We will calculate the chi-square statistic, degrees of freedom, and the p-value. To calculate the chi-square statistic, we need the observed counts as well as the expected counts. To ﬁnd the observed counts, we use the observed percentages. For example, 18.7% of 712 = 0.187(712) = 133. observed counts: expected counts: Blue Orange Green Yellow Red Brown 133 170.9 139 113.9 133 142.4 108 92.6 96 92.6 103 99.6 χ2 = (observed − expected)2 expected = (133 − 170.9)2 170.9 + (133 − 142.4)2 142.4 + · · · + (108 − 92.6)2 92.6 + (96 − 92.6)2 92.6 =8.41 + 0.62 + 5.53 + 0.12 + 2.56 + 0.12 =17.36 Because there are six colors, the degrees of freedom is 6 − 1 = 5. In a chi-square test, the p-value is always the area to the right of the chi-square statistic. Here, the area to the right of 17.36 under the chi-square curve with 5 degrees of freedom is 0.004. Conclude: The p-value of 0.004 is < 0.05, so we reject H0; there is suﬃcient evidence that the distribution of M&M’s does not match the stated distribution on the website in 2008. 336 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA EXAMPLE 6.40 For Wicklin’s sample, which color showed the most prominent diﬀerence from the stated website distribution in 2008? We can compare the website percentages with the observed percentages. However, another approach is to look at the terms used when calculating the chi-square statistic. We note that the largest term, 8.41, corresponds to Blue. This means that the observed number for Blue was, relatively speaking, the farthest from the expected number among all of the colors. This is consistent with the observation that the largest diﬀerence in website percentage and observed percentage is for Blue (24% vs 18.7%). Wicklin observed far fewer Blue M&M’s than would have been expected if the website percentages were still true. 6.3. TESTING FOR GOODNESS OF FIT USING CHI-SQUARE 337 6.3.6 Technology: the chi-square goodness of ﬁt test TI-84: CHI-SQUARE GOODNESS OF FIT TEST Use STAT, TESTS, χ2GOF-Test. 1. Enter the observed counts into list L1 and the expected counts into list L2. 2. Choose STAT. 3. Right arrow to TESTS. 4. Down arrow and choose D:χ2GOF-Test. 5. Leave Observed: L1 and Expected: L2. 6. Enter the degrees of freedom after df: 7. Choose Calculate and hit ENTER, which returns: χ2 p df chi-square test statistic p-value degrees of freedom TI-83: Unfortunately the TI-83 does not have this test built in. To carry out the test manually, make list L3 = (L1 - L2)2 / L2 and do 1-Var-Stats on L3. The sum of L3 will correspond to the value of χ2 for this test. CASIO FX-9750GII: CHI-SQUARE GOODNESS OF FIT TEST 1. Navigate to STAT (MENU button, then hit the 2 button or select STAT). 2. Enter the observed counts into a list (e.g. List 1) and the expected counts into list (e.g. List 2). 3. Choose the TEST option (F3 button). 4. Choose the CHI option (F3 button). 5. Choose the GOF option (F1 button). 6. Adjust the Observed and Expected lists to the corresponding list numbers from Step 2. 7. Enter the degrees of freedom, df. 8. Specify a list where the contributions to the test statistic will be reported using CNTRB. This list number should be diﬀerent from the others. 9. Hit the EXE button, which returns chi-square test statistic p-value degrees of freedom list showing the test statistic contributions χ2 p df CNTRB GUIDED PRACTICE 6.41 Use the table below and a calculator to ﬁnd the χ2-statistic and p-value for chi-square goodness of ﬁt test.33 observed counts: expected counts: Blue Orange Green Yellow Red Brown 133 170.9 139 113.9 133 142.4 103 99.6 108 92.6 96 92.6 33Enter the observed counts into L1 and the expected counts into L2. the GOF test. Make sure that Observed: is L1 and Expected: is L2. Let df: be 5. You should ﬁnd that χ2 = 17.36 and p-value = 0.004. 338 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA Section summary The inferential procedures we saw in the ﬁrst two sections of this chapter are based on the test statistic following a normal distribution. In this section, we introduced a new distribution called the chi-square distribution. • While a normal distribution is deﬁned by its mean and standard deviation, the chi-square distribution is deﬁned by just one parameter called degrees of freedom. • For a chi-square distribution, as the degrees of freedom increases: the center increases, the spread increases, and the shape becomes more symmetric and more normal.34 • When we want to see if a model is a good ﬁt for observed data or if data is representative of a particular population, we can use a χ2χ2χ2 goodness of ﬁt test. This test is used when there is one variable with multiple categories (bins) that can be arranged in a one-way table. • In a chi-square goodness of ﬁt test, we calculate a χ2χ2χ2-statistic, which is a measure of how far the observed counts in the sample are from the expected counts, relative to the expected counts, under the null hypothesis. χ2 = (observed − expected)2 . expected – Always use whole numbers (counts) for the observed values, not proportions or percents. – For each category, the expected counts can be found by multiplying the sample size by the expected proportion under the null hypothesis. Expected counts do not need to be integers. • A larger χ2 represents greater deviation between the observed values and the expected values, relative to the expected values. For a ﬁxed degrees of freedom, a larger χ2 value leads to a smaller p-value, providing greater evidence against H0. • χ2χ2χ2 tests for a one-way table. When there is one sample and we are comparing the distribution of a categorical variable to a speciﬁed or population distribution, e.g. using sample values to determine if a machine is producing M&M’s with the speciﬁed distribution of color, the hypotheses can often be written as: H0: The distribution of [...] matches the speciﬁed or population distribution. HA: The distribution of [...] doesn’t match the speciﬁed or population distribution. We test these hypotheses at the α signiﬁcance level using a χ2χ2χ2 goodness of ﬁt test. • For the χ2 goodness of ﬁt test, we check the following conditions to verify that the test statistic follows a chi-square distribution. 1. Independence: Data come from a random sample or random process. When sampling without replacement, check that sample size is less than 10% of the population size. 2. Expected counts: All expected counts are ≥ 5. • We calculate the test statistic as follows: test statistic: χ2 = (observed − expected)2 expected ; df = # of categories − 1 • The p-value is the area to the right of the χ2-statistic under the chi-square curve with the appropriate df . • For a χ2 test, the p-value corresponds to the probability of getting a test statistic as large as we got or larger, assuming the null hypothesis is true and assuming the chi-square model holds. 34Technically, however, it is always right skewed. 6.3. TESTING FOR GOODNESS OF FIT USING CHI-SQUARE 339 Exercises 6.25 True or false, Part I. Determine if the statements below are true or false. For each false statement, suggest an alternative wording to make it a true statement. (a) The chi-square distribution, just like the normal distribution, has two parameters, mean and standard deviation. (b) The chi-square distribution is always right skewed, regardless of the value of the degrees of freedom parameter. (c) The chi-square statistic is always greater than or equal to 0. (d) As the degrees of freedom increases, the shape of the chi-square distribution becomes more skewed. 6.26 True or false, Part II. Determine if the statements below are true or false. For each false statement, suggest an alternative wording to make it a true statement. (a) As the degrees of freedom increases, the mean of the chi-square distribution increases. (b) If you found χ2 = 10 with df = 5 you would fail to reject H0 at the 5% signiﬁcance level. (c) When ﬁnding the p-value of a chi-square test, we always shade the tail areas in both tails. (d) As the degrees of freedom increases, the variability of the chi-square distribution decreases. A professor using an open source introductory statistics book predicts 6.27 Open source textbook. that 60% of the students will purchase a hard copy of the book, 25% will print it out from the web, and 15% will read it online. At the end of the semester he asks his students to complete a survey where they indicate what format of the book they used. Of the 126 students, 71 said they bought a hard copy of the book, 30 said they printed it out from the web, and 25 said they read it online. (a) State the hypotheses for testing if the professor’s predictions were inaccurate. (b) How many students did
the professor expect to buy the book, print the book, and read the book exclusively online? (c) List the conditions required for the chi-square goodness of ﬁt test and discuss whether they are satisﬁed. (d) Assume conditions are suﬃciently met. Calculate the chi-square statistic, the degrees of freedom asso- ciated with it, and the p-value. (e) Based on the p-value calculated in part (d), what is the conclusion of the hypothesis test? Interpret your conclusion in this context. 6.28 Barking deer. Microhabitat factors associated with forage and bed sites of barking deer in Hainan Island, China were examined. In this region woods make up 4.8% of the land, cultivated grass plot makes up 14.7%, and deciduous forests make up 39.6%. Of the 426 sites where the deer forage, 4 were categorized as woods, 16 as cultivated grassplot, and 61 as deciduous forests. The table below summarizes these data.35 Woods Cultivated grassplot Deciduous forests Other Total 426 345 61 16 4 (a) Do these data provide convincing evidence that barking deer prefer to forage in certain habitats over others? Conduct an appropriate hypothesis test to answer this research question, and acknowledge any assumptions you had to make to carry out this test. Include all steps of the Identify, Choose, Check, Calculate, Conclude framework. (b) Interpret the calculated p-value in the context of the problem. Photo by Shrikant Rao (http://ﬂic.kr/p/4Xjdkk) CC BY 2.0 license 35Liwei Teng et al. “Forage and bed sites characteristics of Indian muntjac (Muntiacus muntjak) in Hainan Island, China”. In: Ecological Research 19.6 (2004), pp. 675–681. 340 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA 6.4 Chi-square tests in two-way tables We encounter two-way tables in this section, and we learn about two new and closely related chisquare tests. We will answer questions such as the following: • Does the phrasing of the question aﬀect how likely sellers are to disclose problems with a product? • Is gender associated with whether Facebook users know how to adjust their privacy settings? • Is political aﬃliation associated with support for the use of full body scans at airports? Learning objectives 1. Calculate the expected counts and degrees of freedom for a chi-square test involving a two-way table. 2. State and verify whether or not the conditions for a chi-square test for a two-way table are met. 3. Explain the diﬀerence between the chi-square test for homogeneity and chi-square test for independence. 4. Carry out a complete hypothesis test for homogeneity and for independence. 6.4. HOMOGENEITY AND INDEPENDENCE IN TWO-WAY TABLES 341 6.4.1 Introduction to two-way tables Google is constantly running experiments to test new search algorithms. For example, Google might test three algorithms using a sample of 10,000 google.com search queries. Figure 6.10 shows an example of 10,000 queries split into three algorithm groups.36 The group sizes were speciﬁed before the start of the experiment to be 5000 for the current algorithm and 2500 for each test algorithm. Search algorithm Counts current 5000 test 1 2500 test 2 2500 Total 10000 Figure 6.10: Experiment breakdown of test subjects into three search groups. EXAMPLE 6.42 What is the ultimate goal of the Google experiment? What are the null and alternative hypotheses, in regular words? The ultimate goal is to see whether there is a diﬀerence in the performance of the algorithms. The hypotheses can be described as the following: H0: The algorithms each perform equally well. HA: The algorithms do not perform equally well. In this experiment, the explanatory variable is the search algorithm. However, an outcome variable is also needed. This outcome variable should somehow reﬂect whether the search results align with the user’s interests. One possible way to quantify this is to determine whether (1) there was no new, related search, and the user clicked one of the links provided, or (2) there was a new, related search performed by the user. Under scenario (1), we might think that the user was satisﬁed with the search results. Under scenario (2), the search results probably were not relevant, so the user tried a second search. Figure 6.11 provides the results from the experiment. These data are very similar to the count data in Section 6.3. However, now the diﬀerent combinations of two variables are binned in a twoway table. In examining these data, we want to evaluate whether there is strong evidence that at least one algorithm is performing better than the others. To do so, we apply a chi-square test to this two-way table. The ideas of this test are similar to those ideas in the one-way table case. However, degrees of freedom and expected counts are computed a little diﬀerently than before. No new search New search Total Search algorithm test 1 1749 751 2500 current 3511 1489 5000 test 2 1818 682 2500 Total 7078 2922 10000 Figure 6.11: Results of the Google search algorithm experiment. WHAT IS SO DIFFERENT ABOUT ONE-WAY TABLES AND TWO-WAY TABLES? A one-way table describes counts for each outcome in a single variable. A two-way table describes counts for combinations of outcomes for two variables. When we consider a two-way table, we often would like to know, are these variables related in any way? The hypothesis test for this Google experiment is really about assessing whether there is statistically signiﬁcant evidence that the choice of the algorithm aﬀects whether a user performs a second search. In other words, the goal is to check whether the three search algorithms perform diﬀerently. 36Google regularly runs experiments in this manner to help improve their search engine. It is entirely possible that if you perform a search and so does your friend, that you will have diﬀerent search results. While the data presented in this section resemble what might be encountered in a real experiment, these data are simulated. 342 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA 6.4.2 Expected counts in two-way tables EXAMPLE 6.43 From the experiment, we estimate the proportion of users who were satisﬁed with their initial search (no new search) as 7078/10000 = 0.7078. If there really is no diﬀerence among the algorithms and 70.78% of people are satisﬁed with the search results, how many of the 5000 people in the “current algorithm” group would be expected to not perform a new search? About 70.78% of the 5000 would be satisﬁed with the initial search: 0.7078 × 5000 = 3539 users That is, if there was no diﬀerence between the three groups, then we would expect 3539 of the current algorithm users not to perform a new search. GUIDED PRACTICE 6.44 Using the same rationale described in Example 6.43, about how many users in each test group would not perform a new search if the algorithms were equally helpful?37 We can compute the expected number of users who would perform a new search for each group using the same strategy employed in Example 6.43 and Guided Practice 6.44. These expected counts were used to construct Figure 6.12, which is the same as Figure 6.11, except now the expected counts have been added in parentheses. Search algorithm No new search New search Total current 3511 1489 5000 (3539) (1461) test 1 1749 751 2500 (1769.5) (730.5) test 2 1818 682 2500 (1769.5) (730.5) Total 7078 2922 10000 Figure 6.12: The observed counts and the (expected counts). The examples and exercises above provided some help in computing expected counts. In general, expected counts for a two-way table may be computed using the row totals, column totals, and the table total. For instance, if there was no diﬀerence between the groups, then about 70.78% of each column should be in the ﬁrst row: 0.7078 × (column 1 total) = 3539 0.7078 × (column 2 total) = 1769.5 0.7078 × (column 3 total) = 1769.5 Looking back to how the fraction 0.7078 was computed – as the fraction of users who did not perform a new search (7078/10000) – these three expected counts could have been computed as row 1 total table total row 1 total table total row 1 total table total (column 1 total) = 3539 (column 2 total) = 1769.5 (column 3 total) = 1769.5 This leads us to a general formula for computing expected counts in a two-way table when we would like to test whether there is strong evidence of an association between the column variable and row variable. 37We would expect 0.7078 ∗ 2500 = 1769.5. It is okay that this is a fraction. 6.4. HOMOGENEITY AND INDEPENDENCE IN TWO-WAY TABLES 343 COMPUTING EXPECTED COUNTS IN A TWO-WAY TABLE To identify the expected count for the ith row and jth column, compute Expected Countrow i, col j = (row i total) × (column j total) table total 6.4.3 The chi-square test for homogeneity in two-way tables The chi-square test statistic for a two-way table is found the same way it is found for a one-way table. For each table count, compute General formula Row 1, Col 1 Row 1, Col 2 ... Row 2, Col 3 (observed count − expected count)2 expected count (3511 − 3539)2 3539 (1749 − 1769.5)2 1769.5 ... (682 − 730.5)2 730.5 = 0.222 = 0.237 = 3.220 Adding the computed value for each cell gives the chi-square test statistic χ2: χ2 = 0.222 + 0.237 + · · · + 3.220 = 6.120 Just like before, this test statistic follows a chi-square distribution. However, the degrees of freedom is computed a little diﬀerently for a two-way table.38 For two way tables, the degrees of freedom is equal to df = (number of rows - 1) × (number of columns - 1) In our example, the degrees of freedom is df = (2 − 1) × (3 − 1) = 2 If the null hypothesis is true (i.e. the algorithms are equally useful), then the test statistic χ2 = 6.12 closely follows a chi-square distribution with 2 degrees of freedom. Using this information, we can compute the p-value for the test, which is depicted in Figure 6.13. COMPUTING DEGREES OF FREEDOM FOR A TWO-WAY TABLE When using the chi-square test to a two-way table, we use df = (R − 1) × (C − 1) where R is the number of rows in the table and C is the number of columns. USE TWO-PROPORT
ION METHODS FOR 2-BY-2 CONTINGENCY TABLES When analyzing 2-by-2 contingency tables, use the two-proportion methods introduced in Section 6.2. 38Recall: in the one-way table, the degrees of freedom was the number of groups minus 1. 344 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA Figure 6.13: Computing the p-value for the Google hypothesis test. CONDITIONS FOR THE CHI-SQUARE TEST FOR HOMOGENEITY There are two conditions that must be checked before performing a chi-square test for homogeneity. If these conditions are not met, this test should not be used. Independence. Data should be come from multiple independent random samples or from a randomized experiment with multiple treatments. Data can then be organized into a twoway table. When sampling without replacement, the sample size should be less than 10% of the population size for each sample. Large expected counts. All of the cells in the two-way table must have at least 5 expected cases under the assumption that the null hypothesis is true. EXAMPLE 6.45 Compute the p-value and draw a conclusion about whether the search algorithms have diﬀerent performances. Here, found that the degrees of freedom for this 3 × 2 table is 2. The p-value corresponds to the area under the chi-square curve with 2 degrees of freedom to the right of χ2 = 6.120. Using a calculator, we ﬁnd that the p-value = 0.047. Using an α = 0.05 signiﬁcance level, we reject H0. That is, the data provide convincing evidence that there is some diﬀerence in performance among the algorithms. Notice that the conclusion of the test is that there is some diﬀerence in performance among the algorithms. This chi-square test does not tell us which algorithm performed better than the others. To answer this question, we could compare the relevant proportions or construct bar graphs. The proportion that resulted in the new search can be calculated as current: 1489 5000 = 0.298 test 1: 751 2500 = 0.300 test 2: 682 2500 = 0.136. This suggests that the test 2 algorithm performed better than the current algorithm and test 1 algorithm, because it led to fewer new searches; however, to formally test this speciﬁc claim we would need to use a test that includes a multiple comparisons correction, which is beyond the scope of this textbook. A careful reader may have noticed that when there are exactly 2 random samples or treatments and the counts can be arranged in a 2 × 2 table, both a chi-square test for homogeneity and a 2-proportion Z-test could apply. In this case, the chi-square test for homogeneity and the two-sided 2-proportion Z-test are equivalent, meaning that they produce the same p-value.39 39Sometimes the success-failure condition for the Z-test is weakened to require the number of successes and failures to be at least 5, making it consistent with the chi-square condition that expected counts must at least 5. 051015 6.4. HOMOGENEITY AND INDEPENDENCE IN TWO-WAY TABLES 345 χ2χ2χ2 TEST FOR HOMOGENEITY When there are multiple samples or treatments and we are comparing the distribution of a categorical variable across several groups, e.g. comparing the distribution of rural/urban/suburban dwellers among 4 states, Identify: Identify the hypotheses and the signiﬁcance level, α. H0: The distribution of [...] is the same for each population/treatment. HA: The distribution of [...] is not the same for each population/treatment. Choose: Choose the correct test procedure and identify it by name. Here we use a χ2χ2χ2 test for homogeneity. Check: Check that the test statistic follows a chi-square distribution. 1. Independence: Data come from multiple random samples or from a randomized experiment with multiple treatments. When sampling without replacement, the sample size should be less than 10% of the population size for each sample. 2. Expected counts: All expected counts are ≥ 5 (calculate and record expected counts). Calculate: Calculate the χ2-statistic, df , and p-value. test statistic: χ2 = (observed − expected)2 expected df = (# of rows − 1) × (# of columns − 1) p-value = (area to the right of χ2-statistic with the appropriate df ) Conclude: Compare the p-value to α, and draw a conclusion in context. If the p-value is < α, reject H0; there is suﬃcient evidence that [HA in context]. If the p-value is > α, do not reject H0; there is not suﬃcient evidence that [HA in context]. 346 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA EXAMPLE 6.46 In an experiment, each individual was asked to be a seller of an iPod (a product commonly used to store music on before smart phones). The participant received $10 + 5% of the sale price for participating. The iPod they were selling had frozen twice in the past inexplicitly but otherwise worked ﬁne. Unbeknownst to the participants who were the sellers in the study, the buyers were collaborating with the researchers to evaluate the inﬂuence of diﬀerent questions on the likelihood of getting the sellers to disclose the past issues with the iPod. The scripted buyers started with “Okay, I guess I’m supposed to go ﬁrst. So you’ve had the iPod for 2 years ...” and ended with one of three questions: • General: What can you tell me about it? • Positive Assumption: It doesn’t have any problems, does it? • Negative Assumption: What problems does it have? The outcome variable is whether the participant discloses or hides the problem with the iPod. Question Type General Positive Assump. Negative Assump. Response Disclose Hide Total 2 71 73 23 50 73 36 37 73 Does the phrasing of the question aﬀect how likely individuals are to disclose the problems with the iPod? Carry out an appropriate test at the 0.05 signiﬁcance level. Identify: We will test the following hypotheses at the α = 0.05 signiﬁcance level. H0: The likelihood of disclosing the problem is the same for each question type. HA: The likelihood of disclosing the problem is not the same for each question type. Choose: We want to know if the distribution of disclose/hide is the same for each of the three question types, so we want a chi-square test for homogeneity. Check: This is an experiment in which there were three randomly allocated treatments. Here a treatment corresponds to a question type. All values in the table of expected counts are ≥ 5. Table of expected counts: Question Type General Positive Assump. Negative Assump. Response Disclose Hide 20.3 52.7 20.3 52.7 20.3 52.7 Calculate: Using technology, we get χ2 = 40.1 df = (# of rows − 1) × (# of columns − 1) = 2 × 1 = 2 The p-value is the area under the chi-square curve with 2 degrees of freedom to the right of χ2 = 40.1. Thus, the p-value is almost 0. Conclude: Because the p-value ≈ 0 < α, we reject H0. We have strong evidence that the likelihood of disclosing the problem is not the same for each question type. GUIDED PRACTICE 6.47 If an error was made in the test in the previous example, would it have been a Type I error or a Type II error?40 40In this test, the p-value was less than α, so we rejected H0. If H0 is in fact true, and we reject it, that would be committing a Type I error. We could not have made a Type II error, because a Type II error involves not rejecting H0. 6.4. HOMOGENEITY AND INDEPENDENCE IN TWO-WAY TABLES 347 6.4.4 The chi-square test for independence in two-way tables Often, instead of having separate random samples or treatments, we have just one sample and we want to look at the association between two variables. When these two variables are categorical, we can arrange the responses in a two-way table. In Chapter 3 we looked at independence in the context of probability. Here we look at independence in the context of inference. We want to know if any observed association is due to random chance or if there is evidence of a real association in the population that the sample was taken from. To answer this, we use a chi-square test for independence. The chi-square test for independence applies when there is only one random sample and there are two categorical variables. The null claim is always that the two variables are independent, while the alternate claim is that the variables are dependent. EXAMPLE 6.48 Figure 6.14 summarizes the results of a Pew Research poll. A random sample of adults in the U.S. was taken, and each was asked whether they approved or disapproved of the job being done by President Obama, Democrats in Congress, and Republicans in Congress. The results are shown in Figure 6.14. We would like to determine if the three groups and the approval ratings are associated. What are appropriate hypotheses for such a test? H0: The group and their ratings are independent. (There is no diﬀerence in approval ratings between the three groups.) HA: The group and their ratings are dependent. (There is some diﬀerence in approval ratings between the three groups, e.g. perhaps Obama’s approval diﬀers from Democrats in Congress.) Obama Democrats Republicans Congress Approve Disapprove Total 842 616 1458 736 646 1382 541 842 1383 Total 2119 2104 4223 Figure 6.14: Pew Research poll results of a March 2012 poll. CONDITIONS FOR THE CHI-SQUARE TEST FOR INDEPENDENCE There are two conditions that must be checked before performing a chi-square test for independence. If these conditions are not met, this test should not be used. Independence. The data must be arrived at by taking one random sample. When sampling without replacement from a ﬁnite population, the sample size should be less than 10% of the population size. After the data is collected, it is separated and categorized according to two variables and can be organized into a two-way table. Large expected counts. All of the cells in the two-way table must have at least 5 expected cases assuming the null hypothesis is true. 348 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA EXAMPLE 6.49 First, we observe that the data came from a random sample of adults in the U.S. and that the population size for adults in the U.S. is much larger than 10 times the sample size. Next, let’s compute the expected values that correspond to
Figure 6.14, if the null hypothesis is true, that is, if group and rating are independent. The expected count for row one, column one is found by multiplying the row one total (2119) and column one total (1458), then dividing by the table total (4223): 2119×1458 = 731.6. Similarly for the ﬁrst column and the second row: 2104×1458 4223 = 726.4. Repeating this process, we get the expected counts: 4223 Approve Disapprove Obama Congr. Dem. Congr. Rep. 731.6 726.4 694.0 689.0 693.5 688.5 The table above gives us the number we would expect for each of the six combinations if group and rating were really independent. Because all of the expected counts are at least 5 and there is one random sample, we can carry out the chi-square test for independence. The chi-square test for independence and the chi-square test for homogeneity both involve counts in a two-way table. The chi-square statistic and the degrees of freedom are calculated in the same way. EXAMPLE 6.50 Calculate the chi-square statistic. We calculate (obs−exp)2 chi-square test statistic. exp for each of the six cells in the table. Adding the results of each cell gives the χ2 = (obs − exp)2 exp = (842 − 731.6)2 731.6 =16.7 + · · · = 106.4 + · · · EXAMPLE 6.51 Find the p-value for the test and state the appropriate conclusion. We must ﬁrst ﬁnd the degrees of freedom for this chi-square test. Because there are 2 rows and 3 columns, the degrees of freedom is df = (2 − 1) × (3 − 1) = 2. We ﬁnd the area to the right of χ2 = 106.4 under the chi-square curve with df = 2. The p-value is extremely small, much less than 0.01, so we reject H0. We have evidence that the three groups and their approval ratings are dependent. 6.4. HOMOGENEITY AND INDEPENDENCE IN TWO-WAY TABLES 349 χ2χ2χ2 TEST FOR INDEPENDENCE When there is one sample and we are looking for association or dependence between two categorical variables, e.g. testing for an association between gender and political party, Identify: Identify the hypotheses and the signiﬁcance level, α. H0: [variable 1] and [variable 2] are independent. HA: [variable 1] and [variable 2] are dependent. Choose: Choose the correct test procedure and identify it by name. Here we use a χ2χ2χ2 test for independence. Check: Check that the test statistic follows a chi-square distribution. 1. Independence: Data come from one random sample. If sampling without replacement, check that the sample size is less than 10% of the population size. 2. Expected counts: All expected counts are ≥ 5 (calculate and record expected counts). Calculate: Calculate the χ2-statistic, df , and p-value. test statistic: χ2 = (observed − expected)2 expected df = (# of rows − 1) × (# of columns − 1) p-value = (area to the right of χ2-statistic with the appropriate df ) Conclude: Compare the p-value to α, and draw a conclusion in context. If the p-value is < α, reject H0; there is suﬃcient evidence that [HA in context]. If the p-value is > α, do not reject H0; there is not suﬃcient evidence that [HA in context]. 350 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA EXAMPLE 6.52 A 2021 Pew Research poll asked a random sample of U.S. residents their generation and whether they have personally taken action to help address climate change within the last year. The data are shown below. Response Gen Z Generation Millenial Gen X Boomer & older Total Took action Didn’t take action Total 912 3,160 3,518 6,074 13,664 620 2,275 2,709 4,798 10,402 292 885 809 1,276 3,262 We can see that the percent in the sample from each generation that took action vary: 32% for Gen Z, 28% for Millenial, 23% for Gen X, and 21% for Boomer & older. However, could this be due to random variation based on who happened to end up in the sample? Carry out an appropriate test at the 0.05 signiﬁcance level to see if there is an association between generation and taking action to help address climate change. Identify: We will test the following hypotheses at the α = 0.05 signiﬁcance level. H0: Generation and taking action to help address climate change are independent. HA: Generation and taking action to help address climate change are dependent. Choose: Two variables were recorded on the respondents: generation and whether or not they have taken action to help address climate change within the last year. We want to know if these variables are associated / dependent, so we will carry out a chi-square test for independence. Check: According to the problem, there was one random sample taken. We note that the population of U.S. residents is much larger than 10 times the sample size of 13,664. Also, all values in the table of expected counts are ≥ 5. Table of expected counts: Gen Z Generation Millenial Gen X Boomer & older Response Took action Didn’t take action 217.72 754.39 839.85 1450.00 694.28 2405.60 2678.10 4624.00 Calculate: Using technology, we get χ2 = 91.9. The degrees of freedom for this test is given by: df = (# of rows − 1) × (# of columns − 1) = 3 × 1 = 3. The p-value is the area under the chi-square curve with 3 degrees of freedom to the right of χ2 = 91.9. Thus, the p-value = 8.46x10−20 ≈ 0. Conclude: Because the p-value ≈ 0 < α, we reject H0. We have suﬃcient evidence that generation and taking action to help address climate change are dependent. GUIDED PRACTICE 6.53 In context, interpret the p-value of the test in the previous example.41 41The p-value in this test corresponds to the area to the right of χ2 = 91.9 under the chi-square curve with 3 degrees of freedom. Assuming the probability model is true and assuming the null hypothesis is true, i.e. that generation and response really are independent, there is close to a 0% probability of getting a χ2-statistic as large or larger than 91.9. Equivalently, it is the probability of our observed counts being this diﬀerent from the expected counts, relative to the expected counts, if the null is true and the model holds. Because the p-value is so small, we reject the null hypothesis. 6.4. HOMOGENEITY AND INDEPENDENCE IN TWO-WAY TABLES 351 6.4.5 Technology: the chi-square test for two-way tables TI-83/84: ENTERING DATA INTO A TWO-WAY TABLE 1. Hit 2ND x−1 (i.e. MATRIX). 2. Right arrow to EDIT. 3. Hit 1 or ENTER to select matrix A. 4. Enter the dimensions by typing #rows, ENTER, #columns, ENTER. 5. Enter the data from the two-way table. TI-83/84: CHI-SQUARE TEST FOR HOMOGENEITY AND INDEPENDENCE Use STAT, TESTS, χ2-Test. 1. First enter two-way table data as described in the previous box. 2. Choose STAT. 3. Right arrow to TESTS. 4. Down arrow and choose C:χ2-Test. 5. Down arrow, choose Calculate, and hit ENTER, which returns χ2 p df chi-square test statistic p-value degrees of freedom TI-83/84: CHI-SQUARE TEST FOR HOMOGENEITY AND INDEPENDENCE TI-83/84: Finding the expected counts 1. First enter two-way table data as described previously. 2. Carry out the chi-square test for homogeneity or independence as described in previous box. 3. Hit 2ND x−1 (i.e. MATRIX). 4. Right arrow to EDIT. 5. Hit 2 to see matrix B. This matrix contains the expected counts. 352 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA CASIO FX-9750GII: CHI-SQUARE TEST FOR HOMOGENEITY AND INDEPENDENCE 1. Navigate to STAT (MENU button, then hit the 2 button or select STAT). 2. Choose the TEST option (F3 button). 3. Choose the CHI option (F3 button). 4. Choose the 2WAY option (F2 button). 5. Enter the data into a matrix: • Hit MAT (F2 button). • Navigate to a matrix you would like to use (e.g. Mat C) and hit EXE. • Specify the matrix dimensions: m is for rows, n is for columns. • Enter the data. • Return to the test page by hitting EXIT twice. 6. Enter the Observed matrix that was used by hitting MAT (F1 button) and the matrix letter (e.g. C). 7. Enter the Expected matrix where the expected values will be stored (e.g. D). 8. Hit the EXE button, which returns χ2 p df chi-square test statistic p-value degrees of freedom 9. To see the expected values of the matrix, go to MAT (F6 button) and select the corre- sponding matrix. GUIDED PRACTICE 6.54 The table from Figure 6.14 is reproduced below. Use a calculator to ﬁnd the expected values and the χ2-statistic, df , and p-value for the chi-square test for independence.42 Obama Democrats Republicans Congress Approve Disapprove Total 842 616 1458 736 646 1382 541 842 1383 Total 2119 2104 4223 42First create a 2 × 3 matrix with the data. The ﬁnal summaries should be χ2 = 106.4, p-value is p = 8.06 × 10−24 ≈ 0, and df = 2. Below is the matrix of expected values: Approve Disapprove Obama Congr. Dem. Congr. Rep. 731.59 726.41 693.45 688.55 693.96 689.04 6.4. HOMOGENEITY AND INDEPENDENCE IN TWO-WAY TABLES 353 Section summary • When there are two categorical variables, rather than one, the data can be arranged in a two-way table. • When working with a two-way table, the expected count for each row,column combination is calculated as: expected count = (row total)×(column total) . table total • When categorical data are arranged in a two way table, use the χ2 test for homogeneity or the χ2 test for independence. These tests are almost identical; the diﬀerences lie in the data collection method and in the hypotheses. • When there are multiple random samples or treatments and we are comparing the distribution of a categorical variable across several groups, e.g. comparing the distribution of rural/urban/suburban dwellers among 4 states, the hypotheses can be written as follows: H0: The distribution of [...] is the same for each population/treatment. HA: The distribution of [...] is not the same for each population/treatment. We test these hypotheses at the α signiﬁcance level using a χ2χ2χ2 test for homogeneity. • When there is one random sample and we are looking for association or dependence between two categorical variables, e.g. testing for an association between gender and political party, the hypotheses can be written as: H0: [variable 1] and [variable 2] are independent. HA: [variable 1] and [variable 2] are dependent. We test these hypotheses at the α signiﬁ
cance level using a χ2χ2χ2 test for independence. • In addition to the independence/random condition, all expected counts must be at least 5 for the test statistic to follow a chi-square distribution. • The chi-square statistic and associated df are found as follows: test statistic: χ2 = (observed − expected)2 expected df = (# of rows − 1)(# of cols − 1) • The p-value is the area to the right of χ2-statistic under the chi-square curve with the appro- priate df . 354 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA Exercises 6.29 Quitters. Does being part of a support group aﬀect the ability of people to quit smoking? A county health department enrolled 300 smokers in a randomized experiment. 150 participants were randomly assigned to a group that used a nicotine patch and met weekly with a support group; the other 150 received the patch and did not meet with a support group. At the end of the study, 40 of the participants in the patch plus support group had quit smoking while only 30 smokers had quit in the other group. (a) Create a two-way table presenting the results of this study. (b) Answer each of the following questions under the null hypothesis that being part of a support group does not aﬀect the ability of people to quit smoking, and indicate whether the expected values are higher or lower than the observed values. i. How many subjects in the “patch + support” group would you expect to quit? ii. How many subjects in the “patch only” group would you expect to not quit? 6.30 Full body scan, Part II.A news article reports that “Americans have diﬀering views on two potentially inconvenient and invasive practices that airports could implement to uncover potential terrorist attacks.” This news piece was based on a survey conducted among a random sample of 1,137 adults nationwide, where one of the questions on the survey was “Some airports are now using ‘full-body’ digital x-ray machines to electronically screen passengers in airport security lines. Do you think these new x-ray machines should or should not be used at airports?” Below is a summary of responses based on party aﬃliation.43 The diﬀerences in each political group may be due to chance. Complete the following computations under the null hypothesis of independence between an individual’s party aﬃliation and his support of full-body scans. It may be useful to ﬁrst add on an extra column for row totals before proceeding with the computations. Party Aﬃliation Republican Democrat Answer Should Should not Don’t know/No answer Total 264 38 16 318 299 55 15 369 Independent 351 77 22 450 (a) How many Republicans would you expect to not support the use of full-body scans? (b) How many Democrats would you expect to support the use of full- body scans? (c) How many Independents would you expect to not know or not answer? A survey asked 827 randomly sampled registered voters in California “Do 6.31 Offshore drilling. you support? Or do you oppose? Drilling for oil and natural gas oﬀ the Coast of California? Or do you not know enough to say?” Below is the distribution of responses, separated based on whether or not the respondent has a college degree.44 Complete a chi-square test for these data to test whether there is an association between opinions regarding oﬀshore drilling for oil and having a college degree. Include all steps of the Identify, Choose, Check, Calculate, Conclude framework. College Degree Yes 154 Support Oppose 180 Do not know 104 438 Total No 132 126 131 389 43S. Condon. “Poll: 4 in 5 Support Full-Body Airport Scanners”. In: CBS News (2010). 44Survey USA, Election Poll #16804, data collected July 8-11, 2010. 6.4. HOMOGENEITY AND INDEPENDENCE IN TWO-WAY TABLES 355 6.32 Parasitic worm. Lymphatic ﬁlariasis is a disease caused by a parasitic worm. Complications of the disease can lead to extreme swelling and other complications. Here we consider results from a randomized experiment that compared three diﬀerent drug treatment options to clear people of the this parasite, which people are working to eliminate entirely. The results for the second year of the study are given below:45 Clear at Year 2 Not Clear at Year 2 Three drugs Two drugs Two drugs annually 52 31 42 2 24 14 (a) Set up hypotheses for evaluating whether there is any diﬀerence in the performance of the treatments, and also check conditions. (b) Statistical software was used to run a chi-square test, which output: X 2 = 23.7 df = 2 p-value = 7.2e-6 Use these results to evaluate the hypotheses from part (a), and provide a conclusion in the context of the problem. 45Christopher King et al. “A Trial of a Triple-Drug Treatment for Lymphatic Filariasis”. In: New England Journal of Medicine 379 (2018), pp. 1801–1810. 356 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA Chapter highlights Calculating a conﬁdence interval or a test statistic and p-value are generally done with statistical software. It is important, then, to focus not on the calculations, but rather on 1. choosing the correct procedure 2. understanding when the procedures do or do not apply, and 3. interpreting the results. Choosing the correct procedure requires understanding the type of data and the method of data collection. All of the inference procedures in Chapter 6 are for categorical variables. Here we list the ﬁve tests encountered in this chapter and when to use them. • 1-proportion Z-test – 1 random sample, a yes/no variable – Compare the sample proportion to a ﬁxed / hypothesized proportion. • 2-proportion Z-test – 2 independent random samples or randomly allocated treatments – Compare two populations or treatments to each other with respect to one yes/no variable; e.g. comparing the proportion over age 65 in two distinct populations. • χ2χ2χ2 goodness of ﬁt test – 1 random sample, a categorical variable (generally at least three categories) – Compare the distribution of a categorical variable to a ﬁxed or known population distri- bution; e.g. looking at distribution of color among M&M’s. • χ2χ2χ2 test for homogeneity: – 2 or more independent random samples or randomly allocated treatments – Compare the distribution of a categorical variable across several populations or treatments; e.g. party aﬃliation over various years, or patient improvement compared over 3 treatments. • χ2χ2χ2 test for independence – 1 random sample, 2 categorical variables – Determine if, in a single population, there is an association between two categorical variables; e.g. grade level and favorite class. Even when the data and data collection method correspond to a particular test, we must verify that conditions are met to see if the assumptions of the test are reasonable. All of the inferential procedures of this chapter require some type of random sample or process. In addition, the 1-proportion Z-test/interval and the 2-proportion Z-test/interval require that the success-failure condition is met and the three χ2 tests require that all expected counts are at least 5. Finally, understanding and communicating the logic of a test and being able to accurately interpret a conﬁdence interval or p-value are essential. For a refresher on this, review Chapter 5: Foundations for inference. 6.4. HOMOGENEITY AND INDEPENDENCE IN TWO-WAY TABLES 357 Chapter exercises 6.33 Active learning. A teacher wanting to increase the active learning component of her course is concerned about student reactions to changes she is planning to make. She conducts a survey in her class, asking students whether they believe more active learning in the classroom (hands on exercises) instead of traditional lecture will helps improve their learning. She does this at the beginning and end of the semester and wants to evaluate whether students’ opinions have changed over the semester. Can she used the methods we learned in this chapter for this analysis? Explain your reasoning. 6.34 Website experiment. The OpenIntro website occasionally experiments with design and link placement. We conducted one experiment testing three diﬀerent placements of a download link for this textbook on the book’s main page to see which location, if any, led to the most downloads. The number of site visitors included in the experiment was 701 and is captured in one of the response combinations in the following table: Download No Download Position 1 Position 2 Position 3 13.8% 14.6% 12.1% 18.3% 18.5% 22.7% (a) Calculate the actual number of site visitors in each of the six response categories. (b) Each individual in the experiment had an equal chance of being in any of the three experiment groups. However, we see that there are slightly diﬀerent totals for the groups. Is there any evidence that the groups were actually imbalanced? Carry out an appropriate test and include all steps of the ICCCC framework. (c) Complete an appropriate hypothesis test to check whether there is evidence that there is a higher rate of site visitors clicking on the textbook link in any of the three groups. Include all steps of the Identify, Choose, Check, Calculate, Conclude framework. 6.35 Shipping holiday gifts. A local news survey asked 500 randomly sampled Los Angeles residents which shipping carrier they prefer to use for shipping holiday gifts. The table below shows the distribution of responses by age group as well as the expected counts for each cell (shown in parentheses). Shipping Method 18-34 Age 35-54 55+ USPS UPS FedEx Something else Not sure Total 72 52 31 7 3 (81) (53) (21) (5) (5) 97 76 24 6 6 (102) (68) (27) (7) (5) 76 34 9 3 4 (62) (41) (16) (4) (3) 165 209 126 Total 245 162 64 16 13 500 (a) State the null and alternative hypotheses for testing for independence of age and preferred shipping method for holiday gifts among Los Angeles residents. (b) Are the conditions for inference using a chi-square test satisﬁed? 6.36 The Civil War. A national survey conducted among a simple random sample of 1,507 adults shows that 56% of Americans think the Civil War is still relevant to American politics and political life.46 (a) Condu
ct a hypothesis test to determine if these data provide strong evidence that the majority of the Americans think the Civil War is still relevant. (b) Interpret the p-value in this context. (c) Calculate a 90% conﬁdence interval for the proportion of Americans who think the Civil War is still relevant. Interpret the interval in this context, and comment on whether or not the conﬁdence interval agrees with the conclusion of the hypothesis test. 46Pew Research Center Publications, Civil War at 150: Still Relevant, Still Divisive, data collected between March 30 - April 3, 2011. 358 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA 6.37 College smokers. who smoke. Out of a random sample of 200 students from this university, 40 students smoke. We are interested in estimating the proportion of students at a university (a) Calculate a 95% conﬁdence interval for the proportion of students at this university who smoke, and interpret this interval in context. (Reminder: Check conditions.) (b) If we wanted the margin of error to be no larger than 2% at a 95% conﬁdence level for the proportion of students who smoke, how big of a sample would we need? It is believed that large doses of acetaminophen (the active 6.38 Acetaminophen and liver damage. ingredient in over the counter pain relievers like Tylenol) may cause damage to the liver. A researcher wants to conduct a study to estimate the proportion of acetaminophen users who have liver damage. For participating in this study, he will pay each subject $20 and provide a free medical consultation if the patient has liver damage. (a) If he wants to limit the margin of error of his 98% conﬁdence interval to 2%, what is the minimum amount of money he needs to set aside to pay his subjects? (b) The amount you calculated in part (a) is substantially over his budget so he decides to use fewer subjects. How will this aﬀect the width of his conﬁdence interval? 6.39 Life after college. We are interested in estimating the proportion of graduates at a mid-sized university who found a job within one year of completing their undergraduate degree. Suppose we conduct a survey and ﬁnd out that 348 of the 400 randomly sampled graduates found jobs. The graduating class under consideration included over 4500 students. (a) Describe the population parameter of interest. What is the value of the point estimate of this parameter? (b) Check if the conditions for constructing a conﬁdence interval based on these data are met. (c) Calculate a 95% conﬁdence interval for the proportion of graduates who found a job within one year of completing their undergraduate degree at this university, and interpret it in the context of the data. (d) What does “95% conﬁdence” mean? (e) Now calculate a 99% conﬁdence interval for the same parameter and interpret it in the context of the data. (f) Compare the widths of the 95% and 99% conﬁdence intervals. Which one is wider? Explain. 6.40 Diabetes and unemployment. A Gallup poll surveyed Americans about their employment status and whether or not they have diabetes. The survey results indicate that 1.5% of the 47,774 employed (full or part time) and 2.5% of the 5,855 unemployed 18-29 year olds have diabetes.47 (a) Create a two-way table presenting the results of this study. (b) State appropriate hypotheses to test. (c) The sample diﬀerence is about 1%. If we completed the hypothesis test, we would ﬁnd that the p-value is very small (about 0), meaning the diﬀerence is statistically signiﬁcant. Use this result to explain the diﬀerence between statistically signiﬁcant and practically signiﬁcant ﬁndings. 6.41 Rock-paper-scissors. Rock-paper-scissors is a hand game played by two or more people where players choose to sign either rock, paper, or scissors with their hands. For your statistics class project, you want to evaluate whether players choose between these three options randomly, or if certain options are favored above others. You ask two friends to play rock-paper-scissors and count the times each option is played. The following table summarizes the data: Rock Paper 43 21 Scissors 35 Use these data to evaluate whether players choose between these three options randomly, or if certain options are favored above others. Make sure to clearly outline each step of your analysis, and interpret your results in context of the data and the research question. 47Gallup Wellbeing, Employed Americans in Better Health Than the Unemployed, data collected Jan. 2, 2011 - May 21, 2012. 6.4. HOMOGENEITY AND INDEPENDENCE IN TWO-WAY TABLES 359 6.42 2010 Healthcare Law. On June 28, 2012 the U.S. Supreme Court upheld the much debated 2010 healthcare law, declaring it constitutional. A Gallup poll released the day after this decision indicates that 46% of 1,012 randomly sampled Americans agree with this decision. At a 95% conﬁdence level, this sample has a 3% margin of error. Based on this information, determine if the following statements are true or false, and explain your reasoning.48 (a) We are 95% conﬁdent that between 43% and 49% of Americans in this sample support the decision of the U.S. Supreme Court on the 2010 healthcare law. (b) We are 95% conﬁdent that between 43% and 49% of Americans support the decision of the U.S. Supreme Court on the 2010 healthcare law. (c) If we considered many random samples of 1,012 Americans, and we calculated the sample proportions of those who support the decision of the U.S. Supreme Court, 95% of those sample proportions will be between 43% and 49%. (d) The margin of error at a 90% conﬁdence level would be higher than 3%. 6.43 Browsing on the mobile device. A survey of 2,254 randomly selected American adults indicates that 17% of cell phone owners browse the internet exclusively on their phone rather than a computer or other device.49 (a) According to an online article, a report from a mobile research company indicates that 38 percent of Chinese mobile web users only access the internet through their cell phones.50 Conduct a hypothesis test to determine if these data provide strong evidence that the proportion of Americans who only use their cell phones to access the internet is diﬀerent than the Chinese proportion of 38%. (b) Interpret the p-value in this context. (c) Calculate a 95% conﬁdence interval for the proportion of Americans who access the internet on their cell phones, and interpret the interval in this context. 6.44 Which chi-square test? Part 1. Consider each of the following tables. Determine (i) if a goodness of ﬁt test, test for homogeneity, or test for independence is more appropriate, and (ii) how many degrees of freedom should be used for the test. (a) (b) Favorite Animal Count Red Panda Koala Otter Fennec Fox Hedgehog 22 7 13 25 38 Favorite Kid Food Pizza Tacos Mac and Cheese Chicken or Veggie Nuggets Broccoli Count 167 48 171 74 2 (c) Freshman Sophomore Other Rushing Not 275 14 392 5 725 7 (d) Commute Time Count ≤ 10 minutes 11-30 minutes 31-60 minutes > 60 minutes 198 130 48 29 6.45 Which chi-square test? Part 2. Consider each of the following planned studies. Determine (i) if a goodness of ﬁt test, test for homogeneity, or test for independence is more appropriate, and (ii) how many degrees of freedom should be used for the test. (a) A state is conducting a study to better understand pay for tradespeople in the state’s three largest cities. In each city, the state will take a random sample of tradespeople and estimate the proportion who made at least $100,000 in each of the cities. In their ﬁnal report, they would also like to note whether that proportion varies across the three cities. (b) A particular gene has 3 variants that can be found in proportions p1 = 0.15, p2 = 0.60, and p3 = 0.25 in the general population. Scientists suspect diﬀerent variants of this gene might indicate an elevated risk for a particular genetic disease, and one way to evaluate this is to see if the general population distribution is the same in patients with the disease. The scientists will sample 450 patients with the disease and identify which variant each patient has. (c) A candy company produces candy pieces in 5 diﬀerent colors that are mixed into bags. The colors should be in the following proportions: 15% green, 22% orange, 20% yellow, 24% red, and 19% purple. As a quality control check, the company randomly samples 1500 candy pieces and wants to determine if the target proportions match those of the observed distribution. 48Gallup, Americans Issue Split Decision on Healthcare Ruling, data collected June 28, 2012. 49Pew Internet, Cell Internet Use 2012, data collected between March 15 - April 13, 2012. 50S. Chang. “The Chinese Love to Use Feature Phone to Access the Internet”. In: M.I.C Gadget (2012). 360 Chapter 7 Inference for numerical data 7.1 Inference for a mean with the t-distribution 7.2 Inference with paired data 7.3 Inference for the difference of two means 361 Chapter 5 introduced a framework for statistical inference based on conﬁdence intervals and hypothesis tests. Chapter 6 summarized inference procedures for categorical data (counts and proportions), using the normal distribution and the chi-square distribution. In this chapter, we focus on inference procedures for numerical data and we encounter a new distribution. In each case, the inference ideas remain the same: 1. Determine which point estimate or test statistic is useful. 2. Identify an appropriate distribution for the point estimate or test statistic. 3. Apply the ideas from Chapter 5 using the distribution from step 2. Each section in Chapter 7 explores a new situation: a single mean (7.1), a mean of diﬀerences (7.2); and a diﬀerence of means (7.3). For videos, slides, and other resources, please visit www.openintro.org/ahss 362 CHAPTER 7. INFERENCE FOR NUMERICAL DATA 7.1 Inference for a mean with the ttt-distribution In this section, we turn our attention to numerical variables and answer questions such as the following: • How well can we estimate the mean income of people in
a certain city, county, or state? • What is the average mercury content in various types of ﬁsh? • Are people’s run times getting faster or slower, on average? • How does the sample size aﬀect the expected error in our estimates? • When is it reasonable to model the sample mean ¯x using a normal distribution, and when will we need to use a new distribution, known as the t-distribution? Learning objectives 1. Understand the relationship between a t-distribution and a normal distribution, and explain why we use a t-distribution for inference on a mean. 2. State and verify whether or not the conditions for inference for a mean based on the tdistribution are met. Understand when it is necessary to look at the distribution of the sample data. 3. Know the degrees of freedom associated with a one-sample t-procedure. 4. Carry out a complete hypothesis test for a single mean. 5. Carry out a complete conﬁdence interval procedure for a single mean. 6. Find the minimum sample size needed to estimate a mean with C% conﬁdence and a margin of error no greater than a certain value. 7.1.1 Using a normal distribution for inference when σσσ is known In Section 4.2 we saw that the distribution of a sample mean is normal if the population is normal or if the sample size is at least 30. In these problems, we used the population mean and population standard deviation to ﬁnd a Z-score. However, in the case of inference, these values will be unknown. In rare circumstances we may know the standard deviation of a population, even though we do not know its mean. For example, in some industrial processes, the mean may be known to shift over time, while the standard deviation of the process remains the same. In these cases, we can use the normal model as the basis for our inference procedures. We use ¯x as our point estimate for µ and the SD formula for a sample mean calculated in Section 4.2: σ¯x = σ√ n . That leads to a conﬁdence interval and a test statistic as follows: CI: ¯x ± z σ √ n ¯x − null value σ√ n Z = What happens if we do not know the population standard deviation σ, as is usually the case? The best we can do is use the sample standard deviation, denoted by s, to estimate the population standard deviation. SE = s √ n 7.1. INFERENCE FOR A MEAN WITH THE T -DISTRIBUTION 363 However, when we do this we run into a problem: when carrying out our inference procedures, we will be trying to estimate two quantities: both the mean and the standard deviation. Looking at the SD and SE formulas, we can make some important observations that will give us a hint as to what will happen when we use s instead of σ. • For a given population, σ is a ﬁxed number and does not vary. • s, the standard deviation of a sample, will vary from one sample to the next and will not be exactly equal to σ. • The larger the sample size n, the better the estimate s will tend to be for σ. For this reason, the normal model still works well when the sample size is large. For smaller sample sizes, we run into a problem: our use of s, which is used when computing the standard error, tends to add more variability to our test statistic. It is this extra variability that leads us to a new distribution: the t-distribution. 7.1.2 Introducing the ttt-distribution When we use the sample standard deviation s in place of the population standard deviation σ to standardize the sample mean, we get an entirely new distribution - one that is similar to the normal distribution, but has greater spread. This distribution is known as the t-distribution. A t-distribution, shown as a solid line in Figure 7.1, has a bell shape. However, its tails are thicker than the normal model’s. We can see that a greater proportion of the area under the t-distribution is beyond 2 standard units from 0 than under the normal distribution. These extra thick tails are exactly the correction we need to resolve the problem of a poorly estimated standard deviation. Figure 7.1: Comparison of a t-distribution (solid line) and a normal distribution (dotted line). The t-distribution, always centered at zero, has a single parameter: degrees of freedom. The degrees of freedom (df ) describes the precise form of the bell-shaped t-distribution. Several t-distributions are shown in Figure 7.2. When there are more degrees of freedom, the t-distribution looks more like the standard normal distribution. Figure 7.2: The larger the degrees of freedom, the more closely the t-distribution resembles the standard normal distribution. −4−2024−4−2024normalt, df = 8t, df = 4t, df = 2t, df = 1 364 CHAPTER 7. INFERENCE FOR NUMERICAL DATA DEGREES OF FREEDOM The degrees of freedom describes the shape of the t-distribution. The larger the degrees of freedom, the more closely the distribution resembles the standard normal distribution. When the degrees of freedom is large, about 30 or more, the t-distribution is nearly indistinguishable from the normal distribution. In Section 7.1.4, we will see how degrees of freedom relates to sample size. We will ﬁnd it useful to become familiar with the t-distribution, because it plays a very similar role to the normal distribution during inference. We use a ttt-table, partially shown in Figure 7.3, in place of the normal probability table when the population standard deviation is unknown, especially when the sample size is small. A larger table is presented in Appendix C.3. df one tail 1 2 3 ... 17 18 19 20 ... 1000 0.100 3.078 1.886 1.638 ... 1.333 1.330 1.328 1.325 ... 1.282 ∞ 1.282 80% 0.050 6.314 2.920 2.353 ... 1.740 1.734 1.729 1.725 ... 1.646 1.645 90% 0.025 12.71 4.303 3.182 ... 2.110 2.101 2.093 2.086 ... 1.962 1.960 95% 0.010 31.82 6.965 4.541 ... 2.567 2.552 2.539 2.528 ... 2.330 2.326 98% 0.005 63.66 9.925 5.841 2.898 2.878 2.861 2.845 2.581 2.576 99% Conﬁdence level C Figure 7.3: An abbreviated look at the t-table. Each row represents a diﬀerent t-distribution. The columns describe the cutoﬀs for speciﬁc tail areas. The row with df = 18 has been highlighted. Each row in the t-table represents a t-distribution with diﬀerent degrees of freedom. The columns correspond to tail probabilities. For instance, if we know we are working with the tdistribution with df = 18, we can examine row 18, which is highlighted in Figure 7.3. If we want the value in this row that identiﬁes the cutoﬀ for an upper tail of 10%, we can look in the column where one tail is 0.100. This cutoﬀ is 1.33. If we had wanted the cutoﬀ for the lower 10%, we would use -1.33. Just like the normal distribution, all t-distributions are symmetric. EXAMPLE 7.1 What proportion of the t-distribution with 18 degrees of freedom falls below -2.10? Just like a normal probability problem, we ﬁrst draw a picture as shown in Figure 7.4 and shade the area below -2.10. To ﬁnd this area, we identify the appropriate row: df = 18. Then we identify the column containing the absolute value of -2.10; it is the third column. Because we are looking for just one tail, we examine the top line of the table, which shows that a one tail area for a value in the third row corresponds to 0.025. That is, 2.5% of the distribution falls below -2.10. EXAMPLE 7.2 For the t-distribution with 18 degrees of freedom, what percent of the curve is contained between -1.330 and +1.330? Using row df = 18, we ﬁnd 1.330 in the table. The area in each tail is 0.100 for a total of 0.200, which leaves 0.800 in the middle between -1.33 and +1.33. This corresponds to the 80%, which can be found at the very bottom of that column. 7.1. INFERENCE FOR A MEAN WITH THE T -DISTRIBUTION 365 Figure 7.4: The t-distribution with 18 degrees of freedom. The area below -2.10 has been shaded. Figure 7.5: Left: The t-distribution with 3 degrees of freedom, with the area farther than 3.182 units from 0 shaded. Right: The t-distribution with 20 degrees of freedom, with the area above 1.65 shaded. EXAMPLE 7.3 For the t-distribution with 3 degrees of freedom, as shown in the left panel of Figure 7.5, what should the value of t be so that 95% of the area of the curve falls between -t and +t? We can look at the column in the t-table that says 95% along the bottom row and trace it up to row df = 3 to ﬁnd that t = 3.182. EXAMPLE 7.4 A t-distribution with 20 degrees of freedom is shown in the right panel of Figure 7.5. Estimate the proportion of the distribution falling above 1.65. We identify the row in the t-table using the degrees of freedom: df = 20. Then we look for 1.65; it is not listed. It falls between the ﬁrst and second columns. Since these values bound 1.65, their tail areas will bound the tail area corresponding to 1.65. We identify the one tail area of the ﬁrst and second columns, 0.050 and 0.10, and we conclude that between 5% and 10% of the distribution is more than 1.65 standard deviations above the mean. If we like, we can identify the precise area using statistical software: 0.0573. When the desired degrees of freedom is not listed on the table, choose a conservative value: round the degrees of freedom down, i.e. move up to the previous row listed. Another option is to use a calculator or statistical software to get a precise answer. −4−2024−4−2024−4−2024 366 CHAPTER 7. INFERENCE FOR NUMERICAL DATA 7.1.3 Technology: ﬁnding area under the ttt-distribution It is possible to ﬁnd areas under a t-distribution on a calculator. TI-84: FINDING AREA UNDER THE T-CURVE Use 2ND VARS, tcdf to ﬁnd an area/proportion/probability between two t-scores or to the left or right of a t-score. 1. Choose 2ND VARS (i.e. DISTR). 2. Choose 6:tcdf. 3. Enter the lower (left) t-score and the upper (right) t-score. • If ﬁnding just a lower tail area, set lower to -100. • For an upper tail area, set upper to 100. 4. Enter the degrees of freedom after df:. 5. Down arrow, choose Paste, and hit ENTER. TI-83: Do steps 1-2, then enter the lower bound, upper bound, degrees of freedom, e.g. tcdf(2, 100, 5), and hit ENTER. CASIO FX-9750GII: FINDING AREA UNDER THE T-DISTRIBUTION 1. Navigate to STAT (MENU, then hit 2
). 2. Select DIST (F5), then t (F2), and then tcd (F2). 3. If needed, set Data to Variable (Var option, which is F2). 4. Enter the Lower t-score and the Upper t-score. Set the degrees of freedom (df). • If ﬁnding just a lower tail area, set Lower to -100. • For an upper tail area, set Upper to 100. 5. Hit EXE, which will return the area probability (p) along with the t-scores for the lower and upper bounds. GUIDED PRACTICE 7.5 Use a calculator to ﬁnd the area to the right of t = 3 under the t-distribution with 35 degrees of freedom.1 GUIDED PRACTICE 7.6 Without doing any calculations, will the area to the right of Z = 3 under the standard normal curve be greater than, less than, or equal to the area to the right of t = 3 with 35 degrees of freedom?2 1Because we want to shade to the right of t = 3, we let lower = 3. There is no upper bound, so use a large value such as 100 for upper. Let df = 35. The area is 0.0025 or 0.25%. 2Because the t-distribution has greater spread and thicker tails than the normal distribution, we would expect the upper tail area to the right of Z = 3 to be less than the upper tail area to the right of t = 3. One can conﬁrm that the area to the right of Z = 3 is 0.0013, which is less than 0.0025. With a smaller degrees of freedom, this diﬀerence would be even more pronounced. Try it! 7.1. INFERENCE FOR A MEAN WITH THE T -DISTRIBUTION 367 7.1.4 Checking conditions for inference on a mean using the ttt-distribution Using the t-distribution for inference on a mean requires that the theoretical sampling distribution for the sample mean ¯x is nearly normal. In practice, we check whether this assumption is reasonable by verifying that certain conditions are met. Independence. Observations can be considered independent when the data are collected from a random process, such as rolling a die, or from a random sample. Without a random sample or process, the standard error formula would not apply, and it is unclear to what population the inference would apply. Recall that when sampling without replacement from a ﬁnite population, the observations can be considered independent when sampling less than 10% of the population. Sample size / nearly normal population. We saw in Section 4.2 that in order for the sampling distribution for a sample mean to be nearly normal, we also need the sample to be drawn from a nearly normal population or we need the sample size to be at least 30 (n ≥ 30). What should we do when the sample size is small and we are not sure whether the population distribution is nearly normal? In this case, the best we can do is look at the data for excessive skew. If the data are very skewed or have obvious outliers, this suggests that the sample did not come from a nearly normal population. However, if the data do not show obvious skew or outliers, then the idea of a nearly normal population is generally considered reasonable. Note that by looking at a small data set, we cannot prove that the population distribution is nearly normal. However, the data can suggest to us whether the population distribution being nearly normal is an unreasonable assumption. THE NORMALITY CONDITION WITH SMALL SAMPLES If the sample is small and there is strong skew or extreme outliers in the data, the population from which the sample was drawn may not be nearly normal. Ideally, we use a graph of the data to check for strong skew or outliers. When the full data set is not available, summary statistics can also be used. For larger samples, it is less necessary to check for skew in the data. If the sample size is 30 or more, it is no longer necessary that the population distribution be nearly normal. When the sample size is large, the Central Limit Theorem tells us that the sampling distribution for the sample mean will be nearly normal regardless of the distribution of the population. 7.1.5 One-sample ttt-interval for a mean Dolphins are at the top of the oceanic food chain, which causes dangerous substances such as mercury to concentrate in their organs and muscles. This is an important problem for both dolphins and other animals, like humans, who eat them. We would like to create a conﬁdence interval to estimate the average mercury content in dolphin muscles. We will use a sample of 19 Risso’s dolphins from the Taiji area in Japan. The data are summarized in Figure 7.7. Because we are estimating a mean, we would like to construct a t-interval, but ﬁrst we must check whether the conditions for using a t-interval are met. We will start by assuming that the sample of 19 Risso’s dolphins constitutes a random sample. Next, we note that the sample size is small (less than 30), and we do not know whether the distribution of mercury content for all dolphins is nearly normal. Therefore, we must look at the data. Since we do not have all of the data to graph, we look at the summary statistics provided in Figure 7.7. These summary statistics do not suggest any strong skew or outliers; all observations are within 2.5 standard deviations of the mean. Based on this evidence, we believe it is reasonable that the population distribution of mercury content in dolphins could be nearly normal. 368 CHAPTER 7. INFERENCE FOR NUMERICAL DATA Figure 7.6: A Risso’s dolphin. —————————– Photo by Mike Baird (www.bairdphotos.com). CC BY 2.0 license. n 19 ¯x 4.4 s minimum maximum 2.3 1.7 9.2 Figure 7.7: Summary of mercury content in the muscle of 19 Risso’s dolphins from the Taiji area. Measurements are in µg/wet g (micrograms of mercury per wet gram of muscle). With both conditions met, we will construct a 95% conﬁdence interval. Recall that a conﬁdence interval has the following form: point estimate ± critical value × SE of estimate n. What The point estimate is the sample mean and the SE of the sample mean is given by s/ do we use for the critical value? Since we are using the t-distribution, we use a t-table to ﬁnd the critical value. We denote the critical value t. √ • For a 95% conﬁdence interval, we want to ﬁnd the cutoﬀ t such that 95% of the t-distribution is between -t and t. • Using the t-table on page 364, we look at the row that corresponds to the degrees of freedom and the column that corresponds to the conﬁdence level. DEGREES OF FREEDOM FOR A SINGLE SAMPLE If the sample has n observations and we are examining a single mean, then we use the tdistribution with df = n − 1 degrees of freedom. 7.1. INFERENCE FOR A MEAN WITH THE T -DISTRIBUTION 369 EXAMPLE 7.7 Calculate a 95% conﬁdence interval for the average mercury content in dolphin muscles based on this sample. Recall that n = 19, ¯x = 4.4 µg/wet g, and s = 2.3 µg/wet g. To ﬁnd the critical value t we use the t-distribution with n − 1 degrees of freedom. The sample size is 19, so df = 19 − 1 = 18 degrees of freedom. Using the t-table with row df = 18 and column corresponding to a 95% conﬁdence level, we get t = 2.10. The point estimate is the sample mean ¯x and the standard error of a sample mean is given by s√ n . Now we have all the pieces we need to calculate a 95% conﬁdence interval for the average mercury content in dolphin muscles. point estimate ± critical value × SE of estimate s √ n 2.3 √ 19 4.4 ± 2.10 × ¯x ± t × df = n − 1 df = 18 = (3.29, 5.51) EXAMPLE 7.8 How do we interpret this 95% conﬁdence interval? To what population is it applicable? A random sample of Risso’s dolphins was taken from the Taiji area in Japan. The mercury content in the muscles of other types of dolphins and from dolphins from other regions may vary. Therefore, we can only make an inference to Risso’s dolphins from this area. We are 95% conﬁdent the true average mercury content in the muscles of Risso’s dolphins in the Taiji area of Japan is between 3.29 and 5.51 µg/wet gram. 370 CHAPTER 7. INFERENCE FOR NUMERICAL DATA CONSTRUCTING A CONFIDENCE INTERVAL FOR A MEAN To carry out a complete conﬁdence interval procedure to estimate a single mean µ, Identify: Identify the parameter and the conﬁdence level, C%. The parameter will be an unknown population mean, e.g. the true mean (or average) mercury content in Risso’s dolphins. Choose: Choose the appropriate interval procedure and identify it by name. To estimate a single mean we use a 1-sample ttt-interval. Check: Check conditions for the sampling distribution for ¯x to be nearly normal. 1. Independence: Data come from a random sample or random process. When sampling without replacement, check that sample size is less than 10% of the population size. 2. Large sample or normal population: n ≥ 30 or the population distr. is nearly normal. If the sample size is less than 30 and the population distribution is unknown, check for strong skew or outliers in the data. If neither is found, the condition that the population distribution is nearly normal is considered reasonable. Calculate: Calculate the conﬁdence interval and record it in interval form. point estimate ± t × SE of estimate, df = n − 1 point estimate: the sample mean ¯x s√ n SE of estimate: t: use a t-table at row df = n − 1 and conﬁdence level C% ( , ) Conclude: Interpret the interval and, if applicable, draw a conclusion in context. . A conclusion Here, we are C% conﬁdent that the true mean of [...] is between depends upon whether the interval is entirely above, is entirely below, or contains the value of interest. and 7.1. INFERENCE FOR A MEAN WITH THE T -DISTRIBUTION 371 EXAMPLE 7.9 The FDA’s webpage provides some data on mercury content of ﬁsh. Based on a sample of 15 croaker white ﬁsh (Paciﬁc), a sample mean and standard deviation were computed as 0.287 and 0.069 ppm (parts per million), respectively. The 15 observations ranged from 0.18 to 0.41 ppm. Construct an appropriate 95% conﬁdence interval for the true average mercury content of croaker white ﬁsh (Paciﬁc). Is there evidence that the average mercury content is greater than 0.275 ppm? Use the ﬁve step framework to organize your work. Identify: The parameter of interest is the true mean mercury content in croaker white ﬁsh (Pac
iﬁc). We want to estimate this at the 95% conﬁdence level. Choose: Because the parameter to be estimated is a single mean, we will use a 1-sample t-interval. Check: We must check that the sampling distribution for the mean can be modeled using a normal distribution. We will assume that the sample constitutes a random sample of less than 10% of all croaker white ﬁsh (Paciﬁc) and that independence is reasonable. The sample size n is small, but there are no obvious outliers; all observations are within 2 standard deviations of the mean. If there is skew, it is not too great. Therefore we think it is reasonable that the population distribution of mercury content in croaker white ﬁsh (Paciﬁc) could be nearly normal. Calculate: We will calculate the interval: point estimate ± t × SE of estimate The point estimate is the sample mean: ¯x = 0.287 The SE of the sample mean is: n = 0.069√ s√ 15 We ﬁnd t for the one-sample case using the t-table at row df = n − 1 and conﬁdence level C%. For a 95% conﬁdence level and df = 15 − 1 = 14, t = 2.145. So the 95% conﬁdence interval is given by: 0.287 ± 2.145 × 0.069 √ 15 df = 14 0.287 ± 2.145 × 0.0178 = (0.249, 0.325) Conclude: We are 95% conﬁdent that the true average mercury content of croaker white ﬁsh (Paciﬁc) is between 0.249 and 0.325 ppm. Because the interval contains 0.275 as well as values less than 0.275, we do not have evidence that the true average mercury content is greater than 0.275 ppm. EXAMPLE 7.10 Based on the interval calculated in Example 7.9 above, can we say that 95% of croaker white ﬁsh (Paciﬁc) have mercury content between 0.249 and 0.325 ppm? No. The interval estimates the average amount of mercury with 95% conﬁdence. It is not trying to capture 95% of the values. 372 CHAPTER 7. INFERENCE FOR NUMERICAL DATA 7.1.6 Technology: the 1-sample ttt-interval TI-83/84: 1-SAMPLE T-INTERVAL Use STAT, TESTS, TInterval. 1. Choose STAT. 2. Right arrow to TESTS. 3. Down arrow and choose 8:TInterval. 4. Choose Data if you have all the data or Stats if you have the mean and standard deviation. • If you choose Data, let List be L1 or the list in which you entered your data (don’t forget to enter the data!) and let Freq be 1. • If you choose Stats, enter the mean, SD, and sample size. 5. Let C-Level be the desired conﬁdence level. 6. Choose Calculate and hit ENTER, which returns: the conﬁdence interval the sample mean the sample SD the sample size ( , ) ¯x Sx n CASIO FX-9750GII: 1-SAMPLE T-INTERVAL 1. Navigate to STAT (MENU button, then hit the 2 button or select STAT). 2. If necessary, enter the data into a list. 3. Choose the INTR option (F3 button), t (F2 button), and 1-S (F1 button). 4. Choose either the Var option (F2) or enter the data in using the List option. 5. Specify the interval details: • Conﬁdence level of interest for C-Level. • If using the Var option, enter the summary statistics. If using List, specify the list and leave Freq value at 1. 6. Hit the EXE button, which returns Left, Right ¯x sx n ends of the conﬁdence interval sample mean sample standard deviation sample size GUIDED PRACTICE 7.11 Use a calculator to ﬁnd a 95% conﬁdence interval for the mean mercury content in croaker white ﬁsh (Paciﬁc). The sample size was 15, and the sample mean and standard deviation were computed as 0.287 and 0.069 ppm (parts per million), respectively.3 . 3Choose TInterval or equivalent. We do not have all the data, so choose Stats on a TI or Var on a Casio. Enter ¯x and Sx. Note: Sx is the sample standard deviation (0.069), not the SE. Let n = 15 and C-Level = 0.95. This should give the interval (0.249, 0.325). 7.1. INFERENCE FOR A MEAN WITH THE T -DISTRIBUTION 373 7.1.7 Choosing a sample size when estimating a mean In Section 6.1.5, we looked at sample size considerations when estimating a proportion. We take the same approach when estimating a mean. Recall that the margin of error is measured as the distance between the point estimate and the upper or lower bound of the conﬁdence interval. We want to estimate a mean with a particular conﬁdence level while putting an upper bound on the margin of error. What is the smallest sample size that will satisfy these conditions? For a one-sample t-interval, the margin of error, M E, is given by M E = t × s√ n . The challenge in this case is that we need to know n to ﬁnd t. But n is precisely what we are attempting to solve for! Fortunately, in most cases we will have a reasonable estimate for the population standard deviation and the desired n will be large, so we can use M E = z × σ√ n , making it easier to solve for n. EXAMPLE 7.12 Blood pressure oscillates with the beating of the heart, and the systolic pressure is deﬁned as the peak pressure when a person is at rest. The standard deviation of systolic blood pressure for people in the U.S. is about 25 mmHg (millimeters of mercury). How large of a sample is necessary to estimate the average systolic blood pressure of people in a particular town with a margin of error no greater than 4 mmHg using a 95% conﬁdence level? For this problem, we want to ﬁnd the sample size n so that the margin of error, M E, is less than or equal to 4 mmHg. We start by writing the following inequality: z × σ √ n ≤ 4 For a 95% conﬁdence level, the critical value z = 1.96. Our best estimate for the population standard deviation is σ = 25. We substitute in these two values and we solve for n. 1.96 × 1.96 × 1.96 × 25 4 25 √ n 25 4 2 ≤ 4 √ n ≤ ≤ n 150.06 ≤ n n = 151 The minimum sample size that meets the condition is 151. We round up because the sample size must be an integer and it must be greater than or equal to 150.06. IDENTIFY A SAMPLE SIZE FOR A PARTICULAR MARGIN OF ERROR To estimate the minimum sample size required to achieve a margin of error less than or equal to m, with C% conﬁdence, we set up an inequality as follows: z σ √ n ≤ m z depends on the desired conﬁdence level and σ is the standard deviation associated with the population. We solve for the sample size, n. Sample size computations are helpful in planning data collection, and they require careful forethought. 374 CHAPTER 7. INFERENCE FOR NUMERICAL DATA 7.1.8 Hypothesis testing for a mean Is the typical U.S. runner getting faster or slower over time? Technological advances in shoes, training, and diet might suggest runners would be faster. An opposing viewpoint might say that with the average body mass index on the rise, people tend to run slower. In fact, all of these components might be inﬂuencing run time. We consider this question in the context of the Cherry Blossom Race, which is a 10-mile race in Washington, DC each spring. The average time for all runners who ﬁnished the Cherry Blossom Race in 2006 was 93.3 minutes (93 minutes and about 18 seconds). We want to determine using data from 100 participants in the 2017 Cherry Blossom Race whether runners in this race are getting faster or slower, versus the other possibility that there has been no change. Figure 7.8 shows run times for 100 randomly selected participants. Figure 7.8: A histogram of time for the sample of 2017 Cherry Blossom Race participants. EXAMPLE 7.13 What are appropriate hypotheses for this context? We know that the average run time for all runners in 2006 was 93.3 minutes. We have a sample of times from the 2017 race. We are interested in whether the average run time has changed, so we will use a two-sided HA. Let µ represent the average 10-mile run time of all participants in 2017, which is unknown to us. H0: µ = 93.3 minutes. The average run time of all participants in 2017 was 93.3 min. HA: µ = 93.3 minutes. The average run time of all participants in 2017 was not 93.3 min. The data come from a random sample from a large population, so the observations are independent. Do we need to check for skew in the data? No – with a sample size of 100, well over 30, the Central Limit Theorem tells us that the sampling distribution for ¯x will be nearly normal. With independence satisﬁed and slight skew not a concern for this large of a sample, we can proceed with performing a hypothesis test using the t-distribution. The sample mean and sample standard deviation of the 100 runners from the 2017 Cherry Blossom Race are 97.3 and 17.0 minutes, respectively. We want to know whether the observed sample mean of 97.3 is far enough away from 93.3 to provide convincing evidence of a real diﬀerence, or if it is within the realm of expected variation for a sample of size 100. Time (Minutes)Frequency608010012014005101520 7.1. INFERENCE FOR A MEAN WITH THE T -DISTRIBUTION 375 To answer this question we will ﬁnd the test statistic and p-value for the hypothesis test. Since we will be using a sample standard deviation in our calculation of the test statistic, we will need to use a t-distribution, just as we did with conﬁdence intervals for a mean. We call the test statistic a T -statistic. It has the same general form as a Z-statistic. T = point estimate − null value SE of estimate As we saw before, when carrying out inference on a single mean, the degrees of freedom is given by n − 1. THE T-STATISTIC The T-statistic (or T-score) is analogous to a Z-statistic (or Z-score). Both represent how many standard errors the observed value is from the null value. EXAMPLE 7.14 Calculate the test statistic, degrees of freedom, and p-value for this test. Here, our point estimate is the sample mean, ¯x = 97.3 minutes. The SE of the sample mean is given by s√ n , so the SE of estimate = 17.0√ 100 = 1.7 minutes. T = 97.3 − 93.3 1.7 = 2.35 df = 100 − 1 = 99 Using a calculator, we ﬁnd that the area above 2.35 under the t-distribution with 99 degrees of freedom is 0.01. Because this is a two-tailed test, we double this. So the p-value = 2 × 0.01 = 0.02. EXAMPLE 7.15 Does the data provide suﬃcient evidence that the average Cherry Blossom Run time in 2017 is diﬀerent than in 2006? This depends upon the desired signiﬁcance level. Since the p-value = 0.02 < 0.05, there is suﬃcient evidence at the 5% signiﬁ
cance level. However, as the p-value of 0.02 > 0.01, there is not suﬃcient evidence at the 1% signiﬁcance level. EXAMPLE 7.16 Would you expect the hypothesized value of 93.3 to fall inside or outside of a 95% conﬁdence interval? What about a 99% conﬁdence interval? Because the hypothesized value of 93.3 was rejected by the two-sided α = 0.05 test, we would expect it to be outside the 95% conﬁdence interval. However, because the hypothesized value of 93.3 was not rejected by the two-sided α = 0.01 test, we would expect it to fall inside the (wider) 99% conﬁdence interval. 376 CHAPTER 7. INFERENCE FOR NUMERICAL DATA HYPOTHESIS TEST FOR A MEAN To carry out a complete hypothesis test to test the claim that a single mean µ is equal to a null value µ0, Identify: Identify the hypotheses and the signiﬁcance level, α. H0: µ = µ0 HA: µ = µ0; HA: µ > µ0; or HA: µ < µ0 Choose: Choose the appropriate test procedure and identify it by name. To test hypotheses about a single mean we use a 1-sample ttt-test. Check: Check conditions for the sampling distribution for ¯x to be nearly normal. 1. Independence: Data come from a random sample or random process. When sampling without replacement, check that sample size is less than 10% of the population size. 2. Large sample or normal population: n ≥ 30 or the population distr. is nearly normal. - If the sample size is less than 30 and the population distribution is unknown, check for strong skew or outliers in the data. If neither is found, then the condition that the population is nearly normal is considered reasonable. Calculate: Calculate the t-statistic, df , and p-value. T = point estimate − null value SE of estimate , df = n − 1 point estimate: the sample mean ¯x s√ n SE of estimate: null value: µ0 p-value = (based on the t-statistic, the df , and the direction of HA) Conclude: Compare the p-value to α, and draw a conclusion in context. If the p-value is < α, reject H0; there is suﬃcient evidence that [HA in context]. If the p-value is > α, do not reject H0; there is not suﬃcient evidence that [HA in context]. 7.1. INFERENCE FOR A MEAN WITH THE T -DISTRIBUTION 377 EXAMPLE 7.17 Recall the example involving the mercury content in croaker white ﬁsh (Paciﬁc). Based on a sample of size 15, a sample mean and standard deviation were computed as 0.287 and 0.069 ppm (parts per million), respectively. Carry out an appropriate test to determine if 0.25 is a reasonable value for the average mercury content of croaker white ﬁsh (Paciﬁc). Use the ﬁve step method to organize your work. Identify: We will test the following hypotheses at the α = 0.05 signiﬁcance level. H0: µ = 0.25 HA: µ = 0.25 The mean mercury content is not 0.25 ppm. Choose: Because we are hypothesizing about a single mean we choose the 1-sample t-test. Check: The conditions were checked previously, namely – the data come from a random sample of less than 10% of the population of all croaker white ﬁsh (Paciﬁc), and because n is less than 30, we veriﬁed that there is no strong skew or outliers in the data, so the assumption that the population distribution of mercury is nearly normally distributed is reasonable. Calculate: We will calculate the t-statistic and the p-value. T = point estimate − null value SE of estimate The point estimate is the sample mean: ¯x = 0.287 The SE of the sample mean is: n = 0.069√ s√ 15 = 0.0178 The null value is the value hypothesized for the parameter in H0, which is 0.25. For the 1-sample t-test, df = n − 1. T = 0.287 − 0.25 0.0178 = 2.07 df = 15 − 1 = 14 Because HA is a two-tailed test ( = ), the p-value corresponds to the area to the right of t = 2.07 plus the area to the left of t = −2.07 under the t-distribution with 14 degrees of freedom. The p-value = 2 × 0.029 = 0.058. Conclude: The p-value of 0.058 > 0.05, so we do not reject the null hypothesis. We do not have suﬃcient evidence that the average mercury content in croaker white ﬁsh (Paciﬁc) is not 0.25. GUIDED PRACTICE 7.18 Recall that the 95% conﬁdence interval for the average mercury content in croaker white ﬁsh was (0.249, 0.325). Discuss whether the conclusion of the hypothesis test in the previous example is consistent or inconsistent with the conclusion of the conﬁdence interval.4 4It is consistent because 0.25 is located (just barely) inside the conﬁdence interval, so it is considered a reasonable value. Our hypothesis test did not reject the hypothesis that µ = 0.25, also implying that it is a reasonable value. Note that the p-value was just over the cutoﬀ of 0.05. This is consistent with the value of 0.25 being just inside the conﬁdence interval. Also note that the hypothesis test did not prove that µ = 0.25. The value 0.25 is just one of many reasonable values for the true mean. 378 CHAPTER 7. INFERENCE FOR NUMERICAL DATA 7.1.9 Technology: the 1-sample ttt-test TI-83/84: 1-SAMPLE T-TEST Use STAT, TESTS, T-Test. 1. Choose STAT. 2. Right arrow to TESTS. 3. Down arrow and choose 2:T-Test. 4. Choose Data if you have all the data or Stats if you have the mean and standard deviation. 5. Let µ0 be the null or hypothesized value of µ. • If you choose Data, let List be L1 or the list in which you entered your data (don’t forget to enter the data!) and let Freq be 1. • If you choose Stats, enter the mean, SD, and sample size. 6. Choose =, <, or > to correspond to HA. 7. Choose Calculate and hit ENTER, which returns: t T-statistic p-value p the sample mean ¯x Sx n the sample standard deviation the sample size CASIO FX-9750GII: 1-SAMPLE T-TEST 1. Navigate to STAT (MENU button, then hit the 2 button or select STAT). 2. If necessary, enter the data into a list. 3. Choose the TEST option (F3 button). 4. Choose the t option (F2 button). 5. Choose the 1-S option (F1 button). 6. Choose either the Var option (F2) or enter the data in using the List option. 7. Specify the test details: • Specify the sidedness of the test using the F1, F2, and F3 keys. • Enter the null value, µ0. • If using the Var option, enter the summary statistics. If using List, specify the list and leave Freq values at 1. 8. Hit the EXE button, which returns alternative hypothesis ¯x sx n t T-statistic p-value p sample mean sample standard deviation sample size GUIDED PRACTICE 7.19 The average time for all runners who ﬁnished the Cherry Blossom Run in 2006 was 93.3 minutes. In 2017, the average time for 100 randomly selected participants was 97.3, with a standard deviation of 17.0 minutes. Use a calculator to ﬁnd the T -statistic and p-value for the appropriate test to see if the average time for the participants in 2017 is diﬀerent than it was in 2006.5 5Choose T-Test or equivalent. Let µ0 be 93.3. ¯x is 97.3, Sx is 17.0, and n = 100. Choose = to correspond to HA. We get t = 2.353 and the p-value p = 0.021. 7.1. INFERENCE FOR A MEAN WITH THE T -DISTRIBUTION 379 Section summary • The t-distribution. – When calculating a test statistic for a mean, using the sample standard deviation in place of the population standard deviation gives rise to a new distribution called the t-distribution. – As the sample size and degrees of freedom increase, s becomes a more stable estimate of σ, and the corresponding t-distribution has smaller spread. – As the degrees of freedom go to ∞, the t-distribution approaches the normal distribution. This is why we can use the t-table at df = ∞ to ﬁnd the value of z. • When carrying out inference for a single mean, we use the t-distribution with n − 1 degrees of freedom. • When there is one sample and the parameter of interest is a single mean: – Estimate µ at the C% conﬁdence level using a 1-sample ttt-interval. – Test H0: µ = µ0 at the α signiﬁcance level using a 1-sample ttt-test. • The one-sample t-interval and t-test require that the sampling distribution for ¯x be nearly normal. For this reason we must check that the following conditions are met. 1. Independence: The data come from a random sample or random process. When sampling without replacement, check that the sample size is less than 10% of the population size. 2. Large sample or normal population: n ≥ 30 or population distribution is nearly normal. - If the sample size is less than 30 and the population distribution is unknown, check for strong skew or outliers in the data. If neither is found, then the condition that the population distribution is nearly normal is considered reasonable. • When the conditions are met, we calculate the conﬁdence interval and the test statistic as we did in the previous chapter, except that we use t for the critical value and we use T for the test statistic. Conﬁdence interval: point estimate ± t × SE of estimate Test statistic: T = point estimate − null value SE of estimate Here the point estimate is the sample mean: ¯x. The SE of estimate is the SE of the sample mean: s√ n . The degrees of freedom is given by df = n − 1. • To calculate the minimum sample size required to estimate a mean with C% conﬁdence and a margin of error no greater than m, we set up an inequality as follows: z σ √ n ≤ m z depends on the desired conﬁdence level and σ is the standard deviation associated with the population. We solve for the sample size, n. Always round the answer up to the next integer, since n refers to a number of people or things. 380 CHAPTER 7. INFERENCE FOR NUMERICAL DATA Exercises 7.1 Identify the critical ttt. A random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical t-value (t) for the given sample size and conﬁdence level. (a) n = 6, CL = 90% (b) n = 21, CL = 98% (c) n = 29, CL = 95% (d) n = 12, CL = 99% 7.2 ttt-distribution. The ﬁgure on the right shows three unimodal and symmetric curves: the standard normal (z) distribution, the t-distribution with 5 degrees of freedom, and the t-distribution with 1 degree of freedom. Determine which is which, and explain your reasoning. 7.3 Find the p-value, Part I. A random sample is selected from an approximately normal population wit
h an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at α = 0.05. (a) n = 11, T = 1.91 (b) n = 17, T = −3.45 (c) n = 7, T = 0.83 (d) n = 28, T = 2.13 7.4 Find the p-value, Part II. A random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at α = 0.01. (a) n = 26, T = 2.485 (b) n = 18, T = 0.5 7.5 Working backwards, Part I. A 95% conﬁdence interval for a population mean, µ, is given as (18.985, 21.015). This conﬁdence interval is based on a simple random sample of 36 observations. Calculate the sample mean and standard deviation. Assume that all conditions necessary for inference are satisﬁed. Use the t-distribution in any calculations. 7.6 Working backwards, Part II. A 90% conﬁdence interval for a population mean is (65, 77). The population distribution is approximately normal and the population standard deviation is unknown. This conﬁdence interval is based on a simple random sample of 25 observations. Calculate the sample mean, the margin of error, and the sample standard deviation. −4−2024soliddasheddotted 7.1. INFERENCE FOR A MEAN WITH THE T -DISTRIBUTION 381 7.7 Sleep habits of New Yorkers. New York is known as “the city that never sleeps”. A random sample of 25 New Yorkers were asked how much sleep they get per night. Statistical summaries of these data are shown below. The point estimate suggests New Yorkers sleep less than 8 hours a night on average. Evaluate the claim that New York is the city that never sleeps keeping in mind that, despite this claim, the true average number of hours New Yorkers sleep could be less than 8 hours or more than 8 hours. n 25 ¯x 7.73 s min max 9.78 6.17 0.77 (a) Write the hypotheses in symbols and in words. (b) Check conditions, then calculate the test statistic, T , and the associated degrees of freedom. (c) Find and interpret the p-value in this context. Drawing a picture may be helpful. (d) What is the conclusion of the hypothesis test? (e) If you were to construct a 90% conﬁdence interval that corresponded to this hypothesis test, would you expect 8 hours to be in the interval? 7.8 Heights of adults. Researchers studying anthropometry collected body girth measurements and skeletal diameter measurements, as well as age, weight, height and gender, for 507 physically active individuals. The histogram below shows the sample distribution of heights in centimeters.6 Min Q1 Median Mean SD Q3 Max 147.2 163.8 170.3 171.1 9.4 177.8 198.1 (a) What is the point estimate for the average height of active individuals? What about the median? (b) What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR? (c) Is a person who is 1m 80cm (180 cm) tall considered unusually tall? And is a person who is 1m 55cm (155cm) considered unusually short? Explain your reasoning. (d) The researchers take another random sample of physically active individuals. Would you expect the mean and the standard deviation of this new sample to be the ones given above? Explain your reasoning. (e) The sample means obtained are point estimates for the mean height of all active individuals, if the sample of individuals is equivalent to a simple random sample. What measure do we use to quantify the variability of such an estimate? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample. 6G. Heinz et al. “Exploring relationships in body dimensions”. In: Journal of Statistics Education 11.2 (2003). Height150160170180190200020406080100 382 CHAPTER 7. INFERENCE FOR NUMERICAL DATA 7.9 Find the mean. You are given the following hypotheses: H0 : µ = 60 HA : µ = 60 We know that the sample standard deviation is 8 and the sample size is 20. For what sample mean would the p-value be equal to 0.05? Assume that all conditions necessary for inference are satisﬁed. 7.10 ttt vs. zzz. For a given conﬁdence level, t than z∗ aﬀects the width of the conﬁdence interval. df is larger than z. Explain how t∗ df being slightly larger Georgianna claims that in a small city renowned for its music school, the average 7.11 Play the piano. child takes less than 5 years of piano lessons. We have a random sample of 20 children from the city, with a mean of 4.6 years of piano lessons and a standard deviation of 2.2 years. (a) Evaluate Georgianna’s claim using a hypothesis test and include all steps of the Identify, Choose, Check, Calculate, Conclude framework. (b) Construct a 95% conﬁdence interval for the number of years students in this city take piano lessons and include all steps of the Identify, Choose, Check, Calculate, Conclude framework. (c) Do your results from the hypothesis test and the conﬁdence interval agree? Explain your reasoning. 7.12 Auto exhaust and lead exposure. Researchers interested in lead exposure due to car exhaust sampled the blood of 52 police oﬃcers subjected to constant inhalation of automobile exhaust fumes while working traﬃc enforcement in a primarily urban environment. The blood samples of these oﬃcers had an average lead concentration of 124.32 µg/l and a SD of 37.74 µg/l; a previous study of individuals from a nearby suburb, with no history of exposure, found an average blood level concentration of 35 µg/l.7 (a) Write down the hypotheses that would be appropriate for testing if the police oﬃcers appear to have been exposed to a diﬀerent concentration of lead. (b) Explicitly state and check all conditions necessary for inference on these data. (c) Regardless of your answers in part (b), test the hypothesis that the downtown police oﬃcers have a higher lead exposure than the group in the previous study. Interpret your results in context. A market researcher wants to evaluate car insurance savings at a 7.13 Car insurance savings. competing company. Based on past studies he is assuming that the standard deviation of savings is $100. He wants to collect data such that he can get a margin of error of no more than $10 at a 95% conﬁdence level. How large of a sample should he collect? 7.14 SAT scores. The standard deviation of SAT scores for students at a particular Ivy League college is 250 points. Two statistics students, Raina and Luke, want to estimate the average SAT score of students at this college as part of a class project. They want their margin of error to be no more than 25 points. (a) Raina wants to use a 90% conﬁdence interval. How large a sample should she collect? (b) Luke wants to use a 99% conﬁdence interval. Without calculating the actual sample size, determine whether his sample should be larger or smaller than Raina’s, and explain your reasoning. (c) Calculate the minimum required sample size for Luke. 7WI Mortada et al. “Study of lead exposure from automobile exhaust as a risk for nephrotoxicity among traﬃc policemen.” In: American journal of nephrology 21.4 (2000), pp. 274–279. 7.2. INFERENCE WITH PAIRED DATA 383 7.2 Inference with paired data When we have two observations on each person or each case, we can answer questions such as the following: • Do students do better on reading or writing sections of standardized tests? • How do the number of days with temperature above 90°F compare between 1948 and 2018? • Are Amazon textbook prices lower than the college bookstore prices? If so, how much lower, on average? Learning objectives 1. Distinguish between paired and unpaired data. 2. Recognize that inference procedures with paired data use the same one-sample t-procedures as in the previous section, and that these procedures are applied using the diﬀerences of the paired observations. 3. Carry out a complete hypothesis test with paired data. 4. Carry out a complete conﬁdence interval procedure with paired data. 7.2.1 Paired observations and samples In the previous edition of this textbook, we found that Amazon prices were, on average, lower than those of the UCLA Bookstore for UCLA courses in 2010. It’s been several years, and many stores have adapted to the online market, so we wondered, how is the UCLA Bookstore doing today? We sampled 201 UCLA courses. Of those, 68 required books that could be found on Amazon. A portion of the data set from these courses is shown in Figure 7.9, where prices are in U.S. dollars. subject course number 1 American Indian Studies M10 2 Anthropology 3 Arts and Architecture ... ... 67 Korean 68 2 10 ... 1 M10 Jewish Studies bookstore 47.97 14.26 13.50 ... 24.96 35.96 amazon 47.45 13.55 12.53 ... 23.79 32.40 price diﬀerence 0.52 0.71 0.97 ... 1.17 3.56 Figure 7.9: Five cases of the textbooks data set. Each textbook has two corresponding prices in the data set: one for the UCLA Bookstore and one for Amazon. Therefore, each textbook price from the UCLA bookstore has a natural correspondence with a textbook price from Amazon. When two sets of observations have this special correspondence, they are said to be paired. PAIRED DATA Two sets of observations are paired if each observation in one set has a special correspondence or connection with exactly one observation in the other data set. 384 CHAPTER 7. INFERENCE FOR NUMERICAL DATA Figure 7.10: Histogram of the diﬀerence in price for each book sampled. These data are very strongly skewed. Explore this data set on Tableau Public . To analyze paired data, it is often useful to look at the diﬀerence in outcomes of each pair of observations. In the textbook data set, we look at the diﬀerences in prices, which is represented as the diff variable. Here, for each book, the diﬀerences are taken as UCLA Bookstore price − Amazon price It is important that we always subtract using a consistent order; here Amazon prices are always subtracted from UCLA prices. A histogram of these diﬀerences is shown in Figure 7.10. Using diﬀerences between paired observations
is a common and useful way to analyze paired data. GUIDED PRACTICE 7.20 The ﬁrst diﬀerence shown in Figure 7.9 is computed as: 47.97 − 47.45 = 0.52. What does this diﬀerence tell us about the price for this textbook on Amazon versus the UCLA bookstore?8 7.2.2 Hypothesis tests for a mean of differences To analyze a paired data set, we simply analyze the diﬀerences. We can use the same t- distribution techniques we applied in the last section. ndiﬀ 68 ¯xdiﬀ 3.58 sdiﬀ 13.42 Figure 7.11: Summary statistics for the price diﬀerences. There were 68 books, so there are 68 diﬀerences. 8The diﬀerence is taken as UCLA Bookstore price − Amazon price. Because the diﬀerence is positive, it tells us that the UCLA Bookstore price was greater for this textbook. In fact, it was $0.52, or 52 cents, more expensive at the UCLA bookstore than on Amazon. UCLA Bookstore Price − Amazon Price (USD)Frequency−$20$0$20$40$60$800102030 7.2. INFERENCE WITH PAIRED DATA 385 We will set up and implement a hypothesis test to determine whether, on average, there is a diﬀerence in textbook prices between Amazon and the UCLA bookstore. We are considering two scenarios: there is no diﬀerence in prices or there is some diﬀerence in prices. H0: µdiﬀ = 0. On average, there is no diﬀerence in textbook prices. HA: µdiﬀ = 0. On average, there is some diﬀerence in textbook prices. Can the t-distribution be used for this application? The observations are based on a random sample from a large population, so independence is reasonable. While the distribution of the data is very strongly skewed, we do have n = 68 observations. This sample size is large enough that we do not have to worry about whether the population distribution for diﬀerence in price might be nearly normal or not. Because the conditions are satisﬁed, we can use the t-distribution to this setting. We compute the standard error associated with ¯xdiﬀ using the standard deviation of the dif- ferences (sdiﬀ = 13.42) and the number of diﬀerences (ndiﬀ = 68): SE¯xdiﬀ = sdiﬀ √ ndiﬀ = 13.42 √ 68 = 1.63 Next we compute the test statistic. The point estimate is the observed value of ¯xdiﬀ . The null value is the value hypothesized under the null hypothesis. Here, the null hypothesis is that the true mean of the diﬀerences is 0. T = point estimate − null value SE of estimate = 3.58 − 0 1.63 = 2.20 The degrees of freedom are df = 68 − 1 = 67. To visualize the p-value, the sampling distribution for ¯xdiﬀ is drawn as though H0 is true. This is shown in Figure 7.12. Because this is a two-sided test, the p-value corresponds to the area in both tails. Using statistical software, we ﬁnd the area in the tails to be 0.0312. Because the p-value of 0.0312 is less than 0.05, we reject the null hypothesis. We have evidence that, on average, there is a diﬀerence in textbook prices. In particular, we can say that, on average, Amazon prices are lower than the UCLA Bookstore prices for UCLA course textbooks. Figure 7.12: Sampling distribution for the mean diﬀerence in book prices, if the true average diﬀerence is zero. EXAMPLE 7.21 We have evidence to conclude Amazon is, on average, less expensive. Does this mean that UCLA students should always buy their books on Amazon? No. The fact that Amazon is, on average, less expensive, does not imply that it is less expensive for every textbook. Examining the distribution shown in Figure 7.10, we see that there are certainly a handful of cases where Amazon prices are much lower than the UCLA Bookstore’s, which suggests it is worth checking Amazon or other online sites before purchasing. However, in many cases the Amazon price is above what the UCLA Bookstore charges, and most of the time the price isn’t that diﬀerent. For reference, this is a very diﬀerent result from what we (the authors) had seen in a similar data set from 2010. At that time, Amazon prices were almost uniformly lower than those of the UCLA Bookstore’s and by a large margin, making the case to use Amazon over the UCLA Bookstore quite compelling at that time. m0 = 0xdiff = 3.58 386 CHAPTER 7. INFERENCE FOR NUMERICAL DATA HYPOTHESIS TEST FOR A MEAN OF DIFFERENCES To carry out a complete hypothesis test to test the claim that a mean of diﬀerences µdiﬀ is equal to 0, Identify: Identify the hypotheses and the signiﬁcance level, α. H0: µdiﬀ = 0 HA: µdiﬀ = 0; HA: µdiﬀ > 0; or HA: µdiﬀ < 0 Choose: Choose the appropriate test procedure and identify it by name. To test hypotheses about a mean of diﬀerences we use a 1-sample ttt-test with paired data. Check: Check conditions for the sampling distribution for ¯xdiﬀ to be nearly normal. 1. Independence: Data come from one random sample (with paired data) or from a randomized matched pairs experiment. When sampling without replacement, check that the sample size is less than 10% of the population size. 2. Large sample or normal population: ndiﬀ ≥ 30 or population of diﬀs is nearly normal. - If the number of diﬀerences is less than 30 and the distribution of the population of diﬀerences is unknown, check for strong skew or outliers in the sample diﬀerences. If neither is found, then the condition that the population of diﬀerences is nearly normal is considered reasonable. Calculate: Calculate the t-statistic, df , and p-value. T = point estimate − null value SE of estimate , df = ndiﬀ − 1 point estimate: the sample mean of diﬀerences ¯xdiﬀ SE of estimate: sdiﬀ√ ndiﬀ null value: 0 p-value = (based on the t-statistic, the df , and the direction of HA) Conclude: Compare the p-value to α, and draw a conclusion in context. If the p-value is < α, reject H0; there is suﬃcient evidence that [HA in context]. If the p-value is > α, do not reject H0; there is not suﬃcient evidence that [HA in context]. 7.2. INFERENCE WITH PAIRED DATA 387 EXAMPLE 7.22 An SAT preparation company claims that its students’ scores improve by over 100 points on average after their course. A consumer group would like to evaluate this claim, and they collect data on a random sample of 30 students who took the class. Each of these students took the SAT before and after taking the company’s course, so we have a diﬀerence in scores for each student. We will examine these diﬀerences x1 = 57, x2 = 133, ..., x30 = 140. The distribution of the diﬀerences has a mean of 135.9, a standard deviation of 82.2, and is shown below. Do the data provide convincing evidence to back up the company’s claim? Use the ﬁve step framework to organize your work. Identify: We will test the following hypotheses at the α = 0.05 level: H0: µdiﬀ = 100. Student scores improve by 100 points, on average. HA: µdiﬀ > 100. Student scores improve by more than 100 points, on average. Here, diﬀ = SAT score after course - SAT score before course. Choose: Because we have paired data and the parameter to be estimated is a mean of diﬀerences, we will use a 1-sample t-test with paired data. Check: We have a random sample of students and have paired data on them. We will assume that this sample of size 30 represents less than 10% of the total population of such students. Finally, the number of diﬀerences is ndiﬀ = 30 ≥ 30, so we can proceed with the 1-sample t-test. Calculate: We will calculate the test statistic, df , and p-value. T = point estimate − null value SE of estimate The point estimate is the sample mean of diﬀerences: ¯xdiﬀ = 135.9 The SE of the sample mean of diﬀerences is: = 15.0 sdiﬀ√ ndiﬀ = 82.2√ 30 Because this is a one-sample t-test, the degrees of freedom is ndiﬀ − 1. T = 135.9 − 100 82.2√ 30 = 135.9 − 100 15.0 = 2.4 df = 30 − 1 = 29 The p-value is the area to the right of 2.4 under the t-distribution with 29 degrees of freedom. The p-value = 0.012. Conclude: p-value = 0.012 < α so we reject the null hypothesis. The data provide convincing evidence to support the company’s claim that students’ scores improve by more than 100 points, on average, following the class. GUIDED PRACTICE 7.23 Because we found evidence to support the company’s claim, does this mean that a student will score more than 100 points higher on the SAT if they take the class than if they do not take the class?9 9No. First, this is an observational study, so we cannot make a causal conclusion. Maybe SAT test takers tend to improve their score over time even if they don’t take this SAT class. Second, the test considers the average. It does not imply that each student improved. With a sample standard deviation of 82.2 and a mean of 135.9, some students did worse after the SAT class, as shown in the histogram in Example 7.22. Differences−10001002003000510 388 CHAPTER 7. INFERENCE FOR NUMERICAL DATA 7.2.3 Technology: the 1-sample ttt-test with paired data When carrying out a 1-sample t-test with paired data, make sure to use the sample diﬀerences or the summary statistics for the diﬀerences. TI-83/84: 1-SAMPLE T-TEST WITH PAIRED DATA Use STAT, TESTS, T-Test. 1. Choose STAT. 2. Right arrow to TESTS. 3. Down arrow and choose 2:T-Test. 4. Choose Data if you have all the data or Stats if you have the mean and standard deviation. 5. Let µ0 be the null or hypothesized value of µdiﬀ . • If you choose Data, let List be L3 or the list in which you entered the diﬀerences, and let Freq be 1. • If you choose Stats, enter the mean, SD, and sample size of the diﬀerences. 6. Choose =, <, or > to correspond to HA. 7. Choose Calculate and hit ENTER, which returns: t p ¯x Sx n T-statistic p-value the sample mean of the diﬀerences the sample SD of the diﬀerences the sample size of the diﬀerences CASIO FX-9750GII: 1-SAMPLE T-TEST WITH PAIRED DATA 1. Compute the diﬀerences of the paired observations. 2. Using the computed diﬀerences, follow the instructions for a 1-sample t-test. 7.2.4 Conﬁdence intervals for the mean of a difference In the previous examples, we carried out a 1-sample t-test with paired data, where the null hypothesis was that the true mean of diﬀerences is zero. Sometimes we want to estimate the true mean of diﬀerences with a conﬁdence interval, and we use a 1-sample t-inter
val with paired data. Consider again the table summarizing data on: UCLA Bookstore price − Amazon price, for each of the 68 books sampled. ndiﬀ 68 ¯xdiﬀ 3.58 sdiﬀ 13.42 Figure 7.13: Summary statistics for the price diﬀerences. There were 68 books, so there are 68 diﬀerences. We construct a 95% conﬁdence interval for the average price diﬀerence between books at the UCLA Bookstore and on Amazon. Conditions have already veriﬁed, namely, that we have paired data from a random sample and that the number of diﬀerences is at least 30. We must ﬁnd the critical value, t. Since df = 67 is not on the t-table, round the df down to 60 to get a t of 2.00 for 95% conﬁdence. (See Section 7.2.5 for how to get a more precise interval using a calculator.) 7.2. INFERENCE WITH PAIRED DATA 389 Plugging the t value, point estimate, and standard error into the conﬁdence interval formula, we get: point estimate ± t × SE of estimate → 3.58 ± 2.00 × 13.42 √ 68 → (0.33, 6.83) We are 95% conﬁdent that the UCLA bookstore is, on average, between $0.33 and $6.83 more expensive than Amazon for UCLA course books. This interval does not contain zero, so it is consistent with the earlier hypothesis test that rejected the null hypothesis that the average diﬀerence was 0. Because our interval is entirely above 0, we have evidence that the true average diﬀerence is greater than zero. Unlike the hypothesis test, the conﬁdence interval gives us a good idea of how much more expensive the UCLA bookstore might be, on average. EXAMPLE 7.24 Based on the interval, can we say that 95% of the books cost between $0.33 and $6.83 more at the UCLA Bookstore than on Amazon? No. This interval is attempting to estimate the average diﬀerence with 95% conﬁdence. It is not attempting to capture 95% of the values. A quick look at Figure 7.10 shows that much less than 95% of the diﬀerences fall between $0.32 and $6.84. CONSTRUCTING A CONFIDENCE INTERVAL FOR A MEAN OF DIFFERENCES To carry out a complete conﬁdence interval procedure to estimate a mean of diﬀerences µdiﬀ , Identify: Identify the parameter and the conﬁdence level, C%. The parameter will be a mean of diﬀerences, e.g. county population (year 2018 − year 2017). the true mean of the diﬀerences in Choose: Choose the appropriate interval procedure and identify it by name. To estimate a mean of diﬀerences we use a 1-sample ttt-interval with paired data. Check: Check conditions for the sampling distribution for ¯xdiﬀ to be nearly normal. 1. Independence: Data come from one random sample (with paired data) or from a randomized matched pairs experiment. When sampling without replacement, check that the sample size is less than 10% of the population size. 2. Large sample or normal population: ndiﬀ ≥ 30 or population of diﬀs is nearly normal. - If the number of diﬀerences is less than 30 and the distribution of the population of diﬀerences is unknown, check for strong skew or outliers in the sample diﬀerences. If neither is found, then the condition that the population of diﬀerences is nearly normal is considered reasonable. Calculate: Calculate the conﬁdence interval and record it in interval form. point estimate ± t × SE of estimate, df : ndiﬀ − 1 point estimate: the sample mean of diﬀerences ¯xdiﬀ SE of estimate: sdiﬀ√ ndiﬀ t: use a t-table at row df = ndiﬀ − 1 and conﬁdence level C% ( , ) Conclude: Interpret the interval and, if applicable, draw a conclusion in context. We are C% conﬁdent that the true mean of the diﬀerences in [...] and . If applicable, draw a conclusion based on whether the interval is entirely above, is is between entirely below, or contains the value 0. 390 CHAPTER 7. INFERENCE FOR NUMERICAL DATA EXAMPLE 7.25 An SAT preparation company claims that its students’ scores improve by over 100 points on average after their course. A consumer group would like to evaluate this claim, and they collect data on a random sample of 30 students who took the class. Each of these students took the SAT before and after taking the company’s course, so we have a diﬀerence in scores for each student. We will examine these diﬀerences x1 = 57, x2 = 133, ..., x30 = 140 as a sample to evaluate the company’s claim. The distribution of the diﬀerences, shown in Figure 7.14, has a mean of 135.9 and a standard deviation of 82.2. Construct a conﬁdence interval to estimate the true average increase in SAT after taking the company’s course. Is there evidence at the 95% conﬁdence level that students score an average of more than 100 points higher after the class? Use the ﬁve step framework to organize your work. Identify: The parameter we want to estimate is µdiﬀ , the true change in SAT score after taking the company’s course. Here, diﬀ = SAT score after course − SAT score before course. We will estimate this parameter at the 95% conﬁdence level. Choose: Because we have paired data and the parameter to be estimated is a mean of diﬀerences, we will use a 1-sample t-interval with paired data. Check: We have a random sample of students with paired observations on them. We will assume that these 30 students represent less than 10% of the total number of such students. Finally, the number of diﬀerences is ndiﬀ = 30 ≥ 30, so we can proceed with the 1-sample t-interval. Calculate: We will calculate the conﬁdence interval as follows. point estimate ± t × SE of estimate The point estimate is the sample mean of diﬀerences: ¯xdiﬀ = 135.9 The SE of the sample mean of diﬀerences is: sdiﬀ√ ndiﬀ = 82.2√ 30 = 15.0 We ﬁnd t for the one-sample case using the t-table at row df = n − 1 and conﬁdence level C%. For a 95% conﬁdence level and df = 30 − 1 = 29, t = 2.045. The 95% conﬁdence interval is given by: 135.9 ± 2.045 × 82.2 √ 30 135.9 ± 2.045 × 15.0 = (105.2, 166.6) df = 29 Conclude: We are 95% conﬁdent that the true average increase in SAT score following the company’s course is between 105.2 points to 166.6 points. There is suﬃcient evidence that students score greater than 100 points higher, on average, after the company’s course because the entire interval is above 100. GUIDED PRACTICE 7.26 Based on the interval (105.2, 166.6), calculated previously, can we say that 95% of student scores increased between 105.2 and 166.6 points after taking the company’s course?10 10No. This interval is attempting to capture the average increase. It is not attempting to capture 95% of the increases. Looking at Figure 7.14, we see that only a small percent had increases between 105.2 and 166.6. 7.2. INFERENCE WITH PAIRED DATA 391 Figure 7.14: SAT score after course minus the SAT score before course. 7.2.5 Technology: the 1-sample ttt-interval with paired data When carrying out a 1-sample t-interval with paired data, make sure to use the sample diﬀer- ences or the summary statistics for the diﬀerences. TI-83/84: 1-SAMPLE T-INTERVAL WITH PAIRED DATA Use STAT, TESTS, TInterval. 1. Choose STAT. 2. Right arrow to TESTS. 3. Down arrow and choose 8:TInterval. 4. Choose Data if you have all the data or Stats if you have the mean and standard deviation. • If you choose Data, let List be L3 or the list in which you entered the diﬀerences (don’t forget to enter the diﬀerences!) and let Freq be 1. • If you choose Stats, enter the mean, SD, and sample size of the diﬀerences. 5. Let C-Level be the desired conﬁdence level. 6. Choose Calculate and hit ENTER, which returns: ( , ) ¯x Sx n the conﬁdence interval for the mean of the diﬀerences the sample mean of the diﬀerences the sample SD of the diﬀerences the number of diﬀerences in the sample CASIO FX-9750GII: 1-SAMPLE T-INTERVAL WITH PAIRED DATA 1. Compute the diﬀerences of the paired observations. 2. Using the computed diﬀerences, follow the instructions for a 1-sample t-interval. GUIDED PRACTICE 7.27 In our UCLA textbook example, we had 68 diﬀerences of paired observations. Because df = 67 was not on our t-table, we rounded the df down to 60. This gave us a 95% conﬁdence interval (0.325, 6.834). Use a calculator to ﬁnd the more exact 95% conﬁdence interval based on 67 degrees of freedom. How diﬀerent is it from the one we calculated based on 60 degrees of freedom?11 ndiﬀ 68 ¯xdiﬀ 3.58 sdiﬀ 13.42 11Choose TInterval or equivalent. We do not have all the data, so choose Stats on a TI or Var on a Casio. Enter ¯x = 3.58 and Sx = 13.42. Let n = 68 and C-Level = 0.95. This should give the interval (0.332, 6.828). The intervals are equivalent when rounded to two decimal places. Differences−10001002003000510 392 CHAPTER 7. INFERENCE FOR NUMERICAL DATA Section summary • Paired data can come from a random sample or a matched pairs experiment. With paired data, we are often interested in whether the diﬀerence is positive, negative, or zero. For example, the diﬀerence of paired data from a matched pairs experiment tells us whether one treatment did better, worse, or the same as the other treatment for each subject. • We use the notation ¯xdiﬀ to represent the mean of the sample diﬀerences. Likewise, sdiﬀ is the standard deviation of the sample diﬀerences, and ndiﬀ is the number of sample diﬀerences. • To carry out inference with paired data, we ﬁrst ﬁnd all of the sample diﬀerences. Then, we perform a one-sample procedure using the diﬀerences. For this reason, the conﬁdence interval and hypothesis test with paired data use the one-sample t-procedures, where the degrees of freedom is given by ndiﬀ − 1. • When there is paired data and the parameter of interest is a mean of diﬀerences: – Estimate µdiﬀ at the C% conﬁdence level using a 1-sample ttt-interval with paired data. – Test H0: µdiﬀ = 0 at the α signiﬁcance level using a 1-sample ttt-test with paired data. • The one-sample t-interval and t-test with paired data require the sampling distribution for ¯xdiﬀ to be nearly normal. For this reason, we must check that the following conditions are met. 1. Independence: Data should come from one random sample (with paired observations) or from a randomized matched pairs experiment. If sampling without replacement, check that the sample
size is less than 10% of the population size. 2. Large sample or normal population: ndiﬀ ≥ 30 or population of diﬀerences nearly normal. - If the number of diﬀerences is less than 30 and it is not known that the population of diﬀerences is nearly normal, we argue that the population of diﬀerences could be nearly normal if there is no strong skew or outliers in the sample diﬀerences. • When the conditions are met, we calculate the conﬁdence interval and the test statistic as we did in the previous section. Here, our data is a list of diﬀerences. Conﬁdence interval: point estimate ± t × SE of estimate Test statistic: T = point estimate − null value SE of estimate Here the point estimate is the mean of sample diﬀerences: ¯xdiﬀ . The SE of estimate is the SE of a mean of sample diﬀerences: sdiﬀ√ ndiﬀ . The degrees of freedom is given by df = ndiﬀ − 1. 7.2. INFERENCE WITH PAIRED DATA 393 Exercises 7.15 Air quality. Air quality measurements were collected in a random sample of 25 country capitals in 2013, and then again in the same cities in 2014. We would like to use these data to compare average air quality between the two years. Should we use a paired or non-paired test? Explain your reasoning. 7.16 True / False: paired. Determine if the following statements are true or false. If false, explain. (a) In a paired analysis we ﬁrst take the diﬀerence of each pair of observations, and then we do inference on these diﬀerences. (b) Two data sets of diﬀerent sizes cannot be analyzed as paired data. (c) Consider two sets of data that are paired with each other. Each observation in one data set has a natural correspondence with exactly one observation from the other data set. (d) Consider two sets of data that are paired with each other. Each observation in one data set is subtracted from the average of the other data set’s observations. 7.17 Paired or not? Part I. In each of the following scenarios, determine if the data are paired. (a) Compare pre- (beginning of semester) and post-test (end of semester) scores of students. (b) Assess gender-related salary gap by comparing salaries of randomly sampled men and women. (c) Compare artery thicknesses at the beginning of a study and after 2 years of taking Vitamin E for the same group of patients. (d) Assess eﬀectiveness of a diet regimen by comparing the before and after weights of subjects. 7.18 Paired or not? Part II. In each of the following scenarios, determine if the data are paired. (a) We would like to know if Intel’s stock and Southwest Airlines’ stock have similar rates of return. To ﬁnd out, we take a random sample of 50 days, and record Intel’s and Southwest’s stock on those same days. (b) We randomly sample 50 items from Target stores and note the price for each. Then we visit Walmart and collect the price for each of those same 50 items. (c) A school board would like to determine whether there is a diﬀerence in average SAT scores for students at one high school versus another high school in the district. To check, they take a simple random sample of 100 students from each high school. 7.19 Global warming, Part I. Let’s consider a limited set of climate data, examining temperature diﬀerences in 1948 vs 2018. We sampled 197 locations from the National Oceanic and Atmospheric Administration’s (NOAA) historical data, where the data was available for both years of interest. We want to know: were there more days with temperatures exceeding 90°F in 2018 or in 1948?12 The diﬀerence in number of days exceeding 90°F (number of days in 2018 - number of days in 1948) was calculated for each of the 197 locations. The average of these diﬀerences was 2.9 days with a standard deviation of 17.2 days. We are interested in determining whether these data provide strong evidence that there were more days in 2018 that exceeded 90°F from NOAA’s weather stations. (a) Is there a relationship between the observations collected in 1948 and 2018? Or are the observations in the two groups independent? Explain. (b) Write hypotheses for this research in symbols and in words. (c) Check the conditions required to complete this test. A histogram of the diﬀerences is given to the right. (d) Calculate the test statistic and ﬁnd the p-value. (e) Use α = 0.05 to evaluate the test, and interpret your conclusion in context. (f) What type of error might we have made? Explain in context what the error means. (g) Based on the results of this hypothesis test, would you expect a conﬁdence interval for the average diﬀerence between the number of days exceeding 90°F from 1948 and 2018 to include 0? Explain your reasoning. 12NOAA, www.ncdc.noaa.gov/cdo-web/datasets, April 24, 2019. Differences in Number of Days−60−40−20020406001020304050−60−40−200204060 394 CHAPTER 7. INFERENCE FOR NUMERICAL DATA 7.20 High School and Beyond, Part I. The National Center of Education Statistics conducted a survey of high school seniors, collecting test data on reading, writing, and several other subjects. Here we examine a simple random sample of 200 students from this survey. Side-by-side box plots of reading and writing scores as well as a histogram of the diﬀerences in scores are shown below. (a) Is there a clear diﬀerence in the average reading and writing scores? (b) Are the reading and writing scores of each student independent of each other? (c) Create hypotheses appropriate for the following research question: is there an evident diﬀerence in the average scores of students in the reading and writing exam? (d) Check the conditions required to complete this test. (e) The average observed diﬀerence in scores is ¯xread−write = −0.545, and the standard deviation of the diﬀerences is 8.887 points. Do these data provide convincing evidence of a diﬀerence between the average scores on the two exams? (f) What type of error might we have made? Explain what the error means in the context of the application. (g) Based on the results of this hypothesis test, would you expect a conﬁdence interval for the average diﬀerence between the reading and writing scores to include 0? Explain your reasoning. 7.21 Global warming, Part II. We considered the change in the number of days exceeding 90°F from 1948 and 2018 at 197 randomly sampled locations from the NOAA database in Exercise 7.19. The mean and standard deviation of the reported diﬀerences are 2.9 days and 17.2 days. Calculate a 90% conﬁdence interval for the average diﬀerence between number of days exceeding 90°F between 1948 and 2018. Does the conﬁdence interval provide convincing evidence that there were more days exceeding 90°F in 2018 than in 1948 at NOAA stations? Include all steps of the Identify, Choose, Check, Calculate, Conclude framework. 7.22 High school and beyond, Part II. We considered the diﬀerences between the reading and writing scores of a random sample of 200 students who took the High School and Beyond Survey in Exercise 7.20. The mean and standard deviation of the diﬀerences are ¯xread−write = −0.545 and 8.887 points. (a) Calculate a 95% conﬁdence interval for the average diﬀerence between the reading and writing scores of all students. (b) Interpret this interval in context. (c) Does the conﬁdence interval provide convincing evidence that there is a real diﬀerence in the average scores? Explain. yscoresreadwrite20406080Differences in scores (read − write)−20−1001020010203040 7.3. INFERENCE FOR THE DIFFERENCE OF TWO MEANS 395 7.3 Inference for the difference of two means Often times we wish to compare two groups to each other to answer questions such as the following: • Does treatment using embryonic stem cells (ESCs) help improve heart function following a heart attack? • Is there convincing evidence that newborns from mothers who smoke have a diﬀerent average birth weight than newborns from mothers who don’t smoke? • Is there statistically signiﬁcant evidence that one variation of an exam is harder than another variation? • Are faculty willing to pay someone named “John” more than someone named “Jennifer”? If so, how much more? Learning objectives 1. Determine when it is appropriate to use a one-sample t-procedure versus a two-sample t- procedure. 2. State and verify whether or not the conditions for inference on the diﬀerence of two means using the t-distribution are met. 3. Be able to use a calculator or other software to ﬁnd the degrees of freedom associated with a two-sample t-procedure. 4. Carry out a complete conﬁdence interval procedure for the diﬀerence of two means. 5. Carry out a complete hypothesis test for the diﬀerence of two means. 396 CHAPTER 7. INFERENCE FOR NUMERICAL DATA 7.3.1 Sampling distribution for the difference of two means (review) In this section we are interested in comparing the means of two independent groups. We want to estimate how far apart µ1 and µ2 are and test whether their diﬀerence is zero or not. Before we perform inference for the diﬀerence of means, let’s review the sampling distribution for ¯x1 − ¯x2, which will be used as the point estimate for µ1 − µ2. We know from Section 4.3 that when the independence condition is satisﬁed, the sampling distribution for ¯x1 − ¯x2 is centered on µ1 − µ2 and has standard deviation of σ¯x1−¯x2 = σ2 1 n1 + σ2 2 n2 When the individual population standard deviations are unknown, we estimate the standard deviation of ¯x1 − ¯x2 using the Standard Error, abbreviated SE, by plugging in the sample standard deviations as our best guesses of the population standard deviations: SE¯x1−¯x2 = s2 1 n1 + s2 2 n2 The diﬀerence of two sample means ¯x1 − ¯x2 follows a nearly normal distribution when certain conditions are met. First, the sampling distribution for each sample mean must be nearly normal, and second, the observations must be independent, both within and between groups. Under these two conditions, the sampling distribution for ¯x1 − ¯x2 may be well approximated using the normal model. 7.3.2 Checking conditions for inference on a difference of means When comparing two means, we carry out infere
nce on a diﬀerence of means, µ1 − µ2. We will use the t-distribution just as we did when carrying out inference on a single mean. In order to use the t-distribution, we need the sampling distribution for ¯x1 − ¯x2 to be nearly normal. We check whether this assumption is reasonable by verifying the following conditions. Independence. Observations can be considered independent when the data are collected from two independent random samples or from a randomized experiment with two treatments. Randomly assigning subjects to treatments is equivalent to randomly assigning treatments to subjects. When sampling without replacement, the observations can be considered independent when the sample size is less than 10% of the population size for both samples. Sample size / nearly normal population. Each population distribution should be nearly normal or each sample size should be at least 30. As before, if the sample sizes are small and the population distributions are not known to be nearly normal, we look at the data for excessive skew or outliers. If we do not ﬁnd excessive skew or outliers in either group, the assumption that the populations are nearly normal to be reasonable is typically considered reasonable. 7.3. INFERENCE FOR THE DIFFERENCE OF TWO MEANS 397 7.3.3 Conﬁdence intervals for a difference of means What’s in a name? Are employers more likely to oﬀer interviews or higher pay to prospective employees when the name on a resume suggests the candidate is a man versus a woman? This is a challenging question to tackle, because employers are inﬂuenced by many aspects of a resume. Thinking back to Chapter 1 on data collection, we could imagine a host of confounding factors associated with name and gender. How could we possibly isolate just the factor of name? We would need an experiment in which name was the only variable and everything else was held constant. Researchers at Yale carried out precisely this experiment. Their results were published in the Proceedings of the National Academy of Sciences (PNAS). The researchers sent out resumes to faculty at academic institutions for a lab manager position. The resumes were identical, except that on half of them the applicant’s name was John and on the other half, the applicant’s name was Jennifer. They wanted to see if faculty, speciﬁcally faculty trained in conducting scientiﬁcally objective research, held implicit gender biases. Unlike in the matched pairs scenario, each faculty member received only one resume. We are interested in comparing the mean salary oﬀered to John relative to the mean salary oﬀered to Jennifer. Instead of taking the average of a set of diﬀerences, we ﬁnd the average of each group separately and take their diﬀerence. Let ¯x1 : mean salary oﬀered to John ¯x2 : mean salary oﬀered to Jennifer We will use ¯x1 − ¯x2 as our point estimate for µ1 − µ2. The data is given in the table below. Name John Jennifer n 63 64 ¯x $30,238 $26,508 s $5567 $7247 We can calculate the diﬀerence as ¯x1 − ¯x2 = 30, 238 − 26, 508 = 3730. EXAMPLE 7.28 Interpret the point estimate 3730. Why might we want to construct a conﬁdence interval? The average salary oﬀered to John was $3,730 higher than the average salary oﬀered to Jennifer. Because there is randomness in which faculty ended up in the John group and which faculty ended up in the Jennifer group, we want to see if the diﬀerence of $3,730 is beyond what could be expected by random variation. In order to answer this, we will ﬁrst want to calculate the SE for the diﬀerence of sample means. EXAMPLE 7.29 Calculate and interpret the SE for the diﬀerence of sample means. SE = s2 1 n1 + s2 2 n2 = (5567)2 63 + (7247)2 64 = 1151 Using samples of size n1 = 63 and n2 = 64, the typical error when using ¯x1 − ¯x2 to estimate µ1 − µ2, the real diﬀerence in mean salary that the faculty would oﬀer John versus Jennifer, is $1151. We see that the diﬀerence of sample means of $3,730 is more than 3 SE above 0, which makes us think that the diﬀerence being 0 is unreasonable. We would like to construct a 95% conﬁdence interval for the theoretical diﬀerence in mean salary that would be oﬀered to John versus Jennifer. For this, we need the degrees of freedom associated with a two-sample t-interval. 398 CHAPTER 7. INFERENCE FOR NUMERICAL DATA For the one-sample t-procedure, the degrees of freedom is given by the simple expression n − 1, where n is the sample size. For the two-sample t-procedures, however, there is a complex formula for calculating the degrees of freedom, which is based on the two sample sizes and the two sample standard deviations. In practice, we ﬁnd the degrees of freedom using software or a calculator (see Section 7.3.4). If this is not possible, the alternative is to use the smaller of n1 − 1 and n2 − 1. DEGREES OF FREEDOM FOR TWO-SAMPLE T-PROCEDURES Use statistical software or a calculator to compute the degrees of freedom for two-sample t-procedures. If this is not possible, use the smaller of n1 − 1 and n2 − 1. EXAMPLE 7.30 Verify that conditions are met for a two-sample t-test. Then, construct the 95% conﬁdence interval for the diﬀerence of means. We noted previously that this is an experiment and that the two treatments (name Jennifer and name John) were randomly assigned. Also, both sample sizes are well over 30, so the distribution of ¯x1 − ¯x2 is nearly normal. Using a calculator, we ﬁnd that df = 118.1. Since 118.1 is not on the t-table, we round the degrees of freedom down to 100.13 Using a t-table at row df = 100 with 95% conﬁdence, we get a t = 1.984. We calculate the conﬁdence interval as follows. point estimate ± t × SE of estimate 3730 ± 1.984 × 1151 = 3730 ± 2284 = (1446, 6014) Based on this interval, we are 95% conﬁdent that the true diﬀerence in mean salary that these faculty would oﬀer John versus Jennifer is between $1,495 and $6,055. That is, we are 95% conﬁdent that the mean salary these faculty would oﬀer John for a lab manager position is between $1,446 and $6,014 more than the mean salary they would oﬀer Jennifer for the position. The results of these studies and others like it are alarming and disturbing.14 One aspect that makes this bias so diﬃcult to address is that the experiment, as well-designed as it was, cannot send us much signal about which faculty are discriminating. Each faculty member received only one of the resumes. A faculty member that oﬀered “Jennifer” a very low salary may have also oﬀered “John” a very low salary. We might imagine an experiment in which each faculty received both resumes, so that we could compare how much they would oﬀer a Jennifer versus a John. However, the matched pairs scenario is clearly not possible in this case, because what makes the experiment work is that the resumes are exactly the same except for the name. An employer would notice something ﬁshy if they received two identical resumes. It is only possible to say that overall, the faculty were willing to oﬀer John more money for the lab manager position than Jennifer. Finding proof of bias for individual cases is a persistent challenge in enforcing anti-discrimination laws. 13Using technology, we get a more precise interval, based on 118.1 df : (1461, 5999). 14A similar study sent out identical resumes with diﬀerent names to investigate the importance of perceived race. Resumes with a name commonly perceived to be for a White person (e.g. Emily) were 50% more likely to receive a callback than the same resume with a name commonly perceived to be for a Black person (e.g. Lakisha). More information is given in Appendix B – see the resume data set. 7.3. INFERENCE FOR THE DIFFERENCE OF TWO MEANS 399 CONSTRUCTING A CONFIDENCE INTERVAL FOR THE DIFFERENCE OF TWO MEANS To carry out a complete conﬁdence interval procedure to estimate the diﬀerence of two means µ1 − µ2, Identify: Identify the parameter and the conﬁdence level, C%. The parameter will be a diﬀerence of means, e.g. the true diﬀerence in mean cholesterol reduction (mean treatment A − mean treatment B). Choose: Choose the appropriate interval procedure and identify it by name. To estimate the diﬀerence of means we use a 2-sample ttt-interval. Check: Check conditions for the sampling distribution for ¯x1 − ¯x2 to be nearly normal. 1. Independence: Data come from 2 independent random samples or from a randomized experiment with 2 treatments. When sampling without replacement, check that the sample size is less than 10% of the population size for each sample. 2. Large samples or normal populations: n1 ≥ 30 and n2 ≥ 30 or both population distributions are nearly normal. - If the sample sizes are less than 30 and the population distributions are unknown, check for strong skew or outliers in either data set. If neither is found, the condition that both population distributions are nearly normal is considered reasonable. Calculate: Calculate the conﬁdence interval and record it in interval form. point estimate ± t × SE of estimate, df : use calculator or other technology point estimate: the diﬀerence of sample means ¯x1 − ¯x2 SE of estimate: + s2 2 n2 t: use a t-table at row df and conﬁdence level C% s2 1 n1 ( , ) Conclude: Interpret the interval and, if applicable, draw a conclusion in context. . If We are C% conﬁdent that the true diﬀerence in mean [...] applicable, draw a conclusion based on whether the interval is entirely above, is entirely below, or contains the value 0. is between and 400 CHAPTER 7. INFERENCE FOR NUMERICAL DATA EXAMPLE 7.31 An instructor decided to run two slight variations of the same exam. Prior to passing out the exams, she shuﬄed the exams together to ensure each student received a random version. Summary statistics for how students performed on these two exams are shown in Figure 7.31. Anticipating complaints from students who took Version B, she would like to evaluate whether the diﬀerence observed in the groups is so large that it provides convincing evidence that Version B was more diﬃcult (on average) than Version A. U
se a 95% conﬁdence interval to estimate the diﬀerence in average score: version A - version B. Version A B n 30 30 ¯x 79.4 74.1 s min max 100 45 100 32 14 20 Identify: The parameter we want to estimate is µ1 − µ2, which is the true average score under Version A − the true average score under Version B. We will estimate this parameter at the 95% conﬁdence level. Choose: Because we are comparing two means, we will use a 2-sample t-interval. Check: The data was collected from a randomized experiment with two treatments: Version A and Version B of test. The 10% condition does not need to be checked here since we are not sampling from a population. There were 30 students in each group, so the condition that both group sizes are at least 30 is met. Calculate: We will calculate the conﬁdence interval as follows. point estimate ± t × SE of estimate The point estimate is the diﬀerence of sample means: ¯x1 − ¯x2 = 79.4 − 74.1 = 5.3 The SE of a diﬀerence of sample means is: s2 1 n1 + s2 2 n2 = 142 30 + 202 30 = 4.46 In order to ﬁnd the critical value t, we must ﬁrst ﬁnd the degrees of freedom. Using a calculator, we ﬁnd df = 51.9. We round down to 50, and using a t-table at row df = 50 and conﬁdence level 95%, we get t = 2.009. The 95% conﬁdence interval is given by: (79.4 − 74.1) ± 2.009 × 142 30 + 202 30 df = 51.9 5.3 ± 2.009 × 4.46 = (−3.66, 14.26) Conclude: We are 95% conﬁdent that the true diﬀerence in average score between Version A and Version B is between -3.66 and 14.26 points. Because the interval contains both positive and negative values, the data do not convincingly show that one exam version is more diﬃcult than the other, and the teacher should not be convinced that she should add points to the Version B exam scores. 7.3. INFERENCE FOR THE DIFFERENCE OF TWO MEANS 401 7.3.4 Technology: the 2-sample ttt-interval TI-83/84: 2-SAMPLE T-INTERVAL Use STAT, TESTS, 2-SampTInt. 1. Choose STAT. 2. Right arrow to TESTS. 3. Down arrow and choose 0:2-SampTTInt. 4. Choose Data if you have all the data or Stats if you have the means and standard deviations. • If you choose Data, let List1 be L1 or the list that contains sample 1 and let List2 be L2 or the list that contains sample 2 (don’t forget to enter the data!). Let Freq1 and Freq2 be 1. • If you choose Stats, enter the mean, SD, and sample size for sample 1 and for sample 2. 5. Let C-Level be the desired conﬁdence level and let Pooled be No. 6. Choose Calculate and hit ENTER, which returns: the conﬁdence interval degrees of freedom mean of sample 1 mean of sample 2 ( , ) df ¯x1 ¯x2 Sx1 Sx2 n1 n2 SD of sample 1 SD of sample 2 size of sample 1 size of sample 2 CASIO FX-9750GII: 2-SAMPLE T-INTERVAL 1. Navigate to STAT (MENU button, then hit the 2 button or select STAT). 2. If necessary, enter the data into a list. 3. Choose the INTR option (F4 button). 4. Choose the t option (F2 button). 5. Choose the 2-S option (F2 button). 6. Choose either the Var option (F2) or enter the data in using the List option. 7. Specify the test details: • Conﬁdence level of interest for C-Level. • If using the Var option, enter the summary statistics for each group. If using List, specify the lists and leave Freq values at 1. • Choose whether to pool the data or not. 8. Hit the EXE button, which returns Left, Right df ¯x1, ¯x2 sx1, sx2 n1, n2 ends of the conﬁdence interval degrees of freedom sample means sample standard deviations sample sizes 402 CHAPTER 7. INFERENCE FOR NUMERICAL DATA GUIDED PRACTICE 7.32 Use the data below and a calculator to ﬁnd a 95% conﬁdence interval for the diﬀerence in average scores between Version A and Version B of the exam from the previous example.15 Version A B n 30 30 ¯x 79.4 74.1 s min max 100 45 100 32 14 20 7.3.5 Hypothesis testing for the difference of two means Four cases from a data set called ncbirths, which represents mothers and their newborns in North Carolina, are shown in Figure 7.15. We are particularly interested in two variables: weight and smoke. The weight variable represents the weights of the newborns and the smoke variable describes which mothers smoked during pregnancy. We would like to know, is there convincing evidence that newborns from mothers who smoke have a diﬀerent average birth weight than newborns from mothers who don’t smoke? The smoking group includes a random sample of 50 cases and the nonsmoking group contains a random sample of 100 cases, represented in Figure 7.16. fAge mAge weeks weight 5.00 5.88 8.13 male ... ... female 9.25 NA NA 19 ... 45 sex female female 37 36 41 ... 36 13 14 15 ... 50 smoke nonsmoker nonsmoker smoker nonsmoker 1 2 3 ... 150 Figure 7.15: Four cases from the ncbirths data set. The value “NA”, shown for the ﬁrst two entries of the ﬁrst variable, indicates pieces of data that are missing. Figure 7.16: The top panel represents birth weights for infants whose mothers smoked. The bottom panel represents the birth weights for infants whose mothers who did not smoke. The distributions exhibit moderate-to-strong and strong skew, respectively. 15Choose 2-SampTInt or equivalent. Because we have the summary statistics rather than all of the data, choose Stats. Let ¯x1=79.41, Sx1=14, n1=30, ¯x2=74.1, Sx2 = 20, and n2 = 30. The interval is (−3.6, 14.2) with df = 51.9. Newborn Weights (lbs) From Mothers Who Smoked0246810Newborn Weights (lbs) From Mothers Who Did Not Smoke0246810 7.3. INFERENCE FOR THE DIFFERENCE OF TWO MEANS 403 EXAMPLE 7.33 Set up appropriate hypotheses to evaluate whether there is a relationship between a mother smoking and average birth weight. Let µ1 represent the mean for mothers that did smoke and µ2 represent the mean for mothers that did not smoke. We will take the diﬀerence as: smoker − nonsmoker. The null hypothesis represents the case of no diﬀerence between the groups. H0: µ1 − µ2 = 0. There is no diﬀerence in average birth weight for newborns from mothers who did and did not smoke. HA: µ1 − µ2 = 0. There is some diﬀerence in average newborn weights from mothers who did and did not smoke. We check the two conditions necessary to use the t-distribution to the diﬀerence in sample means. (1) Because the data come from a sample, we need there to be two independent random samples. In fact, there was only one random sample, but it is reasonable that the two groups here are independent of each other, so we will consider the assumption of independence reasonable. (2) The sample sizes of 50 and 100 are well over 30, so we do not worry about the distributions of the original populations. Since both conditions are satisﬁed, the diﬀerence in sample means may be modeled using a t-distribution. mean st. dev. samp. size smoker 6.78 1.43 50 nonsmoker 7.18 1.60 100 Figure 7.17: Summary statistics for the ncbirths data set. EXAMPLE 7.34 We will use the summary statistics in Figure 7.17 for this exercise. (a) What is the point estimate of the population diﬀerence, µ1 − µ2? (b) Compute the standard error of the point estimate from part (a). (a) The point estimate is the diﬀerence of sample means: ¯x1 − ¯x2 = 6.78 − 7.18 = −0.40 pounds. (b) The standard error for a diﬀerence of sample means is calculated analogously to the standard deviation for a diﬀerence of sample means. SE = s2 1 n1 + s2 2 n2 = 1.432 50 + 1.602 100 = 0.26 pounds EXAMPLE 7.35 Compute the test statistic. We have already found the point estimate and the SE of estimate. The null hypothesis is that the two means are equal, or that their diﬀerence equals 0. The null value for the diﬀerence, therefore is 0. We now have everything we need to compute the test statistic. T = point estimate − null value SE of estimate = −0.40 − 0 0.26 = −1.54 404 CHAPTER 7. INFERENCE FOR NUMERICAL DATA EXAMPLE 7.36 Draw a picture to represent the p-value for this hypothesis test, then calculate the p-value. To depict the p-value, we draw the distribution of the point estimate as though H0 were true and shade areas representing at least as much evidence against H0 as what was observed. Both tails are shaded because it is a two-sided test. We saw previously that the degrees of freedom can be found using software or using the smaller of n1 − 1 and n2 − 1. If we use 50 − 1 = 49 degrees of freedom, we ﬁnd that the area in the lower tail is 0.065. The p-value is twice this, or 2 × 0.065 = 0.130. See Section 7.3.6 for a shortcut to compute the degrees of freedom and p-value on a calculator. EXAMPLE 7.37 What can we conclude from this p-value? Use a signiﬁcance level of α = 0.05. This p-value of 0.130 is larger the signiﬁcance level of 0.05, so we do not reject the null hypothesis. There is not suﬃcient evidence to say there is a diﬀerence in average birth weight of newborns from North Carolina mothers who did smoke during pregnancy and newborns from North Carolina mothers who did not smoke during pregnancy. EXAMPLE 7.38 Does the conclusion to Example 7.35 mean that smoking and average birth weight are unrelated? Not necessarily. It is possible that there is some diﬀerence but that we did not detect it. The result must be considered in light of other evidence and research. In fact, larger data sets do tend to show that women who smoke during pregnancy have smaller newborns. GUIDED PRACTICE 7.39 If we made an error in our conclusion, which type of error could we have made: Type I or Type II?16 GUIDED PRACTICE 7.40 If we made a Type II Error and there is a diﬀerence, what could we have done diﬀerently in data collection to be more likely to detect the diﬀerence?17 16Since we did not reject H0, it is possible that we made a Type II Error. It is possible that there is some diﬀerence but that we did not detect it. 17We could have collected more data. If the sample sizes are larger, we tend to have a better shot at ﬁnding a diﬀerence if one exists. In other words, increasing the sample size increases the power of the test. mn-ms = 0obs. diff 7.3. INFERENCE FOR THE DIFFERENCE OF TWO MEANS 405 HYPOTHESIS TEST FOR THE DIFFERENCE OF TWO MEANS To carry out a complete hypothesis t
est to test the claim that two means µ1 and µ2 are equal to each other, Identify: Identify the hypotheses and the signiﬁcance level, α. H0: µ1 = µ2 HA: µ1 = µ2; HA: µ1 > µ2; or HA: µ1 < µ2 Choose: Choose the appropriate test procedure and identify it by name. To test hypotheses about the diﬀerence of means we use a 2-sample ttt-test. Check: Check conditions for the sampling distribution for ¯x1 − ¯x2 to be nearly normal. 1. Independence: Data come from 2 independent random samples or from a randomized experiment with 2 treatments. When sampling without replacement, check that the sample size is less than 10% of the population size for each sample. 2. Large samples or normal populations: n1 ≥ 30 and n2 ≥ 30 or both population distributions are nearly normal. - If the sample sizes are less than 30 and the population distributions are unknown, check for excessive skew or outliers in either data set. If neither is found, the condition that both population distributions are nearly normal is considered reasonable. Calculate: Calculate the t-statistic, df , and p-value. T = point estimate − null value SE of estimate df : use calculator or other technology point estimate: the diﬀerence of sample means ¯x1 − ¯x2 SE of estimate: s2 1 n1 + s2 2 n2 p-value = (based on the t-statistic, the df , and the direction of HA) Conclude: Compare the p-value to α, and draw a conclusion in context. If the p-value is < α, reject H0; there is suﬃcient evidence that [HA in context]. If the p-value is > α, do not reject H0; there is not suﬃcient evidence that [HA in context]. 406 CHAPTER 7. INFERENCE FOR NUMERICAL DATA EXAMPLE 7.41 Do embryonic stem cells (ESCs) help improve heart function following a heart attack? The following table and ﬁgure summarize results from an experiment to test ESCs in sheep that had a heart attack. ESCs control n 9 9 ¯x 3.50 -4.33 s 5.17 2.76 Each of these sheep was randomly assigned to the ESC or control group, and the change in their hearts’ pumping capacity was measured. A positive value generally corresponds to increased pumping capacity, which suggests a stronger recovery. The sample data is also graphed. Use the given information and an appropriate statistical test to answer the research question. Identify: Let µ1 be the mean percent change for sheep that receive ESC and let µ2 be the mean percent change for sheep in the control group. We will use an α = 0.05 signiﬁcance level. H0: µ1 − µ2 = 0. The stem cells do not improve heart pumping function. HA: µ1 − µ2 > 0. The stem cells do improve heart pumping function. Choose: Because we are hypothesizing about a diﬀerence of means we choose the 2-sample t-test. Check: The data come from a randomized experiment with two treatment groups: ESC and control. Because this is an experiment, we do not need to check the 10% condition. The group sizes are small, but the data show no excessive skew or outliers, so the assumption that the population distributions are nearly normal is reasonable. Calculate: We will calculate the t-statistic and the p-value. T = point estimate − null value SE of estimate The point estimate is the diﬀerence of sample means: ¯x1 − ¯x2 = 3.50 − (−4.33) = 7.83 The SE of a diﬀerence of sample means: s2 1 n1 + s2 2 n2 = (5.17)2 9 + (2.76)2 9 = 1.95 T = 3.50 − (−4.33) − 0 (5.17)2 9 + (2.76)2 9 = 7.83 − 0 1.95 = 4.01 Because HA is an upper tail test ( > ), the p-value corresponds to the area to the right of t = 4.01 with the appropriate degrees of freedom. Using a calculator, we ﬁnd get df = 12.2 and p-value = 8.4 × 10−4 = 0.00084. Conclude: The p-value is much less than 0.05, so we reject the null hypothesis. There is suﬃcient evidence that embryonic stem cells improve the heart’s pumping function in sheep that have suﬀered a heart attack. frequency−10−5051015Embryonic stem cell transplantPercent change in heart pumping function0123frequency−10−50510150123Control (no treatment)Percent change in heart pumping function 7.3. INFERENCE FOR THE DIFFERENCE OF TWO MEANS 407 7.3.6 Technology: the 2-sample ttt-test TI-83/84: 2-SAMPLE T-TEST Use STAT, TESTS, 2-SampTTest. 1. Choose STAT. 2. Right arrow to TESTS. 3. Choose 4:2-SampTTest. 4. Choose Data if you have all the data or Stats if you have the means and standard deviations. • If you choose Data, let List1 be L1 or the list that contains sample 1 and let List2 be L2 or the list that contains sample 2 (don’t forget to enter the data!). Let Freq1 and Freq2 be 1. • If you choose Stats, enter the mean, SD, and sample size for sample 1 and for sample 2 5. Choose =, <, or > to correspond to HA. 6. Let Pooled be NO. 7. Choose Calculate and hit ENTER, which returns: T-statistic t p-value p degrees of freedom df ¯x1 mean of sample 1 ¯x2 mean of sample 2 Sx1 Sx2 n1 n2 SD of sample 1 SD of sample 2 size of sample 1 size of sample 2 CASIO FX-9750GII: 2-SAMPLE T-TEST 1. Navigate to STAT (MENU button, then hit the 2 button or select STAT). 2. If necessary, enter the data into a list. 3. Choose the TEST option (F3 button). 4. Choose the t option (F2 button). 5. Choose the 2-S option (F2 button). 6. Choose either the Var option (F2) or enter the data in using the List option. 7. Specify the test details: • Specify the sidedness of the test using the F1, F2, and F3 keys. • If using the Var option, enter the summary statistics for each group. If using List, specify the lists and leave Freq values at 1. • Choose whether to pool the data or not. 8. Hit the EXE button, which returns µ1 t p df µ2 alt. hypothesis T-statistic p-value degrees of freedom ¯x1, ¯x2 sx1, sx2 n1, n2 sample means sample standard deviations sample sizes 408 CHAPTER 7. INFERENCE FOR NUMERICAL DATA GUIDED PRACTICE 7.42 Use the data below and a calculator to ﬁnd the test statistics and p-value for a one-sided test, testing whether there is evidence that embryonic stem cells (ESCs) help improve heart function for sheep that have experienced a heart attack.18 ESCs control n 9 9 ¯x 3.50 -4.33 s 5.17 2.76 18Choose 2-SampTTest or equivalent. Because we have the summary statistics rather than all of the data, choose Stats. Let ¯x1=3.50, Sx1=5.17, n1=9, ¯x2=-4.33, Sx2 = 2.76, and n2 = 9. We get t = 4.01, and the p-value p = 8.4 × 10−4 = 0.00084. The degrees of freedom for the test is df = 12.2. 7.3. INFERENCE FOR THE DIFFERENCE OF TWO MEANS 409 Section summary • This section introduced inference for a diﬀerence of means, which is distinct from inference for a mean of diﬀerences. To calculate a diﬀerence of means, ¯x1 − ¯x2, we ﬁrst calculate the mean of each group, then we take the diﬀerence between those two numbers. To calculate a mean of diﬀerence, ¯xdiﬀ , we ﬁrst calculate all of the diﬀerences, then we ﬁnd the mean of those diﬀerences. • Inference for a diﬀerence of means is based on the t-distribution. The degrees of freedom is complicated to calculate and we rely on a calculator or other software to calculate this.19 • When there are two samples or treatments and the parameter of interest is a diﬀerence of means: – Estimate µ1 − µ2 at the C% conﬁdence level using a 2-sample ttt-interval. – Test H0: µ1 − µ2 = 0 at the α signiﬁcance level using a 2-sample ttt-test. • The 2-sample t-test and t-interval require the sampling distribution for ¯x1 − ¯x2 to be nearly normal. For this reason we must check that the following conditions are met. 1. Independence: The data should come from 2 independent random samples or from a randomized experiment with 2 treatments. When sampling without replacement, check that the sample size is less than 10% of the population size for each sample. 2. Large samples or normal populations: n1 ≥ 30 and n2 ≥ 30 or both population distribu- tions are nearly normal. - If the sample sizes are less than 30 and it is not known that both population distributions are nearly normal, check for excessive skew or outliers in the data. If neither exists, the condition that both population distributions could be nearly normal is considered reasonable. • When the conditions are met, we calculate the conﬁdence interval and the test statistic as follows. Conﬁdence interval: point estimate ± t × SE of estimate Test statistic: T = point estimate − null value SE of estimate Here the point estimate is the diﬀerence of sample means: ¯x1 − ¯x2. s2 1 n1 The SE of estimate is the SE of a diﬀerence of sample means: + s2 2 n2 . Find and record the df using a calculator or other software. 19If this is not available, one can use df = min(n1 − 1, n2 − 1). 410 CHAPTER 7. INFERENCE FOR NUMERICAL DATA Exercises 7.23 Diamonds, Part I. Prices of diamonds are determined by what is known as the 4 Cs: cut, clarity, color, and carat weight. The prices of diamonds go up as the carat weight increases, but the increase is not smooth. For example, the diﬀerence between the size of a 0.99 carat diamond and a 1 carat diamond is undetectable to the naked human eye, but the price of a 1 carat diamond tends to be much higher than the price of a 0.99 diamond. In this question we use two random samples of diamonds, 0.99 carats and 1 carat, each sample of size 23, and compare the average prices of the diamonds. In order to be able to compare equivalent units, we ﬁrst divide the price for each diamond by 100 times its weight in carats. That is, for a 0.99 carat diamond, we divide the price by 99. For a 1 carat diamond, we divide the price by 100. The distributions and some sample statistics are shown below.20 Conduct a hypothesis test to evaluate if there is a diﬀerence between the average standardized prices of 0.99 and 1 carat diamonds. Include all steps of the Identify, Choose, Check, Calculate, Conclude framework. Mean SD n 0.99 carats $44.51 $13.32 23 1 carat $56.81 $16.13 23 7.24 Diamonds, Part II. In Exercise 7.23, we discussed diamond prices (standardized by weight) for diamonds with weights 0. 99 carats and 1 carat. See the table for summary statistics, and then construct a 95% conﬁdence interval for the diﬀerence in means between the standardized prices of 0.99 and 1 car
at diamonds. Include all steps of the Identify, Choose, Check, Calculate, Conclude framework. Mean SD n 0.99 carats $44.51 $13.32 23 1 carat $56.81 $16.13 23 Chicken farming is a multi-billion dollar industry, and any 7.25 Chicken diet and weight, Part I. methods that increase the growth rate of young chicks can reduce consumer costs while increasing company proﬁts, possibly by millions of dollars. An experiment was conducted to measure and compare the eﬀectiveness of various feed supplements on the growth rate of chickens. Newly hatched chicks were randomly allocated into six groups, and each group was given a diﬀerent feed supplement. Below are some summary statistics from this data set along with box plots showing the distribution of weights by feed type.21 casein horsebean linseed meatmeal soybean sunﬂower Mean 323.58 160.20 218.75 276.91 246.43 328.92 SD 64.43 38.63 52.24 64.90 54.13 48.84 n 12 10 12 11 14 12 (a) Describe the distributions of weights of chickens that were fed linseed and horsebean. (b) Do these data provide strong evidence that the average weights of chickens that were fed linseed and horsebean are diﬀerent? Use a 5% signiﬁcance level. (c) What type of error might we have committed? Explain. (d) Would your conclusion change if we used α = 0.01? 20H. Wickham. ggplot2: elegant graphics for data analysis. Springer New York, 2009. 21Chicken Weights by Feed Type, from the datasets package in R.. Point price (in dollars)0.99 carats1 carat20406080Weight (in grams)caseinhorsebeanlinseedmeatmealsoybeansunflower100150200250300350400lll 7.3. INFERENCE FOR THE DIFFERENCE OF TWO MEANS 411 7.26 Fuel efﬁciency of manual and automatic cars, Part I. Each year the US Environmental Protection Agency (EPA) releases fuel economy data on cars manufactured in that year. Below are summary statistics on fuel eﬃciency (in miles/gallon) from random samples of cars with manual and automatic transmissions. Do these data provide strong evidence of a diﬀerence between the average fuel eﬃciency of cars with manual and automatic transmissions in terms of their average city mileage?22 City MPG Automatic Manual Mean SD n 16.12 3.58 26 19.85 4.51 26 7.27 Chicken diet and weight, Part II. Casein is a common weight gain supplement for humans. Does it have an eﬀect on chickens? Using data provided in Exercise 7.25, test the hypothesis that the average weight of chickens that were fed casein is diﬀerent than the average weight of chickens that were fed soybean. If your hypothesis test yields a statistically signiﬁcant result, discuss whether or not the higher average weight of chickens can be attributed to the casein diet. Conditions for inference were checked in Exercise 7.25. 7.28 Fuel efﬁciency of manual and automatic cars, Part II. The table provides summary statistics on highway fuel economy of the same 52 cars from Exercise 7.26. Use these statistics to calculate a 98% conﬁdence interval for the diﬀerence between average highway mileage of manual and automatic cars, and interpret this interval in the context of the data.23 Hwy MPG Automatic Manual Mean SD n 22.92 5.29 26 27.88 5.01 26 22U.S. Department of Energy, Fuel Economy Data, 2012 Dataﬁle. 23U.S. Department of Energy, Fuel Economy Data, 2012 Dataﬁle. City MPGautomaticmanual152535Hwy MPGautomaticmanual152535 412 CHAPTER 7. INFERENCE FOR NUMERICAL DATA 7.29 Prison isolation experiment, Part I. Subjects from Central Prison in Raleigh, NC, volunteered for an experiment involving an “isolation” experience. The goal of the experiment was to ﬁnd a treatment that reduces subjects’ psychopathic deviant T scores. This score measures a person’s need for control or their rebellion against control, and it is part of a commonly used mental health test called the Minnesota Multiphasic Personality Inventory (MMPI) test. The experiment had three treatment groups: (1) Four hours of sensory restriction plus a 15 minute “therapeutic” tape advising that professional help is available. (2) Four hours of sensory restriction plus a 15 minute “emotionally neutral” tape on training hunting dogs. (3) Four hours of sensory restriction but no taped message. Forty-two subjects were randomly assigned to these treatment groups, and an MMPI test was administered before and after the treatment. Distributions of the diﬀerences between pre and post treatment scores (pre - post) are shown below, along with some sample statistics. Use this information to independently test the eﬀectiveness of each treatment. Make sure to clearly state your hypotheses, check conditions, and interpret results in the context of the data.24 Mean SD n Tr 1 Tr 2 2.86 6.21 7.94 12.3 14 14 Tr 3 -3.21 8.57 14 7.30 True / False: comparing means. Determine if the following statements are true or false, and explain your reasoning for statements you identify as false. (a) As the degrees of freedom increases, the t-distribution approaches normality. (b) We use a pooled standard error for calculating the standard error of the diﬀerence between means when sample sizes of groups are equal to each other. 24Prison isolation experiment, stat.duke.edu/resources/datasets/prison-isolation. Treatment 1020400246Treatment 2−20−1001020024Treatment 3−20−100024 7.3. INFERENCE FOR THE DIFFERENCE OF TWO MEANS 413 Chapter highlights We’ve reviewed a wide set of inference procedures over the last 3 chapters. Let’s revisit each and discuss the similarities and diﬀerences among them. The following conﬁdence intervals and tests are structurally the same – they all involve inference on a population parameter, where that parameter is a proportion, a diﬀerence of proportions, a mean, a mean of diﬀerences, or a diﬀerence of means. • 1-proportion z-test/interval • 2-proportion z-test/interval • 1-sample t-test/interval • 1-sample t-test/interval with paired data • 2-sample t-test/interval The above inferential procedures all involve a point estimate, a standard error of the estimate, and an assumption about the shape of the sampling distribution for the point estimate. From Chapter 6, the χ2 tests and their uses are as follows: • χ2 goodness of ﬁt - compares a categorical variable to a known/ﬁxed distribution. • χ2 test for homogeneity - compares a categorical variable across multiple groups. • χ2 test for independence - looks for association between two categorical variables. χ2 is a measure of overall deviation between observed values and expected values. These tests stand apart from the others because when using χ2 there is not a parameter of interest. For this reason there are no conﬁdence intervals using χ2. Also, for χ2 tests, the hypotheses are usually written in words, because they are about the distribution of one or more categorical variables, not about a single parameter. While formulas and conditions vary, all of these procedures follow the same basic logic and process. • For a conﬁdence interval, identify the parameter to be estimated and the conﬁdence level. For a hypothesis test, identify the hypotheses to be tested and the signiﬁcance level. • Choose the correct procedure. • Check that both conditions for its use are met. • Calculate the conﬁdence interval or the test statistic and p-value, as well as the df if applicable. • Interpret the results and draw a conclusion based on the data. For a summary of these hypothesis test and conﬁdence interval procedures (including one more that we will encounter in Section 8.4), see the Inference Guide in Appendix D.3. 414 CHAPTER 7. INFERENCE FOR NUMERICAL DATA Chapter exercises 7.31 Gaming and distracted eating, Part I. A group of researchers are interested in the possible eﬀects of distracting stimuli during eating, such as an increase or decrease in the amount of food consumption. To test this hypothesis, they monitored food intake for a group of 44 patients who were randomized into two equal groups. The treatment group ate lunch while playing solitaire, and the control group ate lunch without any added distractions. Patients in the treatment group ate 52.1 grams of biscuits, with a standard deviation of 45.1 grams, and patients in the control group ate 27.1 grams of biscuits, with a standard deviation of 26.4 grams. Do these data provide convincing evidence that the average food intake (measured in amount of biscuits consumed) is diﬀerent for the patients in the treatment group? Assume that conditions for inference are satisﬁed.25 7.32 Gaming and distracted eating, Part II. The researchers from Exercise 7.31 also investigated the eﬀects of being distracted by a game on how much people eat. The 22 patients in the treatment group who ate their lunch while playing solitaire were asked to do a serial-order recall of the food lunch items they ate. The average number of items recalled by the patients in this group was 4. 9, with a standard deviation of 1.8. The average number of items recalled by the patients in the control group (no distraction) was 6.1, with a standard deviation of 1.8. Do these data provide strong evidence that the average number of food items recalled by the patients in the treatment and control groups are diﬀerent? 7.33 Sample size and pairing. Determine if the following statement is true or false, and if false, explain your reasoning: If comparing means of two groups with equal sample sizes, always use a paired test. 7.34 College credits. A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor decides to randomly sample 100 students by using the registrar’s database of students. The histogram below shows the distribution of the number of credits taken by these students. Sample statistics for this distribution are also provided. Min Q1 Median Mean SD Q3 Max 8 13 14 13.65 1.91 15 18 (a) What is the point estimate for the average number of credits taken per semester by students at this college? What about the median? (b) What is the point estimate for the standard deviation of the number of credits taken per semester b
y students at this college? What about the IQR? (c) Is a load of 16 credits unusually high for this college? What about 18 credits? Explain your reasoning. (d) The college counselor takes another random sample of 100 students and this time ﬁnds a sample mean of 14.02 units. Should she be surprised that this sample statistic is slightly diﬀerent than the one from the original sample? Explain your reasoning. (e) The sample means given above are point estimates for the mean number of credits taken by all students at that college. What measures do we use to quantify the variability of this estimate? Compute this quantity using the data from the original sample. 25R.E. Oldham-Cooper et al. “Playing a computer game during lunch aﬀects fullness, memory for lunch, and later snack intake”. In: The American Journal of Clinical Nutrition 93.2 (2011), p. 308. Number of creditsFrequency8101214161801020 7.3. INFERENCE FOR THE DIFFERENCE OF TWO MEANS 415 7.35 Hen eggs. The distribution of the number of eggs laid by a certain species of hen during their breeding period has a mean of 35 eggs with a standard deviation of 18.2. Suppose a group of researchers randomly samples 45 hens of this species, counts the number of eggs laid during their breeding period, and records the sample mean. They repeat this 1,000 times, and build a distribution of sample means. (a) What is this distribution called? (b) Would you expect the shape of this distribution to be symmetric, right skewed, or left skewed? Explain your reasoning. (c) Calculate the variability of this distribution and state the appropriate term used to refer to this value. (d) Suppose the researchers’ budget is reduced and they are only able to collect random samples of 10 hens. The sample mean of the number of eggs is recorded, and we repeat this 1,000 times, and build a new distribution of sample means. How will the variability of this new distribution compare to the variability of the original distribution? 7.36 Forest management. Forest rangers wanted to better understand the rate of growth for younger trees in the park. They took measurements of a random sample of 50 young trees in 2009 and again measured those same trees in 2019. The data below summarize their measurements, where the heights are in feet: 2009 12.0 3.5 50 2019 Diﬀerences 24.5 9.5 50 12.5 7.2 50 ¯x s n Construct a 99% conﬁdence interval for the average growth of (what had been) younger trees in the park over 2009-2019. 7.37 Exclusive relationships. A survey conducted on a reasonably random sample of 203 undergraduates asked, among many other questions, about the number of exclusive relationships these students have been in. The histogram below shows the distribution of the data from this sample. The sample average is 3.2 with a standard deviation of 1.97. Estimate the average number of exclusive relationships Duke students have been in using a 90% conﬁdence interval and interpret this interval in context. Check any conditions required for inference, and note any assumptions you must make as you proceed with your calculations and conclusions. Number of exclusive relationships0246810020406080100 416 CHAPTER 7. INFERENCE FOR NUMERICAL DATA 7.38 Age at ﬁrst marriage, Part I. The National Survey of Family Growth conducted by the Centers for Disease Control gathers information on family life, marriage and divorce, pregnancy, infertility, use of contraception, and men’s and women’s health. One of the variables collected on this survey is the age at ﬁrst marriage. The histogram below shows the distribution of ages at ﬁrst marriage of 5,534 randomly sampled women between 2006 and 2010. The average age at ﬁrst marriage among these women is 23.44 with a standard deviation of 4.72.26 Estimate the average age at ﬁrst marriage of women using a 95% conﬁdence interval, and interpret this interval in context. Discuss any relevant assumptions. 7.39 Online communication. A study suggests that the average college student spends 10 hours per week communicating with others online. You believe that this is an underestimate and decide to collect your own sample for a hypothesis test. You randomly sample 60 students from your dorm and ﬁnd that on average they spent 13.5 hours a week communicating with others online. A friend of yours, who oﬀers to help you with the hypothesis test, comes up with the following set of hypotheses. Indicate any errors you see. H0 : ¯x < 10 hours HA : ¯x > 13.5 hours 7.40 Age at ﬁrst marriage, Part II. Exercise 7.38 presents the results of a 2006 - 2010 survey showing that the average age of women at ﬁrst marriage is 23.44. Suppose a social scientist thinks this value has changed since the survey was taken. Below is how she set up her hypotheses. Indicate any errors you see. H0 : ¯x = 23.44 years old HA : ¯x = 23.44 years old 26Centers for Disease Control and Prevention, National Survey of Family Growth, 2010. Age at first marriage101520253035404502004006008001000 7.3. INFERENCE FOR THE DIFFERENCE OF TWO MEANS 417 7.41 Friday the 13th, Part I. In the early 1990’s, researchers in the UK collected data on traﬃc ﬂow, number of shoppers, and traﬃc accident related emergency room admissions on Friday the 13th and the previous Friday, Friday the 6th. The histograms below show the distribution of number of cars passing by a speciﬁc intersection on Friday the 6th and Friday the 13th for many such date pairs. Also given are some sample statistics, where the diﬀerence is the number of cars on the 6th minus the number of cars on the 13th.27 6th 128,385 7,259 10 13th 126,550 7,664 10 Diﬀ. 1,835 1,176 10 ¯x s n (a) Are there any underlying structures in these data that should be considered in an analysis? Explain. (b) What are the hypotheses for evaluating whether the number of people out on Friday the 6th is diﬀerent than the number out on Friday the 13th? (c) Check conditions to carry out the hypothesis test from part (b). (d) Calculate the test statistic and the p-value. (e) What is the conclusion of the hypothesis test? (f) Interpret the p-value in this context. (g) What type of error might have been made in the conclusion of your test? Explain. The Friday the 13th study reported in Exercise 7.41 also provides data 7.42 Friday the 13th, Part II. on traﬃc accident related emergency room admissions. The distributions of these counts from Friday the 6th and Friday the 13th are shown below for six such paired dates along with summary statistics. You may assume that conditions for inference are met. 6th 7.5 3.33 6 13th 10.83 3.6 6 diﬀ -3.33 3.01 6 Mean SD n (a) Conduct a hypothesis test to evaluate if there is a diﬀerence between the average numbers of traﬃc accident related emergency room admissions between Friday the 6th and Friday the 13th. (b) Calculate a 95% conﬁdence interval for the diﬀerence between the average numbers of traﬃc accident related emergency room admissions between Friday the 6th and Friday the 13th. (c) The conclusion of the original study states, “Friday 13th is unlucky for some. The risk of hospital admission as a result of a transport accident may be increased by as much as 52%. Staying at home is recommended.” Do you agree with this statement? Explain your reasoning. 27T.J. Scanlon et al. “Is Friday the 13th Bad For Your Health?” In: BMJ 307 (1993), pp. 1584–1586. Friday the 6th12000013000014000001234Friday the 13th1200001300001400000123Difference020004000012345Friday the 6th510012Friday the 13th510012Difference−50012 418 Chapter 8 Introduction to linear regression 8.1 Line ﬁtting, residuals, and correlation 8.2 Fitting a line by least squares regression 8.3 Transformations for skewed data 8.4 Inference for the slope of a regression line 419 Linear regression is a very powerful statistical technique. Many people have some familiarity with regression just from reading the news, where graphs with straight lines are overlaid on scatterplots. Linear models can be used to see trends and to make predictions. For videos, slides, and other resources, please visit www.openintro.org/ahss 420 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION 8.1 Line ﬁtting, residuals, and correlation In this section, we investigate bivariate data. We examine criteria for identifying a linear model and introduce a new bivariate summary called correlation. We answer questions such as the following: • How do we quantify the strength of the linear association between two numerical variables? • What does it mean for two variables to have no association or to have a nonlinear association? • Once we ﬁt a model, how do we measure the error in the model’s predictions? Learning objectives 1. Distinguish between the data point y and the predicted value ˆy based on a model. 2. Calculate a residual and draw a residual plot. 3. Interpret the standard deviation of the residuals. 4. Interpret the correlation coeﬃcient and estimate it from a scatterplot. 5. Know and apply the properties of the correlation coeﬃcient. 8.1.1 Fitting a line to data Requests from twelve separate buyers were simultaneously placed with a trading company to purchase Target Corporation stock (ticker TGT, April 26th, 2012). We let x be the number of stocks to purchase and y be the total cost. Because the cost is computed using a linear formula, the linear ﬁt is perfect, and the equation for the line is: y = 5 + 57.49x. If we know the number of stocks purchased, we can determine the cost based on this linear equation with no error. Additionally, we can say that each additional share of the stock cost $57.49 and that there was a $5 fee for the transaction. Figure 8.1: Total cost of a trade against number of shares purchased. llllllllllll051015202530050010001500Number of Target Corporation stocks to purchase0102030050010001500Total cost of the shares (dollars) 8.1. LINE FITTING, RESIDUALS, AND CORRELATION 421 Perfect linear relationships are unrealistic in almost any natural process. For example, if we took family income (x), this value would provide some useful info
rmation about how much ﬁnancial support a college may oﬀer a prospective student (y). However, the prediction would be far from perfect, since other factors play a role in ﬁnancial support beyond a family’s income. It is rare for all of the data to fall perfectly on a straight line. Instead, it’s more common for data to appear as a cloud of points, such as those shown in Figure 8.2. In each case, the data fall around a straight line, even if none of the observations fall exactly on the line. The ﬁrst plot shows a relatively strong downward linear trend, where the remaining variability in the data around the line is minor relative to the strength of the relationship between x and y. The second plot shows an upward trend that, while evident, is not as strong as the ﬁrst. The last plot shows a very weak downward trend in the data, so slight we can hardly notice it. In each of these examples, we can consider how to draw a “best ﬁt line”. For instance, we might wonder, should we move the line up or down a little, or should we tilt it more or less? As we move forward in this chapter, we will learn diﬀerent criteria for line-ﬁtting, and we will also learn about the uncertainty associated with estimates of model parameters. Figure 8.2: Three data sets where a linear model may be useful even though the data do not all fall exactly on the line. We will also see examples in this chapter where ﬁtting a straight line to the data, even if there is a clear relationship between the variables, is not helpful. One such case is shown in Figure 8.3 where there is a very strong relationship between the variables even though the trend is not linear. Figure 8.3: A linear model is not useful in this nonlinear case. These data are from an introductory physics experiment. −50050−50050500100015000100002000002040−2000200400lllllllllllllllllllllllllAngle of Incline (Degrees)Distance Traveled (m)0510150306090Best fitting line is flat (!) 422 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION 8.1.2 Using linear regression to predict possum head lengths Brushtail possums are a marsupial that lives in Australia. A photo of one is shown in Figure 8.4. Researchers captured 104 of these animals and took body measurements before releasing the animals back into the wild. We consider two of these measurements: the total length of each possum, from head to tail, and the length of each possum’s head. Figure 8.5 shows a scatterplot for the head length and total length of the 104 possums. Each point represents a single point from the data. Figure 8.4: The common brushtail possum of Australia. ———————————— Photo by Peter Firminger on Flickr: http://ﬂic.kr/p/6aPTn CC BY 2.0 license. Figure 8.5: A scatterplot showing head length against total length for 104 brushtail possums. A point representing a possum with head length 94.1 mm and total length 89 cm is highlighted. Total length (cm)Head length (mm)7580859095859095100l 8.1. LINE FITTING, RESIDUALS, AND CORRELATION 423 The head and total length variables are associated: possums with an above average total length also tend to have above average head lengths. While the relationship is not perfectly linear, it could be helpful to partially explain the connection between these variables with a straight line. We want to describe the relationship between the head length and total length variables in the possum data set using a line. In this example, we will use the total length, x, to explain or predict a possum’s head length, y. When we use x to predict y, we usually call x the explanatory variable or predictor variable, and we call y the response variable. We could ﬁt the linear relationship by eye, as in Figure 8.6. The equation for this line is ˆy = 41 + 0.59x A “hat” on y is used to signify that this is a predicted value, not an observed value. We can use this line to discuss properties of possums. For instance, the equation predicts a possum with a total length of 80 cm will have a head length of ˆy = 41 + 0.59(80) = 88.2 The value ˆy may be viewed as an average: the equation predicts that possums with a total length of 80 cm will have an average head length of 88.2 mm. The value ˆy is also a prediction: absent further information about an 80 cm possum, this is our best prediction for a the head length of a single 80 cm possum. 424 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION 8.1.3 Residuals Residuals are the leftover variation in the response variable after ﬁtting a model. Each observation will have a residual, and three of the residuals for the linear model we ﬁt for the possum data are shown in Figure 8.6. If an observation is above the regression line, then its residual, the vertical distance from the observation to the line, is positive. Observations below the line have negative residuals. One goal in picking the right linear model is for these residuals to be as small as possible. Figure 8.6: A reasonable linear model was ﬁt to represent the relationship between head length and total length. Let’s look closer at the three residuals featured in Figure 8.6. The observation marked by an “×” has a small, negative residual of about -1; the observation marked by “+” has a large residual of about +7; and the observation marked by “” has a moderate residual of about -4. The size of a residual is usually discussed in terms of its absolute value. For example, the residual for “” is larger than that of “×” because \| − 4\| is larger than \| − 1\|. RESIDUAL: DIFFERENCE BETWEEN OBSERVED AND EXPECTED The residual for a particular observation (x, y) is the diﬀerence between the observed response and the response we would predict based on the model: residual = observed y − predicted y = y − ˆy We typically identify ˆy by plugging x into the model. Total Length (cm)Head Length (mm)7580859095859095100 8.1. LINE FITTING, RESIDUALS, AND CORRELATION 425 EXAMPLE 8.1 The linear ﬁt shown in Figure 8.6 is given as ˆy = 41 + 0.59x. Based on this line, compute and interpret the residual of the observation (77.0, 85.3). This observation is denoted by “×” on the plot. Recall that x is the total length measured in cm and y is head length measured in mm. We ﬁrst compute the predicted value based on the model: ˆy = 41 + 0.59x = 41 + 0.59(77.0) = 86.4 Next we compute the diﬀerence of the actual head length and the predicted head length: residual = y − ˆy = 85.3 − 86.4 = − 1.1 The residual for this point is -1.1 mm, which is very close to the visual estimate of -1 mm. For this particular possum with total length of 77 cm, the model’s prediction for its head length was 1.1 mm too high. GUIDED PRACTICE 8.2 If a model underestimates an observation, will the residual be positive or negative? What about if it overestimates the observation?1 GUIDED PRACTICE 8.3 Compute the residual for the observation (95.5, 94.0), denoted by “” in the ﬁgure, using the linear model: ˆy = 41 + 0.59x.2 Residuals are helpful in evaluating how well a linear model ﬁts a data set. We often display the residuals in a residual plot such as the one shown in Figure 8.7. Here, the residuals are calculated for each x value, and plotted versus x. For instance, the point (85.0, 98.6) had a residual of 7.45, so in the residual plot it is placed at (85.0, 7.45). Creating a residual plot is sort of like tipping the scatterplot over so the regression line is horizontal. From the residual plot, we can better estimate the standard deviation of the residuals, often denoted by the letter s. The standard deviation of the residuals tells us typical size of the residuals. As such, it is a measure of the typical deviation between the y values and the model predictions. In other words, it tells us the typical prediction error using the model.3 1If a model underestimates an observation, then the model estimate is below the actual. The residual, which is the actual observation value minus the model estimate, must then be positive. The opposite is true when the model overestimates the observation: the residual is negative. 2First compute the predicted value based on the model, then compute the residual. ˆy = 41 + 0.59x = 41 + 0.59(95.50) = 97.3 residual = y − ˆy = 94.0 − 97.3 = −3.3 The residual is -3.3, so the model overpredicted the head length for this possum by 3.3 mm. 3The standard deviation of the residuals is calculated as: s = (yi−ˆy)2 n−2 . 426 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION EXAMPLE 8.4 Estimate the standard deviation of the residuals for predicting head length from total length using the line: ˆy = 41 + 0.59x using Figure 8.7. Also, interpret the quantity in context. To estimate this graphically, we use the residual plot. The approximate 68, 95 rule for standard deviations applies. Approximately 2/3 of the points are within ± 2.5 and approximately 95% of the points are within ± 5, so 2.5 is a good estimate for the standard deviation of the residuals. The typical error when predicting head length using this model is about 2.5 mm. Figure 8.7: Left: Scatterplot of head length versus total length for 104 brushtail possums. Three particular points have been highlighted. Right: Residual plot for the model shown in left panel. STANDARD DEVIATION OF THE RESIDUALS The standard deviation of the residuals, often denoted by the letter s, tells us the typical error in the predictions using the regression model. It can be estimated from a residual plot. Total Length (cm)Head Length (mm)75808590958590951007580859095−505Total Length (cm)Residuals7580859095−505 8.1. LINE FITTING, RESIDUALS, AND CORRELATION 427 EXAMPLE 8.5 One purpose of residual plots is to identify characteristics or patterns still apparent in data after ﬁtting a model. Figure 8.8 shows three scatterplots with linear models in the ﬁrst row and residual plots in the second row. Can you identify any patterns remaining in the residuals? In the ﬁrst data set (ﬁrst column), the residuals show no obvious patterns. The residuals appear to be scattered randomly around the dashed line that represents 0. The second data set sho
ws a pattern in the residuals. There is some curvature in the scatterplot, which is more obvious in the residual plot. We should not use a straight line to model these data. Instead, a more advanced technique should be used. The last plot shows very little upwards trend, and the residuals also show no obvious patterns. It is reasonable to try to ﬁt a linear model to the data. However, it is unclear whether there is statistically signiﬁcant evidence that the slope parameter is diﬀerent from zero. The slope of the sample regression line is not zero, but we might wonder if this could be due to random variation. We will address this sort of scenario in Section 8.4. Figure 8.8: Sample data with their best ﬁtting lines (top row) and their corresponding residual plots (bottom row). xxyg$residualsxyg$residuals 428 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION 8.1.4 Describing linear relationships with correlation When a linear relationship exists between two variables, we can quantify the strength and direction of the linear relation with the correlation coeﬃcient, or just correlation for short. Figure 8.9 shows eight plots and their corresponding correlations. Figure 8.9: Sample scatterplots and their correlations. The ﬁrst row shows variables with a positive relationship, represented by the trend up and to the right. The second row shows variables with a negative trend, where a large value in one variable is associated with a low value in the other. Only when the relationship is perfectly linear is the correlation either −1 or 1. If the linear relationship is strong and positive, the correlation will be near +1. If it is strong and negative, it will be near −1. If there is no apparent linear relationship between the variables, then the correlation will be near zero. CORRELATION MEASURES THE STRENGTH OF A LINEAR RELATIONSHIP Correlation, which always takes values between -1 and 1, describes the direction and strength of the linear relationship between two numerical variables. The strength can be strong, moderate, or weak. We compute the correlation using a formula, just as we did with the sample mean and standard deviation. Formally, we can compute the correlation for observations (x1, y1), (x2, y2), ..., (xn, yn) using the formula r = 1 n − 1 xi − ¯x yi − ¯y sx sy where ¯x, ¯y, sx, and sy are the sample means and standard deviations for each variable. This formula is rather complex, and we generally perform the calculations on a computer or calculator. We can note, though, that the computation involves taking, for each point, the product of the Z-scores that correspond to the x and y values. r = 0.33yr = 0.69yr = 0.98yr = 1.00r = −0.08yr = −0.64yr = −0.92yr = −1.00 8.1. LINE FITTING, RESIDUALS, AND CORRELATION 429 EXAMPLE 8.6 Take a look at Figure 8.6 on page 424. How would the correlation between head length and total body length of possums change if head length were measured in cm rather than mm? What if head length were measured in inches rather than mm? Here, changing the units of y corresponds to multiplying all the y values by a certain number. This would change the mean and the standard deviation of y, but it would not change the correlation. To see this, imagine dividing every number on the vertical axis by 10. The units of y are now in cm rather than in mm, but the graph has remain exactly the same. The units of y have changed, by the relative distance of the y values about the mean are the same; that is, the Z-scores corresponding to the y values have remained the same. CHANGING UNITS OF XXX AND YYY DOES NOT AFFECT THE CORRELATION The correlation, r, between two variables is not dependent upon the units in which the variables are recorded. Correlation itself has no units. Correlation is intended to quantify the strength of a linear trend. Nonlinear trends, even when strong, sometimes produce correlations that do not reﬂect the strength of the relationship; see three such examples in Figure 8.10. Figure 8.10: Sample scatterplots and their correlations. In each case, there is a strong relationship between the variables. However, the correlation is not very strong, and the relationship is not linear. GUIDED PRACTICE 8.7 It appears no straight line would ﬁt any of the datasets represented in Figure 8.10. Try drawing nonlinear curves on each plot. Once you create a curve for each, describe what is important in your ﬁt.4 4We’ll leave it to you to draw the lines. In general, the lines you draw should be close to most points and reﬂect overall trends in the data. r = −0.23yr = 0.31yr = 0.50 430 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION EXAMPLE 8.8 Consider the four scatterplots in Figure 8.11. In which scatterplot is the correlation between x and y the strongest? All four data sets have the exact same correlation of r = 0.816 as well as the same equation for the best ﬁt line! This group of four graphs, known as Anscombe’s Quartet, remind us that knowing the value of the correlation does not tell us what the corresponding scatterplot looks like. It is always important to ﬁrst graph the data. Investigate Anscombe’s Quartet in Desmos: https://www.desmos.com/calculator/paknt6oneh. Figure 8.11: Four scatterplots from Desmos with best ﬁt line drawn in. 8.1. LINE FITTING, RESIDUALS, AND CORRELATION 431 Section summary • In Chapter 2 we introduced a bivariate display called a scatterplot, which shows the relationship between two numerical variables. When we use x to predict y, we call x the explanatory variable or predictor variable, and we call y the response variable. • A linear model for bivariate numerical data can be useful for prediction when the association between the variables follows a constant, linear trend. Linear models should not be used if the trend between the variables is curved. • When we write a linear model, we use ˆy to indicate that it is the model or the prediction. The value ˆy can be understood as a prediction for y based on a given x, or as an average of the y values for a given x. • The residual is the error between the true value and the modeled value, computed as y − ˆy. The order of the diﬀerence matters, and the sign of the residual will tell us if the model overpredicted or underpredicted a particular data point. • The symbol s in a linear model is used to denote the standard deviation of the residuals, and it measures the typical prediction error by the model. • A residual plot is a scatterplot with the residuals on the vertical axis. The residuals are often plotted against x on the horizontal axis, but they can also be plotted against y, ˆy, or other variables. Two important uses of a residual plot are the following. – Residual plots help us see patterns in the data that may not have been apparent in the scatterplot. – The standard deviation of the residuals is easier to estimate from a residual plot than from the original scatterplot. • Correlation, denoted with the letter r, measures the strength and direction of a linear rela- tionship. The following are some important facts about correlation. – The value of r is always between −1 and 1, inclusive, with an r = −1 indicating a perfect negative relationship (points fall exactly along a line that has negative slope) and an r = 1 indicating a perfect positive relationship (points fall exactly along a line that has positive slope). – An r = 0 indicates no linear association between the variables, though there may well exist a quadratic or other type of association. – Just like Z-scores, the correlation has no units. Changing the units in which x or y are measured does not aﬀect the correlation. – Correlation is sensitive to outliers. Adding or removing a single point can have a big eﬀect on the correlation. – As we learned previously, correlation is not causation. Even a very strong correlation cannot prove causation; only a well-designed, controlled, randomized experiment can prove causation. 432 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION Exercises 8.1 Visualize the residuals. The scatterplots shown below each have a superimposed regression line. If we were to construct a residual plot (residuals versus x) for each, describe what those plots would look like. 8.2 Trends in the residuals. Shown below are two plots of residuals remaining after ﬁtting a linear model to two diﬀerent sets of data. Describe important features and determine if a linear model would be appropriate for these data. Explain your reasoning. 8.3 Identify relationships, Part I. For each of the six plots, identify the strength of the relationship (e.g. weak, moderate, or strong) in the data and whether ﬁtting a linear model would be reasonable. llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(a)llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(b)llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(a)llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(b)lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(a)lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(b)lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(c)lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(d)lllllllllllllllllllllllllllllllllllllllll
llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(e)lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(f) 8.1. LINE FITTING, RESIDUALS, AND CORRELATION 433 8.4 Identify relationships, Part II. For each of the six plots, identify the strength of the relationship (e.g. weak, moderate, or strong) in the data and whether ﬁtting a linear model would be reasonable. 8.5 Exams and grades. The two scatterplots below show the relationship between ﬁnal and mid-semester exam grades recorded during several years for a Statistics course at a university. (a) Based on these graphs, which of the two exams has the strongest correlation with the ﬁnal exam grade? Explain. (b) Can you think of a reason why the correlation between the exam you chose in part (a) and the ﬁnal exam is higher? lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(a)lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(b)lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(c)lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(d)lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(e)lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(f)Exam 1Final Exam406080100406080100Exam 2Final Exam406080100406080100 434 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION 8.6 Spouses, Part I. The Great Britain Oﬃce of Population Census and Surveys once collected data on a random sample of 170 married women in Britain, recording the age (in years) and heights (converted here to inches) of the women and their spouses.5 The scatterplot on the left shows the spouse’s age plotted against the woman’s age, and the plot on the right shows spouse’s height plotted against the woman’s height. (a) Describe the relationship between the ages of women in the sample and their spouses’ ages. (b) Describe the relationship between the heights of women in the sample and their spouses’ heights. (c) Which plot shows a stronger correlation? Explain your reasoning. (d) Data on heights were originally collected in centimeters, and then converted to inches. Does this conversion aﬀect the correlation between heights of women in the sample and their spouses’ heights? 8.7 Match the correlation, Part I. Match each correlation to the corresponding scatterplot. (a) r = −0.7 (b) r = 0.45 (c) r = 0.06 (d) r = 0.92 8.8 Match the correlation, Part II. Match each correlation to the corresponding scatterplot. (a) r = 0.49 (b) r = −0.48 (c) r = −0.03 (d) r = −0.85 8.9 Speed and height. 1,302 UCLA students were asked to ﬁll out a survey where they were asked about their height, fastest speed they have ever driven, and gender. The scatterplot on the left displays the relationship between height and fastest speed, and the scatterplot on the right displays the breakdown by gender in this relationship. (a) Describe the relationship between height and fastest speed. (b) Why do you think these variables are positively associated? (c) What role does gender play in the relationship between height and fastest driving speed? 5D.J. Hand. A handbook of small data sets. Chapman & Hall/CRC, 1994. Woman's age (in years)Spouse's age (in years)204060204060Woman's height (in inches)Spouse's height (in inches)6065707555606570(1)(2)(3)(4)(1)(2)(3)(4)Height (in inches)Fastest speed (in mph)607080050100150llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllHeight (in inches)Fastest speed (in mph)607080050100150lfemalemale 8.1. LINE FITTING, RESIDUALS, AND CORRELATION 435 8.10 Guess the correlation. Eduardo and Rosie are both collecting data on number of rainy days in a year and the total rainfall for the year. Eduardo records rainfall in inches and Rosie in centimeters. How will their correlation coeﬃcients compare? 8.11 The Coast Starlight, Part I. The Coast Starlight Amtrak train runs from Seattle to Los Angeles. The scatterplot below displays the distance between each stop (in miles) and the amount of time it takes to travel from one stop to another (in minutes). (a) Describe the relationship between distance and travel time. (b) How would the relationship change if travel time was instead measured in hours, and distance was instead measured in kilometers? (c) Correlation between travel time (in miles) and distance (in minutes) is r = 0.636. What is the correlation between travel time (in kilometers) and distance (in hours)? 8.12 Crawling babies, Part I. A study conducted at the University of Denver investigated whether babies take longer to learn to crawl in cold months, when they are often bundled in clothes that restrict their movement, than in warmer months.6 Infants born during the study year were split into twelve groups, one for each birth month. We consider the average crawling age of babies in each group against the average temperature when the babies are six months old (that’s when babies often begin trying to crawl). Temperature is measured in degrees Fahrenheit (◦F) and age is measured in weeks. (a) Describe the relationship between temperature and crawling age. (b) How would the relationship change if temperature was measured in degrees Celsius (◦C) and age was measured in months? (c) The correlation between temperature in ◦F and age in weeks was r = −0.70. If we converted the temperature to ◦C and age to months, what would the correlation be? 6J.B. Benson. “Season of birth and onset of locomotion: Theoretical and methodological implications”. In: Infant behavior and development 16.1 (1993), pp. 69–81. issn: 0163-6383. llllllllllllllllDistance (miles)Travel Time (minutes)010020030060120180240300360llllllllllll3040506070293031323334Temperature (F)Avg. crawling age (weeks) 436 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION 8.13 Body measurements, Part I. Researchers studying anthropometry collected body girth measurements and skeletal diameter measurements, as well as age, weight, height and gender for 507 physically active individuals.7 The scatterplot below shows the relationship between height and shoulder girth (over deltoid muscles), both measured in centimeters. (a) Describe the relationship between shoulder girth and height. (b) How would the relationship change if shoulder girth was measured in inches while the units of height remained in centimeters? 8.14 Body measurements, Part II. The scatterplot below shows the relationship between weight measured in kilograms and hip girth measured in centimeters from the data described in Exercise 8.13. (a) Describe the relationship between hip girth and weight. (b) How would the relationship change if weight was measured in pounds while the units for hip girth remained in centimeters? 8.15 Correlation, Part I. spouses if the set of women always married someone who was What would be the correlation between the ages of a set of women and their (a) 3 years younger than themselves? (b) 2 years older than themselves? (c) half as old as themselves? 8.16 Correlation, Part II. What would be the correlation between the annual salaries of males and females at a company if for a certain type of position men always made (a) $5,000 more than women? (b) 25% more than women? (c) 15% less than women? 7G. Heinz et al. “Exploring relationships in body dimensions”. In: Journal of Statistics Education 11.2 (2003). 90100110120130150160170180190200Shoulder girth (cm)Height (cm)8090100110120130406080100Hip girth (cm)Weight (kg) 8.2. FITTING A LINE BY LEAST SQUARES REGRESSION 437 8.2 Fitting a line by least squares regression In this section, we answer the following questions: • How well can we predict ﬁnancial aid based on family income for a particular college? • How does one ﬁnd, interpret, and apply the least squares regression line? • How do we measure the ﬁt of a model and compare diﬀerent models to each other? • Why do models sometimes make predictions that are ridiculous or impossible? Learning objectives 1. Calculate the slope and y-intercept of the least squares regression line using the relevant summary statistics. Interpret these quantities in context. 2. Understand why the least squares regression line is called the least squares regression line. 3. Interpret the explained variance R2. 4. Understand the concept of extrapolation and why it is dangerous. 5. Identify outliers and inﬂuential points in a scatterplot. 8.2.1 An objective measure for ﬁnding the best line Fitting linear models by eye is open to criticism since it is based on an individual preference. In this section, we use least squares regression as a more rigorous approach. This section considers family income and gift aid data from a random sample of ﬁfty students in the freshman class of Elmhurst College in Illinois. Gift aid is ﬁnancial aid that does not need to be paid back, as opposed to a loan. A scatterplot of the data is shown in Figure 8.12 along with two linear ﬁts. The lines follow a negative trend in the data; students who have higher family incomes tended to have lower gift aid from the university. Figure 8.12: Gift aid and family income for a random sample of 50 freshman students from Elmhurst College. Two li
nes are ﬁt to the data, the solid line being the least squares line. Family Income ($1000s)0501001502002500102030Gift Aid From University ($1000s) 438 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION We begin by thinking about what we mean by “best”. Mathematically, we want a line that has small residuals. Perhaps our criterion could minimize the sum of the residual magnitudes: \|y1 − ˆy1\| + \|y2 − ˆy2\| + · · · + \|yn − ˆyn\| which we could accomplish with a computer program. The resulting dashed line shown in Figure 8.12 demonstrates this ﬁt can be quite reasonable. However, a more common practice is to choose the line that minimizes the sum of the squared residuals: (y1 − ˆy1)2 + (y2 − ˆy2)2 + · · · + (yn − ˆyn)2 The line that minimizes the sum of the squared residuals is represented as the solid line in Figure 8.12. This is commonly called the least squares line. Both lines seem reasonable, so why do data scientists prefer the least squares regression line? One reason is that it is easier to compute by hand and in most statistical software. Another, and more compelling, reason is that in many applications, a residual twice as large as another residual is more than twice as bad. For example, being oﬀ by 4 is usually more than twice as bad as being oﬀ by 2. Squaring the residuals accounts for this discrepancy. In Figure 8.13, we imagine the squared error about a line as actual squares. The least squares regression line minimizes the sum of the areas of these squared errors. In the ﬁgure, the sum of the squared error is 4 + 1 + 1 = 6. There is no other line about which the sum of the squared error will be smaller. Figure 8.13: A visualization of least squares regression using Desmos. Try out this and other interactive Desmos activities at openintro.org/ahss/desmos. 8.2. FITTING A LINE BY LEAST SQUARES REGRESSION 439 8.2.2 Finding the least squares line For the Elmhurst College data, we could ﬁt a least squares regression line for predicting gift aid based on a student’s family income and write the equation as: aid = a + b × family income Here a is the y-intercept of the least squares regression line and b is the slope of the least squares regression line. a and b are both statistics that can be calculated from the data. In the next section we will consider the corresponding parameters that they statistics attempt to estimate. We can enter all of the data into a statistical software package and easily ﬁnd the values of a and b. However, we can also calculate these values by hand, using only the summary statistics. • The slope of the least squares line is given by b = r sy sx where r is the correlation between the variables x and y, and sx and sy are the sample standard deviations of x, the explanatory variable, and y, the response variable. • The point of averages (¯x, ¯y) is always on the least squares line. Plugging this point in for x and y in the least squares equation and solving for a gives ¯y = a + b¯x a = ¯y − b¯x FINDING THE SLOPE AND INTERCEPT OF THE LEAST SQUARES REGRESSION LINE The least squares regression line for predicting y based on x can be written as: ˆy = a + bx. We ﬁrst ﬁnd b, the slope, and then we solve for a, the y-intercept. b = r sy sx ¯y = a + b¯x GUIDED PRACTICE 8.9 Figure 8.14 shows the sample means for the family income and gift aid as $101,800 and $19,940, respectively. Plot the point (101.8, 19.94) on Figure 8.12 to verify it falls on the least squares line (the solid line).8 family income, in $1000s (“x”) ¯x = 101.8 sx = 63.2 mean sd gift aid, in $1000s (“y”) ¯y = 19.94 sy = 5.46 r = −0.499 Figure 8.14: Summary statistics for family income and gift aid. 8If you need help ﬁnding this location, draw a straight line up from the x-value of 100 (or thereabout). Then draw a horizontal line at 20 (or thereabout). These lines should intersect on the least squares line. 440 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION EXAMPLE 8.10 Using the summary statistics in Figure 8.14, ﬁnd the equation of the least squares regression line for predicting gift aid based on family income. b = r sy sx = (−0.499) 5.46 63.2 a = ¯y − b¯x = 19.94 − (−0.0431)(101.8) = 24.3 = −0.0431 ˆy = 24.3 − 0.0431x or aid = 24.3 − 0.0431 × family income EXAMPLE 8.11 Say we wanted to predict a student’s family income based on the amount of gift aid that they received. Would this least squares regression line be the following? aid = 24.3 − 0.0431 × family income No. The equation we found was for predicting aid, not for predicting family income. We would have to calculate a new regression line, letting y be family income and x be aid. This would give us: b = r sy sx = (−0.499) 63.2 5.46 a = ¯y − b¯x = 19.94 − (−5.776)(101.8) = 607.9 = −5.776 ˆy = 607.3 − 5.776x or family income = 607.3 − 5.776 × aid We mentioned earlier that a computer is usually used to compute the least squares line. A summary table based on computer output is shown in Figure 8.15 for the Elmhurst College data. The ﬁrst column of numbers provides estimates for b0 and b1, respectively. Compare these to the result from Example 8.2.2. (Intercept) family income Estimate 24.3193 -0.0431 Std. Error 1.2915 0.0108 t value Pr(>\|t\|) 0.0000 0.0002 18.83 -3.98 Figure 8.15: Summary of least squares ﬁt for the Elmhurst College data. Compare the parameter estimates in the ﬁrst column to the results of Guided Practice 8.2.2. EXAMPLE 8.12 Examine the second, third, and fourth columns in Figure 8.15. Can you guess what they represent? We’ll look at the second row, which corresponds to the slope. The ﬁrst column, Estimate = -0.0431, tells us our best estimate for the slope of the population regression line. We call this point estimate b. The second column, Std. Error = 0.0108, is the standard error of this point estimate. The third column, t value = -3.98, is the T test statistic for the null hypothesis that the slope of the population regression line = 0. The last column, Pr(>\|t\|) = 0.0002, is the p-value for this two-sided T -test. We will get into more of these details in Section 8.4. 8.2. FITTING A LINE BY LEAST SQUARES REGRESSION 441 EXAMPLE 8.13 Suppose a high school senior is considering Elmhurst College. Can she simply use the linear equation that we have found to calculate her ﬁnancial aid from the university? No. Using the equation will provide a prediction or estimate. However, as we see in the scatterplot, there is a lot of variability around the line. While the linear equation is good at capturing the trend in the data, there will be signiﬁcant error in predicting an individual student’s aid. Additionally, the data all come from one freshman class, and the way aid is determined by the university may change from year to year. 8.2.3 Interpreting the coefﬁcients of a regression line Interpreting the coeﬃcients in a regression model is often one of the most important steps in the analysis. EXAMPLE 8.14 The slope for the Elmhurst College data for predicting gift aid based on family income was calculated as -0.0431. Intepret this quantity in the context of the problem. You might recall from an algebra course that slope is change in y over change in x. Here, both x and y are in thousands of dollars. So if x is one unit or one thousand dollars higher, the line will predict that y will change by 0.0431 thousand dollars. In other words, for each additional thousand dollars of family income, on average, students receive 0.0431 thousand, or $43.10 less in gift aid. Note that a higher family income corresponds to less aid because the slope is negative. EXAMPLE 8.15 The y-intercept for the Elmhurst College data for predicting gift aid based on family income was calculated as 24.3. Intepret this quantity in the context of the problem. The intercept a describes the predicted value of y when x = 0. The predicted gift aid is 24.3 thousand dollars if a student’s family has no income. The meaning of the intercept is relevant to this application since the family income for some students at Elmhurst is $0. In other applications, the intercept may have little or no practical value if there are no observations where x is near zero. Here, it would be acceptable to say that the average gift aid is 24.3 thousand dollars among students whose family have 0 dollars in income. INTERPRETING COEFFICIENTS IN A LINEAR MODEL • The slope, b, describes the average increase or decrease in the y variable if the explanatory variable x is one unit larger. • The y-intercept, a, describes the predicted outcome of y if x = 0. The linear model must be valid all the way to x = 0 for this to make sense, which in many applications is not the case. 442 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION GUIDED PRACTICE 8.16 In the previous chapter, we encountered a data set that compared the price of new textbooks for UCLA courses at the UCLA Bookstore and on Amazon. We ﬁt a linear model for predicting price at UCLA Bookstore from price on Amazon and we get: where x is the price on Amazon and y is the price at the UCLA bookstore. Interpret the coeﬃcients in this model and discuss whether the interpretations make sense in this context.9 ˆy = 1.86 + 1.03x GUIDED PRACTICE 8.17 Can we conclude that if Amazon raises the price of a textbook by 1 dollar, the UCLA Bookstore will raise the price of the textbook by $1.03?10 EXERCISE CAUTION WHEN INTERPRETING COEFFICIENTS OF A LINEAR MODEL • The slope tells us only the average change in y for each unit change in x; it does not tell us how much y might change based on a change in x for any particular individual. Moreover, in most cases, the slope cannot be interpreted in a causal way. • When a value of x = 0 doesn’t make sense in an application, then the interpretation of the y-intercept won’t have any practical meaning. 8.2.4 Extrapolation is treacherous When those blizzards hit the East Coast this winter, it proved to my satisfaction that global warming was a fraud. That snow was freezing cold. But in an alarming trend, temperatures this spring have risen. Consider this: On February
6th it was 10 degrees. Today it hit almost 80. At this rate, by August it will be 220 degrees. So clearly folks the climate debate rages on. Stephen Colbert April 6th, 2010 11 Linear models can be used to approximate the relationship between two variables. However, these models have real limitations. Linear regression is simply a modeling framework. The truth is almost always much more complex than our simple line. For example, we do not know how the data outside of our limited window will behave. 9The y-intercept is 1.86 and the units of y are in dollars. This tells us that when a textbook costs 0 dollars on Amazon, the predicted price of the textbook at the UCLA Bookstore is 1.86 dollars. This does not make sense as Amazon does not sell any $0 textbooks. The slope is 1.03, with units (dollars)/(dollars). On average, for every extra dollar that a book costs on Amazon, it costs an extra 1.03 dollars at the UCLA Bookstore. This interpretation does make sense in this context. 10No. The slope describes the overall trend. This is observational data; a causal conclusion cannot be drawn. Remember, a causal relationship can only be concluded by a well-designed randomized, controlled experiment. Additionally, there may be large variation in the points about the line. The slope does not tell us how much y might change based on a change in x for a particular textbook. 11www.cc.com/video-clips/l4nkoq/ 8.2. FITTING A LINE BY LEAST SQUARES REGRESSION 443 EXAMPLE 8.18 Use the model aid = 24.3 − 0.0431 × family income to estimate the aid of another freshman student whose family had income of $1 million. Recall that the units of family income = 1000: family income are in $1000s, so we want to calculate the aid for aid = 24.3 − 0.0431 × family income aid = 24.3 − 0.431(1000) = −18.8 The model predicts this student will have -$18,800 in aid (!). Elmhurst College cannot (or at least does not) require any students to pay extra on top of tuition to attend. Using a model to predict y-values for x-values outside the domain of the original data is called extrapolation. Generally, a linear model is only an approximation of the real relationship between two variables. If we extrapolate, we are making an unreliable bet that the approximate linear relationship will be valid in places where it has not been analyzed. 8.2.5 Using R2R2R2 to describe the strength of a ﬁt We evaluated the strength of the linear relationship between two variables earlier using the correlation, r. However, it is more common to explain the ﬁt of a model using R2, called Rsquared or the explained variance. If provided with a linear model, we might like to describe how closely the data cluster around the linear ﬁt. Figure 8.16: Gift aid and family income for a random sample of 50 freshman students from Elmhurst College, shown with the least squares regression line (ˆy) and the average line (¯y). We are interested in how well a model accounts for or explains the location of the y values. The R2 of a linear model describes how much smaller the variance (in the y direction) about the regression line is than the variance about the horizontal line ¯y. For example, consider the Elmhurst College data, shown in Figure 8.16. The variance of the response variable, aid received, is s2 aid = 29.8. However, if we apply our least squares line, then this model reduces our uncertainty in predicting aid using a student’s family income. The variability in the residuals describes how much variation remains after using the model: s2 = 22.4. We could say that the reduction in the variance was: RES aid − s2 s2 s2 aid RES = 29.8 − 22.4 29.8 = 7.5 29.8 = 0.25 Family Income ($1000s)0501001502002500102030Gift Aid From University ($1000s) 444 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION If we used the simple standard deviation of the residuals, this would be exactly R2. However, the standard way of computing the standard deviation of the residuals is slightly more sophisticated.12 To avoid any trouble, we can instead use a sum of squares method. If we call the sum of the squared errors about the regression line SSRes and the sum of the squared errors about the mean SSM , we can deﬁne R2 as follows: R2 = SSM − SSRes SSM = 1 − SSRes SSM (a) (b) Figure 8.17: (a) The regression line is equivalent to ¯y; R2 = 0. (b) The regression line passes through all of the points; R2 = 1. Try out this and other interactive Desmos activities at openintro.org/ahss/desmos. GUIDED PRACTICE 8.19 Using the formula for R2, conﬁrm that in Figure 8.17 (a), R2 = 0 and that in Figure 8.17 (b), R2 = 1.13 R2R2R2 IS THE EXPLAINED VARIANCE R2 is always between 0 and 1, inclusive. It tells us the proportion of variation in the y values that is explained by a regression model. The higher the value of R2, the better the model “explains” the response variable. The value of R2 is, in fact, equal to r2, where r is the correlation. This means that r = ± Use this fact to answer the next two practice problems. √ R2. GUIDED PRACTICE 8.20 If a linear model has a very strong negative relationship with a correlation of -0.97, how much of the variation in the response variable is explained by the linear model?14 GUIDED PRACTICE 8.21 If a linear model has an R2 or explained variance of 0.94, what is the correlation?15 12In computing the standard deviation of the residuals, we divide by n − 2 rather than by n − 1 to account for the n − 2 degrees of freedom. 13(a) SSRes = SSM = (−1)2 + (2)2 + (−1)2 = 6, so R2 = 1 − 6 14R2 = (−0.97)2 = 0.94 or 94%. 94% of the variation in y is explained by the linear model. 15We take the square root of R2 and get 0.97, but we must be careful, because r could be 0.97 or -0.97. Without 6 = 0. (b) R2 = 1 − 0 8 . knowing the slope or seeing the scatterplot, we have no way of knowing if r is positive or negative. 8.2. FITTING A LINE BY LEAST SQUARES REGRESSION 445 8.2.6 Technology: linear correlation and regression Get started quickly with this Desmos LinReg Calculator (available at openintro.org/ahss/desmos). Calculator instructions TI-84: FINDING aaa, bbb, R2R2R2, AND rrr FOR A LINEAR MODEL Use STAT, CALC, LinReg(a + bx). 1. Choose STAT. 2. Right arrow to CALC. 3. Down arrow and choose 8:LinReg(a+bx). • Caution: choosing 4:LinReg(ax+b) will reverse a and b. 4. Let Xlist be L1 and Ylist be L2 (don’t forget to enter the x and y values in L1 and L2 before doing this calculation). 5. Leave FreqList blank. 6. Leave Store RegEQ blank. 7. Choose Calculate and hit ENTER, which returns: a, the y-intercept of the best ﬁt line b, the slope of the best ﬁt line a b r2 R2, the explained variance r, the correlation coeﬃcient r TI-83: Do steps 1-3, then enter the x list and y list separated by a comma, e.g. LinReg(a+bx) L1, L2, then hit ENTER. 446 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION WHAT TO DO IF R2R2R2 AND rrr DO NOT SHOW UP ON A TI-83/84 If r2 and r do now show up when doing STAT, CALC, LinReg, the diagnostics must be turned on. This only needs to be once and the diagnostics will remain on. 1. Hit 2ND 0 (i.e. CATALOG). 2. Scroll down until the arrow points at DiagnosticOn. 3. Hit ENTER and ENTER again. The screen should now say: DiagnosticOn Done WHAT TO DO IF A TI-83/84 RETURNS: ERR: DIM MISMATCH This error means that the lists, generally L1 and L2, do not have the same length. 1. Choose 1:Quit. 2. Choose STAT, Edit and make sure that the lists have the same number of entries. CASIO FX-9750GII: FINDING aaa, bbb, R2R2R2, AND rrr FOR A LINEAR MODEL 1. Navigate to STAT (MENU button, then hit the 2 button or select STAT). 2. Enter the x and y data into 2 separate lists, e.g. x values in List 1 and y values in List 2. Observation ordering should be the same in the two lists. For example, if (5, 4) is the second observation, then the second value in the x list should be 5 and the second value in the y list should be 4. 3. Navigate to CALC (F2) and then SET (F6) to set the regression context. • To change the 2Var XList, navigate to it, select List (F1), and enter the proper list number. Similarly, set 2Var YList to the proper list. 4. Hit EXIT. 5. Select REG (F3), X (F1), and a+bx (F2), which returns: a, the y-intercept of the best ﬁt line b, the slope of the best ﬁt line r, the correlation coeﬃcient R2, the explained variance a b r r2 MSe Mean squared error, which you can ignore If you select ax+b (F1), the a and b meanings will be reversed. 8.2. FITTING A LINE BY LEAST SQUARES REGRESSION 447 GUIDED PRACTICE 8.22 The data set loan50, introduced in Chapter 1, contains information on randomly sampled loans oﬀered through Lending Club. A subset of the data matrix is shown in Figure 8.18. Use a calculator to ﬁnd the equation of the least squares regression line for predicting loan amount from total income based on this subset.16 total income 59000 60000 75000 75000 254000 67000 28800 loan amount 22000 6000 25000 6000 25000 6400 3000 1 2 3 4 5 6 7 Figure 8.18: Sample of data from loan50. 16a = 6497 and b = 0.0774, therefore ˆy = 6497 + 0.0774x. 448 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION 8.2.7 Types of outliers in linear regression Outliers in regression are observations that fall far from the “cloud” of points. These points are especially important because they can have a strong inﬂuence on the least squares line. EXAMPLE 8.23 There are six plots shown in Figure 8.19 along with the least squares line and residual plots. For each scatterplot and residual plot pair, identify any obvious outliers and note how they inﬂuence the least squares line. Recall that an outlier is any point that doesn’t appear to belong with the vast majority of the other points. (1) There is one outlier far from the other points, though it only appears to slightly inﬂuence the line. (2) There is one outlier on the right, though it is quite close to the least squares line, which suggests it wasn’t very inﬂuential. (3) There is one point far away from the cloud, and this outlier appears to pull the least squares line up on the right; examine how t
he line around the primary cloud doesn’t appear to ﬁt very well. (4) There is a primary cloud and then a small secondary cloud of four outliers. The secondary cloud appears to be inﬂuencing the line somewhat strongly, making the least squares line ﬁt poorly almost everywhere. There might be an interesting explanation for the dual clouds, which is something that could be investigated. (5) There is no obvious trend in the main cloud of points and the outlier on the right appears to largely control the slope of the least squares line. (6) There is one outlier far from the cloud, however, it falls quite close to the least squares line and does not appear to be very inﬂuential. Examine the residual plots in Figure 8.19. You will probably ﬁnd that there is some trend in the main clouds of (3) and (4). In these cases, the outliers inﬂuenced the slope of the least squares lines. In (5), data with no clear trend were assigned a line with a large trend simply due to one outlier (!). LEVERAGE Points that fall horizontally away from the center of the cloud tend to pull harder on the line, so we call them points with high leverage. Points that fall horizontally far from the line are points of high leverage; these points can strongly inﬂuence the slope of the least squares line. If one of these high leverage points does appear to actually invoke its inﬂuence on the slope of the line – as in cases (3), (4), and (5) of Example 8.23 – then we call it an inﬂuential point. Usually we can say a point is inﬂuential if, had we ﬁtted the line without it, the inﬂuential point would have been unusually far from the least squares line. It is tempting to remove outliers. Don’t do this without a very good reason. Models that ignore exceptional (and interesting) cases often perform poorly. For instance, if a ﬁnancial ﬁrm ignored the largest market swings – the “outliers” – they would soon go bankrupt by making poorly thought-out investments. DON’T IGNORE OUTLIERS WHEN FITTING A FINAL MODEL If there are outliers in the data, they should not be removed or ignored without a good reason. Whatever ﬁnal model is ﬁt to the data would not be very helpful if it ignores the most exceptional cases. 8.2. FITTING A LINE BY LEAST SQUARES REGRESSION 449 Figure 8.19: Six plots, each with a least squares line and residual plot. All data sets have at least one outlier. (1)(2)(3)(4)(5)(6) 450 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION 8.2.8 Categorical predictors with two levels (special topic) Categorical variables are also useful in predicting outcomes. Here we consider a categorical predictor with two levels (recall that a level is the same as a category). We’ll consider eBay auctions for a video game, Mario Kart for the Nintendo Wii, where both the total price of the auction and the condition of the game were recorded. Here we want to predict total price based on game condition, which takes values used and new. A plot of the auction data is shown in Figure 8.20. Figure 8.20: Total auction prices for the game Mario Kart, divided into used (x = 0) and new (x = 1) condition games with the least squares regression line shown. To incorporate the game condition variable into a regression equation, we must convert the categories into a numerical form. We will do so using an indicator variable called cond new, which takes value 1 when the game is new and 0 when the game is used. Using this indicator variable, the linear model may be written as price = α + β × cond new The ﬁtted model is summarized in Figure 8.21, and the model with its parameter estimates is given as price = 42.87 + 10.90 × cond new For categorical predictors with two levels, the linearity assumption will always be satisﬁed. However, we must evaluate whether the residuals in each group are approximately normal with equal variance. Based on Figure 8.20, both of these conditions are reasonably satisﬁed. (Intercept) cond new Estimate 42.87 10.90 Std. Error 0.81 1.26 t value Pr(>\|t\|) 0.0000 0.0000 52.67 8.66 Figure 8.21: Least squares regression summary for the Mario Kart data. EXAMPLE 8.24 Interpret the two parameters estimated in the model for the price of Mario Kart in eBay auctions. The intercept is the estimated price when cond new takes value 0, i.e. when the game is in used condition. That is, the average selling price of a used version of the game is $42.87. The slope indicates that, on average, new games sell for about $10.90 more than used games. INTERPRETING MODEL ESTIMATES FOR CATEGORICAL PREDICTORS. The estimated intercept is the value of the response variable for the ﬁrst category (i.e. the category corresponding to an indicator value of 0). The estimated slope is the average change in the response variable between the two categories. 0(used)1(new)3040506070Total Priceprice = 42.87 + 10.90 cond_new 8.2. FITTING A LINE BY LEAST SQUARES REGRESSION 451 Section summary • We deﬁne the best ﬁt line as the line that minimizes the sum of the squared residuals (errors) about the line. That is, we ﬁnd the line that minimizes (y1 − ˆy1)2 +(y2 − ˆy2)2 +· · ·+(yn − ˆyn)2 = (yi − ˆyi)2. We call this line the least squares regression line. • We write the least squares regression line in the form: ˆy = a + bx, and we can calculate a and b based on the summary statistics as follows: b = r sy sx and a = ¯y − b¯x. • Interpreting the slope and y-intercept of a linear model – The slope, b, describes the average increase or decrease in the y variable if the explanatory variable x is one unit larger. – The y-intercept, a, describes the average or predicted outcome of y if x = 0. The linear model must be valid all the way to x = 0 for this to make sense, which in many applications is not the case. • Two important considerations about the regression line – The regression line provides estimates or predictions, not actual values. It is important to know how large s, the standard deviation of the residuals, is in order to know about how much error to expect in these predictions. – The regression line estimates are only reasonable within the domain of the data. Predicting y for x values that are outside the domain, known as extrapolation, is unreliable and may produce ridiculous results. • Using R2 to assess the ﬁt of the model – R2, called R-squared or the explained variance, is a measure of how well the model explains or ﬁts the data. R2 is always between 0 and 1, inclusive, or between 0% and 100%, inclusive. The higher the value of R2, the better the model “ﬁts” the data. – The R2 for a linear model describes the proportion of variation in the y variable that is explained by the regression line. – R2 applies to any type of model, not just a linear model, and can be used to compare the ﬁt among various models. √ √ – The correlation r = − R2. The value of R2 is always positive and cannot tell us the direction of the association. If ﬁnding r based on R2, make sure to use either the scatterplot or the slope of the regression line to determine the sign of r. R2 or r = • When a residual plot of the data appears as a random cloud of points, a linear model is generally appropriate. If a residual plot of the data has any type of pattern or curvature, such as a ∪-shape, a linear model is not appropriate. • Outliers in regression are observations that fall far from the “cloud” of points. • An inﬂuential point is a point that has a big eﬀect or pull on the slope of the regression line. Points that are outliers in the x direction will have more pull on the slope of the regression line and are more likely to be inﬂuential points. 452 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION Exercises 8.17 Units of regression. Consider a regression predicting weight (kg) from height (cm) for a sample of adult males. What are the units of the correlation coeﬃcient, the intercept, and the slope? 8.18 Which is higher? Determine if I or II is higher or if they are equal. Explain your reasoning. For a regression line, the uncertainty associated with the slope estimate, b, is higher when I. there is a lot of scatter around the regression line or II. there is very little scatter around the regression line 8.19 Over-under, Part I. Suppose we ﬁt a regression line to predict the shelf life of an apple based on its weight. For a particular apple, we predict the shelf life to be 4.6 days. The apple’s residual is -0.6 days. Did we over or under estimate the shelf-life of the apple? Explain your reasoning. 8.20 Over-under, Part II. Suppose we ﬁt a regression line to predict the number of incidents of skin cancer per 1,000 people from the number of sunny days in a year. For a particular year, we predict the incidence of skin cancer to be 1.5 per 1,000 people, and the residual for this year is 0.5. Did we over or under estimate the incidence of skin cancer? Explain your reasoning. 8.21 Tourism spending. The Association of Turkish Travel Agencies reports the number of foreign tourists visiting Turkey and tourist spending by year.17 Three plots are provided: scatterplot showing the relationship between these two variables along with the least squares ﬁt, residuals plot, and histogram of residuals. (a) Describe the relationship between number of tourists and spending. (b) What are the explanatory and response variables? (c) Why might we want to ﬁt a regression line to these data? (d) Do the data meet the conditions required for ﬁtting a least squares line? In addition to the scatterplot, use the residual plot and histogram to answer this question. 17Association of Turkish Travel Agencies, Foreign Visitors Figure & Tourist Spendings By Years. 050001500025000050001000015000Number of tourists (thousands)Spending (million $)Number of tourists (thousands)Residuals050001500025000−100001000Residuals−1500−7500750150001020 8.2. FITTING A LINE BY LEAST SQUARES REGRESSION 453 8.22 Nutrition at Starbucks, Part I. The scatterplot below shows the relationship between the number of calories and amount of carbohydrates (in grams) Starbucks food menu items contain.18 S
ince Starbucks only lists the number of calories on the display items, we are interested in predicting the amount of carbs a menu item has based on its calorie content. (a) Describe the relationship between number of calories and amount of carbohydrates (in grams) that Starbucks food menu items contain. (b) In this scenario, what are the explanatory and response variables? (c) Why might we want to ﬁt a regression line to these data? (d) Do these data meet the conditions required for ﬁtting a least squares line? Exercise 8.11 introduces data on the Coast Starlight Amtrak train 8.23 The Coast Starlight, Part II. that runs from Seattle to Los Angeles. The mean travel time from one stop to the next on the Coast Starlight is 129 mins, with a standard deviation of 113 minutes. The mean distance traveled from one stop to the next is 108 miles with a standard deviation of 99 miles. The correlation between travel time and distance is 0.636. (a) Write the equation of the regression line for predicting travel time. (b) Interpret the slope and the intercept in this context. (c) Calculate R2 of the regression line for predicting travel time from distance traveled for the Coast Starlight, and interpret R2 in the context of the application. (d) The distance between Santa Barbara and Los Angeles is 103 miles. Use the model to estimate the time it takes for the Starlight to travel between these two cities. (e) It actually takes the Coast Starlight about 168 mins to travel from Santa Barbara to Los Angeles. Calculate the residual and explain the meaning of this residual value. (f) Suppose Amtrak is considering adding a stop to the Coast Starlight 500 miles away from Los Angeles. Would it be appropriate to use this linear model to predict the travel time from Los Angeles to this point? 8.24 Body measurements, Part III. Exercise 8.13 introduces data on shoulder girth and height of a group of individuals. The mean shoulder girth is 107.20 cm with a standard deviation of 10.37 cm. The mean height is 171.14 cm with a standard deviation of 9.41 cm. The correlation between height and shoulder girth is 0.67. (a) Write the equation of the regression line for predicting height. (b) Interpret the slope and the intercept in this context. (c) Calculate R2 of the regression line for predicting height from shoulder girth, and interpret it in the context of the application. (d) A randomly selected student from your class has a shoulder girth of 100 cm. Predict the height of this student using the model. (e) The student from part (d) is 160 cm tall. Calculate the residual, and explain what this residual means. (f) A one year old has a shoulder girth of 56 cm. Would it be appropriate to use this linear model to predict the height of this child? 18Source: Starbucks.com, collected on March 10, 2011, www.starbucks.com/menu/nutrition. CaloriesCarbs (grams)10020030040050020406080CaloriesResiduals100200300400500−20020Residuals−40−20020400510152025 454 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION 8.25 Murders and poverty, Part I. per million from percentage living in poverty in a random sample of 20 metropolitan areas. The following regression output is for predicting annual murders s = Pr(>\|t\|) 0.001 0.000 adj = 68.89% R2 (Intercept) poverty% 5.512 Estimate -29.901 2.559 Std. Error 7.789 0.390 t value -3.839 6.562 R2 = 70.52% (a) Write out the linear model. (b) Interpret the intercept. (c) Interpret the slope. (d) Interpret R2. (e) Calculate the correlation coeﬃcient. 8.26 Cats, Part I. The following regression output is for predicting the heart weight (in g) of cats from their body weight (in kg). The coeﬃcients are estimated using a dataset of 144 domestic cats. s = Pr(>\|t\|) 0.607 0.000 adj = 64.41% R2 (Intercept) body wt 1.452 Estimate -0.357 4.034 Std. Error 0.692 0.250 t value -0.515 16.119 R2 = 64.66% (a) Write out the linear model. (b) Interpret the intercept. (c) Interpret the slope. (d) Interpret R2. (e) Calculate the correlation coeﬃcient. 8.27 Outliers, Part I. Identify the outliers in the scatterplots shown below, and determine what type of outliers they are. Explain your reasoning. llllllllllllllllllllPercent in PovertyAnnual Murders per Million14%18%22%26%10203040Body weight (kg)Heart weight (g)2.02.53.03.54.05101520(a)(b)(c) 8.2. FITTING A LINE BY LEAST SQUARES REGRESSION 455 8.28 Outliers, Part II. Identify the outliers in the scatterplots shown below and determine what type of outliers they are. Explain your reasoning. 8.29 Urban homeowners, Part I. The scatterplot below shows the percent of families who own their home vs. the percent of the population living in urban areas.19 There are 52 observations, each corresponding to a state in the US. Puerto Rico and District of Columbia are also included. (a) Describe the relationship between the percent of families who own their home and the percent of the population living in urban areas. (b) The outlier at the bottom right corner is District of Columbia, where 100% of the population is considered urban. What type of an outlier is this observation? 8.30 Crawling babies, Part II. Exercise 8.12 introduces data on the average monthly temperature during the month babies ﬁrst try to crawl (about 6 months after birth) and the average ﬁrst crawling age for babies born in a given month. A scatterplot of these two variables reveals a potential outlying month when the average temperature is about 53◦F and average crawling age is about 28.5 weeks. Does this point have high leverage? Is it an inﬂuential point? 19United States Census Bureau, 2010 Census Urban and Rural Classiﬁcation and Urban Area Criteria and Housing Characteristics: 2010. (a)(b)(c)Percent Urban Population40%60%80%100%45%55%65%75%Percent Own Their Home 456 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION 8.3 Transformations for skewed data County population size among the counties in the US is very strongly right skewed. Can we apply a transformation to make the distribution more symmetric? How would such a transformation aﬀect the scatterplot and residual plot when another variable is graphed against this variable? In this section, we will see the power of transformations for very skewed data. Learning objectives 1. See how a log transformation can bring symmetry to an extremely skewed variable. 2. Recognize that data can often be transformed to produce a linear relationship, and that this transformation often involves log of the y-values and sometimes log of the x-values. 3. Use residual plots to assess whether a linear model for transformed data is reasonable. 8.3.1 Introduction to transformations EXAMPLE 8.25 Consider the histogram of county populations shown in Figure 8.22(a), which shows extreme skew. What isn’t useful about this plot? Nearly all of the data fall into the left-most bin, and the extreme skew obscures many of the potentially interesting details in the data. There are some standard transformations that may be useful for strongly right skewed data where much of the data is positive but clustered near zero. A transformation is a rescaling of the data using a function. For instance, a plot of the logarithm (base 10) of county populations results in the new histogram in Figure 8.22(b). This data is symmetric, and any potential outliers appear much less extreme than in the original data set. By reigning in the outliers and extreme skew, transformations like this often make it easier to build statistical models against the data. Transformations can also be applied to one or both variables in a scatterplot. A scatterplot of the population change from 2010 to 2017 against the population in 2010 is shown in Figure 8.23(a). In this ﬁrst scatterplot, it’s hard to decipher any interesting patterns because the population variable is so strongly skewed. However, if we apply a log10 transformation to the population variable, as shown in Figure 8.23(b), a positive association between the variables is revealed. While ﬁtting a line to predict population change (2010 to 2017) from population (in 2010) does not seem reasonable, ﬁtting a line to predict population from log10(population) does seem reasonable. original observation) and inverse ( Transformations other than the logarithm can be useful, too. For instance, the square root √ original observation ) are commonly used by data scientists. Com( mon goals in transforming data are to see the data structure diﬀerently, reduce skew, assist in modeling, or straighten a nonlinear relationship in a scatterplot. 1 8.3. TRANSFORMATIONS FOR SKEWED DATA 457 (a) (b) Figure 8.22: (a) A histogram of the populations of all US counties. (b) A histogram of log10-transformed county populations. For this plot, the x-value corresponds to the power of 10, e.g. “4” on the x-axis corresponds to 104 = 10,000. (a) (b) Figure 8.23: (a) Scatterplot of population change against the population before the change. (b) A scatterplot of the same data but where the population size has been log-transformed. Population in 2010 (m = millions)0m2m4m6m8m10m050010001500200025003000Frequencylog10(Population in 2010)23456705001000FrequencyPopulation Change from 2010 to 2017−20%0%20%40%0m2m4m6m8m10mPopulation in 2010 (m = millions)log10(Population in 2010)234567−20%0%20%40%Population Change from 2010 to 2017 458 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION 8.3.2 Transformations to achieve linearity Figure 8.24: Variable y is plotted against x. A nonlinear relationship is evident by the ∪-pattern shown in the residual plot. The curvature is also visible in the original plot. EXAMPLE 8.26 Consider the scatterplot and residual plot in Figure 8.24. The regression output is also provided. Is the linear model ˆy = −52.3564 + 2.7842x a good model for the data? The regression equation is y = -52.3564 + 2.7842 x Predictor Constant x Coef -52.3564 2.7842 SE Coef 7.2757 0.1768 T -7.196 15.752 P 3e-08 < 2e-16 S = 13.76 R-Sq = 88.26% R-Sq(adj) = 87.91% We can note the R2 value is fairly large. However, this alone does not mean that the mo
del is good. Another model might be much better. When assessing the appropriateness of a linear model, we should look at the residual plot. The ∪-pattern in the residual plot tells us the original data is curved. If we inspect the two plots, we can see that for small and large values of x we systematically underestimate y, whereas for middle values of x, we systematically overestimate y. The curved trend can also be seen in the original scatterplot. Because of this, the linear model is not appropriate, and it would not be appropriate to perform a t-test for the slope because the conditions for inference are not met. However, we might be able to use a transformation to linearize the data. Regression analysis is easier to perform on linear data. When data are nonlinear, we sometimes transform the data in a way that makes the resulting relationship linear. The most common transformation is log of the y values. Sometimes we also apply a transformation to the x values. We generally use the residuals as a way to evaluate whether the transformed data are more linear. If so, we can say that a better model has been found. y50100150−25025xResiduals2030405060 8.3. TRANSFORMATIONS FOR SKEWED DATA 459 EXAMPLE 8.27 Using the regression output for the transformed data, write the new linear regression equation. The regression equation is log(y) = 1.722540 + 0.052985 x Predictor Constant x Coef 1.722540 0.052985 SE Coef 0.056731 0.001378 T 30.36 38.45 P < 2e-16 < 2e-16 S = 0.1073 R-Sq = 97.82% R-Sq(adj) = 97.75% The linear regression equation can be written as: log(y) = 1.723 + 0.053x Figure 8.25: A plot of log(y) against x. The residuals don’t show any evident patterns, which suggests the transformed data is well-ﬁt by a linear model. GUIDED PRACTICE 8.28 Which of the following statements are true? There may be more than one.20 (a) There is an apparent linear relationship between x and y. (b) There is an apparent linear relationship between x and log(y). (c) The model provided by Regression I (ˆy = −52.3564 + 2.7842x) yields a better ﬁt. (d) The model provided by Regression II ( log(y) = 1.723 + 0.053x) yields a better ﬁt. 20Part (a) is false since there is a nonlinear (curved) trend in the data. Part (b) is true. Since the transformed data shows a stronger linear trend, it is a better ﬁt, i.e. Part (c) is false, and Part (d) is true. log(y)345−0.20.00.2xResiduals2030405060 460 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION Section summary • A transformation is a rescaling of the data using a function. When data are very skewed, a log transformation often results in more symmetric data. • Regression analysis is easier to perform on linear data. When data are nonlinear, we sometimes transform the data in a way that results in a linear relationship. The most common transformation is log of the y-values. Sometimes we also apply a transformation to the x-values. • To assess the model, we look at the residual plot of the transformed data. If the residual plot of the original data has a pattern, but the residual plot of the transformed data has no pattern, a linear model for the transformed data is reasonable, and the transformed model provides a better ﬁt than the simple linear model. 8.3. TRANSFORMATIONS FOR SKEWED DATA 461 Exercises 8.31 Used trucks. The scatterplot below shows the relationship between year and price (in thousands of $) of a random sample of 42 pickup trucks. Also shown is a residuals plot for the linear model for predicting price from year. (a) Describe the relationship between these two variables and comment on whether a linear model is appro- priate for modeling the relationship between year and price. (b) The scatterplot below shows the relationship between logged (natural log) price and year of these trucks, as well as the residuals plot for modeling these data. Comment on which model (linear model from earlier or logged model presented here) is a better ﬁt for these data. (c) The output for the logged model is given below. Interpret the slope in context of the data. (Intercept) Year Estimate -271.981 0.137 Std. Error 25.042 0.013 t value Pr(>\|t\|) 0.000 -10.861 0.000 10.937 1995200020055101520YearPrice (thousand $)199520002005−505YearResiduals1995200020051.01.52.02.53.0Yearlog(price)199520002005−1.0−0.50.00.51.0YearResiduals 462 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION 8.32 Income and hours worked. The scatterplot below shows the relationship between income and years worked for a random sample of 787 Americans. Also shown is a residuals plot for the linear model for predicting income from hours worked. The data come from the 2012 American Community Survey.21 (a) Describe the relationship between these two variables and comment on whether a linear model is appro- priate for modeling the relationship between year and price. (b) The scatterplot below shows the relationship between logged (natural log) income and hours worked, as well as the residuals plot for modeling these data. Comment on which model (linear model from earlier or logged model presented here) is a better ﬁt for these data. (c) The output for the logged model is given below. Interpret the slope in context of the data. (Intercept) hrs work Estimate 1.017 0.058 Std. Error 0.113 0.003 t value Pr(>\|t\|) 0.000 0.000 9.000 21.086 21United States Census Bureau. Summary File. 2012 American Community Survey. U.S. Census Bureau’s American Community Survey Oﬃce, 2013. Web. 0204060801000100200300400Hours workedIncome (thousand $)020406080100−400−2000200400Hours workedResiduals020406080100−20246Hours workedlog(income)020406080100−4−2024Hours workedResiduals 8.4. INFERENCE FOR THE SLOPE OF A REGRESSION LINE 463 8.4 Inference for the slope of a regression line Here we encounter our last conﬁdence interval and hypothesis test procedures, this time for making inferences about the slope of the population regression line. We can use this to answer questions such as the following: • Is the unemployment rate a signiﬁcant linear predictor for the loss of the President’s party in the House of Representatives? • On average, how much less in college gift aid do students receive when their parents earn an additional $1000 in income? Learning objectives 1. Recognize that the slope of the sample regression line is a point estimate and has an associated standard error. 2. Be able to read the results of computer regression output and identify the quantities needed for inference for the slope of the regression line, speciﬁcally the slope of the sample regression line, the SE of the slope, and the degrees of freedom. 3. State and verify whether or not the conditions are met for inference on the slope of the regression line based using the t-distribution. 4. Carry out a complete conﬁdence interval procedure for the slope of the regression line. 5. Carry out a complete hypothesis test for the slope of the regression line. 6. Distinguish between when to use the t-test for the slope of a regression line and when to use the 1-sample t-test for a mean of diﬀerences. 8.4.1 The role of inference for regression parameters Previously, we found the equation of the regression line for predicting gift aid from family income at Elmhurst College. The slope, b, was equal to −0.0431. This is the slope for our sample data. However, the sample was taken from a larger population. We would like to use the slope computed from our sample data to estimate the slope of the population regression line. The equation for the population regression line can be written as µy = α + βx Here, α and β represent two model parameters, namely the y-intercept and the slope of the true or population regression line. (This use of α and β have nothing to do with the α and β we used previously to represent the probability of a Type I Error and Type II Error!) The parameters α and β are estimated using data. We can look at the equation of the regression line calculated from a particular data set: ˆy =a + bx and see that a and b are point estimates for α and β, respectively. If we plug in the values of a and b, the regression equation for predicting gift aid based on family income is: ˆy = 24.3193 − 0.0431x 464 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION The slope of the sample regression line, −0.0431, is our best estimate for the slope of the population regression line, but there is variability in this estimate since it is based on a sample. A diﬀerent sample would produce a somewhat diﬀerent estimate of the slope. The standard error of the slope tells us the typical variation in the slope of the sample regression line and the typical error in using this slope to estimate the slope of the population regression line. We would like to construct a 95% conﬁdence interval for β, the slope of the population regression line. As with means, inference for the slope of a regression line is based on the t-distribution. INFERENCE FOR THE SLOPE OF A REGRESSION LINE Inference for the slope of a regression line is based on the t-distribution with n − 2 degrees of freedom, where n is the number of paired observations. Once we verify that conditions for using the t-distribution are met, we will be able to construct the conﬁdence interval for the slope using a critical value t based on n − 2 degrees of freedom. We will use a table of the regression summary to ﬁnd the point estimate and standard error for the slope. 8.4.2 Conditions for the least squares line Conditions for inference in the context of regression can be more complicated than when dealing with means or proportions. Inference for parameters of a regression line involves the following assumptions: Linearity. The true relationship between the two variables follows a linear trend. We check whether this is reasonable by examining whether the data follows a linear trend. If there is a nonlinear left panel of Figure 8.26), an advanced regression method from another book or trend (e.g. later course should be applied. Nearly normal residuals. For each x-value, the residuals sho
uld be nearly normal. When this assumption is found to be unreasonable, it is usually because of outliers or concerns about inﬂuential points. An example which suggestions non-normal residuals is shown in the second panel of Figure 8.26. If the sample size n ≥ 30, then this assumption is not necessary. Constant variability. The variability of points around the true least squares line is constant for all values of x. An example of non-constant variability is shown in the third panel of Figure 8.26. Independent. The observations are independent of one other. The observations can be considered independent when they are collected from a random sample or randomized experiment. Be careful of data collected sequentially in what is called a time series. An example of data collected in such a fashion is shown in the fourth panel of Figure 8.26. We see in Figure 8.26, that patterns in the residual plots suggest that the assumptions for regression inference are not met in those four examples. In fact, identifying nonlinear trends in the data, outliers, and non-constant variability in the residuals are often easier to detect in a residual plot than in a scatterplot. We note that the second assumption regarding nearly normal residuals is particularly diﬃcult to assess when the sample size is small. We can make a graph, such as a histogram, of the residuals, but we cannot expect a small data set to be nearly normal. All we can do is to look for excessive skew or outliers. Outliers and inﬂuential points in the data can be seen from the residual plot as well as from a histogram of the residuals. CONDITIONS FOR INFERENCE ON THE SLOPE OF A REGRESSION LINE 1. The data is collected from a random sample or randomized experiment. 2. The residual plot appears as a random cloud of points and does not have any patterns or signiﬁcant outliers that would suggest that the linearity, nearly normal residuals, constant variability, or independence assumptions are unreasonable. 8.4. INFERENCE FOR THE SLOPE OF A REGRESSION LINE 465 Figure 8.26: Four examples showing when the inference methods in this chapter are insuﬃcient to apply to the data. In the left panel, a straight line does not ﬁt the data. In the second panel, there are outliers; two points on the left are relatively distant from the rest of the data, and one of these points is very far away from the line. In the third panel, the variability of the data around the line increases with larger values of x. In the last panel, a time series data set is shown, where successive observations are highly correlated. Figure 8.27: Left: Scatterplot of gift aid versus family income for 50 freshmen at Elmhurst college. Right: Residual plot for the model shown in left panel. 8.4.3 Constructing a conﬁdence interval for the slope of a regression line We would like to construct a conﬁdence interval for the slope of the regression line for predicting gift aid based on family income for all freshmen at Elmhurst college. Do conditions seem to be satisﬁed? We recall that the 50 freshmen in the sample were randomly chosen, so the observations are independent. Next, we need to look carefully at the scatterplot and the residual plot. ALWAYS CHECK CONDITIONS Do not blindly apply formulas or rely on regression output; always ﬁrst look at a scatterplot or a residual plot. If conditions for ﬁtting the regression line are not met, the methods presented here should not be applied. The scatterplot seems to show a linear trend, which matches the fact that there is no curved trend apparent in the residual plot. Also, the standard deviation of the residuals is mostly constant for diﬀerent x values and there are no outliers or inﬂuential points. There are no patterns in the residual plot that would suggest that a linear model is not appropriate, so the conditions are reasonably met. We are now ready to calculate the 95% conﬁdence interval. xxyg$residualsxyg$residualsxyg$residualsFamily Income ($1000s)0501001502002500102030Gift aid from university ($1000s)Family Income ($1000s)050100150200250−12−8−404812Residuals 466 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION (Intercept) family income Estimate 24.3193 -0.0431 Std. Error 1.2915 0.0108 t value Pr(>\|t\|) 0.0000 0.0002 18.83 -3.98 Figure 8.28: Summary of least squares ﬁt for the Elmhurst College data, where we are predicting gift aid by the university based on the family income of students. EXAMPLE 8.29 Construct a 95% conﬁdence interval for the slope of the regression line for predicting gift aid from family income at Elmhurst college. As usual, the conﬁdence interval will take the form: point estimate ± critical value × SE of estimate The point estimate for the slope of the population regression line is the slope of the sample regression line: −0.0431. The standard error of the slope can be read from the table as 0.0108. Note that we do not need to divide 0.0108 by the square root of n or do any further calculations on 0.0108; 0.0108 is the SE of the slope. Note that the value of t given in the table refers to the test statistic, not to the critical value t. To ﬁnd t we can use a t-table. Here n = 50, so df = 50 − 2 = 48. Using a t-table, we round down to row df = 40 and we estimate the critical value t = 2.021 for a 95% conﬁdence level. The conﬁdence interval is calculated as: −0.0431 ± 2.021 × 0.0108 = (−0.065, −0.021) t using exactly 48 degrees of freedom is equal to 2.01 and gives the same interval of Note: (−0.065, −0.021). EXAMPLE 8.30 Intepret the conﬁdence interval in context. What can we conclude? We are 95% conﬁdent that the slope of the population regression line, the true average change in gift aid for each additional $1000 in family income, is between −$0.065 thousand dollars and −$0.021 thousand dollars. That is, we are 95% conﬁdent that, on average, when family income is $1000 higher, gift aid is between $21 and $65 lower. Because the entire interval is negative, we have evidence that the slope of the population regression In other words, we have evidence that there is a signiﬁcant negative linear line is less than 0. relationship between gift aid and family income. 8.4. INFERENCE FOR THE SLOPE OF A REGRESSION LINE 467 CONSTRUCTING A CONFIDENCE INTERVAL FOR THE SLOPE OF REGRESSION LINE To carry out a complete conﬁdence interval procedure to estimate the slope of the population regression line β, Identify: Identify the parameter and the conﬁdence level, C%. the slope of the The parameter will be a slope of the population regression line, e.g. population regression line relating air quality index to average rainfall per year for each city in the United States. Choose: Choose the correct interval procedure and identify it by name. To estimate the slope of a regression model we use a ttt-interval for the slope. Check: Check conditions for using a t-interval for the slope. 1. Independence: Data should come from a random sample or randomized experiment. If sampling without replacement, check that the sample size is less than 10% of the population size. 2. Linearity: Check that the scatterplot does not show a curved trend and that the residual plot shows no ∪-shape pattern. 3. Constant variability: Use the residual plot to check that the standard deviation of the residuals is constant across all x-values. 4. Normality: The population of residuals is nearly normal or the sample size is ≥ 30. If the sample size is less than 30 check for strong skew or outliers in the sample residuals. If neither is found, then the condition that the population of residuals is nearly normal is considered reasonable. Calculate: Calculate the conﬁdence interval and record it in interval form. point estimate ± t × SE of estimate, df = n − 2 point estimate: the slope b of the sample regression line SE of estimate: SE of slope (ﬁnd using computer output) t: use a t-distribution with df = n − 2 and conﬁdence level C% ( , ) Conclude: Interpret the interval and, if applicable, draw a conclusion in context. We are C% conﬁdent that the true slope of the regression line, the average change in [y] for each unit increase in [x], is between . If applicable, draw a conclusion based on whether the interval is entirely above, is entirely below, or contains the value 0. and 468 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION Figure 8.29: Left: Scatterplot of head length versus total length for 104 brushtail possums. Right: Residual plot for the model shown in left panel. EXAMPLE 8.31 The regression summary below shows statistical software output from ﬁtting the least squares length for 104 brushtail possums. The regression line for predicting head length from total scatterplot and residual plot are shown above. Predictor Constant total length Coef 42.70979 0.57290 SE Coef 5.17281 0.05933 T 8.257 9.657 P 5.66e-13 4.68e-16 S = 2.595 R-Sq = 47.76% R-Sq(adj) = 47.25% Construct a 95% conﬁdence interval for the slope of the regression line. evidence that there is a positive, linear relationship between head length and total length? Is there convincing Identify: The parameter of interest is the slope of the population regression line for predicting head length from body length. We want to estimate this at the 95% conﬁdence level. Choose: Because the parameter to be estimated is the slope of a regression line, we will use the t-interval for the slope. Check: These data come from a random sample. The residual plot shows no pattern so a linear model seems reasonable. The residual plot also shows that the residuals have constant standard deviation. Finally, n = 104 ≥ 30 so we do not have to check for skew in the residuals. All four conditions are met. Calculate: We will calculate the interval: point estimate ± t × SE of estimate We read the slope of the sample regression line and the corresponding SE from the table. The point estimate is b = 0.57290. The SE of the slope is 0.05933, which can be found next to the slope of 0.57290. The degrees of freedom is df = n − 2 = 104 − 2 = 102. As before, we ﬁnd the critic
al value t using a t-table (the t value is not the same as the T -statistic for the hypothesis test). Using the t-table at row df = 100 (round down since 102 is not on the table) and conﬁdence level 95%, we get t = 1.984. So the 95% conﬁdence interval is given by: 0.57290 ± 1.984 × 0.05933 (0.456, 0.691) Conclude: We are 95% conﬁdent that the slope of the population regression line is between 0.456 and 0.691. That is, we are 95% conﬁdent that the true average increase in head length for each additional cm in total length is between 0.456 mm and 0.691 mm. Because the interval is entirely above 0, we do have evidence of a positive linear association between the head length and body length for brushtail possums. Total Length (cm)Head Length (mm)7580859095859095100Total Length (cm)Residuals7580859095−505 8.4. INFERENCE FOR THE SLOPE OF A REGRESSION LINE 469 8.4.4 Midterm elections and unemployment Elections for members of the United States House of Representatives occur every two years, coinciding every four years with the U.S. Presidential election. The set of House elections occurring during the middle of a Presidential term are called midterm elections. In America’s two-party system, one political theory suggests the higher the unemployment rate, the worse the President’s party will do in the midterm elections. To assess the validity of this claim, we can compile historical data and look for a connection. We consider every midterm election from 1898 to 2018, with the exception of those elections during the Great Depression. Figure 8.30 shows these data and the least-squares regression line: % change in House seats for President’s party = −7.36 − 0.89 × (unemployment rate) We consider the percent change in the number of seats of the President’s party (e.g. percent change in the number of seats for Republicans in 2018) against the unemployment rate. Examining the data, there are no clear deviations from linearity, the constant variance condition, or the normality of residuals. While the data are collected sequentially, a separate analysis was used to check for any apparent correlation between successive observations; no such correlation was found. Figure 8.30: The percent change in House seats for the President’s party in each election from 1898 to 2018 plotted against the unemployment rate. The two points for the Great Depression have been removed, and a least squares regression line has been ﬁt to the data. Explore this data set on Tableau Public . Percent Change in Seats ofPresident's Party in House of Rep.Percent Unemploymentllllllllllll4%8%12%−30%−20%−10%0%10%McKinley1898Reagan1982Clinton1994Bush2002Obama2010Trump2018lDemocratRepublican 470 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION GUIDED PRACTICE 8.32 The data for the Great Depression (1934 and 1938) were removed because the unemployment rate was 21% and 18%, respectively. Do you agree that they should be removed for this investigation? Why or why not?22 There is a negative slope in the line shown in Figure 8.30. However, this slope (and the yintercept) are only estimates of the parameter values. We might wonder, is this convincing evidence that the “true” linear model has a negative slope? That is, do the data provide strong evidence that the political theory is accurate? We can frame this investigation as a statistical hypothesis test: H0: β = 0. The true linear model has slope zero. HA: β < 0. The true linear model has a slope less than zero. The higher the unemployment, the greater the loss for the President’s party in the House of Representatives. We would reject H0 in favor of HA if the data provide strong evidence that the slope of the population regression line is less than zero. To assess the hypotheses, we identify a standard error for the estimate, compute an appropriate test statistic, and identify the p-value. Before we calculate these quantities, how good are we at visually determining from a scatterplot when a slope is signiﬁcantly less than or greater than 0? And why do we tend to use a 0.05 signiﬁcance level as our cutoﬀ? Try out the following activity which will help answer these questions. TESTING FOR THE SLOPE USING A CUTOFF OF 0.05 What does it mean to say that the slope of the population regression line is signiﬁcantly greater than 0? And why do we tend to use a cutoﬀ of α = 0.05? This 5-minute interactive task will explain: www.openintro.org/why05 22We will provide two considerations. Each of these points would have very high leverage on any least-squares regression line, and years with such high unemployment may not help us understand what would happen in other years where the unemployment is only modestly high. On the other hand, these are exceptional cases, and we would be discarding important information if we exclude them from a ﬁnal analysis. 8.4. INFERENCE FOR THE SLOPE OF A REGRESSION LINE 471 8.4.5 Understanding regression output from software The residual plot shown in Figure 8.31 shows no pattern that would indicate that a linear model is inappropriate. Therefore we can carry out a test on the population slope using the sample slope as our point estimate. Just as for other point estimates we have seen before, we can compute a standard error and test statistic for b. The test statistic T follows a t-distribution with n − 2 degrees of freedom. Figure 8.31: The residual plot shows no pattern that would indicate that a linear model is inappropriate. Explore this data set on Tableau Public . HYPOTHESIS TESTS ON THE SLOPE OF THE REGRESSION LINE Use a t-test with n − 2 degrees of freedom when performing a hypothesis test on the slope of a regression line. We will rely on statistical software to compute the standard error and leave the explanation of how this standard error is determined to a second or third statistics course. Figure 8.32 shows software output for the least squares regression line in Figure 8.30. The row labeled unemp represents the information for the slope, which is the coeﬃcient of the unemployment variable. (Intercept) unemp Estimate -7.3644 -0.8897 Std. Error 5.1553 0.8350 t value Pr(>\|t\|) 0.1646 0.2961 -1.43 -1.07 Figure 8.32: Least squares regression summary for the percent change in seats of president’s party in House of Reprepsentatives based on percent unemployment. ResidualsPercent Unemploymentllllllllllll4%8%12%−10%0%10%lDemocratRepublican 472 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION Figure 8.33: The distribution shown here is the sampling distribution for b, if the null hypothesis was true. The shaded tail represents the p-value for the hypothesis test evaluating whether there is convincing evidence that higher unemployment corresponds to a greater loss of House seats for the President’s party during a midterm election. EXAMPLE 8.33 What do the ﬁrst column of numbers in the regression summary represent? The entries in the ﬁrst column represent the least squares estimates for the y-intercept and slope, a and b respectively. Using this information, we could write the equation for the least squares regression line as ˆy = −7.3644 − 0.8897x where y in this case represents the percent change in the number of seats for the president’s party, and x represents the unemployment rate. We previously used a test statistic T for hypothesis testing in the context of means. Regression is very similar. Here, the point estimate is b = −0.8897. The SE of the estimate is 0.8350, which is given in the second column, next to the estimate of b. This SE represents the typical error when using the slope of the sample regression line to estimate the slope of the population regression line. The null value for the slope is 0, so we now have everything we need to compute the test statistic. We have: T = point estimate − null value SE of estimate = −0.8897 − 0 0.8350 = −1.07 This value corresponds to the T -score reported in the regression output in the third column along the unemp row. EXAMPLE 8.34 In this example, the sample size n = 27. hypothesis test. Identify the degrees of freedom and p-value for the The degrees of freedom for this test is n − 2, or df = 27 − 2 = 25. We could use a table or a calculator to ﬁnd the probability of a value less than -1.07 under the t-distribution with 25 degrees of freedom. However, the two-side p-value is given in Figure 8.32, next to the corresponding tstatistic. Because we have a one-sided alternative hypothesis, we take half of this. The p-value for the test is 0.2961 2 = 0.148. −2.62−1.74−0.8700.871.742.62 8.4. INFERENCE FOR THE SLOPE OF A REGRESSION LINE 473 Because the p-value is so large, we do not reject the null hypothesis. That is, the data do not provide convincing evidence that a higher unemployment rate is associated with a larger loss for the President’s party in the House of Representatives in midterm elections. DON’T CARELESSLY USE THE P-VALUE FROM REGRESSION OUTPUT The last column in regression output often lists p-values for one particular hypothesis: a twosided test where the null value is zero. If your test is one-sided and the point estimate is in the direction of HA, then you can halve the software’s p-value to get the one-tail area. If neither of these scenarios match your hypothesis test, be cautious about using the software output to obtain the p-value. HYPOTHESIS TEST FOR THE SLOPE OF REGRESSION LINE To carry out a complete hypothesis test for the claim that there is no linear relationship between two numerical variables, i.e. that β = 0, Identify: Identify the hypotheses and the signiﬁcance level, α. H0: β = 0 HA: β = 0; HA: β > 0; or HA: β < 0 Choose: Choose the correct test procedure and identify it by name. To test hypotheses about the slope of a regression model we use a ttt-test for the slope. Check: Check conditions for using a t-test for the slope. 1. Independence: Data should come from a random sample or randomized experiment. If sampling without replacement, check that the sample size is less than 10% of the population size. 2. Linearity: Check that
the scatterplot does not show a curved trend and that the residual plot shows no ∪-shape pattern. 3. Constant variability: Use the residual plot to check that the standard deviation of the residuals is constant across all x-values. 4. Normality: The population of residuals is nearly normal or the sample size is ≥ 30. If the sample size is less than 30 check for strong skew or outliers in the sample residuals. If neither is found, then the condition that the population of residuals is nearly normal is considered reasonable. Calculate: Calculate the t-statistic, df , and p-value. T = point estimate − null value SE of estimate , df = n − 2 point estimate: the slope b of the sample regression line SE of estimate: SE of slope (ﬁnd using computer output) null value: 0 p-value = (based on the t-statistic, the df , and the direction of HA) Conclude: Compare the p-value to α, and draw a conclusion in context. If the p-value is < α, reject H0; there is suﬃcient evidence that [HA in context]. If the p-value is > α, do not reject H0; there is not suﬃcient evidence that [HA in context]. 474 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION EXAMPLE 8.35 The regression summary below shows statistical software output from ﬁtting the least squares regression line for predicting gift aid based on family income for 50 randomly selected freshman students at Elmhurst College. The scatterplot and residual plot were shown in Figure 8.27. Predictor Constant family income Coef 24.31933 -0.04307 SE Coef 1.29145 0.01081 T 18.831 -3.985 P < 2e-16 0.000229 S = 4.783 R-Sq = 24.86% R-Sq(adj) = 23.29% Do these data provide convincing evidence that there is a negative, linear relationship between family income and gift aid? Carry out a complete hypothesis test at the 0.05 signiﬁcance level. Use the ﬁve step framework to organize your work. Identify: We will test the following hypotheses at the α = 0.05 signiﬁcance level. H0: β = 0. There is no linear relationship. HA: β < 0. There is a negative linear relationship. Here, β is the slope of the population regression line for predicting gift aid from family income at Elmhurst College. Choose: Because the hypotheses are about the slope of a regression line, we choose the t-test for a slope. Check: The data come from a random sample of less than 10% of the total population of freshman students at Elmhurst College. The lack of any pattern in the residual plot indicates that a linear model is reasonable. Also, the residual plot shows that the residuals have constant variance. Finally, n = 50 ≥ 30 so we do not have to worry too much about any skew in the residuals. All four conditions are met. Calculate: We will calculate the t-statistic, degrees of freedom, and the p-value. T = point estimate − null value SE of estimate We read the slope of the sample regression line and the corresponding SE from the table. The point estimate is: b = −0.04307. The SE of the slope is: SE = 0.01081. T = −0.04307 − 0 0.01081 = −3.985 Because HA uses a less than sign (<), meaning that it is a lower-tail test, the p-value is the area to the left of t = −3.985 under the t-distribution with 50 − 2 = 48 degrees of freedom. The p-value = 1 2 (0.000229) ≈ 0.0001. Conclude: The p-value of 0.0001 is < 0.05, so we reject H0; there is suﬃcient evidence that there is a negative linear relationship between family income and gift aid at Elmhurst College. Guided Practice 8.36 In context, interpret the p-value from the previous example.23 23Assuming that the probability model is true and assuming that the null hypothesis is true, i.e. there really is no linear relationship between family income and gift aid at Elmhurst College, there is only a 0.0001 chance of getting a test statistic this small or smaller (HA uses a <, so the p-value represents the area in the left tail). Because this value is so small, we reject the null hypothesis. 8.4. INFERENCE FOR THE SLOPE OF A REGRESSION LINE 475 8.4.6 Technology: the ttt-test/interval for the slope We generally rely on regression output from statistical software programs to provide us with the necessary quantities: b and SE of b. However we can also ﬁnd the test statistic, p-value, and conﬁdence interval using Desmos or a handheld calculator. Get started quickly with this Desmos T-Test/Interval Calculator (available at openintro.org/ahss/desmos). For instructions on implementing the T-Test/Interval on the TI or Casio, see the Graphing Calculator Guides at openintro.org/ahss. 8.4.7 Which inference procedure to use for paired data? In Section 7.2.4, we looked at a set of paired data involving the price of textbooks for UCLA courses at the UCLA Bookstore and on Amazon. The left panel of Figure 8.34 shows the diﬀerence in price (UCLA Bookstore − Amazon) for each book. Because we have two data points on each textbook, it also makes sense to construct a scatterplot, as seen in the right panel of Figure 8.34. Figure 8.34: Left: histogram of the diﬀerence (UCLA Bookstore price - Amazon price) for each book sampled. Right: scatterplot of Amazon Price versus UCLA Bookstore price. EXAMPLE 8.37 What additional information does the scatterplot provide about the price of textbooks at UCLA Bookstore and on Amazon? With a scatterplot, we see the relationship between the variables. We can see that when UCLA Bookstore price is larger, Amazon price also tends to be larger. We can consider the strength of the correlation and we can draw the linear regression equation for predicting Amazon price from UCLA Bookstore price. UCLA Bookstore Price − Amazon Price (USD)Frequency−$20$0$20$40$60$800102030UCLA Bookstore Price$0$50$100$150$200$0$50$100$150$200Amazon Price 476 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION EXAMPLE 8.38 Which test should we do if we want to check whether: 1. prices for textbooks for UCLA courses are higher at the UCLA Bookstore than on Amazon 2. there is a signiﬁcant, positive linear relationship between UCLA Bookstore price and Amazon price? In the ﬁrst case, we are interested in whether the diﬀerences (UCLA Bookstore − Amazon) for all UCLA textbooks are, on average, greater than 0, so we would do a 1-sample t-test for a mean of In the second case, we are interested in whether the slope of the regression line for diﬀerences. predicting Amazon price from UCLA Bookstore price is signiﬁcantly greater than 0, so we would do a t-test for the slope of a regression line. Likewise, a 1-sample t-interval for a mean of diﬀerences would provide an interval of reasonable values for the mean of diﬀerences in textbook price between UCLA Bookstore and Amazon (for all UCLA textbooks), while a t-interval for the slope would provide an interval of reasonable values for the slope of the regression line for predicting Amazon price from UCLA Bookstore price (for all UCLA textbooks). INFERENCE FOR PAIRED DATA A 1-sample t-interval or t-test for a mean of diﬀerences only makes sense when we are asking whether, on average, one variable is greater than, less than or diﬀerent from another (think histogram of the diﬀerences). A t-interval or t-test for the slope of a regression line makes sense when we are interested in the linear relationship between them (think scatterplot). EXAMPLE 8.39 Previously, we looked at the relationship betweeen body length and head length for bushtail possums. We also looked at the relationship between gift aid and family income for freshmen at Elmhurst College. Could we do a 1-sample t-test in either of these scenarios? We have to ask ourselves, does it make sense to ask whether, on average, body length is greater than head length? Similarly, does it make sense to ask whether, on average, gift aid is greater than family income? These don’t seem to be meaningful research questions; a 1-sample t-test for a mean of diﬀerences would not be useful here. GUIDED PRACTICE 8.40 A teacher gives her class a pretest and a posttest. Does this result in paired data? If so, which hypothesis test should she use?24 24Yes, there are two observations for each individual, so there is paired data. The appropriate test depends upon the question she wants to ask. If she is interested in whether, on average, students do better on the posttest than the pretest, should use a 1-sample t-test for a mean of diﬀerences. If she is interested in whether pretest score is a signiﬁcant linear predictor of posttest score, she should do a t-test for the slope. In this situation, both tests could be useful, but which one should be used is dependent on the teacher’s research question. 8.4. INFERENCE FOR THE SLOPE OF A REGRESSION LINE 477 Section summary In Chapter 6, we used a χ2 test for independence to test for association between two categorical variables. In this section, we test for association/correlation between two numerical variables. • We use the slope b as a point estimate for the slope β of the population regression line. The slope of the population regression line is the true increase/decrease in y for each unit increase in x. If the slope of the population regression line is 0, there is no linear relationship between the two variables. • Under certain assumptions, the sampling distribution for b is normal and the distribution of the standardized test statistic using the standard error of the slope follows a ttt-distribution with n − 2 degrees of freedom. • When there is (x, y) data and the parameter of interest is the slope of the population regression line, e.g. the slope of the population regression line relating air quality index to average rainfall per year for each city in the United States: – Estimate β at the C% conﬁdence level using a ttt-interval for the slope. – Test H0: β = 0 at the α signiﬁcance level using a ttt-test for the slope. • The conditions for the t-interval and t-test for the slope of a regression line are the same. 1. Independence: Data come from a random sample or randomized experiment. If sampling without replacement, check that the sample size is less than 10% of the population size. 2. Line
arity: Check that the scatterplot does not show a curved trend and that the residual plot shows no ∪-shape pattern. 3. Constant variability: Use the residual plot to check that the standard deviation of the residuals is constant across all x-values. 3. Normality: The population of residuals is nearly normal or the sample size is ≥ 30. If the sample size is less than 30 check for strong skew or outliers in the sample residuals. If neither is found, then the condition that the population of residuals is nearly normal is considered reasonable. • The conﬁdence interval and test statistic are calculated as follows: Conﬁdence interval: point estimate ± t × SE of estimate, or Test statistic: T = point estimate − null value and p-value SE of estimate point estimate: the slope b of the sample regression line SE of estimate: SE of slope (ﬁnd using computer output) df = n − 2 • The conﬁdence interval for the slope of the population regression line estimates the true average increase in the y-variable for each unit increase in the x-variable. • The t-test for the slope and the 1-sample t-test for a mean of diﬀerences both involve paired, numerical data. However, the t-test for the slope asks if the two variables have a linear relationship, speciﬁcally if the slope of the population regression line is diﬀerent from 0. The 1-sample t-test for a mean of diﬀerences, on the other hand, asks if the two variables are, on average, diﬀerent, speciﬁcally if the mean of the population diﬀerences is not equal to 0. 478 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION Exercises 8.33 Body measurements, Part IV. The scatterplot and least squares summary below show the relationship between weight measured in kilograms and height measured in centimeters of 507 physically active individuals. (Intercept) height Estimate -105.0113 1.0176 Std. Error 7.5394 0.0440 t value -13.93 23.13 Pr(>\|t\|) 0.0000 0.0000 (a) Describe the relationship between height and weight. (b) Write the equation of the regression line. Interpret the slope and intercept in context. (c) Do the data provide strong evidence that an increase in height is associated with an increase in weight? State the null and alternative hypotheses, report the p-value, and state your conclusion. (d) The correlation coeﬃcient for height and weight is 0.72. Calculate R2 and interpret it in context. 8.34 MCU, predict US theater sales. The Marvel Comic Universe movies were an international movie sensation, containing 23 movies at the time of this writing. Here we consider a model predicting an MCU ﬁlm’s gross theater sales in the US based on the ﬁrst weekend sales performance in the US. The data are presented below in both a scatterplot and the model in a regression table. Scientiﬁc notation is used below, e.g. 42.5e6 corresponds to 42.5 × 106. (Intercept) opening weekend us Estimate 42.5e6 2.4361 Std. Error 26.6e6 0.1739 t value Pr(>\|t\|) 0.1251 0.0000 1.60 14.01 (a) Describe the relationship between gross theater sales in the US and ﬁrst weekend sales in the US. (b) Write the equation of the regression line. Inter- pret the slope and intercept in context. (c) Do the data provide strong evidence that higher opening weekend sale is associated with higher gross theater sales? State the null and alternative hypotheses, report the p-value, and state your conclusion. (d) The correlation coeﬃcient for gross sales and ﬁrst weekend sales is 0.950. Calculate R2 and interpret it in context. (e) Suppose we consider a set of all ﬁlms ever released. Do you think the relationship between opening weekend sales and total sales would have as strong of a relationship as what we see with the MCU ﬁlms? Height (cm)Weight (kg)150175200507090110US Opening Weekend Sales$0$100m$200m$300m$0$200m$400m$600m$800mUS Theater Sales 8.4. INFERENCE FOR THE SLOPE OF A REGRESSION LINE 479 8.35 Spouses, Part II. The scatterplot below summarizes womens’ heights and their spouses’ heights for a random sample of 170 married women in Britain, where both partners’ ages are below 65 years. Summary output of the least squares ﬁt for predicting spouse’s height from the woman’s height is also provided in the table. (Intercept) height spouse Estimate 43.5755 0.2863 Std. Error 4.6842 0.0686 t value 9.30 4.17 Pr(>\|t\|) 0.0000 0.0000 (a) Is there strong evidence in this sample that taller women have taller spouses? State the hypotheses and include any information used to conduct the test. (b) Write the equation of the regression line for predicting the height of a woman’s spouse based on the woman’s height. (c) Interpret the slope and intercept in the context of the application. (d) Given that R2 = 0.09, what is the correlation of heights in this data set? (e) You meet a married woman from Britain who is 5’9” (69 inches). What would you predict her spouse’s height to be? How reliable is this prediction? (f) You meet another married woman from Britain who is 6’7” (79 inches). Would it be wise to use the same linear model to predict her spouse’s height? Why or why not? 8.36 Urban homeowners, Part II. Exercise 8.29 gives a scatterplot displaying the relationship between the percent of families that own their home and the percent of the population living in urban areas. Below is a similar scatterplot, excluding District of Columbia, as well as the residuals plot. There were 51 cases. (a) For these data, R2 = 0.28. What is the correlation? How can you tell if it is positive or negative? (b) Examine the residual plot. What do you observe? Is a simple least squares ﬁt appropriate for these data? Woman's height (in inches)Spouse's height (in inches)6065707555606570% Urban population% Who own home4060805560657075−10010 480 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION Exercise 8.25 presents regression output from a model for predicting 8.37 Murders and poverty, Part II. annual murders per million from percentage living in poverty based on a random sample of 20 metropolitan areas. The model output is also provided below. (Intercept) poverty% Estimate -29.901 2.559 Std. Error 7.789 0.390 t value Pr(>\|t\|) 0.001 -3.839 0.000 6.562 s = 5.512 R2 = 70.52% R2 adj = 68.89% (a) What are the hypotheses for evaluating whether poverty percentage is a signiﬁcant predictor of murder rate? (b) State the conclusion of the hypothesis test from part (a) in context of the data. (c) Calculate a 95% conﬁdence interval for the slope of poverty percentage, and interpret it in context of the data. (d) Do your results from the hypothesis test and the conﬁdence interval agree? Explain. 8.38 Babies. Is the gestational age (time between conception and birth) of a low birth-weight baby useful in predicting head circumference at birth? Twenty-ﬁve low birth-weight babies were studied at a Harvard teaching hospital; the investigators calculated the regression of head circumference (measured in centimeters) against gestational age (measured in weeks). The estimated regression line is head circumf erence = 3.91 + 0.78 × gestational age The standard error for the coeﬃcient of gestational age is 0.35. Is there signiﬁcance evidence that gestational age has a positive linear association with head circumference? Use the Identify, Choose, Check, Calculate, Conclude framework and make sure to identify any assumptions used in the test. 8.4. INFERENCE FOR THE SLOPE OF A REGRESSION LINE 481 Chapter highlights This chapter focused on describing the linear association between two numerical variables and ﬁtting a linear model. • The correlation coeﬃcient, r, measures the strength and direction of the linear association between two variables. However, r alone cannot tell us whether data follow a linear trend or whether a linear model is appropriate. • The explained variance, R2, measures the proportion of variation in the y values explained by a given model. Like r, R2 alone cannot tell us whether data follow a linear trend or whether a linear model is appropriate. • Every analysis should begin with graphing the data using a scatterplot in order to see the association and any deviations from the trend (outliers or inﬂuential values). A residual plot helps us better see patterns in the data. • When the data show a linear trend, we ﬁt a least squares regression line of the form: ˆy = a + bx, where a is the y-intercept and b is the slope. It is important to be able to calculate a and b using the summary statistics and to interpret them in the context of the data. • A residual, y − ˆy, measures the error for an individual point. The standard deviation of the residuals, s, measures the typical size of the residuals. • ˆy = a + bx provides the best ﬁt line for the observed data. To estimate or hypothesize about the slope of the population regression line, ﬁrst conﬁrm that the residual plot has no pattern and that a linear model is reasonable, then use a ttt-interval for the slope or a ttt-test for the slope with n − 2 degrees of freedom. In this chapter we focused on simple linear models with one explanatory variable. More complex methods of prediction, such as multiple regression (more than one explanatory variable) and nonlinear regression can be studied in a future course. 482 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION Chapter exercises 8.39 True / False. Determine if the following statements are true or false. If false, explain why. (a) A correlation coeﬃcient of -0.90 indicates a stronger linear relationship than a correlation of 0.5. (b) Correlation is a measure of the association between any two variables. 8.40 Cats, Part II. Exercise 8.26 presents regression output from a model for predicting the heart weight (in g) of cats from their body weight (in kg). The coeﬃcients are estimated using a dataset of 144 domestic cat. The model output is also provided below. Assume that conditions for inference on the slope are met. (Intercept) body wt Estimate -0.357 4.034 Std. Error 0.692 0.250 t value Pr(>\|t\|) 0.607 -0.515 0.000 16.119 s = 1.452 R2 = 64.66% R2 adj = 64.41% (a) What are the hypotheses for evaluat
ing whether body weight is associated with heart weight in cats? (b) State the conclusion of the hypothesis test from part (a) in context of the data. (c) Calculate a 95% conﬁdence interval for the slope of body weight, and interpret it in context of the data. (d) Do your results from the hypothesis test and the conﬁdence interval agree? Explain. 8.41 Nutrition at Starbucks, Part II. Exercise 8.22 introduced a data set on nutrition information on Starbucks food menu items. Based on the scatterplot and the residual plot provided, describe the relationship between the protein content and calories of these menu items, and determine if a simple linear model is appropriate to predict amount of protein from the number of calories. 8.42 Helmets and lunches. The scatterplot shows the relationship between socioeconomic status measured as the percentage of children in a neighborhood receiving reduced-fee lunches at school (lunch) and the percentage of bike riders in the neighborhood wearing helmets (helmet). The average percentage of children receiving reduced-fee lunches is 30.8% with a standard deviation of 26.7% and the average percentage of bike riders wearing helmets is 38.8% with a standard deviation of 16.9%. (a) If the R2 for the least-squares regression line for these data is 72%, what is the correlation between lunch and helmet? (b) Calculate the slope and intercept for the least-squares regression line for these data. (c) Interpret the intercept of the least-squares regression line in the context of the application. (d) Interpret the slope of the least-squares regression line in the context of the application. (e) What would the value of the residual be for a neighborhood where 40% of the children receive reduced-fee lunches and 40% of the bike riders wear helmets? Interpret the meaning of this residual in the context of the application. CaloriesProtein (grams)1002003004005000102030−20020llllllllllllRate of Receiving a Reduced−Fee Lunch0%20%40%60%80%0%20%40%60%Rate of Wearing a Helmet 8.4. INFERENCE FOR THE SLOPE OF A REGRESSION LINE 483 8.43 Match the correlation, Part III. Match each correlation to the corresponding scatterplot. (a) r = −0.72 (b) r = 0.07 (c) r = 0.86 (d) r = 0.99 8.44 Rate my professor. Many college courses conclude by giving students the opportunity to evaluate the course and the instructor anonymously. However, the use of these student evaluations as an indicator of course quality and teaching eﬀectiveness is often criticized because these measures may reﬂect the inﬂuence of non-teaching related characteristics, such as the physical appearance of the instructor. Researchers at University of Texas, Austin collected data on teaching evaluation score (higher score means better) and standardized beauty score (a score of 0 means average, negative score means below average, and a positive score means above average) for a sample of 463 professors.25 The scatterplot below shows the relationship between these variables, and regression output is provided for predicting teaching evaluation score from beauty score. (Intercept) beauty Estimate 4.010 Cell 1 Std. Error 0.0255 0.0322 t value Pr(>\|t\|) 0.0000 157.21 0.0000 4.13 (a) Given that the average standardized beauty score is -0.0883 and average teaching evaluation score is 3.9983, calculate the slope. Alternatively, the slope may be computed using just the information provided in the model summary table. (b) Do these data provide convincing evidence that the slope of the relationship between teaching evaluation and beauty is positive? Explain your reasoning. (c) List the conditions required for linear regression and check if each one is satisﬁed for this model based on the following diagnostic plots. 25Daniel S Hamermesh and Amy Parker. “Beauty in the classroom: Instructors’ pulchritude and putative peda- gogical productivity”. In: Economics of Education Review 24.4 (2005), pp. 369–376. (1)(2)(3)(4)BeautyTeaching evaluation−10122345BeautyResiduals−1012−101Residuals−2−1012050100150Order of data collectionResiduals0100200300400−101 484 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION 8.45 Trees. The scatterplots below show the relationship between height, diameter, and volume of timber in 31 felled black cherry trees. The diameter of the tree is measured 4.5 feet above the ground.26 (a) Describe the relationship between volume and height of these trees. (b) Describe the relationship between volume and diameter of these trees. (c) Suppose you have height and diameter measurements for another black cherry tree. Which of these variables would be preferable to use to predict the volume of timber in this tree using a simple linear regression model? Explain your reasoning. 26Source: R Dataset, stat.ethz.ch/R-manual/R-patched/library/datasets/html/trees.html. lllllllllllllllllllllllllllllllHeight (feet)Volume (cubic feet)60708090020406080lllllllllllllllllllllllllllllllDiameter (inches)Volume (cubic feet)8121620020406080 485 Appendix A Exercise solutions 1 Data collection 1.1 (a) Treatment: 10/43 = 0.23 → 23%. (b) Control: 2/46 = 0.04 → 4%. (c) A higher percentage of patients in the treatment group were pain free 24 hours after receiving acupuncture. (c) It is possible that the observed diﬀerence between the two group percentages is due to chance. 1.3 (a) “Is there an association between air pollution exposure and preterm births?” (b) 143,196 births in Southern California between 1989 and 1993. (c) Measurements of carbon monoxide, nitrogen dioxide, ozone, and particulate matter less than 10µg/m3 (PM10) collected at air-quality-monitoring stations as well as length of gestation. Continous numerical variables. 1.5 (a) “Does explicitly telling children not to cheat aﬀect their likelihood to cheat?”. (b) 160 children between the ages of 5 and 15. (c) Four variables: (1) age (numerical, continuous), (2) sex (categorical), (3) whether they were an only child or not (categorical), (4) whether they cheated or not (categorical). 1.7 Explanatory: acupuncture or not. Response: if the patient was pain free or not. 1.9 (a) 50 × 3 = 150. (b) Four continuous numerical variables: sepal length, sepal width, petal length, and petal width. (c) One categorical variable, species, with three levels: setosa, versicolor, and virginica. 1.11 (a) Airport ownership status (public/private), airport usage status (public/private), latitude, and longitude. (b) Airport ownership status: categorical, not ordinal. Airport usage status: categorical, not ordinal. Latitude: numerical, continuous. Longitude: numerical, continuous. 1.13 (a) Population: all births, sample: 143,196 births between 1989 and 1993 in Southern California. (b) If births in this time span at the geography can be considered to be representative of all births, then the results are generalizable to the population of Southern California. However, since the study is observational the ﬁndings cannot be used to establish causal relationships. 1.15 (a) Population: all asthma patients aged 18-69 who rely on medication for asthma treatment. Sample: 600 such patients. (b) If the patients in this sample, who are likely not randomly sampled, can be considered to be representative of all asthma patients aged 18-69 who rely on medication for asthma treatment, then the results are generalizable to the population deﬁned above. Additionally, since the study is experimental, the ﬁndings can be used to establish causal relationships. 1.17 (a) Observation. (c) Sample statistic (mean). (d) Population parameter (mean). (b) Variable. 1.19 (a) Observational. (b) Use stratiﬁed sampling to randomly sample a ﬁxed number of students, say 10, from each section for a total sample size of 40 students. 1.21 (a) Positive, non-linear, somewhat strong. Countries in which a higher percentage of the population have access to the internet also tend to have higher average life expectancies, however rise in life expectancy trails oﬀ before around 80 years old. (b) Observational. (c) Wealth: countries with individuals who can widely aﬀord the internet can probably also aﬀord basic medical care. (Note: Answers may vary.) 1.23 (a) Simple random sampling is okay. In fact, it’s rare for simple random sampling to not be a reasonable sampling method! (b) The student opinions may vary by ﬁeld of study, so the stratifying by this variable makes sense and would be reasonable. (c) Students of similar ages are probably going to have more similar opinions, and we want clusters to be diverse with respect to the outcome of interest, so this would not be a good approach. (Additional thought: the clusters in this case may also have very diﬀerent numbers of people, which can also create unexpected sample sizes.) 486 APPENDIX A. EXERCISE SOLUTIONS 1.25 (a) The cases are 200 randomly sampled men and women. (b) The response variable is attitude towards a ﬁctional microwave oven. (c) The explanatory variable is dispositional attitude. (d) Yes, the cases are sampled randomly. (e) This is an observational study since there is no random assignment to treatments. (f) No, we cannot establish a causal link between the explanatory and response variables since the study is observational. (g) Yes, the results of the study can be generalized to the population at large since the sample is random. 1.27 (a) Simple random sample. Non-response bias, if only those people who have strong opinions about the survey responds his sample may not be representative of the population. (b) Convenience sample. Under coverage bias, his sample may not be representative of the population since it consists only of his friends. It is also possible that the study will have non-response bias if some choose to not bring back the survey. (c) Convenience sample. This will have a similar issues to handing out surveys to friends. (d) Multi-stage sampling. If the classes are similar to each other with respect to student composition this approach should not introduce bias, other than potential non-response bias. 1.29 (a) Exam
performance. (b) Light level: ﬂuorescent overhead lighting, yellow overhead lighting, no overhead lighting (only desk lamps). (c) Sex: man, woman. 1.31 (a) Experiment. (b) Light level (overhead lighting, yellow overhead lighting, no overhead lighting) and noise level (no noise, construction noise, and human chatter noise). (c) Since the researchers want to ensure equal representation of graduate and undergraduate students, program type will be a blocking variable. 1.33 Need randomization and blinding. One possible outline: (1) Prepare two cups for each participant, one containing regular Coke and the other containing Diet Coke. Make sure the cups are identical and contain equal amounts of soda. Label the cups A (regular) and B (diet). (Be sure to randomize A and B for each trial!) (2) Give each participant the two cups, one cup at a time, in random order, and ask the participant to record a value that indicates how much she liked the beverage. Be sure that neither the participant nor the person handing out the cups knows the identity of the beverage to make this a double-blind experiment. (Answers may vary.) 1.35 (a) Observational study. (b) Dog: Lucy. Cat: Luna. (c) Oliver and Lily. (d) Positive, as the popularity of a name for dogs increases, so does the popularity of that name for cats. 1.37 (a) Experiment. (b) Treatment: 25 grams of chia seeds twice a day, control: placebo. (c) Yes, gender. (d) Yes, single blind since the patients were blinded to the treatment they received. (e) Since this is an experiment, we can make a causal statement. However, since the sample is not random, the causal statement cannot be generalized to the population at large. 1.39 (a) Non-responders may have a diﬀerent response to this question, e.g. parents who returned the surveys likely don’t have diﬃculty spending time with their children. (b) It is unlikely that the women who were reached at the same address 3 years later are a random sample. These missing responders are probably renters (as opposed to homeowners) which means that they might have a lower socio-economic status than the respondents. (c) There is no control group in this study, this is an observational study, and there may be confounding variables, e.g. these people may go running because they are generally healthier and/or do other exercises. controlled 1.41 (a) Randomized experiment. (b) Explanatory: treatment group (categorical, with 3 levels). Response variable: Psychological wellbeing. (c) No, because the participants were volunteers. (d) Yes, because it was an experiment. (e) The statement should say “evidence” instead of “proof”. 1.43 (a) County, state, driver’s race, whether the car was searched or not, and whether the driver was arrested or not. (b) All categorical, non-ordinal. (c) Response: whether the car was searched or not. Explanatory: race of the driver. 487 2 Summarizing data 2.1 (a) There is a weak and positive relationship between age and income. With so few points it is diﬃcult to tell the form of the relationship (linear or not) however the relationship does look somewhat curved. (b) (c) For males as age increases so does income, however this pattern is not apparent for females. 2.3 (a) 0 \| 000003333333 0 \| 7779 1 \| 0011 Legend: (b) 1 \| 0 = 10% (c) (d) 40% (Note: if using only rel. freq. histogram, you can only get an estimate because 7 is in the middle of the bin. Use the dot plot to get a more accurate answer.) 2.5 (a) Positive association: mammals with longer gestation periods tend to live longer as well. (b) Association would still be positive. (c) No, they are not independent. See part (a). 2.7 Both distributions are right skewed and bimodal with modes at 10 and 20 cigarettes; note that people may be rounding their answers to half a pack or a whole pack. The median of each distribution is between 10 and 15 cigarettes. The middle 50% of the data (the IQR) appears to be spread equally in each group and have a width of about 10 to 15. There are potential outliers above 40 cigarettes per day. It appears that respondents who smoke only a few cigarettes (0 to 5) smoke more on the weekdays than on weekends. 2.9 (a) ¯xamtW eekends = 20, ¯xamtW eekdays = 16. (b) samtW eekends = 0, samtW eekdays = 4.18. In this very small sample, higher on weekdays. 2.11 Any 10 employees whose average number of days oﬀ is between the minimum and the mean number of days oﬀ for the entire workforce at this plant. 2.13 (a) Dist 2 has a higher mean since 20 > 13, and a higher standard deviation since 20 is further from the rest of the data than 13. (b) Dist 1 has a higher mean since −20 > −40, and Dist 2 has a higher standard deviation since -40 is farther away from the rest of the data than -20. (c) Dist 2 has a higher mean since all values in this distribution are higher than those in Dist 1, but both distribution have the same standard deviation since they are equally variable around their respective means. (d) Both distributions have the same mean since they’re both centered at 300, but Dist 2 has a higher standard deviation since the observations are farther from the mean than in Dist 1. AgeIncome2030405060020K40K60K80K100K120KMalesAgeIncome2030405060020K40K60K80K100K120KFemalesAgeIncome2030405060020K40K60K80K100K120Kfiber content (% of grams)0.000.020.040.060.080.10llllllllllllllllllllfiber content (% of grams)frequency0.000.020.040.060.080.100.1201234567fiber content (% of grams)relative frequency0.000.020.040.060.080.100.1200.10.20.3 488 APPENDIX A. EXERCISE SOLUTIONS 2.15 (a) About 30. (b) Since the distribution is right skewed the mean is higher than the median. (c) Q1: between 15 and 20, Q3: between 35 and 40, IQR: about 20. (d) Values that are considered to be unusually low or high lie more than 1.5×IQR away from the quartiles. Upper fence: Q3 + 1.5 × IQR = 37.5 + 1.5 × 20 = 67.5; Lower fence: Q1 - 1.5 × IQR = 17.5 + 1.5 × 20 = −12.5; The lowest AQI recorded is not lower than 5 and the highest AQI recorded is not higher than 65, which are both within the fences. Therefore none of the days in this sample would be considered to have an unusually low or high AQI. 2.17 The histogram shows that the distribution is bimodal, which is not apparent in the box plot. The box plot makes it easy to identify more precise values of observations outside of the whiskers. 2.19 (a) The distribution of number of pets per household is likely right skewed as there is a natural boundary at 0 and only a few people have many pets. Therefore the center would be best described by the median, and variability would be best described by the IQR. (b) The distribution of number of distance to work is likely right skewed as there is a natural boundary at 0 and only a few people live a very long distance from work. Therefore the center would be best described by the median, and variability would be best described by the IQR. (c) The distribution of heights of males is likely symmetric. Therefore the center would be best described by the mean, and variability would be best described by the standard deviation. 2.21 (a) The median is a much better measure of the typical amount earned by these 42 people. The mean is much higher than the income of 40 of the 42 people. This is because the mean is an arithmetic average and gets aﬀected by the two extreme observations. The median does not get eﬀected as much since it is robust to outliers. (b) The IQR is a much better measure of variability in the amounts earned by nearly all of the 42 people. The standard deviation gets aﬀected greatly by the two high salaries, but the IQR is robust to these extreme observations. 2.23 (a) The distribution is unimodal and symmetric with a mean of about 25 minutes and a standard deviation of about 5 minutes. There does not appear to be any counties with unusually high or low (b) Answers will vary. There mean travel times. are pockets of longer travel time around DC, Southeastern NY, Chicago, Minneapolis, Los Angeles, and many other big cities. There is also a large section of shorter average commute times that overlap with farmland in the Midwest. Many farmers’ homes are adjacent to their farmland, so their commute would be brief, which may explain why the average commute time for these counties is relatively low. 2.25 (a) 8.85%. (b) 6.94%. (c) 58.86%. (d) 4.56%. 2.27 (a) ZV R = 1.29, ZQR = 0.52. (b) She scored 1.29 standard deviations above the mean on the Verbal Reasoning section and 0.52 standard deviations above the mean on the Quantitative Reasoning section. (c) She did better on the Verbal Reasoning section since her Z-score on that (d) P ercV R = 0.9007 ≈ 90%, section was higher. P ercQR = 0.6990 ≈ 70%. (e) 100% − 90% = 10% did better than her on VR, and 100% − 70% = 30% did better than her on QR. (f) We cannot compare the raw scores since they are on diﬀerent scales. Comparing her percentile scores is more appropriate when comparing her performance to others. (g) Answer to part (b) would not change as Z-scores can be calculated for distributions that are not normal. However, we could not answer parts (c)-(e) since we cannot use the normal probability table to calculate probabilities and percentiles without a normal model. 2.29 (a) Z = 0.84, which corresponds to approximately 159 on QR. (b) Z = −0.52, which corresponds to approximately 147 on VR. 2.31 (a) Z = 1.2, P (Z > 1.2) = 0.1151. (b) Z = −1.28 → 70.6◦F or colder. 2.33 (a) Z = 1.08, P (Z > 1.08) = 0.1401. (b) The answers are very close because only the units were changed. (The only reason why they diﬀer at all because 28◦C is 82.4◦F, not precisely 83◦F.) (c) Since IQR = Q3 − Q1, we ﬁrst need to ﬁnd Q3 and Q1 and take the diﬀerence between the two. Remember that Q3 is the 75th and Q1 is the 25th percentile of a distribution. Q1 = 23.13, Q3 = 26.86, IQR = 26. 86 - 23.13 = 3.73. (a)−1.350(b)01.48(c)0(d)−202VRZ = 1.29QRZ = 0.52 489 2.35 14/20 = 70% are within 1 SD. Within 2 SD: 19/20 = 95%. Within 3 SD: 20/20 = 100%. They follow this rule closely. 2.37 (a)
We see the order of the categories and the relative frequencies in the bar plot. (b) There are no features that are apparent in the pie chart but not in the bar plot. (c) We usually prefer to use a bar plot as we can also see the relative frequencies of the categories in this graph. 2.39 The vertical locations at which the ideological groups break into the Yes, No, and Not Sure categories diﬀer, which indicates that likelihood of supporting the DREAM act varies by political ideology. This suggests that the two variables may be dependent. 7,979 227,571 ≈ 0.035 2.41 (a) (i) False. Instead of comparing counts, we should compare percentages of people in each group who suﬀered cardiovascular problems. (ii) True. (iii) False. Association does not imply causation. We cannot infer a causal relationship based on an observational study. The diﬀerence from part (ii) is subtle. (iv) True. (b) Proportion of all patients who had cardiovascular problems: (c) The expected number of heart attacks in the rosiglitazone group, if having cardiovascular problems and treatment were independent, can be calculated as the number of patients in that group multiplied by the overall cardiovascular problem rate in the study: 67, 593 ∗ 7,979 227,571 ≈ 2370. (d) (i) H0: The treatment and cardiovascular problems are independent. They have no relationship, and the diﬀerence in incidence rates between the rosiglitazone and pioglitazone groups is due to chance. HA: The treatment and cardiovascular problems are not independent. The diﬀerence in the incidence rates between the rosiglitazone and pioglitazone groups is not due to chance and rosiglitazone is associated with an increased risk of serious car(ii) A higher number of padiovascular problems. tients with cardiovascular problems than expected under the assumption of independence would provide support for the alternative hypothesis as this would suggest that rosiglitazone increases the risk of such problems. (iii) In the actual study, we observed 2,593 cardiovascular events in the rosiglitazone group. In the 1,000 simulations under the independence model, we observed somewhat less than 2,593 in every single simulation, which suggests that the actual results did not come from the independence model. That is, the variables do not appear to be independent, and we reject the independence model in favor of the alternative. The study’s results provide convincing evidence that rosiglitazone is associated with an increased risk of cardiovascular problems. 2.43 (a) Decrease: the new score is smaller than the mean of the 24 previous scores. (b) Calculate a weighted mean. Use a weight of 24 for the old mean and 1 for the new mean: (24 × 74 + 1 × 64)/(24 + 1) = 73.6. (c) The new score is more than 1 standard deviation away from the previous mean, so increase. 2.45 No, we would expect this distribution to be right skewed. There are two reasons for this: (1) there is a natural boundary at 0 (it is not possible to watch less than 0 hours of TV), (2) the standard deviation of the distribution is very large compared to the mean. 2.47 The distribution of ages of best actress winners are right skewed with a median around 30 years. The distribution of ages of best actress winners is also right skewed, though less so, with a median around 40 years. The diﬀerence between the peaks of these distributions suggest that best actress winners are typically younger than best actor winners. The ages of best actress winners are more variable than the ages of best actor winners. There are potential outliers on the higher end of both of the distributions. 2.49 (b) Z = 10−7.44 2.51 (a) Z = 5.5−7.44 1.33 = −1.49; P (Z < −1.49) = 0.068. Approximately 6.8% of the newborns were of 1.33 = 1.925. Uslow birth weight. ing a lower bound of 2 and an upper bound of 5, we get P (Z > 1.925) = 0.027. Approximately 2.7% of the newborns weighed over 10 pounds. (c) Approximately 2.7% of the newborns weighed over 10 pounds. Because there were 23,419 of them, about 0.027 × 23419 ≈ 632 weighed greater than 10 pounds. (d) Because we have the percentile, this is the inverse problem. To get the Z-score, use the inverse normal option with 0.90 to get Z = 1.28. Then solve for x in 1.28 = x−7.44 to get x = 9.15. To be at the 90th 1.33 percentile among this group, a newborn would have to weigh 9.15 pounds. 2.53 (a) 93.94%. (b) 93.53%. (c) 80.49 miles/hour. (d) 70.54%. l60708090Scores 490 APPENDIX A. EXERCISE SOLUTIONS 3 Probability (d) 0.11/0.33 ≈ 0.33. nation. (b) 0.60 + 0.20 − 0.18 = 0.62. (c) 0.18/0.20 = 0.9. (e) No, otherwise the answers to (c) and (d) would be the same. (f) 0.06/0.34 ≈ 0.18. 3.17 (a) 0.3. (b) 0.3. (c) 0.3. (d) 0.3 × 0.3 = 0.09. (e) Yes, the population that is being sampled from is identical in each draw. 3.19 (a) 2/9 ≈ 0.22. (b) 3/9 ≈ 0.33. (c) 3 9 ≈ 0.067. (d) No, e.g. in this exercise, removing one chip meaningfully changes the probability of what might be drawn next. 3.21 P (1leggings, 2jeans, 3jeans) = 5 22 = 0.0173. However, the person with leggings could have come 2nd or 3rd, and these each have this same probability, so 3 × 0.0173 = 0.0519. 23 × 6 24 × 7 10 × 2 3.23 (a) 3.25 0.0714. Even when a patient tests positive for lupus, there is only a 7.14% chance that he actually has lupus. House may be right. (b) 0.84 3.27 (a) P(pass) = 0.5, but it should be 0.16. (b) P(pass) = 0.2, instead of 0.16. (c) P(pass) = 0.17, instead of 0.16. 3.1 (a) False. (b) False. There are red face cards. card cannot be both a face card and an ace. These are independent trials. (c) True. A 3.3 (a) 10 tosses. Fewer tosses mean more variability in the sample fraction of heads, meaning there’s a better chance of getting at least 60% heads. (b) 100 tosses. More ﬂips means the observed proportion of heads would often be closer to the average, 0.50, and therefore also above 0.40. (c) 100 tosses. With more ﬂips, the observed proportion of heads would often be closer to the average, 0.50. (d) 10 tosses. Fewer ﬂips would increase variability in the fraction of tosses that are heads. 3.5 (a) 0.510 = 0.00098. (b) 0.510 = 0.00098. (c) P (at least one tails) = 1 − P (no tails) = 1 − (0.510) ≈ 1 − 0.001 = 0.999. 3.7 (a) No, there are voters who are both independent and swing voters. (b) (c) Each Independent voter is either a swing voter or not. Since 35% of voters are Independents and 11% are both Independent and swing voters, the other 24% must not be swing voters. (d) 0.47. (e) 0.53. (f) P(Independent) × P(swing) = 0.35 × 0.23 = 0.08, which does not equal P(Independent and swing) = 0.11, so the events are dependent. 3.9 (a) If the class is not graded on a curve, they are independent. If graded on a curve, then neither independent nor disjoint – unless the instructor will only give one A, which is a situation we will ignore in parts (b) and (c). (b) They are probably not independent: if you study together, your study habits would be related, which suggests your course performances are also related. (c) No. See the answer to part (a) when the course is not graded on a curve. More generally: if two things are unrelated (independent), then one occurring does not preclude the other from occurring. 3.11 (a) 280/792 = 0.354. (b) 445/792 = 0.562. (c) 231/792 = 0.292, (d) 0.354 × 0.562 = 0.199 = 0.292. The events are not independent, so you cannot just multiply the unconditional probabilities. 3.13 (a) No, but we could if A and B are independent. (b-i) 0.21. (b-ii) 0.79. (b-iii) 0.3. (c) No, because 0.1 = 0.21, where 0.21 was the value computed under independence from part (a). (d) 0.143. 3.15 (a) No, 0.18 of respondents fall into this combi- 241112IndependentSwingCan constructbox plots?Passed?yes, 0.8Yes, 0.860.80.86 = 0.688No, 0.140.80.14 = 0.112no, 0.2Yes, 0.650.20.65 = 0.13No, 0.350.20.35 = 0.07Lupus?Resultyes, 0.02positive, 0.980.020.98 = 0.0196negative, 0.020.020.02 = 0.0004no, 0.98positive, 0.260.980.26 = 0.2548negative, 0.740.980.74 = 0.7252 3.29 (a) Starting at row 3 of the random number table, we will read across the table two digits at a time. If the random number is between 00-15, the car will fail the pollution test. If the number is between 1699, the car will pass the test. (Answers may vary.) (b) Fleet 1: 18-52-97-32-85-95-29 → P-P-P-P-P-P-P → ﬂeet passes Fleet 2: 14-96-06-67-17-49-59 → F-P-F-P-P-P-P → ﬂeet fails Fleet 3: 05-33-67-97-58-11-81 → F-P-P-P-P-F-P → ﬂeet fails Fleet 4: 23-81-83-21-71-08-50 → P-P-P-P-P-F-P → ﬂeet fails Fleet 5: 82-84-39-31-83-14-34 → P-P-P-P-P-F-P → ﬂeet fails estimate: 4 / 5 = 0.80 (c) P(at least one car fails in a ﬂeet of seven) = 1 - P(no cars fail) = 1 − (0.84)7 = 0.705. 3.31 (a) Mean: $3 ∗ 0.5 + $5 ∗ 0.3 + $10 ∗ 0.15 + $25 ∗ 0.05 = $5.75. (b) To compute the SD, it is easier to ﬁrst compute the variance: (3 − 5.75)2 ∗ 0.5 + (5 − 5.75)2 ∗ 0.3 + (10 − 5.75)2 ∗ 0.15 + (25 − 5.75)2 ∗ 0.05 = 25.1875. The SD is then the square root of this value: $5.02. 3.33 (a) E(X) = 3.59. SD(X) = 9.64. (b) E(X) = -1.41. SD(X) = 9.64. (c) No, the expected net proﬁt is negative, so on average you expect to lose money. 3.35 5% increase in value. 3.37 E = -0.0526. SD = 0.9986. 3.39 (a) Let X represent the amount of lemonade in the pitcher, Y represent the amount of lemonade in a glass, and W represent the amount left over after. Then, µW = E(X − Y ) = 64 − 12 = 52 (b) σW = √ SD(X)2 + SD(Y )2 = 4 = 2 = P (Z > −1) = (c) P (W > 50) = P Z > 50−52 1 − 0.1587 = 0.8413 1.7322 + 12 ≈ √ 2 6, though specifying such details would be necessary. 3.45 (a) 0.8752×0.125 = 0.096. (b) µ = 8, σ = 7.48. 491 3.47 If p is the probability of a success, then the mean of a Bernoulli random variable X is given by µ = E[X] = P (X = 0) × 0 + P (X = 1) × 1 = (1 − p.49 (a) 5 + 5 (d) 5 = 5. (b) 5 + 5 = 10 + 5 + 1 = 16. = 5. (c) 5 = 10. 3 1 4 3 4 5 3.51 (a) Binomial conditions are met: (1) Independent trials: In a random sample, whether or not one 18-20 year old has consumed alcohol does not depend on whether or not another one has. (2) Fixed number of trials: n = 10. (
3) Only two outcomes at each trial: Consumed or did not consume alcohol. (4) Probability of a success is the same for each trial: p = 0.697. (b) 0.203. (c) 0.203. (d) 0.167. (e) 0.997. 3.53 (a) 1 − 0.753 = 0.5781. (b) 0.1406. (c) 0.4219. (d) 1 − 0.253 = 0.9844. 3.55 (a) µ = 35, σ = 3.24 (b) Z = 45−35 3.24 = 3.09. 45 is more than 3 standard deviations away from the mean, we can assume that it is an unusual observation. Therefore yes, we would be surprised. (c) Using the normal approximation, 0.0010. With 0.5 correction, 0.0017. 3.57 (a) Invalid. Sum is greater than 1. (b) Valid. Probabilities are between 0 and 1, and they sum to 1. In this class, every student gets a C. (c) Invalid. Sum is less than 1. (d) Invalid. There is a negative probability. (e) Valid. Probabilities are between 0 and 1, and they sum to 1. (f) Invalid. There is a negative probability. 3.59 0.8247. 3.41 (a) The combined scores follow a normal distribution with µcombined = 304 and σcombined = 10.38. Then, P(combined score > 320) is approximately 0.06. (b) Z=1.28 (using calculator or table). Then we set 1.28 = x−304 10.38 and ﬁnd x ≈ 317. 3.43 (a) No. The cards are not independent. For example, if the ﬁrst card is an ace of clubs, that implies the second card cannot be an ace of clubs. Additionally, there are many possible categories, which would need to be simpliﬁed. (b) No. There are six events under consideration. The Bernoulli distribution allows for only two events or categories. Note that rolling a die could be a Bernoulli trial if we simplify to two events, e.g. rolling a 6 and not rolling a 3.61 (a) E = $3.90. SD = $0.34. (b) E = $27.30. SD = $0.89. 3.63 (a) 13. (b) No, these 27 students are not a random sample from the university’s student population. For example, it might be argued that the proportion of smokers among students who go to the gym at 9 am on a Saturday morning would be lower than the proportion of smokers in the university as a whole. 3.65 0 wins (-$3): 0.1458. 1 win (-$1): 0.3936. 2 wins (+$1): 0.3543. 3 wins (+$3): 0.1063. HIV?Resultyes, 0.259positive, 0.9970.2590.997 = 0.2582negative, 0.0030.2590.003 = 0.0008no, 0.741positive, 0.0740.7410.074 = 0.0548negative, 0.9260.7410.926 = 0.6862 492 APPENDIX A. EXERCISE SOLUTIONS 4 Distributions of random variables 4.1 (a) Each observation in each of the distributions represents the sample proportion (ˆp) from samples of size n = 20, n = 100, and n = 500, respectively. (b) The centers for all three distributions are at 0.95, the true population parameter. When n is small, the distribution is skewed to the left and not smooth. As n increases, the variability of the distribution (standard deviation) decreases, and the shape of the distribution becomes more unimodal and symmetric. 4.3 (a) False. Doesn’t satisfy success-failure condition. (b) True. The success-failure condition is not satisﬁed. In most samples we would expect ˆp to be close to 0.08, the true population proportion. While ˆp can be much above 0.08, it is bound below by 0, suggesting it would take on a right skewed shape. Plotting the sampling distribution would conﬁrm this suspicion. (c) False. SE ˆp = 0.0243, and ˆp = 0.12 is only 0.12−0.08 0.0243 = 1.65 SEs away from the mean, which would not be considered unusual. (d) True. ˆp = 0.12 is 2.32 standard errors away from the mean, which is often considered unusual. (e) False. Decreases the SE by a factor of 1/ 4.5 (a) SD ˆp = p(1 − p)/n = 0.0707. This describes the typical distance that the sample proportion will deviate from the true proportion, p = 0.5. (b) ˆp approximately follows N (0.5, 0.0707). Z = (0.55 − 0.50)/0.0707 ≈ 0.71. This corresponds to an upper tail of about 0.2389. That is, P (ˆp > 0.55) ≈ 0.24. √ 2. 4.7 (a) First we need to check that the necessary conditions are met. There are 200 × 0.08 = 16 expected successes and 200 × (1 − 0.08) = 184 expected failures, therefore the success-failure condition is met. Then the binomial distribution can be approximated by N (µ = 16, σ = 3.84). P (X < 12) = P (Z < −1.04) = 0.1492. (b) Since the successfailure condition is met the sampling distribution for ˆp ∼ N (µ = 0.08, σ = 0.0192). P (ˆp < 0.06) = P (Z < −1.04) = 0.1492. (c) As expected, the two answers are the same. 4.9 The sampling distribution is the distribution of sample proportions from samples of the same size randomly sampled from the same population. As the same size increases, the shape of the sampling distribution (when p = 0.1) will go from being rightskewed to being more symmetric and resembling the normal distribution. With larger sample sizes, the spread of the sampling distribution gets smaller. Regardless of the sample size, the center of the sampling distribution is equal to the true mean of that population, provided the sampling isn’t biased. 4.11 (a) The distribution is unimodal and strongly right skewed with a median between 5 and 10 years old. Ages range from 0 to slightly over 50 years old, and the middle 50% of the distribution is roughly between 5 and 15 years old. There are potential outliers on the higher end. (b) When the sample size is small, the sampling distribution is right skewed, just like the population distribution. As the sample size increases, the sampling distribution gets more unimodal, symmetric, and approaches normality. The variability also decreases. This is consistent with the Central Limit Theorem. 4.13 (a) Right skewed. There is a long tail on the higher end of the distribution but a much shorter tail on the lower end. (b) Less than, as the median would be less than the mean in a right skewed distribution. (c) We should not. (d) Even though the population distribution is not normal, the conditions for inference are reasonably satisﬁed, with the possible exception of skew. If the skew isn’t very strong (we should ask to see the data), then we can use the Central Limit Theorem to estimate this probability. For now, we’ll assume the skew isn’t very strong, though the description suggests it is at least moderate to 60): Z = 2.58 → strong. Use N (1.3, SD¯x = 0.3/ 0.0049. (e) It would decrease it by a factor of 1/ 2. √ √ 4.15 The centers are the same in each plot, and each data set is from a nearly normal distribution, though the histograms may not look very normal since each represents only 100 data points. The only way to tell which plot corresponds to which scenario is to examine the variability of each distribution. Plot B is the most variable, followed by Plot A, then Plot C. This means Plot B will correspond to the original data, Plot A to the sample means with size 5, and Plot C to the sample means with size 25. 4.17 (a) Z = −3.33 → 0.0004. (b) The population SD is known and the data are nearly normal, so the sample mean will be nearly normal with distribution N (µ, σ/ n), i.e. N (2.5, 0.0095). (c) Z = −10.54 → ≈ 0. (d) See below: √ (e) We could not estimate (a) without a nearly normal population distribution. We also could not estimate (c) since the sample size is not suﬃcient to yield a nearly normal sampling distribution if the population distribution is not nearly normal. 2.412.442.472.502.532.562.59PopulationSampling (n = 10) 493 √ √ 4.25 (a) µ¯x1 = 15, σ¯x1 = 20/ 50 = 2.8284. √ (b) µ¯x2 = 20, σ¯x1 = 10/ 30 = 1.8257. (c) µ¯x2−¯x1 = 20/ + 10/ 20 − 15 = 5, σ¯x2−¯x1 = = 3.3665. (d) Think of ¯x1 and ¯x2 as being random variables, and we are considering the standard deviation of the diﬀerence of these two random variables, so we square each standard deviation, add them together, and then take the square root of the sum: 502 302 √ SD¯x2−¯x1 = SD2 ¯x2 + SD2 ¯x1 4.27 Want to ﬁnd the probability that there will be 1,787 or more enrollees. Using the normal approximation, with µ = np = 2, 500 × 0.7 = 1750 √ and σ = np(1 − p) = 2, 500 × 0.7 × 0.3 ≈ 23, Z = 1.61, and P (Z > 1.61) = 0.0537. With a 0.5 correction: 0.0559. 4.29 Z = 1.56, P (Z > 1.56) = 0.0594, i.e. 6%. 4.31 This is the same as checking that the average bag weight of the 10 bags is greater than 46 lbs. SD¯x = 3.2√ 1.012 = 0.988; 10 P (z > 0.988) = 0.162 = 16.2%. = 1.012; z = 46−45 4.33 First we need to check that the necessary conditions are met. There are 100 × 0.360 = 36.0 expected successes and 100×(1−0.360) = 64.0 expected failures, therefore the success-failure condition is met. Calculate using either (1) the normal approximation to the binomial distribution or (2) the sampling distribution of ˆp. (1) The binomial distribution can be approximated by N (µ = 0.360, σ = 4.8). P (X ≥ 35) = P (Z > −0.208) = 0.5823. (2) The sampling distribution of ˆp ∼ N (µ = 0.360, σ = 0.048). P (ˆp > 0.35) = P (Z > −0.208) = 0.5823. n = ˆp(1− ˆp) (e) Compute the SE using ˆp = 0.127 in place of p: 0.127(1−0.127) 212 = 0.023. (f) The SE ≈ standard error is the standard deviation of ˆp. A value of 0.10 would be about one standard error away from the observed value, which would not represent a very uncommon deviation. (Usually beyond about 2 standard errors is a good rule of thumb.) The engineer should not be surprised. (g) Recomputed standard 0.1(1−0.1) 212 = 0.021. This error using p = 0.1: SE = value isn’t very diﬀerent, which is typical when the standard error is computed using relatively similar proportions (and even sometimes when those proportions are quite diﬀerent!). 4.19 (a) We cannot use the normal model for this calculation, but we can use the histogram. About 500 songs are shown to be longer than 5 minutes, so the probability is about 500/3000 = 0.167. (b) Two diﬀerent answers are reasonable. Option 1Since the population distribution is only slightly skewed to the right, even a small sample size will yield a nearly normal sampling distribution. We also know that the songs are sampled randomly and the sample size is less than 10% of the population, so the length of one song in the sample is independent of another. We are looking for the probability that the total length of 15 songs is more than 60 minutes, which means that the average song should last at least
60/15 = 4 min√ 15, Z = 1.31 → 0.0951. utes. Using SD¯x = 1.63/ Option 2Since the population distribution is not normal, a small sample size may not be suﬃcient to yield a nearly normal sampling distribution. Therefore, we cannot estimate the probability using the tools we (c) We can now be conﬁdent have learned so far. that the conditions are satisﬁed. Z = 0.92 → 0.1788. 4.21 (a) SD¯x = 25√ = 2.89. (b) Z = 1.73, which 75 indicates that the two values are not unusually distant from each other when accounting for the uncertainty in John’s point estimate. 4.23 (a) µ ˆpN E = 0.01. σ ˆpN E = 0.0031. (b) µ ˆpN Y = 0.06. σ ˆpN Y = 0.0075. (c) µ ˆpN Y − ˆpN B = 0.06 − 0.01 = 0.05. σ ˆpN Y − ˆpN B = 0.0081. (d) We can think of ˆpN E and ˆpN Y as being random variables, and we are considering the standard deviation of the diﬀerence of these two random variables, so we square each standard deviation, add them together, and then take the square root of the sum: SD ˆpN Y − ˆpN E = SD2 ˆpN Y + SD2 ˆpN E 5 Foundations for inference 5.1 (a) Mean. Each student reports a numerical value: a number of hours. (b) Mean. Each student reports a number, which is a percentage, and we can average over these percentages. (c) Proportion. Each student reports Yes or No, so this is a categorical variable and we use a proportion. (d) Mean. Each student reports a number, which is a percentage like in part (b). (e) Proportion. Each student reports whether or not s/he expects to get a job, so this is a categorical variable and we use a proportion. 5.3 (a) The sample is from all computer chips manufactured at the factory during the week of production. We might be tempted to generalize the population to represent all weeks, but we should exercise caution here since the rate of defects may change over time. (b) The fraction of computer chips manufactured at the factory during the week of production that had defects. (c) Estimate the parameter using the data: ˆp = 27 (d) Standard error (or SE). 212 = 0.127. 494 APPENDIX A. EXERCISE SOLUTIONS (3) The alternative hypothesis should have a notequals sign, and (4) it should reference the null value, p0 = 0.6, not the observed sample proportion. The correct way to set up these hypotheses is: H0 : p = 0.6 and HA : p = 0.6. 5.17 (a) H0 : punemp = punderemp: The proportions of unemployed and underemployed people who are having relationship problems are equal. HA : punemp = punderemp: The proportions of unemployed and underemployed people who are having relationship problems are diﬀerent. (b) If in fact the two population proportions are equal, the probability of observing at least a 2% diﬀerence between the sample proportions is approximately 0.35. Since this is a high probability we fail to reject the null hypothesis. The data do not provide convincing evidence that the proportion of of unemployed and underemployed people who are having relationship problems are diﬀerent. 5.19 (a) H0: Anti-depressants do not aﬀect the symptoms of Fibromyalgia. HA: Anti-depressants do aﬀect the symptoms of Fibromyalgia (either helping (b) Concluding that anti-depressants or harming). either help or worsen Fibromyalgia symptoms when they actually do neither. (c) Concluding that antidepressants do not aﬀect Fibromyalgia symptoms when they actually do. 5.21 (a) False. Conﬁdence intervals provide a range of plausible values, and sometimes the truth is missed. A 95% conﬁdence interval “misses” about 5% of the time. (b) True. Notice that the description focuses on the true population value. (c) True. The 95% conﬁdence interval is given by: (42.6%, 47.4%), and we can see that 50% is outside of this interval. This means that in a hypothesis test, we would reject the null hypothesis that the proportion is 0.5. (d) False. The standard error describes the uncertainty in the overall estimate from natural ﬂuctuations due to randomness, not the uncertainty corresponding to individuals’ responses. 5.5 (a) Sampling distribution. (b) If the population proportion is in the 5-30% range, the success-failure condition would be satisﬁed and the sampling distribution would be symmetric. (c) We use the standard error to describe the variability: SE = 0.08(1−0.08) 800 = 0.0096. (d) Standard error. (e) The distribution will tend to be more variable when we have fewer observations per sample. n = p(1−p) the that First, general 5.7 Recall formula is point estimate ± z × SE. identify the three diﬀerent values. The point estimate is 45%, z = 1.96 for a 95% conﬁdence level, and SE = 1.2%. Then, plug the values into the formula: 45% ± 1.96 × 1.2% → (42.6%, 47.4%) We are 95% conﬁdent that the proportion of US adults who live with one or more chronic conditions is between 42.6% and 47.4%. 5.9 (a) False. Inference is made on the population parameter, not the point estimate. The point estimate is always in the conﬁdence interval. (b) True. (c) False. The conﬁdence interval is not about a sample mean. (d) False. To be more conﬁdent that we capture the parameter, we need a wider interval. Think about needing a bigger net to be more sure of catching a ﬁsh in a murky lake. (e) True. Optional explanation: This is true since the normal model was used to model the sample mean. The margin of error is half the width of the interval, and the sample mean is the midpoint of the interval. (f) False. In the calculation of the standard error, we divide the standard deviation by the square root of the sample size. To cut the SE (or margin of error) in half, we would need to sample 22 = 4 times the number of people in the initial sample. 5.11 (a) This claim is reasonable, since the entire interval lies above 50%. (b) The value of 70% lies outside of the interval, so we have convincing evidence that the researcher’s conjecture is wrong. (c) A 90% conﬁdence interval will be narrower than a 95% conﬁdence interval. Even without calculating the interval, we can tell that 70% would not fall in the interval, and we would reject the researcher’s conjecture based on a 90% conﬁdence level as well. 5.13 (a) H0 : p = 0.5 (Neither a majority nor minority of students’ grades improved) HA : p = 0.5 (Either a majority or a minority of students’ grades improved) (b) H0 : µ = 15 (The average amount of company time each employee spends not working is 15 minutes for March Madness.) HA : µ = 15 (The average amount of company time each employee spends not working is diﬀerent than 15 minutes for March Madness.) 5.15 (1) The hypotheses should be about the population proportion (p), not the sample proportion. (2) The null hypothesis should have an equal sign. 495 5.23 (a) H0: The restaurant meets food safety and sanitation regulations. HA: The restaurant does not meet food safety and sanitation regulations. (b) The food safety inspector concludes that the restaurant does not meet food safety and sanitation regulations and shuts down the restaurant when the restaurant is actually safe. (c) The food safety inspector concludes that the restaurant meets food safety and sanitation regulations and the restaurant stays open when the restaurant is actually not safe. (d) A Type 1 Error may be more problematic for the restaurant owner since his restaurant gets shut down even though it meets the food safety and sanitation regulations. (e) A Type 2 Error may be more problematic for diners since the restaurant deemed safe by the inspector is actually not. (f) Strong evidence. Diners would rather a restaurant that meet the regulations get shut down than a restaurant that doesn’t meet the regulations not get shut down. 5.25 True. If the sample size gets ever larger, then the standard error will become ever smaller. Eventually, when the sample size is large enough and the standard error is tiny, we can ﬁnd statistically signiﬁcant yet very small diﬀerences between the null value and point estimate (assuming they are not exactly equal). 6 Inference for categorical data 6.1 (a) True. See the reasoning of 6.1(b). (b) True. We take the square root of the sample size in the SE formula. (c) True. The independence and successfailure conditions are satisﬁed. (d) True. The independence and success-failure conditions are satisﬁed. 6.3 (a) False. A conﬁdence interval is constructed to estimate the population proportion, not the sample proportion. (b) True. 95% CI: 82% ± 2%. (c) True. By the deﬁnition of the conﬁdence level. (d) True. Quadrupling the sample size decreases the SE and 4. (e) True. The 95% CI is ME by a factor of 1/ entirely above 50%. √ 6.5 With a random sample, independence is satisﬁed. The success-failure condition is also satisﬁed. M E = z ˆp(1− ˆp) = 0.0397 ≈ 4% 0.56×0.44 600 n = 1.96 6.7 (a) No. The sample only represents students who took the SAT, and this was also an online survey. (b) (0.5289, 0.5711). We are 90% conﬁdent that 53% to 57% of high school seniors who took the SAT are fairly certain that they will participate in a study abroad program in college. (c) 90% of such random samples would produce a 90% conﬁdence in(d) Yes. terval that includes the true proportion. The interval lies entirely above 50%. 6.9 (a) We want to check for a majority (or minority), so we use the following hypotheses: H0 : p = 0.5 HA : p = 0.5 We have a sample proportion of ˆp = 0.55 and a sample size of n = 617 independents. Since this is a random sample, independence is satisﬁed. The success-failure condition is also satisﬁed: 617 × 0.5 and 617 × (1 − 0.5) are both at least 10 (we use the null proportion p0 = 0.5 for this check in a one-proportion hypothesis test). Therefore, we can model ˆp using a normal distribution with a standard error of SE = p(1 − p) n = 0.02 (We use the null proportion p0 = 0.5 to compute the standard error for a one-proportion hypothesis test.) Next, we compute the test statistic: Z = 0.55 − 0.5 0.02 = 2.5 This yields a one-tail area of 0.0062, and a p-value of 2 × 0.0062 = 0.0124. Because the p-value is smaller than 0.05, we reject the null hypothesis. We have strong evidence that the support is
diﬀerent from 0.5, and since the data provide a point estimate above 0.5, we have strong evidence to support this claim by the TV pundit. (b) No. Generally we expect a hypothesis test and a conﬁdence interval to align, so we would expect the conﬁdence interval to show a range of plausible values entirely above 0.5. However, if the conﬁdence level is misaligned (e.g. a 99% conﬁdence level and a α = 0.05 signiﬁcance level), then this is no longer generally true. 6.11 (a) Identify: H0: p = 0.5. HA: p > 0.5. α = 0.05. Choose: 1-proportion Z-test. Check: Independence (random sample, < 10% of population) is satisﬁed, as is the success-failure conditions (using p0 = 0.5, we expect 40 successes and 40 failures). Calculate: Z = 2.91 → p- value = 0.0018. Conclude: Since the p-value < 0.05, we reject the null hypothesis. The data provide strong evidence that the rate of correctly identifying a soda for these people is signiﬁcantly better than just by random guessing. (b) The p-value represents the following conditional probability: P (ˆp > 0.6625 \| p = 0.5). If in fact people cannot tell the diﬀerence between diet and regular soda and they randomly guess, the probability of getting a random sample of 80 people where 66.25% (53/80) or higher identify a soda correctly would be 0.0018. 496 APPENDIX A. EXERCISE SOLUTIONS low, we reject H0. There is strong evidence of a difference in the rates of autism of children of mothers who did and did not use prenatal vitamins during the ﬁrst three months before pregnancy. (b) The title of this newspaper article makes it sound like using prenatal vitamins can prevent autism, which is a causal statement. Since this is an observational study, we cannot make causal statements based on the ﬁndings of the study. A more accurate title would be “Mothers who use prenatal vitamins before pregnancy are found to have children with a lower rate of autism” 6.25 (a) False. The chi-square distribution has one parameter called degrees of freedom. (b) True. (c) True. (d) False. As the degrees of freedom increases, the shape of the chi-square distribution becomes more symmetric. 6.27 (a) H0: The distribution of the format of the book used by the students follows the professor’s predictions. HA: The distribution of the format of the book used by the students does not follow the professor’s predictions. (b) Ehard copy = 126 × 0.60 = 75.6. Eprint = 126 × 0.25 = 31.5. Eonline = 126 × 0.15 = 18.9. (c) Independence: The sample is not random. However, if the professor has reason to believe that the proportions are stable from one term to the next and students are not aﬀecting each other’s study habits, independence could be reasonable. Sample size: All expected counts are at least 5. (d) χ2 = 2.32, df = 2, p-value = 0.313. (e) Since the p-value is large, we fail to reject H0. The data do not provide strong evidence indicating the professor’s predictions were statistically inaccurate. 6.29 (a) Two-way table: Treatment Patch + support group Only patch Total Quit Yes 40 30 70 No 110 120 230 Total 150 150 300 (b-i) Erow1,col1 = (row 1 total)×(col 1 total) table total is lower than the observed value. (b-ii) Erow2,col2 = (row 2 total)×(col 2 total) This is lower than the observed value. table total = 35. This = 115. 6.13 Because a sample proportion (ˆp = 0.55) is available, we use this for the sample size calculations. The margin of error for a 90% conﬁdence interval is 1.6449 × SE = 1.6449 × be less than 0.01, where we use ˆp in place of p: . We want this to p(1−p) n 1.6449 × 0.55(1 − 0.55) n 1.64492 0.55(1 − 0.55) 0.012 ≤ 0.01 ≤ n From this, we get that n must be at least 6697. 6.15 This is not a randomized experiment, and it is unclear whether people would be aﬀected by the behavior of their peers. That is, independence may not hold. Additionally, there are only 5 interventions under the provocative scenario, so the successfailure condition does not hold. Even if we consider a hypothesis test where we pool the proportions, the success-failure condition will not be satisﬁed. Since one condition is questionable and the other is not satisﬁed, the diﬀerence in sample proportions will not follow a nearly normal distribution. 6.17 (a) False. The entire conﬁdence interval is above 0. (b) True. (c) True. (d) True. (e) False. It is simply the negated and reordered values: (-0.06,0.02). 6.19 (a) Standard error: SE = 0.79(1 − 0.79) 347 + 0.55(1 − 0.55) 617 = 0.03 Using z = 1.96, we get: 0.79 − 0.55 ± 1.96 × 0.03 → (0.181, 0.299) We are 95% conﬁdent the proportion of Democrats who support the plan is 18.1% to 29.9% higher than the proportion of Independents who support the plan. (b) True. that 6.21 Identify: Subscript C means control group. Subscript T means truck drivers. H0 : pC = pT . HA : pC = pT . α = 0.05. Choose: 2-proportion Ztest. Check: Independence is satisﬁed (random samples that are independent), as is the success-failure condition, which we check using the pooled proportion (ˆppool = 70/495 = 0.141). Calculate: Z = −1.65 → p-value = 0.0989. Conclude: Since the p-value is > α, we fail to reject H0. The data do not provide strong evidence that the rates of sleep deprivation are diﬀerent for non-transportation workers and truck drivers. 6.23 (a) Subscript V means vitamin group. Subscript N V means no vitamin group. H0 : pV = pN V . H0 : pV = pN V . Independence is satisﬁed (random samples, < 10% of the population), as is the success-failure condition, which we would check using the pooled proportion (ˆppool = 254/483 = 0.53). Z = 2.99 → p-value = 0.0028. Since the p-value is 6.31 Identify: H0: Opinions regarding oﬀshore drilling for oil and having a college degree are independent. HA: Opinions regarding oﬀshore drilling for oil and having a college degree are dependent. α = 0.05. Choose: chi-square test for independence Check: Erow 1,col 1 = 151.5 Erow 2,col 1 = 162.1 Erow 3,col 1 = 124.5 Erow 1,col 2 = 134.5 Erow 2,col 2 = 143.9 Erow 3,col 2 = 110.5 Independence: The sample is random, and from less than 10% of the population, so independence between observations is reasonable. Expected counts: All expected counts are at least 5. Calculate: χ2 = 11.47, df = 2 → p-value = 0.003. Conclude: Since the pvalue < α, we reject H0. There is strong evidence that there is an association between support for oﬀshore drilling and having a college degree. 6.33 No. The samples at the beginning and at the end of the semester are not independent since the survey is conducted on the same students. 6.35 (a) H0: The age of Los Angeles residents is independent of shipping carrier preference variable. HA: The age of Los Angeles residents is associated with the shipping carrier preference variable. (b) The conditions are not satisﬁed since some expected counts are below 5. 6.37 (a) Independence is satisﬁed (random sample), as is the success-failure condition (40 smokers, 160 non-smokers). The 95% CI: (0.145, 0.255). We are 95% conﬁdent that 14.5% to 25.5% of all students at this university smoke. (b) We want zSE to be no larger than 0.02 for a 95% conﬁdence level. We use z = 1.96 and plug in the point estimate ˆp = 0.2 within the SE formula: 1.960.2(1 − 0.2)/n ≤ 0.02. The sample size n should be at least 1,537. 6.39 (a) Proportion of graduates from this university who found a job within one year of graduating. ˆp = 348/400 = 0.87. (b) This is a random sample, so the observations are independent. Success-failure condition is satisﬁed: 348 successes, 52 failures, both 497 well above 10. (c) (0.8371, 0.9029). We are 95% conﬁdent that approximately 84% to 90% of graduates from this university found a job within one year of completing their undergraduate degree. (d) 95% of such random samples would produce a 95% conﬁdence interval that includes the true proportion of students at this university who found a job within one year of graduating from college. (e) (0.8267, 0.9133). Similar interpretation as before. (f) 99% CI is wider, as we are more conﬁdent that the true proportion is within the interval and so need to cover a wider range. 6.41 Use a chi-square goodness of ﬁt test. H0: Each option is equally likely. HA: Some options are preferred over others. Total sample size: 99. Expected counts: (1/3) * 99 = 33 for each option. These are all above 5, so conditions are satisﬁed. df = 3 − 1 = 2 + (21−33)2 and χ2 = (43−33)2 = 7.52 → 33 p-value = 0.023. Since the p-value is less than 5%, we reject H0. The data provide convincing evidence that some options are preferred over others. + (35−33)2 33 33 6.43 (a) H0 : p = 0.38. HA : p = 0.38. Independence (random sample) and the success-failure condition are satisﬁed. Z = −20.5 → p-value ≈ 0. Since the p-value is very small, we reject H0. The data provide strong evidence that the proportion of Americans who only use their cell phones to access the internet is diﬀerent than the Chinese proportion of 38%, and the data indicate that the proportion is lower in the US. (b) If in fact 38% of Americans used their cell phones as a primary access point to the internet, the probability of obtaining a random sample of 2,254 Americans where 17% or less or 59% or more use their only their cell phones to access the internet would be approximately 0. (c) (0.1545, 0.1855). We are 95% conﬁdent that approximately 15.5% to 18.6% of all Americans primarily use their cell phones to browse the internet. 6.45 (a) Since there are 3 independent random samples here, we do a test for homogeneity. df = 2. (b) Goodness of ﬁt test, df = 2. (c) Goodness of ﬁt test, df = 4. 498 APPENDIX A. EXERCISE SOLUTIONS 7 Inference for numerical data 7.1 (a) df = 6 − 1 = 5, t 5 = 2.02 (column with two tails of 0.10, row with df = 5). (b) df = 21 − 1 = 20, t 20 = 2.53 (column with two tails of 0.02, row with df = 20). (d) df = 11, t 11 = 3.11. (c) df = 28, t 28 = 2.05. 7.3 (a) 0.085, do not reject H0. (b) 0.003, reject H0. (c) 0.438, do not reject H0. (d) 0.042, reject H0. 7.5 The mean is the midpoint: ¯x = 20. Identify the margin of error: M E
= 1.015, then use t 35 = 2.03 n in the formula for margin of error and SE = s/ to identify s = 3. √ 7.7 (a) H0: µ = 8 (New Yorkers sleep 8 hrs per night on average.) HA: µ = 8 (New Yorkers sleep less or more than 8 hrs per night on average.) (b) Independence: The sample is random. The min/max suggest there are no concerning outliers. T = −1.75. df = 25 − 1 = 24. (c) p-value = 0.093. If in fact the true population mean of the amount New Yorkers sleep per night was 8 hours, the probability of getting a random sample of 25 New Yorkers where the average amount of sleep is 7.73 hours per night or less (or 8.27 hours or more) is 0.093. (d) Since p-value > 0.05, do not reject H0. The data do not provide strong evidence that New Yorkers sleep more or less than 8 hours per night on average. (e) Yes, since we did not rejected H0. 7.9 T is either -2.09 or 2.09. Then ¯x is one of the following: pothesis and the null value of 5 was in the t-interval. 7.13 If the sample is large, then the margin of error √ n. We want this value to will be about 1.96 × 100/ be less than 10, which leads to n ≥ 384.16, meaning we need a sample size of at least 385 (round up for sample size calculations!). 7.15 Paired, data are recorded in the same cities at two diﬀerent time points. The temperature in a city at one point is not independent of the temperature in the same city at another time point. 7.17 (a) Since it’s the same students at the beginning and the end of the semester, there is a pairing between the datasets, for a given student their beginning and end of semester grades are dependent. (b) Since the subjects were sampled randomly, each observation in the men’s group does not have a special correspondence with exactly one observation in the other (women’s) group. (c) Since it’s the same subjects at the beginning and the end of the study, there is a pairing between the datasets, for a subject student their beginning and end of semester artery thickness are dependent. (d) Since it’s the same subjects at the beginning and the end of the study, there is a pairing between the datasets, for a subject student their beginning and end of semester weights are dependent. −2.09 = 2.09 = ¯x − 60 8√ 20 ¯x − 60 8√ 20 → ¯x = 56.26 → ¯x = 63.74 √ 7.11 (a) Identify: H0: µ = 5. HA: µ < 5. We’ll use α = 0.05. Choose: 1-sample t-test. Check: This is a random sample, so the observations are independent. To proceed, we assume the distribution of years of piano lessons is approximately normal. Calculate: SE = 2.2/ 20 = 0.4919. The test statistic is T = (4.6 − 5)/SE = −0.81. df = 20 − 1 = 19. The one-tail p-value is about 0.21. Conclude: p-value > α = 0.05, so we do not reject H0. That is, we do not have suﬃciently strong evidence to reject Georgianna’s claim. (b) Identify: estimate average hours a child takes piano lessons in this city with 95% conﬁdence. Choose: 1-sample t-interval. Check: same as in part (a). Calculate: Using SE = 0.4919 and t df =19 = 2.093, the conﬁdence interval is (3.57, 5.63). Conclude: We are 95% conﬁdent that the average number of years a child takes piano lessons in this city is 3.57 to 5.63 years. Do not have evidence that average is not 5 because 5 is in the interval. (c) They agree, since we did not reject the null hy- 7.19 (a) For each observation in one data set, there is exactly one specially corresponding observation in the other data set for the same geographic location. The data are paired. (b) H0 : µdiﬀ = 0 (There is no diﬀerence in average number of days exceeding 90°F in 1948 and 2018 for NOAA stations.) HA : µdiﬀ = 0 (There is a diﬀerence.) (c) Locations were randomly sampled, so independence is reasonable. The sample size is at least 30, so we’re just looking for particularly extreme outliers: none are present (the observation oﬀ left in the histogram would be considered a clear outlier, but not a particularly extreme one). Therefore, the conditions are satisﬁed. 1.23 = 2.36 with (d) SE = 17.2/ degrees of freedom df = 197 − 1 = 196. This leads to a one-tail area of 0.0096 and a p-value of about 0.019. (e) Since the p-value is less than 0.05, we reject H0. The data provide strong evidence that NOAA stations observed more 90°F days in 2018 than in 1948. (f) Type 1 Error, since we may have incorrectly rejected H0. This error would mean that NOAA stations did not actually observe a decrease, but the sample we took just so happened to make it appear that this was the case. (g) No, since we rejected H0, which had a null value of 0. 197 = 1.23. T = 2.9−0 √ 7.21 Identify: we want to estimate the average difference in number of days exceeding 90°F for (2018 1948) with 90% conﬁdence. Choose: 1-sample tinterval with paired data. Check: ndif f = 197 ≥ 30 and the locations are randomly sampled. Calculate: t ≈ 1.65. average SEdif f = 1.23 and df = 196. 2.9 ± 1.65 × 1.23 → (0.87, 4.93). Conclude: We are 90% conﬁdent that there was an increase of 0.87 to 4.93 in the average diﬀerence of days that hit 90°F in 2018 relative to 1948 for NOAA stations. We have evidence that the average diﬀerence of days that hit 90°F increased, because the interval is entirely above 0. 7.23 Identify:H0: µ0.99 = µ1 and HA: µ0.99 = µ1; Let α = 0.05. Choose: 2-sample t-test. Check: Independence: Both samples are random and represent less than 10% of their respective populations. Also, we have no reason to think that the 0.99 carats are not independent of the 1 carat diamonds since they are both sampled randomly. Normal populations: The sample distributions are not very skewed, hence we ﬁnd it reasonable that the underlying population distributions are nearly normal. Calculate: T = −2.82, df = 42.5, p-value = 0.007. Conclude: Since p-value < 0.05, reject H0. The data provide convincing evidence that the average standardized price of 0.99 carats and 1 carat diamonds are diﬀerent. 7.25 (a) Chicken fed linseed weighed an average of 218.75 grams while those fed horsebean weighed an average of 160.20 grams. Both distributions are relatively symmetric with no apparent outliers. There is 499 more variability in the weights of chicken fed linseed. (b) H0 : µls = µhb. HA : µls = µhb. We leave the conditions to you to consider. T = 3.02, df = min(11, 9) = 9 → p-value = 0.014. Since p-value < 0.05, reject H0. The data provide strong evidence that there is a signiﬁcant diﬀerence between the average weights of chickens that were fed linseed and horsebean. (c) Type 1 Error, since we rejected H0. (d) Yes, since p-value > 0.01, we would not have rejected H0. 7.27 H0 : µC = µS. HA : µC = µS. T = 3.27, df = 11 → p-value = 0.007. Since p-value < 0.05, reject H0. The data provide strong evidence that the average weight of chickens that were fed casein is different than the average weight of chickens that were fed soybean (with weights from casein being higher). Since this is a randomized experiment, the observed diﬀerence can be attributed to the diet. 7.29 Let µdif f = µpre−post. H0 : µdif f = 0: Treatment has no eﬀect. HA : µdif f = 0: Treatment has an eﬀect on P.D.T. scores, either positive or negative. Conditions: The subjects are randomly assigned to treatments, so independence within and between groups is satisﬁed. All three sample sizes are smaller than 30, so we look for clear outliers. There is a borderline outlier in the ﬁrst treatment group. Since it is borderline, we will proceed, but we should report this caveat with any results. For all three groups: df = 13. T1 = 1.89 → p-value = 0.081, T2 = 1.35 → p-value = 0.200), T3 = −1.40 → (p-value = 0.185). We do not reject the null hypothesis for any of these groups. As earlier noted, there is some uncertainty about if the method applied is reasonable for the ﬁrst group. 7.31 H0 : µT = µC . HA : µT = µC . T = 2.24, df = 21 → p-value = 0.036. Since p-value < 0.05, reject H0. The data provide strong evidence that the average food consumption by the patients in the treatment and control groups are diﬀerent. Furthermore, the data indicate patients in the distracted eating (treatment) group consume more food than patients in the control group. 7.33 False. While it is true that paired analysis requires equal sample sizes, only having the equal sample sizes isn’t, on its own, suﬃcient for doing a paired test. Paired tests require that there be a special correspondence between each pair of observations in the two groups. 7.35 (a) We are building a distribution of sample statistics, in this case the sample mean. Such a distribution is called a sampling distribution. (b) Because we are dealing with the distribution of sample means, we need to check to see if the Central Limit Theorem applies. Our sample size is greater than 30, and we are told that random sampling is employed. With these conditions met, we expect that the dis- 500 APPENDIX A. EXERCISE SOLUTIONS tribution of the sample mean will be nearly normal and therefore symmetric. (c) Because we are dealing with a sampling distribution, we measure its variabil45 = 2.713. ity with the standard error. SE = 18.2/ (d) The sample means will be more variable with the smaller sample size. √ 7.37 Independence: it is a random sample, so we can assume that the students in this sample are independent of each other with respect to number of exclusive relationships they have been in. Notice that there are no students who have had no exclusive relationships in the sample, which suggests some student responses are likely missing (perhaps only positive values were reported). The sample size is at least 30, and there are no particularly extreme outliers, so the normality condition is reasonable. 90% CI: (2.97, 3.43). We are 90% conﬁdent that undergraduate students have been in 2.97 to 3.43 exclusive relationships, on average. 7.39 First, the hypotheses should be about the population mean (µ), not the sample mean. Second, the null hypothesis should have an equal sign and the alternative hypothesis should be about the null hypothesized value, not the observed sample mean. The correct way to se

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