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Chapter 9 © George Hall /Corbis Estimation Using a Single Sample A fﬁrmative action in university admissions is a controversial topic. Some believe that afﬁrmative action programs are no longer needed, whereas others argue that using race and ethnicity as factors in university admissions is necessary to achieve diverse student populations. To assess public opinion on this issue, investigators conducted a survey of 1013 randomly selected U.S. adults. The results are summarized in the article “Poll Finds Sharp Split on Afﬁrmative Action” (San Luis Obispo Tribune, March 8, 2003). The investigators wanted to use the survey data to estimate the true proportion of U.S. adults who believed that programs that give “advantages and preferences to Blacks, Hispanics, and other minorities in hiring, promotions, and college admissions” should be continued. The methods introduced in this chapter will be used to produce the desired estimate. Because the estimate is based only on a sample rather than on a census of all U.S. adults, it is important that this estimate be constructed in a way that also conveys information about the anticipated accuracy. The objective of inferential statistics is to use sample data to decrease our uncertainty about some characteristic of the corresponding population, such as a population mean m or a population proportion p. One way to accomplish this uses the sample data to arrive at a single number that represents a plausible value for the characteristic of interest. Alternatively, an entire range of plausible values for the characteristic can be reported. These two estimation techniques, point estimation and interval estimation, are introduced in this chapter. Improve your understanding and save time! Visit http://www.thomsonedu.com/login where you will ﬁnd: ■ ■ ■ Step-by-step instructions for MINITAB, Excel, TI-83, SPSS, and JMP Video solutions to selected exercises Data sets available for selected examples and exercises ■ ■ Exam-prep pre-tests that build a Personalized Learning Plan based on your results so that you know exactly what to study Help from a live statistics tutor 24 hours a day 475 476 Chapter 9 ■ Estimation Using a Single Sample ........................................................................................................................................ 9.1 Point Estimation The simplest approach to estimating a population characteristic involves using sample data to compute a single number that can be regarded as a plausible value of the characteristic. For example, sample data might suggest that 1000 hr is a plausible value for m, the true mean lifetime for lightbulbs of a particular brand. In a different setting, a sample survey of students at a particular university might lead to the statement that .41 is a plausible value for p, the true proportion of students who favor a fee for recreational facilities. D E F I N I T I O N A point estimate of a population characteristic is a single number that is based on sample data and represents a plausible value of the characteristic. In the examples just given, 1000 is a point estimate of m and .41 is a point estimate of p. The adjective point reﬂects the fact that the estimate corresponds to a single point on the number line. A point estimate is obtained by ﬁrst selecting an appropriate statistic. The estimate is then the value of the statistic for the given sample. For example, the computed value of the sample mean provides a point estimate of a population mean m. .......................................................................................................................................... Example 9.1 Support for Afﬁrmative Action One of the purposes of the survey on afﬁrmative action described in the chapter introduction was to estimate the proportion of the U.S. population who believe that afﬁrmative action programs should be continued. The article reported that 537 of the 1013 people surveyed believed that afﬁrmative action programs should be continued. Let’s use this information to estimate p, where p is the true proportion of all U.S. adults who favor continuing afﬁrmative action programs. With success identiﬁed as a person who believes that afﬁrmative action programs should continue, p is then just the population proportion of successes. The statistic p number of successes in the sample n which is the sample proportion of successes, is an obvious choice for obtaining a point estimate of p. Based on the reported information, the point estimate of p is p 537 .530 1013 That is, based on this random sample, we estimate that 53% of the adults in the United States believe that afﬁrmative action programs should be continued. ■ For purposes of estimating a population proportion p, there is no obvious alternative to the statistic p. In other situations, such as the one illustrated in Example 9.2, there may be several statistics that can be used to obtain an estimate. 9.1 ■ Point Estimation 477 .......................................................................................................................................... Example 9.2 Internet Use by College Students ● The article “Online Extracurricular Activity” (USA Today, March 13, 2000) reported the results of a study of college students conducted by a polling organization called The Student Monitor. One aspect of computer use examined in this study was the number of hours per week spent on the Internet. Suppose that the following observations represent the number of Internet hours per week reported by 20 college students (these data are compatible with summary values given in the article): 4.00 5.00 5.00 5.25 5.50 6.25 6.25 6.50 6.50 7.00 7.25 7.75 8.00 8.00 8.00 8.25 8.50 8.50 9.50 10.50 A dotplot of the data is shown here: 4 5 6 7 8 9 10 Internet time If a point estimate of m, the true mean Internet time per week for college students, is desired, an obvious choice of a statistic for estimating m is the sample mean x. However, there are other possibilities. We might consider using a trimmed mean or even the sample median, because the data set exhibits some symmetry. (If the corresponding population distribution is symmetric, the population mean m and the population median are equal). The three statistics and the resulting estimates of m calculated from the data are gx 141.50 7.075 n 20 7.0 7.25 sample median 7.125 2 112.5 average of middle 7.031 b 10% trimmed mean a 16 observations 16 sample mean x The estimates of the mean Internet time per week for college students differ somewhat from one another. The choice from among them should depend on which statistic tends, on average, to produce an estimate closest to the true value of m. The following subsection discusses criteria for choosing among competing statistics. ■ ■ Choosing a Statistic for Computing an Estimate ......................................... The point of Example 9.2 is that more than one statistic may be reasonable to use to obtain a point estimate of a speciﬁed population characteristic. Loosely speaking, the statistic used should be one that tends to yield an accurate estimate—that is, an estimate close to the value of the population characteristic. Information about the accuracy of estimation for a particular statistic is provided by the statistic’s sampling distribution. Figure 9.1 displays the sampling distributions of three different statistics. The Step-by-step technology instructions available online ● Data set available online 478 Chapter 9 ■ Estimation Using a Single Sample value of the population characteristic, which we refer to as the true value, is marked on the measurement axis. The distribution in Figure 9.1(a) is that of a statistic unlikely to yield an estimate close to the true value. The distribution is centered to the right of the true value, making it very likely that an estimate (a value of the statistic for a particular sample) will be larger than the true value. If this statistic is used to compute an estimate based on a ﬁrst sample, then another estimate based on a second sample, and another estimate based on a third sample, and so on, the long-run average value of these estimates will exceed the true value. Sampling distributions of three different statistics for estimating a population characteristic. Figure 9.1 True value True value (a) (b) True value (c) The sampling distribution of Figure 9.1(b) is centered at the true value. Thus, although one estimate may be smaller than the true value and another may be larger, when this statistic is used many times over with different samples, there will be no long-run tendency to over- or underestimate the true value. Note that even though the sampling distribution is correctly centered, it spreads out quite a bit about the true value. Because of this, some estimates resulting from the use of this statistic will be far above or far below the true value, even though there is no systematic tendency to underestimate or overestimate the true value. In contrast, the mean value of the statistic with the distribution shown in Figure 9.1(c) is equal to the true value of the population characteristic (implying no systematic error in estimation), and the statistic’s standard deviation is relatively small. Estimates based on this third statistic will almost always be quite close to the true value—certainly more often than estimates resulting from the statistic with the sampling distribution shown in Figure 9.1(b). D E F I N I T I O N A statistic whose mean value is equal to the value of the population characteristic being estimated is said to be an unbiased statistic. A statistic that is not unbiased is said to be biased. As an example of a statistic that is biased, consider using the sample range as an estimate of the population range. Because the range of a population is deﬁned as the difference between the largest value in the population and the smallest value, the range for a sample tends to underestimate the population range. This is true because the 9.1 ■ Point Estimation 479 largest value in a sample must be less than or equal to the largest value in the population and the smallest sample value must be greater than or equal to the smallest value in the population. The sample range equals the population range only if the sample includes both the largest and the smallest values in the population; in all other instances, the sample range is smaller than the population range. Thus, msample range population range, implying bias. Let x1, x2, ..., xn represent the values in a random sample. One of the general results concerning the sampling distribution of x, the sample mean, is that mx m. This result says that the x values from all possible random samples of size n center around m, the population mean. For example, if m 100, the distribution is centered at 100, whereas if m 5200, then the distribution is centered at 5200. Therefore, x is an unbiased statistic for estimating m. Similarly, because the sampling distribution of p is centered at p, it follows that p is an unbiased statistic for estimating a population proportion. Using an unbiased statistic that also has a small standard deviation ensures that there will be no systematic tendency to under- or overestimate the value of the population characteristic and that estimates will almost always be relatively close to the true value. Given a choice between several unbiased statistics that could be used for estimating a population characteristic, the best statistic to use is the one with the smallest standard deviation. Consider the problem of estimating a population mean, m. The obvious choice of statistic for obtaining a point estimate of m is the sample mean, x, an unbiased statistic for this purpose. However, when the population distribution is symmetric, x is not the only choice. Other unbiased statistics for estimating m in this case include the sample median and any trimmed mean (with the same number of observations trimmed from each end of the ordered sample). Which statistic should be used? The following facts may be helpful in making a choice. 1. If the population distribution is normal, then x has a smaller standard deviation than any other unbiased statistic for estimating m. However, in this case, a trimmed mean with a small trimming percentage (such as 10%) performs almost as well as x. 2. When the population distribution is symmetric with heavy tails compared to the normal curve, a trimmed mean is a better statistic than x for estimating m. When the population distribution is unquestionably normal, the choice is clear: Use x to estimate m. However, with a heavy-tailed distribution, a trimmed mean gives protection against one or two outliers in the sample that might otherwise have a large effect on the value of the estimate. Now consider estimating another population characteristic, the population variance s2. The sample variance s2 g 1x x2 2 n1 is a good choice for obtaining a point estimate of the population variance s2. It can be shown that s2 is an unbiased statistic for estimating s2; that is, whatever the value of s2, the sampling distribution of s2 is centered at that value. It is precisely for this rea- 480 Chapter 9 ■ Estimation Using a Single Sample son—to obtain an unbiased statistic—that the divisor (n 1) is used. An alternative statistic is the average squared deviation g 1x x2 2 n which one might think has a more natural divisor than s2. However, the average squared deviation is biased, with its values tending to be smaller, on average, than s2. .......................................................................................................................................... Example 9.3 Airborne Times for Flight 448 ● The Bureau of Transportation Statistics provides data on U.S. airline ﬂights. The airborne times (in minutes) for United Airlines ﬂight 448 from Albuquerque to Denver on 10 randomly selected days between January 1, 2003, and March 31, 2003, are 57 54 55 51 56 48 52 51 59 59 2 For these data gx 542, gx 29,498, n 10, and 1g x2 2 n 15422 2 29.498 10 121.6 2 2 a 1x x2 a x Let s2 denote the true variance in airborne time for ﬂight 448. Using the sample variance s2 to provide a point estimate of s2 yields s2 g 1x x2 2 121.6 13.51 n1 9 Using the average squared deviation (with divisor n 10), the resulting point estimate is g 1x x2 2 121.6 12.16 n 10 Because s2 is an unbiased statistic for estimating s2, most statisticians would recommend using the point estimate 13.51. ■ An obvious choice of a statistic for estimating the population standard deviation s is the sample standard deviation s. For the data given in Example 9.3, s 113.51 3.68 Unfortunately, the fact that s2 is an unbiased statistic for estimating s2 does not imply that s is an unbiased statistic for estimating s. The sample standard deviation tends to underestimate slightly the true value of s. However, unbiasedness is not the only criterion by which a statistic can be judged, and there are other good reasons for using s to estimate s. In what follows, whenever we need to estimate s based on a single random sample, we use the statistic s to obtain a point estimate. ● Data set available online 9.1 ■ Exercises ■ Point Estimation 481 9 . 1 – 9 . 1 0 ........................................................................................................... 9.1 ▼ Three different statistics are being considered for estimating a population characteristic. The sampling distributions of the three statistics are shown in the following illustration: 2331 1852 Statistic III Statistic I True value of population characteristic Which statistic would you recommend? Explain your choice. 9.2 Why is an unbiased statistic generally preferred over a biased statistic for estimating a population characteristic? Does unbiasedness alone guarantee that the estimate will be close to the true value? Explain. Under what circumstances might you choose a biased statistic over an unbiased statistic if two statistics are available for estimating a population characteristic? 9.3 Consumption of fast food is a topic of interest to researchers in the ﬁeld of nutrition. The article “Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children” (Pediatrics [2004]: 112–118) reported that 1720 of those in a random sample of 6212 U.S. children indicated that on a typical day they ate fast food. Estimate p, the proportion of children in the U.S. who eat fast food on a typical day. 9.4 ● Data consistent with summary quantities in the article referenced in Exercise 9.3 on total calorie consumption on a particular day are given for a sample of children who did not eat fast food on that day and for a sample of children who did eat fast food on that day. Assume that it is reasonable to regard these samples as representative of the population of children in the United States. 1918 1777 1009 1765 1730 1827 1469 1648 2053 1506 2143 2669 1981 0000 934 875 2328 2207 2434 1811 2267 1250 2526 2117 1195 0000 Fast Food 2523 890 Statistic II Bold exercises answered in back No Fast Food 1758 1511 a. Use the given information to estimate the mean calorie intake for children in the United States on a day when no fast food is consumed. b. Use the given information to estimate the mean calorie intake for children in the United States on a day when fast food is consumed. c. Use the given information to estimate the produce estimates of the standard deviations of calorie intake for days when no fast food is consumed and for days when fast food is consumed. 9.5 Each person in a random sample of 20 students at a particular university was asked whether he or she is registered to vote. The responses (R registered, N not registered) are given here: R R N R N N R R R N R R R R R N R R R N Use these data to estimate p, the true proportion of all students at the university who are registered to vote. 9.6 A study reported in Newsweek (December 23, 1991) involved a sample of 935 smokers. Each individual received a nicotine patch, which delivers nicotine to the bloodstream but at a much slower rate than cigarettes do. Dosage was decreased to 0 over a 12-week period. Suppose that 245 of the subjects were still not smoking 6 months after treatment (this ﬁgure is consistent with information given in the article). Estimate the percentage of all smokers who, when given this treatment, would refrain from smoking for at least 6 months. 9.7 ● The article “Sensory and Mechanical Assessment of the Quality of Frankfurters” (Journal of Texture Studies [1990]: 395–409) reported the following salt content (percentage by weight) for 10 frankfurters: 2.26 2.11 1.64 1.17 1.64 2.36 1.70 2.10 2.19 2.40 a. Use the given data to produce a point estimate of m, the true mean salt content for frankfurters. ● Data set available online but not required ▼ Video solution available 482 Chapter 9 ■ Estimation Using a Single Sample b. Use the given data to produce a point estimate of s2, the variance of salt content for frankfurters. c. Use the given data to produce an estimate of s, the standard deviation of salt content. Is the statistic you used to produce your estimate unbiased? 9.8 ● The following data on gross efﬁciency (ratio of work accomplished per minute to calorie expenditure per minute) for trained endurance cyclists were given in the article “Cycling Efﬁciency Is Related to the Percentage of Type I Muscle Fibers” (Medicine and Science in Sports and Exercise [1992]: 782–88): 18.3 18.9 19.0 20.9 21.4 20.5 20.1 20.1 20.8 20.5 19.9 20.5 20.6 22.1 21.9 21.2 20.5 22.6 22.6 20.5 20.6 22.1 21.9 21.2 a. Assuming that the distribution of gross energy in the population of all endurance cyclists is normal, give a point estimate of m, the population mean gross efﬁciency. b. Making no assumptions about the shape of the population distribution, estimate the proportion of all such cyclists whose gross efﬁciency is at most 20. 9.9 ● A random sample of n 12 four-year-old red pine trees was selected, and the diameter (in inches) of each tree’s main stem was measured. The resulting observations are as follows: 11.3 10.7 12.4 15.2 10.1 12.1 16.2 10.5 11.4 11.0 10.7 12.0 10.1 12.1 16.2 10.5 a. Compute a point estimate of s, the population standard deviation of main stem diameter. What statistic did you use to obtain your estimate? b. Making no assumptions about the shape of the population distribution of diameters, give a point estimate for the Bold exercises answered in back population median diameter. What statistic did you use to obtain the estimate? c. Suppose that the population distribution of diameter is symmetric but with heavier tails than the normal distribution. Give a point estimate of the population mean diameter based on a statistic that gives some protection against the presence of outliers in the sample. What statistic did you use? d. Suppose that the diameter distribution is normal. Then the 90th percentile of the diameter distribution is m 1.28s (so 90% of all trees have diameters less than this value). Compute a point estimate for this percentile. (Hint: First compute an estimate of m in this case; then use it along with your estimate of s from Part (a).) 9.10 ● A random sample of 10 houses in a particular area, each of which is heated with natural gas, is selected, and the amount of gas (in therms) used during the month of January is determined for each house. The resulting observations are as follows: 103 156 118 89 125 147 122 109 138 99 a. Let mJ denote the average gas usage during January by all houses in this area. Compute a point estimate of mJ. b. Suppose that 10,000 houses in this area use natural gas for heating. Let t denote the total amount of gas used by all of these houses during January. Estimate t using the data of Part (a). What statistic did you use in computing your estimate? c. Use the data in Part (a) to estimate p, the proportion of all houses that used at least 100 therms. d. Give a point estimate of the population median usage based on the sample of Part (a). Which statistic did you use? ● Data set available online but not required ▼ Video solution available ........................................................................................................................................ 9.2 Large-Sample Confidence Interval for a Population Proportion In Section 9.1 we saw how to use a statistic to produce a point estimate of a population characteristic. The value of a point estimate depends on which sample, out of all the possible samples, happens to be selected. Different samples usually yield different estimates as a result of chance differences from one sample to another. Because of sampling variability, rarely is the point estimate from a sample exactly equal to the true value of the population characteristic. We hope that the chosen statistic produces an estimate that is close, on average, to the true value. Although a point estimate may 9.2 ■ Large-Sample Conﬁdence Interval for a Population Proportion 483 represent our best single-number guess for the value of the population characteristic, it is not the only plausible value. These considerations suggest the need to indicate in some way a range of plausible values for the population characteristic. A point estimate by itself does not provide this information. As an alternative to a point estimate, we can report an interval of reasonable values for the population characteristic based on the sample data. For example, we might be conﬁdent that for all calls made from AT&T pay phones, the proportion p of calls that are billed to a credit card is in the interval from .53 to .57. The narrowness of this interval implies that we have rather precise information about the value of p. If, with the same high degree of conﬁdence, we could only state that p was between .32 and .74, it would be clear that we had relatively imprecise knowledge of the value of p. D E F I N I T I O N A conﬁdence interval (CI) for a population characteristic is an interval of plausible values for the characteristic. It is constructed so that, with a chosen degree of conﬁdence, the value of the characteristic will be captured between the lower and upper endpoints of the interval. Associated with each conﬁdence interval is a conﬁdence level. The conﬁdence level provides information on how much “conﬁdence” we can have in the method used to construct the interval estimate (not our conﬁdence in any one particular interval). Usual choices for conﬁdence levels are 90%, 95%, and 99%, although other levels are also possible. If we were to construct a 95% conﬁdence interval using the technique to be described shortly, we would be using a method that is “successful” 95% of the time. That is, if this method was used to generate an interval estimate over and over again with different samples, in the long run 95% of the resulting intervals would capture the true value of the characteristic being estimated. Similarly, a 99% conﬁdence interval is one that is constructed using a method that is, in the long run, successful in capturing the true value of the population characteristic 99% of the time. D E F I N I T I O N The conﬁdence level associated with a conﬁdence interval estimate is the success rate of the method used to construct the interval. Many factors inﬂuence the choice of conﬁdence level. These factors will be discussed after we develop the method for constructing conﬁdence intervals. We ﬁrst consider a large-sample conﬁdence interval for a population proportion p. Often an investigator wishes to make an inference about the proportion of individuals or objects in a population that possess a particular property of interest. For example, a university administrator might be interested in the proportion of students who prefer a new web-based computer registration system to the previous registration method. In a different setting, a quality control engineer might be concerned about the proportion of defective parts manufactured using a particular process. 484 Chapter 9 ■ Estimation Using a Single Sample Let p be the proportion of the population that possess the property of interest. Previously, we used the sample proportion p number in the sample that possess the property of interest n to calculate a point estimate of p. We can also use p to form a conﬁdence interval for p. Although a small-sample conﬁdence interval for p can be obtained, our focus is on the large-sample case. The justiﬁcation for the large-sample interval rests on properties of the sampling distribution of the statistic p: 1. The sampling distribution of p is centered at p; that is, mp p. Therefore, p is an unbiased statistic for estimating p. p11 p 2 2. The standard deviation of p is sp n B 3. As long as n is large (np 10 and n(1 p) 10) and the sample size is less than 10% of the population size, the sampling distribution of p is well approximated by a normal curve. The accompanying box summarizes these properties. When n is large, the statistic p has a sampling distribution that is approximately normal with p11 p2 . mean p and standard deviation n B The development of a conﬁdence interval for p is easier to follow if we select a particular conﬁdence level. For a conﬁdence level of 95%, Appendix Table 2, the table of standard normal (z) curve areas, can be used to determine a value z* such that a central area of .95 falls between z* and z*. In this case, the remaining area of .05 is divided equally between the two tails, as shown in Figure 9.2. The total area to the left of the desired z* is .975 (.95 central area .025 area below z*). By locating .9750 in the body of Appendix Table 2, we ﬁnd that the corresponding z critical value is 1.96. Capturing a central area of .95 under the z curve. Figure 9.2 Central area .95 Upper-tail area .025 Lower-tail area .025 z* 1.96 z* 1.96 Generalizing this result to normal distributions other than the standard normal distribution tells us that for any normal distribution, about 95% of the values are within 1.96 standard deviations of the mean. Because (for large random samples) the sampling distribution of p is approximately normal with mean mp p and standard devip11 p 2 , we get the following result. ation sp n B 9.2 ■ Large-Sample Conﬁdence Interval for a Population Proportion 485 When n is large, approximately 95% of all samples of size n will result in a value of p that is p11 p 2 within 1.96sp 1.96 of the true population proportion p. n B If p is within 1.96 B p 1.96 B p11 p2 of p, this means the interval n p11 p 2 n to p 1.96 B p11 p2 n will capture p (and this will happen for 95% of all possible samples). However, if p is p11 p 2 farther away from p than 1.96 (which will happen for about 5% of all n B possible samples), the interval will not include the true value of p. This is shown in Figure 9.3. The population proportion p is captured in the interval from p11 p2 p 1.96 to n B Approximate sampling distribution of p Figure 9.3 π p11 p2 p 1.96 when n B p11 p 2 n B p is within 1.96 1.96 π(1 π) 1.96 ––––––– n π(1 π) ––––––– n of p. p 1.96 π(1 π) ––––––_– n interval captures π ) ( p p 1.96 π(1 π) ––––––_– n ) ( p 1.96 π(1 π) ––––––_– n p p 1.96 π(1 π) ––––––_– n interval does not capture π Because p is within 1.96sp of p 95% of the time, this implies that in repeated sampling, 95% of the time the interval p11 p 2 n B p 1.96 to p 1.96 B p11 p2 n will contain p. p11 p2 must be estimated. As long as the sample size n B p11 p2 p11 p2 . is large, the value of can be used in place of n n B B Since p is unknown, 486 Chapter 9 ■ Estimation Using a Single Sample When n is large, a 95% conﬁdence interval for p is p11 p2 p11 p2 , p 1.96 b n n B B a p 1.96 An abbreviated formula for the interval is p11 p 2 n B p 1.96 p11 p2 p11 p 2 gives the upper endpoint of the interval and p 1.96 n n B B where p 1.96 gives the lower endpoint of the interval. The interval can be used as long as 1. np 10 and n(1 p) 10, 2. the sample size is less than 10% of the population size if sampling is without replacement, 3. the sample can be regarded as a random sample from the population of interest. .......................................................................................................................................... Example 9.4 Afﬁrmative Action Continued Let’s return to the information from the survey on attitudes toward afﬁrmative action (see the chapter introduction and Example 9.1): Total number of people surveyed: 1013 Number who believe afﬁrmative action programs should be continued: 537 A point estimate of p, the true proportion of U.S. adults who believe that afﬁrmative action programs should be continued, is p 537 .530 1013 Because np (1013)(.53) 537 10 and n(1 p) (1013)(.47) 476 10 and the adults in the sample were randomly selected from a large population, the large-sample interval can be used. For a 95% conﬁdence level, a conﬁdence interval for p is p 1.96 B 1.530 2 1.470 2 p11 p2 5.30 1.96 n 1013 B .530 11.962 1.016 2 .530 .031 1.499, .5612 Based on this sample, we can be 95% conﬁdent that p, the true proportion who believe that afﬁrmative action programs should continue, is between .499 and .561. We 9.2 ■ Large-Sample Conﬁdence Interval for a Population Proportion 487 used a method to construct this estimate that in the long run will successfully capture the true value of p 95% of the time. ■ π The 95% conﬁdence interval for p calculated in Example 9.4 is (.499, .561). It is tempting to say that there is a “probability” of .95 that p is between .499 and .561. Do not yield to this temptation! The 95% refers to the percentage of all possible samples resulting in an interval that includes p. In other words, if we take sample after sample from the population and use each one separately to compute a 95% conﬁdence interval, in the long run roughly 95% of these intervals will capture p. Figure 9.4 illustrates this concept for intervals generated from 100 different random samples; 93 of the intervals include p, whereas 7 do not. Any speciﬁc interval, and our interval (.499, .561) in particular, either includes p or it does not (remember, the value of p is ﬁxed but not known to us). We cannot make a chance (probability) statement concerning this particular interval. The conﬁdence level 95% refers to the method used to construct the interval rather than to any particular interval, such as the one we obtained. The formula given for a 95% conﬁdence interval can easily be adapted for other conﬁdence levels. The choice of a 95% conﬁdence level led to the use of the z value 1.96 (chosen to capture a central area of .95 under the standard normal curve) in the formula. Any other conﬁdence level can be obtained by using an appropriate z critical value in place of 1.96. For example, suppose that we wanted to achieve a conﬁdence level of 99%. To obtain a central area of .99, the approximate z critical value would have a cumulative area (area to the left) of .995, as illustrated in Figure 9.5. From Appendix Table 2, we ﬁnd that the corresponding z critical value is 2.58. A 99% conﬁdence interval for p is then obtained by using 2.58 in place of 1.96 in the formula for the 95% conﬁdence interval. * * * * * * * π One hundred 95% conﬁdence intervals for p computed from 100 different random samples (asterisks identify intervals that do not include p). Figure 9.4 Finding the z critical value for a 99% conﬁdence level. Figure 9.5 upper-tail area lower-tail area .01 .005 2 .01 .005 2 central area .99 z* z* Cumulative area .995 488 Chapter 9 ■ Estimation Using a Single Sample The Large-Sample Confidence Interval for P The general formula for a conﬁdence interval for a population proportion p when 1. p is the sample proportion from a random sample, and 2. the sample size n is large (np 10 and n(1 p) 10), and 3. if the sample is selected without replacement, the sample size is small relative to the population size (n is at most 10% of the population size) is p 1z critical value2 p11 p2 n B The desired conﬁdence level determines which z critical value is used. The three most commonly used conﬁdence levels, 90%, 95%, and 99%, use z critical values 1.645, 1.96, and 2.58, respectively. Note: This interval is not appropriate for small samples. It is possible to construct a conﬁdence interval in the small-sample case, but this is beyond the scope of this textbook. Why settle for 95% conﬁdence when 99% conﬁdence is possible? Because the higher conﬁdence level comes with a price tag. The resulting interval is wider than the p11 p2 b whereas the 99% 95% interval. The width of the 95% interval is 2 a 1.96 n B , p11 p2 b . The higher reliability of the 99% interval interval has width 2 a 2.58 n B (where “reliability” is speciﬁed by the conﬁdence level) entails a loss in precision (as indicated by the wider interval). In the opinion of many investigators, a 95% interval is a reasonable compromise between reliability and precision. .......................................................................................................................................... Example 9.5 Dangerous Driving The article “Nine Out of Ten Drivers Admit in Survey to Having Done Something Dangerous” (Knight Ridder Newspapers, July 8, 2005) reported the results of a survey of 1100 drivers. Of those surveyed, 990 admitted to careless or aggressive driving during the previous six months. Assuming that it is reasonable to regard this sample of 1100 as representative of the population of drivers, we can use this information to construct an estimate of p, the true proportion of drivers who have engaged in careless or aggressive driving in the past six months. For this sample p 990 .900 1100 Because np 990 and n(1 p) 110 are both greater than or equal to 10, the sample size is large enough to use the formula for a large-sample conﬁdence interval. A 90% conﬁdence interval for p is then Step-by-step technology instructions available online 9.2 ■ Large-Sample Conﬁdence Interval for a Population Proportion 489 © The Image Bank / Wilfried Krecichwest /Getty Images p 1z critical value2 p11 p2 1.9002 1.1002 .900 1.645 n B 1100 B .900 11.6452 1.0092 .900 .015 1.885, .915 2 Based on these sample data, we can be 90% conﬁdent that the true proportion of drivers who have engaged in careless or aggressive driving in the past six months is between .885 and .915. We have used a method to construct this interval estimate that has a 10% error rate. ■ The conﬁdence level for the z conﬁdence interval for a population proportion is only approximate. That is, when we report a 95% conﬁdence interval for a population proportion, the 95% conﬁdence level implies that we have used a method that produces an interval that includes the actual value of the population proportion 95% of the time in repeated sampling. In fact, because the normal distribution is only an approximation to the sampling distribution of p, the true conﬁdence level may differ somewhat from the reported value. If the conditions (1) np 10 and n(1 p) 10 and (2) n is at most 10% of the population size if sampling without replacement are met, the normal approximation is reasonable and the actual conﬁdence level is usually quite close to the reported level; this is why it is important to check these conditions before computing and reporting a z conﬁdence interval for a population proportion. What should you do if these conditions are not met? If the sample size is too small to satisfy the np and n(1 p) 10 condition, an alternative procedure can be used. onsult a statistician or a more advanced textbook in this case. If the condition that the sample size is less than 10% of the population size when sampling without replacement is not satisﬁed, the z conﬁdence interval tends to be conservative (i.e., it tends to be wider than is necessary to achieve the desired conﬁdence level). In this case, a ﬁnite population correction factor can be used to obtain a more precise interval. Again, it would be wise to consult a statistician or a more advanced textbook. ■ An Alternative to the Large-Sample z Interval .............................................. Investigators have shown that in some instances, even when the sample size conditions of the large-sample z conﬁdence interval for a population proportion are met, the actual conﬁdence level associated with the method may be noticeably different from the reported conﬁdence level. A modiﬁed interval that has an actual conﬁdence level that is closer to the reported conﬁdence level is based on a modiﬁed sample proportion, pmod, the proportion of successes after adding two successes and two failures to the sample. Then pmod is pmod number of successes 2 n4 pmod is used in place of p in the usual conﬁdence interval formula. Properties of this modiﬁed conﬁdence interval are investigated in Activity 9.2 at the end of the chapter. 490 Chapter 9 ■ ■ Estimation Using a Single Sample General Form of a Conﬁdence Interval ........................................................ Many conﬁdence intervals have the same general form as the large-sample z interval for p just considered. We started with a statistic p, from which a point estimate for p was obtained. The standard deviation of this statistic is 2p11 p2/n. This resulted in a conﬁdence interval of the form a point estimate using standard deviation b 1critical value2 a b a speciﬁed statistic of the statistic Because p was unknown, we estimated the standard deviation of the statistic by 2p11 p 2/n, which yielded the interval estimated point estimate using a b 1critical value2 ° standard deviation ¢ a speciﬁed statistic of the statistic For a population characteristic other than p, a statistic for estimating the characteristic is selected. Then (drawing on statistical theory) a formula for the standard deviation of the statistic is given. In practice, it is almost always necessary to estimate this standard deviation (using something analogous to 2p11 p 2/n rather than 2p11 p2/n, for example), so that the interval estimated point estimate using a b 1critical value2 ° standard deviation ¢ a speciﬁed statistic of the statistic is the prototype conﬁdence interval. It is common practice to refer to both the standard deviation of a statistic and the estimated standard deviation of a statistic as the standard error. In this textbook, when we use the term standard error, we mean the estimated standard deviation of a statistic. D E F I N I T I O N The standard error of a statistic is the estimated standard deviation of the statistic. The 95% conﬁdence interval for p is based on the fact that, for approximately p11 p2 95% of all random samples, p is within 1.96 of p. The quantity n B p11 p2 is sometimes called the bound on the error of estimation associated 1.96 n B with a 95% conﬁdence level—we have 95% conﬁdence that the point estimate p is no farther than this quantity from p. 9.2 D ■ Large-Sample Conﬁdence Interval for a Population Proportion 491 E F I N I T I O N If the sampling distribution of a statistic is (at least approximately) normal, the bound on error of estimation, B, associated with a 95% conﬁdence interval is (1.96) # (standard error of the statistic). ■ Choosing the Sample Size ............................................................................... Before collecting any data, an investigator may wish to determine a sample size for which a particular value of the bound on the error is achieved. For example, with p representing the true proportion of students at a university who purchase textbooks over the Internet, the objective of an investigation may be to estimate p to within .05 with 95% conﬁdence. The value of n necessary to achieve this is obtained by equating p11 p 2 .05 to 1.96 and solving for n. n B In general, suppose that we wish to estimate p to within an amount B (the speciﬁed bound on the error of estimation) with 95% conﬁdence. Finding the necessary sample size requires solving the equation p11 p2 n B B 1.96 Solving this equation for n results in n p11 p 2 a 1.96 2 b B Unfortunately, the use of this formula requires the value of p, which is unknown. One possible way to proceed is to carry out a preliminary study and use the resulting data to get a rough estimate of p. In other cases, prior knowledge may suggest a reasonable estimate of p. If there is no reasonable basis for estimating p and a preliminary study is not feasible, a conservative solution follows from the observation that p(1 p) is never larger than .25 (its value when p .5). Replacing p(1 p) with .25, the maximum value, yields n .25 a 1.96 2 b B Using this formula to obtain n gives us a sample size for which we can be 95% conﬁdent that p will be within B of p, no matter what the value of p. The sample size required to estimate a population proportion p to within an amount B with 95% conﬁdence is n p11 p2 a 1.96 2 b B The value of p may be estimated using prior information. In the absence of any such information, using p .5 in this formula gives a conservatively large value for the required sample size (this value of p gives a larger n than would any other value). Chapter 9 492 ■ Estimation Using a Single Sample .......................................................................................................................................... Example 9.6 Snifﬁng Out Cancer Researchers have found biochemical markers of cancer in the exhaled breath of cancer patients, but chemical analysis of breath specimens has not yet proven effective in clinical diagnosis. The authors of the paper “Diagnostic Accuracy of Canine Scent Detection in Early- and Late-Stage Lung and Breast Cancers” (Integrative Cancer Therapies [2006]: 1–10) describe a study to investigate whether dogs can be trained to identify the presence or absence of cancer by snifﬁng breath specimens. Suppose we want to collect data that would allow us to estimate the long-run proportion of accurate identiﬁcations for a particular dog that has completed training. The dog has been trained to lie down when presented with a breath specimen from a cancer patient and to remain standing when presented with a specimen from a person who does not have cancer. How many different breath specimens should be used if we want to estimate the long-run proportion of correct identiﬁcations for this dog to within .05 with 95% conﬁdence? Using a conservative value of p .5 in the formula for required sample size gives n p11 p2 a 1.96 2 1.96 2 b .25 a b 384.16 B .05 Thus, a sample of at least 385 breath specimens should be used. Note that in sample size calculations, we always round up. ■ ■ E x e r c i s e s 9 . 1 1 – 9 . 2 9 .......................................................................................................... 9.11 ▼ For each of the following choices, explain which would result in a wider large-sample conﬁdence interval for p: a. 90% conﬁdence level or 95% conﬁdence level b. n 100 or n 400 9.12 The formula used to compute a large-sample conﬁdence interval for p is p 1z critical value 2 p11 p 2 n B What is the appropriate z critical value for each of the following conﬁdence levels? a. 95% d. 80% b. 90% e. 85% c. 99% Bold exercises answered in back 9.13 The use of the interval p 1z critical value 2 p11 p 2 n B requires a large sample. For each of the following combinations of n and p, indicate whether the given interval would be appropriate. a. n 50 and p .30 b. n 50 and p .05 c. n 15 and p .45 d. n 100 and p .01 e. n 100 and p .70 f. n 40 and p .25 g. n 60 and p .25 h. n 80 and p .10 ● Data set available online but not required ▼ Video solution available 9.2 ■ Large-Sample Conﬁdence Interval for a Population Proportion 493 9.14 Discuss how each of the following factors affects the width of the conﬁdence interval for p: a. The conﬁdence level b. The sample size c. The value of p 9.15 According to an AP-Ipsos poll (June 15, 2005), 42% of 1001 randomly selected adult Americans made plans in May 2005 based on a weather report that turned out to be wrong. a. Construct and interpret a 99% conﬁdence interval for the proportion of Americans who made plans in May 2005 based on an incorrect weather report. b. Do you think it is reasonable to generalize this estimate to other months of the year? Explain. 9.16 The article “Students Increasingly Turn to Credit Cards” (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college freshmen and 48% of college seniors carry a credit card balance from month to month. Suppose that the reported percentages were based on random samples of 1000 college freshmen and 1000 college seniors. a. Construct a 90% conﬁdence interval for the proportion of college freshmen who carry a credit card balance from month to month. b. Construct a 90% conﬁdence interval for the proportion of college seniors who carry a credit card balance from month to month. c. Explain why the two 90% conﬁdence intervals from Parts (a) and (b) are not the same width. 9.17 ▼ The article “CSI Effect Has Juries Wanting More Evidence” (USA Today, August 5, 2004) examines how the popularity of crime-scene investigation television shows is inﬂuencing jurors’ expectations of what evidence should be produced at a trial. In a survey of 500 potential jurors, one study found that 350 were regular watchers of at least one crime-scene forensics television series. a. Assuming that it is reasonable to regard this sample of 500 potential jurors as representative of potential jurors in the United States, use the given information to construct and interpret a 95% conﬁdence interval for the true proportion of potential jurors who regularly watch at least one crime-scene investigation series. b. Would a 99% conﬁdence interval be wider or narrower than the 95% conﬁdence interval from Part (a)? Bold exercises answered in back 9.18 In a survey of 1000 randomly selected adults in the United States, participants were asked what their most favorite and what their least favorite subject was when they were in school (Associated Press, August 17, 2005). In what might seem like a contradiction, math was chosen more often than any other subject in both categories! Math was chosen by 230 of the 1000 as the favorite subject, and it was also chosen by 370 of the 1000 as the least favorite subject. a. Construct a 95% conﬁdence interval for the proportion of U.S. adults for whom math was the favorite subject in school. b. Construct a 95% conﬁdence interval for the proportion of U.S. adults for whom math was the least favorite subject. 9.19 The report “2005 Electronic Monitoring & Surveillance Survey: Many Companies Monitoring, Recording, Videotaping—and Firing—Employees” (American Management Association, 2005) summarized the results of a survey of 526 U.S. businesses. The report stated that 137 of the 526 businesses had ﬁred workers for misuse of the Internet and 131 had ﬁred workers for email misuse. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of businesses in the United States. a. Construct and interpret a 95% conﬁdence interval for the proportion of U.S. businesses that have ﬁred workers for misuse of the Internet. b. What are two reasons why a 90% conﬁdence interval for the proportion of U.S. businesses that have ﬁred workers for misuse of email would be narrower than the 95% conﬁdence interval computed in Part (a). 9.20 In an AP-AOL sports poll (Associated Press, December 18, 2005), 394 of 1000 randomly selected U.S. adults indicated that they considered themselves to be baseball fans. Of the 394 baseball fans, 272 stated that they thought the designated hitter rule should either be expanded to both baseball leagues or eliminated. a. Construct a 95% conﬁdence interval for the proportion of U.S. adults that consider themselves to be baseball fans. b. Construct a 95% conﬁdence interval for the proportion of those who consider themselves to be baseball fans that think the designated hitter rule should be expanded to both leagues or eliminated. c. Explain why the conﬁdence intervals of Parts (a) and ● Data set available online but not required ▼ Video solution available 494 Chapter 9 ■ Estimation Using a Single Sample (b) are not the same width even though they both have a conﬁdence level of 95%. 9.21 The article “Viewers Speak Out Against Reality TV” (Associated Press, September 12, 2005) included the following statement: “Few people believe there’s much reality in reality TV: a total of 82 percent said the shows are either ‘totally made up’ or ‘mostly distorted’.” This statement was based on a survey of 1002 randomly selected adults. Compute and interpret a bound on the error of estimation for the reported percentage. 9.22 One thousand randomly selected adult Americans participated in a survey conducted by the Associated Press (June, 2006). When asked “Do you think it is sometimes justiﬁed to lie or do you think lying is never justiﬁed?” 52% responded that lying was never justiﬁed. When asked about lying to avoid hurting someone’s feelings, 650 responded that this was often or sometimes OK. a. Construct a 90% conﬁdence interval for the proportion of adult Americans who think lying is never justiﬁed. b. Construct a 90% conﬁdence interval for the proportion of adult American who think that it is often or sometimes OK to lie to avoid hurting someone’s feelings. c. Based on the conﬁdence intervals from Parts (a) and (b), comment on the apparent inconsistency in the responses given by the individuals in this sample. ferent countries (San Luis Obispo Tribune, November 21, 2002). The society found that 10% of the participants could not identify their own country on a blank world map. a. Construct a 90% conﬁdence interval for the proportion who can identify their own country on a blank world map. b. What assumptions are necessary for the conﬁdence interval in Part (a) to be valid? c. To what population would it be reasonable to generalize the conﬁdence interval estimate from Part (a)? 9.25 ▼ “Tongue Piercing May Speed Tooth Loss, Researchers Say” is the headline of an article that appeared in the San Luis Obispo Tribune (June 5, 2002). The article describes a study of 52 young adults with pierced tongues. The researchers found receding gums, which can lead to tooth loss, in 18 of the participants. Construct a 95% conﬁdence interval for the proportion of young adults with pierced tongues who have receding gums. What assumptions must be made for use of the z conﬁdence interval to be appropriate? 9.26 USA Today (October 14, 2002) reported that 36% of adult drivers admit that they often or sometimes talk on a cell phone when driving. This estimate was based on data from a sample of 1004 adult drivers, and a bound on the error of estimation of 3.1% was reported. Explain how the given bound on the error can be justiﬁed. 9.23 The article “Doctors Cite Burnout in Mistakes” (San Luis Obispo Tribune, March 5, 2002) reported that many doctors who are completing their residency have ﬁnancial struggles that could interfere with training. In a sample of 115 residents, 38 reported that they worked moonlighting jobs and 22 reported a credit card debt of more than $3000. Suppose that it is reasonable to consider this sample of 115 as a random sample of all medical residents in the United States. a. Construct and interpret a 95% conﬁdence interval for the proportion of U.S. medical residents who work moonlighting jobs. b. Construct and interpret a 90% conﬁdence interval for the proportion of U.S. medical residents who have a credit card debt of more than $3000. c. Give two reasons why the conﬁdence interval in Part (a) is wider than the conﬁdence interval in Part (b). 9.28 In a study of 1710 schoolchildren in Australia (Herald Sun, October 27, 1994), 1060 children indicated that they normally watch TV before school in the morning. (Interestingly, only 35% of the parents said their children watched TV before school!) Construct a 95% conﬁdence interval for the true proportion of Australian children who say they watch TV before school. What assumption about the sample must be true for the method used to construct the interval to be valid? 9.24 The National Geographic Society conducted a study that included 3000 respondents, age 18 to 24, in nine dif- 9.29 ▼ A consumer group is interested in estimating the proportion of packages of ground beef sold at a particular Bold exercises answered in back 9.27 The Gallup Organization conducts an annual survey on crime. It was reported that 25% of all households experienced some sort of crime during the past year. This estimate was based on a sample of 1002 randomly selected adults. The report states, “One can say with 95% conﬁdence that the margin of sampling error is 3 percentage points.” Explain how this statement can be justiﬁed. ● Data set available online but not required ▼ Video solution available 9.3 store that have an actual fat content exceeding the fat content stated on the label. How many packages of ground Bold exercises answered in back ■ Confidence Interval for a Population Mean 495 beef should be tested to estimate this proportion to within .05 with 95% conﬁdence? ● Data set available online but not required ▼ Video solution available ........................................................................................................................................ 9.3 Confidence Interval for a Population Mean In this section, we consider how to use information from a random sample to construct a conﬁdence interval estimate of a population mean, m. We begin by considering the case in which (1) s, the population standard deviation, is known (not realistic, but we will see shortly how to handle the more realistic situation where s is unknown) and (2) the sample size n is large enough for the Central Limit Theorem to apply. In this case, the following three properties about the sampling distribution of x hold: 1. The sampling distribution of x is centered at m, so x is an unbiased statistic for estimating m 1mx m2. s . 2. The standard deviation of x is sx 1n 3. As long as n is large (generally n 30), the sampling distribution of x is approximately normal, even when the population distribution itself is not normal. The same reasoning that was used to develop the large-sample conﬁdence interval for a population proportion p can be used to obtain a conﬁdence interval estimate for m. The One-Sample z Confidence Interval for M The general formula for a conﬁdence interval for a population mean m when 1. x is the sample mean from a random sample, 2. the sample size n is large (generally n 30), and 3. s, the population standard deviation, is known is x 1z critical value2 a s b 1n .......................................................................................................................................... Example 9.7 Cosmic Radiation Cosmic radiation levels rise with increasing altitude, prompting researchers to consider how pilots and ﬂight crews might be affected by increased exposure to cosmic radiation. The paper “Estimated Cosmic Radiation Doses for Flight Personnel” (Space Medicine and Medical Engineering [2002]: 265–269) reported a mean annual cosmic radiation dose of 219 mrems for a sample of ﬂight personnel of Xinjiang Airlines. Suppose that this mean was based on a random sample of 100 ﬂight crew members. 496 Chapter 9 ■ Estimation Using a Single Sample Let m denote the true mean annual cosmic radiation exposure for Xinjiang Airlines ﬂight crew members. Although s, the true population standard deviation, is not usually known, suppose for illustrative purposes that s 35 mrem is known. Because the sample size is large and s is known, a 95% conﬁdence interval for m is x 1z critical value2 a s 35 b 219 11.962 a b 1n 1100 219 6.86 1212.14, 225.862 Based on this sample, plausible values of m, the true mean annual cosmic radiation exposure for Xinjaing Airlines ﬂight crew members, are between 212.14 and 225.86 mrem. A 95% conﬁdence level is associated with the method used to produce this interval estimate. ■ The conﬁdence interval just introduced is appropriate when s is known and n is large, and it can be used regardless of the shape of the population distribution. This is because this conﬁdence interval is based on the Central Limit Theorem, which says that when n is sufﬁciently large, the sampling distribution of x is approximately normal for any population distribution. When n is small, the Central Limit Theorem cannot be used to justify the normality of the x sampling distribution, so the z conﬁdence interval can not be used. One way to proceed in the small-sample case is to make a speciﬁc assumption about the shape of the population distribution and then to use a method of estimation that is valid only under this assumption. The one instance where this is easy to do is when it is reasonable to believe that the population distribution is normal in shape. Recall that for a normal population distribution the sampling distribution of x is normal even for small sample sizes. So, if n is small but the population distribution is normal, the same conﬁdence interval formula just introduced can still be used. If n is small (generally n 30) but it is reasonable to believe that the distribution of values in the population is normal, a conﬁdence interval for m (when s is known) is x 1z critical value2 a s b 1n There are several ways that sample data can be used to assess the plausibility of normality. The two most common ways are to look at a normal probability plot of the sample data (looking for a plot that is reasonably straight) and to construct a boxplot of the data (looking for approximate symmetry and no outliers). ■ Conﬁdence Interval for m When s Is Unknown ............................................. The conﬁdence intervals just developed have an obvious drawback: To compute the interval endpoints, s must be known. Unfortunately, this is rarely the case in practice. We now turn our attention to the situation when s is unknown. The development of the conﬁdence interval in this instance depends on the assumption that the population 9.3 ■ Confidence Interval for a Population Mean 497 distribution is normal. This assumption is not critical if the sample size is large, but it is important when the sample size is small. To understand the derivation of this conﬁdence interval, it is instructive to begin by taking another look at the previous 95% conﬁdence interval. We know that mx m s and sx . Also, when the population distribution is normal, the x distribution is 1n normal. These facts imply that the standardized variable z xm s 1n has approximately a standard normal distribution. Because the interval from 1.96 to 1.96 captures an area of .95 under the z curve, approximately 95% of all samples result in an x value that satisﬁes 1.96 xm 1.96 s 1n Manipulating these inequalities to isolate m in the middle results in the equivalent inequalities: x 1.96 a s s b m x 1.96 a b 1n 1n s b is the lower endpoint of the 95% large-sample conﬁdence 1n s b is the upper endpoint. interval for m, and x 1.96 a 1n If s is unknown, we must use the sample data to estimate s. If we use the sample standard deviation as our estimate, the result is a different standardized variable denoted by t: The term x 1.96 a t xm s 1n The value of s may not be all that close to s, especially when n is small. As a consequence, the use of s in place of s introduces extra variability, so the distribution of t is more spread out than the standard normal (z) distribution. (The value of z varies from sample to sample, because different samples generally result in different x values. There is even more variability in t, because different samples may result in different values of both x and s.) To develop an appropriate conﬁdence interval, we must investigate the probability distribution of the standardized variable t for a sample from a normal population. This requires that we ﬁrst learn about probability distributions called t distributions. ■ t Distributions ...................................................................................................... Just as there are many different normal distributions, there are also many different t distributions. Although normal distributions are distinguished from one another by 498 Chapter 9 ■ Estimation Using a Single Sample their mean m and standard deviation s, t distributions are distinguished by a positive whole number called the number of degrees of freedom (df). There is a t distribution with 1 df, another with 2 df, and so on. I m p o r t a n t P ro p e r t i e s o f t D i s t r i b u t i o n s 1. The t distribution corresponding to any ﬁxed number of degrees of freedom is bell shaped and centered at zero (just like the standard normal (z) distribution). 2. Each t distribution is more spread out than the standard normal (z) distribution. 3. As the number of degrees of freedom increases, the spread of the corresponding t distribution decreases. 4. As the number of degrees of freedom increases, the corresponding sequence of t distributions approaches the standard normal (z) distribution. z curve The properties discussed in the preceding box are illustrated in Figure 9.6, which shows two t curves along with the z curve. t curve for 4 df Appendix Table 3 gives selected critical values for various t distributions. The central areas for which values are tabulated are .80, .90, .95, .98, .99, .998, and .999. To ﬁnd a particular critical value, go down the left margin of the table to the row labeled with the desired number of degrees of freedom. Then move over in that row to the column headed by the desired cen3 2 1 0 1 2 3 tral area. For example, the value in the 12-df row under the column corresponding to central area .95 is 2.18, so 95% of the F i g u r e 9 . 6 Comparison of the z curve and area under the t curve with 12 df lies between 2.18 and 2.18. t curves for 12 df and 4 df. Moving over two columns, we ﬁnd the critical value for central area .99 (still with 12 df) to be 3.06 (see Figure 9.7). Moving down the .99 column to the 20-df row, we see the critical value is 2.85, so the area between 2.85 and 2.85 under the t curve with 20 df is .99. t curve for 12 df Figure 9.7 ues illustrated. t critical val- t curve for 12 df Shaded area .95 2.18 0 2.18 Shaded area .99 3.06 0 3.06 Notice that the critical values increase from left to right in each row of Appendix Table 3. This makes sense because as we move to the right, we capture larger central areas. In each column, the critical values decrease as we move downward, reﬂecting decreasing spread for t distributions with larger degrees of freedom. The larger the number of degrees of freedom, the more closely the t curve resembles the z curve. To emphasize this, we have included the z critical values as the last row of the t table. Furthermore, once the number of degrees of freedom exceeds 30, the critical values change little as the number of degrees of freedom increases. For this reason, Appendix Table 3 jumps from 30 df to 40 df, then to 60 df, then to 120 df, and 9.3 ■ Confidence Interval for a Population Mean 499 ﬁnally to the row of z critical values. If we need a critical value for a number of degrees of freedom between those tabulated, we just use the critical value for the closest df. For df 120, we use the z critical values. Many graphing calculators calculate t critical values for any number of degrees of freedom; so, if you are using such a calculator, it is not necessary to approximate the t critical values as described. ■ One-Sample t Conﬁdence Interval ................................................................ xm is approximately the z (standard 1s/ 1n 2 normal) distribution when n is large led to the z conﬁdence interval when s is known. In the same way, the following proposition provides the key to obtaining a conﬁdence interval when the population distribution is normal but s is unknown. The fact that the sampling distribution of Let x1, x2, ..., xn constitute a random sample from a normal population distribution. Then the probability distribution of the standardized variable t xm s 1n is the t distribution with df n 1. To see how this result leads to the desired conﬁdence interval, consider the case n 25. We use the t distribution with df 24 (n 1). From Appendix Table 3, the interval between 2.06 and 2.06 captures a central area of .95 under the t curve with 24 df. This means that 95% of all samples (with n 25) from a normal population result in values of x and s for which 2.06 xm 2.06 s 1n Algebraically manipulating these inequalities to isolate m yields x 2.06 a s s b m x 2.06 a b 125 125 The 95% conﬁdence interval for m in this situation extends from the lower endpoint s s x 2.06 a b to the upper endpoint x 2.06 a b . This interval can be 125 125 written x 2.06 a s b 125 The differences between this interval and the interval when s is known are the use of the t critical value 2.06 rather than the z critical value 1.96 and the use of the sample standard deviation as an estimate of s. The extra uncertainty that results from estimating s causes the t interval to be wider than the z interval. 500 Chapter 9 ■ Estimation Using a Single Sample If the sample size is something other than 25 or if the desired conﬁdence level is something other than 95%, a different t critical value (obtained from Appendix Table 3) is used in place of 2.06. The One-Sample t Confidence Interval for M The general formula for a conﬁdence interval for a population mean m based on a sample of size n when 1. x is the sample mean from a random sample, 2. the population distribution is normal, or the sample size n is large (generally n 30), and 3. s, the population standard deviation, is unknown is x 1t critical value2 a s b 1n where the t critical value is based on df n 1. Appendix Table 3 gives critical values appropriate for each of the conﬁdence levels 90%, 95%, and 99%, as well as several other less frequently used conﬁdence levels. If n is large (generally n 30), the normality of the population distribution is not critical. However, this conﬁdence interval is appropriate for small n only when the population distribution is (at least approximately) normal. If this is not the case, as might be suggested by a normal probability plot or boxplot, another estimation method should be used. .......................................................................................................................................... Example 9.8 Waiting for Surgery The Cardiac Care Network in Ontario, Canada collected information on the time between the date a patient was recommended for heart surgery and the surgery date for cardiac patients in Ontario (“Wait Times Data Guide,” Ministry of Health and LongTerm Care, Ontario, Canada, 2006). The reported mean waiting time (in days) for samples of patients for two cardiac procedures are given in the accompanying table. (The standard deviations in the table were estimated from information on wait-time variability included in the report.) Surgical Procedure Sample Size Mean Wait Time Standard Deviation Bypass Angiography 539 847 19 18 10 9 If we had access to the raw data (the 539 847 1386 individual wait time observations), we might begin by looking at boxplots. Data consistent with the given 9.3 ■ Confidence Interval for a Population Mean 501 summary quantities were used to generate the boxplots of Figure 9.8. The boxplots for the two surgical procedures are similar. There are outliers in both data sets, which might cause us to question the normality of the two wait-time distributions, but because the sample sizes are large, it is still appropriate to use the t conﬁdence interval. Figure 9.8 Boxplots for Wait time (days) Example 9.8. 60 * * * * * * * * Bypass Angiography 50 40 30 20 10 0 As a next step, we can use the conﬁdence interval of this section to estimate the true mean wait time for each of the two procedures. Let’s ﬁrst focus on the sample of bypass patients. For this group, sample size n 539 sample mean wait time x 19 sample standard deviation s 10 The report referenced here indicated that it is reasonable to regard these data as representative of the Ontario population. So, with m denoting the mean wait time for bypass surgery in Ontario, we can estimate m using a 90% conﬁdence interval. From Appendix Table 3, we use t critical value 1.645 (from the z critical value row because df n 1 538 120, the largest number of degrees of freedom in the table). The 90% conﬁdence interval for m is x 1t critical value2 a s 10 b 19 11.6452 a b 1n 1539 19 .709 118.291, 19.7092 Based on this sample, we are 90% conﬁdent that m is between 18.291 days and 19.709 days. This interval is fairly narrow indicating that our information about the value of m is relatively precise. A graphing calculator or any of the commercially available statistical computing packages can produce t conﬁdence intervals. Conﬁdence interval output from MINITAB for the angiography data is shown here. One-Sample T N Mean 847 18.0000 StDev 9.0000 SE Mean 0.3092 90% CI (17.4908, 18.5092) 502 Chapter 9 ■ Estimation Using a Single Sample The 90% conﬁdence interval for mean wait time for angiography extends from 17.4908 days to 18.5092 days. This interval is narrower than the 90% interval for bypass surgery wait time for two reasons: the sample size is larger (847 rather than 539) and the sample standard deviation is smaller (9 rather than 10). ■ .......................................................................................................................................... Example 9.9 Selﬁsh Chimps? © Alan and Sandy Carey/Getty Images ● The article “Chimps Aren’t Charitable” (Newsday, November 2, 2005) summarized the results of a research study published in the journal Nature. In this study, chimpanzees learned to use an apparatus that dispensed food when either of two ropes was pulled. When one of the ropes was pulled, only the chimp controlling the apparatus received food. When the other rope was pulled, food was dispensed both to the chimp controlling the apparatus and also to a chimp in the adjoining cage. The accompanying data (approximated from a graph in the paper) represent the number of times out of 36 trials that each of seven chimps chose the option that would provide food to both chimps (the “charitable” response). 23 22 21 24 19 20 20 Figure 9.9 is a normal probability plot of these data. The plot is reasonably straight, so it seems plausible that the population distribution is approximately normal. Normal Score 2 1 0 1 Normal probability plot for data of Example 9.9. Figure 9.9 2 19 20 21 22 23 Number of Charitable Responses 24 Calculation of a conﬁdence interval for the population mean number of charitable responses requires x and s. From the given data, we compute x 21.29 Step-by-step technology instructions available online s 1.80 ● Data set available online 9.3 ■ Confidence Interval for a Population Mean 503 The t critical value for a 99% conﬁdence interval based on 6 df is 3.71. The interval is x 1t critical value2 a s 1.80 b 21.29 13.712 a b 1n 17 21.29 2.52 118.77, 23.812 A statistical software package could also have been used to compute the 99% conﬁdence interval. The following is output from SPSS. The slight discrepancy between the hand-calculated interval and the one reported by SPSS occurs because SPSS uses more decimal accuracy in x, s, and t critical values. One-Sample Statistics CharitableResponses N 7 Mean 21.2857 Std. Deviation 1.79947 Std. Error Mean .68014 One-Sample CharitableResponses 99% Conﬁdence Interval of the Difference Lower Upper 18.7642 23.8073 With 99% conﬁdence, we estimate the population mean number of charitable responses (out of 36 trials) to be between 18.77 and 23.81. Remember that the 99% conﬁdence level implies that if the same formula is used to calculate intervals for sample after sample randomly selected from the population, in the long run 99% of these intervals will capture m between the lower and upper conﬁdence limits. Notice that based on this interval, we would conclude that on average chimps choose the charitable option more than half the time (18 out of 36 trials). The Newsday headline “Chimps Aren’t Charitable” was based on additional data from the study indicating that chimps’ charitable behavior was no different when there was another chimp in the adjacent cage than when the adjacent cage was empty. We will revisit this study in Chapter 11 to investigate this further. ■ .......................................................................................................................................... © Stone/Mark Douet /Getty Images E x a m p l e 9 . 1 0 Housework How much time do school-age children spend helping with housework? The article “The Three Corners of Domestic Labor: Mothers’, Fathers’, and Children’s Weekday and Weekend Housework” (Journal of Marriage and the Family [1994]: 657–668) gave information on the number of minutes per weekday spent on housework. The following mean and standard deviation are for a random sample of 26 girls in twoparent families where both parents work full-time: n 26 x 14.0 s 8.6 The authors of the article analyzed these data using methods designed for population distributions that are approximately normal. This assumption appears a bit questionable based on the reported mean and standard deviation (it is impossible to spend less than 0 min per day on housework, so the smallest possible value, 0, is only 1.63 standard deviations below the mean). However, because the authors reported that there were no outliers in the data and because n is relatively close to 30, 504 Chapter 9 ■ Estimation Using a Single Sample we use the t conﬁdence interval formula of this section to compute a 95% conﬁdence interval. Because n 26, df 25, and the appropriate t critical value is 2.06, the conﬁdence interval is then x 1t critical value2 a s 8.6 b 14.0 12.062 a b 1n 126 14.0 3.5 110.5, 17.5 2 Based on the sample data, we believe that the true mean time per weekday spent on housework is between 10.5 and 17.5 min for girls in two-parent families where both parents work. We used a method that has a 5% error rate to construct this interval. We should be somewhat cautious in interpreting this conﬁdence interval because of the concern expressed about the normality of the population distribution. ■ ■ Choosing the Sample Size ............................................................................... When estimating m using a large sample or a small sample from a normal population with known s, the bound B on the error of estimation associated with a 95% conﬁdence interval is B 1.96 a s b 1n Before collecting any data, an investigator may wish to determine a sample size for which a particular value of the bound is achieved. For example, with m representing the average fuel efﬁciency (in miles per gallon, mpg) for all cars of a certain type, the objective of an investigation may be to estimate m to within 1 mpg with 95% conﬁdence. The value of n necessary to achieve this is obtained by setting B 1 and then s b for n. solving 1 1.96 a 1n In general, suppose that we wish to estimate m to within an amount B (the speciﬁed bound on the error of estimation) with 95% conﬁdence. Finding the necessary s b for n. The result is sample size requires solving the equation B 1.96 a 1n n a 1.96s 2 b B Notice that, in general, a large value of s forces n to be large, as does a small value of B. Use of the sample-size formula requires that s be known, but this is rarely the case in practice. One possible strategy for estimating s is to carry out a preliminary study and use the resulting sample standard deviation (or a somewhat larger value, to be conservative) to determine n for the main part of the study. Another possibility is simply to make an educated guess about the value of s and to use that value to calculate n. For a population distribution that is not too skewed, dividing the range (the difference between the largest and the smallest values) by 4 often gives a rough idea of the value of the standard deviation. 9.3 ■ Confidence Interval for a Population Mean 505 The sample size required to estimate a population mean m to within an amount B with 95% conﬁdence is n a 1.96s 2 b B If s is unknown, it may be estimated based on previous information or, for a population that is not too skewed, by using (range)/4. If the desired conﬁdence level is something other than 95%, 1.96 is replaced by the appropriate z critical value (e.g., 2.58 for 99% conﬁdence). .......................................................................................................................................... E x a m p l e 9 . 1 1 Cost of Textbooks The ﬁnancial aid ofﬁce wishes to estimate the mean cost of textbooks per quarter for students at a particular university. For the estimate to be useful, it should be within $20 of the true population mean. How large a sample should be used to be 95% conﬁdent of achieving this level of accuracy? To determine the required sample size, we must have a value for s. The ﬁnancial aid ofﬁce is pretty sure that the amount spent on books varies widely, with most values between $50 and $450. A reasonable estimate of s is then range 450 50 400 100 4 4 4 The required sample size is n a 11.962 11002 2 1.96s 2 b a b 19.82 2 96.04 B 20 Rounding up, a sample size of 97 or larger is recommended. ■ ■ E x e r c i s e s 9.30–9.50 .............................................................................................................. 9.30 Given a variable that has a t distribution with the speciﬁed degrees of freedom, what percentage of the time will its value fall in the indicated region? a. 10 df, between 1.81 and 1.81 b. 10 df, between 2.23 and 2.23 c. 24 df, between 2.06 and 2.06 d. 24 df, between 2.80 and 2.80 e. 24 df, outside the interval from 2.80 to 2.80 f. 24 df, to the right of 2.80 g. 10 df, to the left of 1.81 Bold exercises answered in back 9.31 The formula used to compute a conﬁdence interval for the mean of a normal population when n is small is x 1t critical value 2 s 1n What is the appropriate t critical value for each of the following conﬁdence levels and sample sizes? a. 95% conﬁdence, n 17 b. 90% conﬁdence, n 12 c. 99% conﬁdence, n 24 ● Data set available online but not required ▼ Video solution available 506 Chapter 9 ■ Estimation Using a Single Sample d. 90% conﬁdence, n 25 e. 90% conﬁdence, n 13 f. 95% conﬁdence, n 10 9.32 The two intervals (114.4, 115.6) and (114.1, 115.9) are conﬁdence intervals for m true average resonance frequency (in hertz) for all tennis rackets of a certain type. a. What is the value of the sample mean resonance frequency? b. The conﬁdence level for one of these intervals is 90% and for the other it is 99%. Which is which, and how can you tell? 9.33 ▼ Samples of two different types of automobiles were selected, and the actual speed for each car was determined when the speedometer registered 50 mph. The resulting 95% conﬁdence intervals for true average actual speed were (51.3, 52.7) and (49.4, 50.6). Assuming that the two sample standard deviations are identical, which conﬁdence interval is based on the larger sample size? Explain your reasoning. 9.34 Suppose that a random sample of 50 bottles of a particular brand of cough medicine is selected and the alcohol content of each bottle is determined. Let m denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the sample of 50 results in a 95% conﬁdence interval for m of (7.8, 9.4). a. Would a 90% conﬁdence interval have been narrower or wider than the given interval? Explain your answer. b. Consider the following statement: There is a 95% chance that m is between 7.8 and 9.4. Is this statement correct? Why or why not? c. Consider the following statement: If the process of selecting a sample of size 50 and then computing the corresponding 95% conﬁdence interval is repeated 100 times, 95 of the resulting intervals will include m. Is this statement correct? Why or why not? 9.35 ▼ Acrylic bone cement is sometimes used in hip and knee replacements to ﬁx an artiﬁcial joint in place. The force required to break an acrylic bone cement bond was measured for six specimens under speciﬁed conditions, and the resulting mean and standard deviation were 306.09 Newtons and 41.97 Newtons, respectively. Assuming that it is reasonable to assume that breaking force under these conditions has a distribution that is approximately normal, estimate the true average breaking force for acrylic bone cement under the speciﬁed conditions. Bold exercises answered in back 9.36 The article “The Association Between Television Viewing and Irregular Sleep Schedules Among Children Less Than 3 Years of Age” (Pediatrics [2005]: 851–856) reported the accompanying 95% conﬁdence intervals for average TV viewing time (in hours per day) for three different age groups. Age Group 95% Conﬁdence Interval Less than 12 months 12 to 23 months 24 to 35 months (0.8, 1.0) (1.4, 1.8) (2.1, 2.5) a. Suppose that the sample sizes for each of the three age group samples were equal. Based on the given conﬁdence intervals, which of the age group samples had the greatest variability in TV viewing time? Explain your choice. b. Now suppose that the sample standard deviations for the three age group samples were equal, but that the three sample sizes might have been different. Which of the three age group samples had the largest sample size? Explain your choice. c. The interval (.768, 1.302) is either a 90% conﬁdence interval or a 99% conﬁdence interval for the mean TV viewing time for children less than 12 months old. Is the conﬁdence level for this interval 90% or 99%? Explain your choice. 9.37 Five hundred randomly selected working adults living in Calgary, Canada were asked how long, in minutes, their typical daily commute was (Calgary Herald Trafﬁc Study, Ipsos, September 17, 2005). The resulting sample mean and standard deviation of commute time were 28.5 minutes and 24.2 minutes, respectively. Construct and interpret a 90% conﬁdence interval for the mean commute time of working adult Calgary residents. 9.38 The article “Most Canadians Plan to Buy Treats, Many Will Buy Pumpkins, Decorations and/or Costumes” (Ipsos-Reid, October 24, 2005) summarized results from a survey of 1000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween during 2005. The resulting sample mean and standard deviation were $46.65 and $83.70 respectively. a. Explain how it could be possible for the standard deviation of the anticipated Halloween expense to be larger than the mean anticipated expense. ● Data set available online but not required ▼ Video solution available 9.3 b. Is it reasonable to think that the distribution of the variable anticipated Halloween expense is approximately normal? Explain why or why not. c. Is it appropriate to use the t conﬁdence interval to estimate the mean anticipated Halloween expense for Canadian residents? Explain why or why not. d. If appropriate, construct and interpret a 99% conﬁdence interval for the mean anticipated Halloween expense for Canadian residents. 9.39 Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small commuter airlines must estimate passenger weights. Under the old rule, airlines used 180 lb as a typical passenger weight (including carry-on luggage) in warm months and 185 lb as a typical weight in cold months. The Alaska Journal of Commerce (May 25, 2003) reported that Frontier Airlines conducted a study to estimate average passenger plus carry-on weights. They found an average summer weight of 183 lb and a winter average of 190 lb. Suppose that each of these estimates was based on a random sample of 100 passengers and that the sample standard deviations were 20 lb for the summer weights and 23 lb for the winter weights. a. Construct and interpret a 95% conﬁdence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers. b. Construct and interpret a 95% conﬁdence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers. c. The new FAA recommendations are 190 lb for summer and 195 lb for winter. Comment on these recommendations in light of the conﬁdence interval estimates from Parts (a) and (b). 9.40 “Heinz Plays Catch-up After Under-Filling Ketchup Containers” is the headline of an article that appeared on CNN.com (November 30, 2000). The article stated that Heinz had agreed to put an extra 1% of ketchup into each ketchup container sold in California for a 1-year period. Suppose that you want to make sure that Heinz is in fact fulﬁlling its end of the agreement. You plan to take a sample of 20-oz bottles shipped to California, measure the amount of ketchup in each bottle, and then use the resulting data to estimate the mean amount of ketchup in each bottle. A small pilot study showed that the amount of ketchup in 20-oz bottles varied from 19.9 to 20.3 oz. How many bottles should be included in the sample if you want Bold exercises answered in back ■ Confidence Interval for a Population Mean 507 to estimate the true mean amount of ketchup to within 0.1 oz with 95% conﬁdence? 9.41 ● ▼ Example 9.3 gave the following airborne times for United Airlines ﬂight 448 from Albuquerque to Denver on 10 randomly selected days: 57 54 55 51 56 48 52 51 59 59 a. Compute and interpret a 90% conﬁdence interval for the mean airborne time for ﬂight 448. b. Give an interpretation of the 90% conﬁdence level associated with the interval estimate in Part (a). c. Based on your interval in Part (a), if ﬂight 448 is scheduled to depart at 10 A.M., what would you recommend for the published arrival time? Explain. 9.42 The authors of the paper “Short-Term Health and Economic Beneﬁts of Smoking Cessation: Low Birth Weight” (Pediatrics [1999]: 1312–1320) investigated the medical cost associated with babies born to mothers who smoke. The paper included estimates of mean medical cost for low-birth-weight babies for different ethnic groups. For a sample of 654 Hispanic low-birth-weight babies, the mean medical cost was $55,007 and the standard error (s/ 1n) was $3011. For a sample of 13 Native American low-birth-weight babies, the mean and standard error were $73,418 and $29,577, respectively. Explain why the two standard errors are so different. 9.43 ● ▼ A study of the ability of individuals to walk in a straight line (“Can We Really Walk Straight?” American Journal of Physical Anthropology [1992]: 19–27) reported the following data on cadence (strides per second) for a sample of n 20 randomly selected healthy men: 0.95 0.85 0.92 0.95 0.93 0.86 1.00 0.92 0.85 0.81 0.78 0.93 0.93 1.05 0.93 1.06 1.06 0.96 0.81 0.96 Construct and interpret a 99% conﬁdence interval for the population mean cadence. 9.44 ● Fat contents (in percentage) for 10 randomly selected hot dogs were given in the article “Sensory and Mechanical Assessment of the Quality of Frankfurters” (Journal of Texture Studies [1990]: 395–409). Use the following data to construct a 90% conﬁdence interval for the true mean fat percentage of hot dogs: 25.2 21.3 22.8 17.0 29.8 21.0 25.5 16.0 20.9 19.5 ● Data set available online but not required ▼ Video solution available 508 Chapter 9 ■ Estimation Using a Single Sample 9.45 ● Five students visiting the student health center for a free dental examination during National Dental Hygiene Month were asked how many months had passed since their last visit to a dentist. Their responses were as follows: 6 17 11 22 29 Assuming that these ﬁve students can be considered a random sample of all students participating in the free checkup program, construct a 95% conﬁdence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program. 9.46 The article “First Year Academic Success: A Prediction Combining Cognitive and Psychosocial Variables for Caucasian and African American Students” (Journal of College Student Development [1999]: 599–610) reported that the sample mean and standard deviation for high school grade point average (GPA) for students enrolled at a large research university were 3.73 and 0.45, respectively. Suppose that the mean and standard deviation were based on a random sample of 900 students at the university. a. Construct a 95% conﬁdence interval for the mean high school GPA for students at this university. b. Suppose that you wanted to make a statement about the range of GPAs for students at this university. Is it reasonable to say that 95% of the students at the university have GPAs in the interval you computed in Part (a)? Explain. 9.47 ● ▼ The following data are the calories per half-cup serving for 16 popular chocolate ice cream brands reviewed by Consumer Reports (July 1999): 270 190 150 190 170 160 140 170 160 150 160 110 160 180 290 170 half-cup serving of chocolate ice cream? Explain why or why not. 9.48 The Bureau of Alcohol, Tobacco, and Firearms (BATF) has been concerned about lead levels in California wines. In a previous testing of wine specimens, lead levels ranging from 50 to 700 parts per billion were recorded (San Luis Obispo Telegram Tribune, June 11, 1991). How many wine specimens should be tested if the BATF wishes to estimate the true mean lead level for California wines to within 10 parts per billion with 95% conﬁdence? 9.49 ▼ The article “National Geographic, the Doomsday Machine,” which appeared in the March 1976 issue of the Journal of Irreproducible Results (yes, there really is a journal by that name—it’s a spoof of technical journals!) predicted dire consequences resulting from a nationwide buildup of National Geographic magazines. The author’s predictions are based on the observation that the number of subscriptions for National Geographic is on the rise and that no one ever throws away a copy of National Geographic. A key to the analysis presented in the article is the weight of an issue of the magazine. Suppose that you were assigned the task of estimating the average weight of an issue of National Geographic. How many issues should you sample to estimate the average weight to within 0.1 oz with 95% conﬁdence? Assume that s is known to be 1 oz. 9.50 The formula described in this section for determining sample size corresponds to a conﬁdence level of 95%. What would be the appropriate formula for determining sample size when the desired conﬁdence level is 90%? 98%? Is it reasonable to use the t conﬁdence interval to compute a conﬁdence interval for m, the true mean calories per Bold exercises answered in back ● Data set available online but not required ▼ Video solution available ........................................................................................................................................ 9.4 Interpreting and Communicating the Results of Statistical Analyses The purpose of most surveys and many research studies is to produce estimates of population characteristics. One way of providing such an estimate is to construct and report a conﬁdence interval for the population characteristic of interest. 9.4 ■ ■ Interpreting and Communicating the Results of Statistical Analyses 509 Communicating the Results of a Statistical Analysis .................................... When using sample data to estimate a population characteristic, a point estimate or a conﬁdence interval estimate might be used. Conﬁdence intervals are generally preferred because a point estimate by itself does not convey any information about the accuracy of the estimate. For this reason, whenever you report the value of a point estimate, it is a good idea to also include an estimate of the bound on the error of estimation. Reporting and interpreting a conﬁdence interval estimate requires a bit of care. First, always report both the conﬁdence interval and the conﬁdence level associated with the method used to produce the interval. Then, remember that both the conﬁdence interval and the conﬁdence level should be interpreted. A good strategy is to begin with an interpretation of the conﬁdence interval in the context of the problem and then to follow that with an interpretation of the conﬁdence level. For example, if a 90% conﬁdence interval for p, the proportion of students at a particular university who own a computer, is (.56, .78), we might say interpretation of interval explanation of “90% conﬁdence” interpretation of conﬁdence level We can be 90% conﬁdent that between • 56% and 78% of the students at this university own computers. We have used a method to produce this estimate that is successful in µ capturing the actual population proportion 90% of the time. When providing an interpretation of a conﬁdence interval, remember that the interval is an estimate of a population characteristic and be careful not to say that the interval applies to individual values in the population or to the values of sample statistics. For example, if a 99% conﬁdence interval for m, the mean amount of ketchup in bottles labeled as 12 oz, is (11.94, 11.98), this does not tell us that 99% of 12-oz ketchup bottles contain between 11.94 and 11.98 oz of ketchup. Nor does it tell us that 99% of samples of the same size would have sample means in this particular range. The conﬁdence interval is an estimate of the mean for all bottles in the population of interest. ■ Interpreting the Results of a Statistical Analysis ............................................ Unfortunately, there is no customary way of reporting the estimates of population characteristics in published sources. Possibilities include conﬁdence interval estimate bound on error estimate standard error If the population characteristic being estimated is a population mean, then you may also see sample mean sample standard deviation If the interval reported is described as a conﬁdence interval, a conﬁdence level should accompany it. These intervals can be interpreted just as we have interpreted the conﬁdence intervals in this chapter, and the conﬁdence level speciﬁes the long-run er- 510 Chapter 9 ■ Estimation Using a Single Sample ror rate associated with the method used to construct the interval (e.g., a 95% conﬁdence level speciﬁes a 5% long-run error rate). A form particularly common in news articles is estimate bound on error, where the bound on error is also sometimes called the margin of error. The bound on error reported is usually 2 times the standard deviation of the estimate. This method of reporting is a little more informal than a conﬁdence interval and, if the sample size is reasonably large, is roughly equivalent to reporting a 95% conﬁdence interval. You can interpret these intervals as you would a conﬁdence interval with approximate conﬁdence level of 95%. You must use care in interpreting intervals reported in the form of an estimate standard error. Recall from Section 9.2 that the general form of a conﬁdence interval is estimate (critical value)(standard deviation of the estimate) In journal articles, the estimated standard deviation of the estimate is usually referred to as the standard error. The critical value in the conﬁdence interval formula was determined by the form of the sampling distribution of the estimate and by the conﬁdence level. Note that the reported form, estimate standard error, is equivalent to a conﬁdence interval with the critical value set equal to 1. For a statistic whose sampling distribution is (approximately) normal (such as the mean of a large sample or a large-sample proportion), a critical value of 1 corresponds to an approximate conﬁdence level of about 68%. Because a conﬁdence level of 68% is rather low, you may want to use the given information and the conﬁdence interval formula to convert to an interval with a higher conﬁdence level. When researchers are trying to estimate a population mean, they sometimes report sample mean sample standard deviation. Be particularly careful here. To convert this information into a useful interval estimate of the population mean, you must ﬁrst convert the sample standard deviation to the standard error of the sample mean (by dividing by 1n) and then use the standard error and an appropriate critical value to construct a conﬁdence interval. For example, suppose that a random sample of size 100 is used to estimate the population mean. If the sample resulted in a sample mean of 500 and a sample standard deviation of 20, you might ﬁnd the published results summarized in any of the following ways: 95% conﬁdence interval for the population mean: (496.08, 503.92) mean bound on error: 500 4 mean standard error: 500 2 mean standard deviation: 500 20 ■ What to Look For in Published Data ................................................................ Here are some questions to ask when you encounter interval estimates in research reports. ■ ■ Is the reported interval a conﬁdence interval, mean bound on error, mean standard error, or mean standard deviation? If the reported interval is not a conﬁdence interval, you may want to construct a conﬁdence interval from the given information. What conﬁdence level is associated with the given interval? Is the choice of conﬁdence level reasonable? What does the conﬁdence level say about the long-run error rate of the method used to construct the interval? 9.4 ■ ■ Interpreting and Communicating the Results of Statistical Analyses 511 Is the reported interval relatively narrow or relatively wide? Has the population characteristic been estimated precisely? For example, the article “Use of a Cast Compared with a Functional Ankle Brace After Operative Treatment of an Ankle Fracture” (Journal of Bone and Joint Surgery [2003]: 205–211) compared two different methods of immobilizing an ankle after surgery to repair damage from a fracture. The article includes the following statement: The mean duration (and standard deviation) between the operation and return to work was 6313 days (median, sixty-three days; range, thirty three to ninetyeight days) for the cast group and 6519 days (median, sixty-two days; range, eight to 131 days) for the brace group; the difference was not signiﬁcant. This is an example of a case where we must be careful—the reported intervals are of the form estimate standard deviation. We can use this information to construct a conﬁdence interval for the mean time between surgery and return to work for each method of immobilization. One hundred patients participated in the study, with 50 wearing a cast after surgery and 50 wearing an ankle brace (random assignment was used to assign patients to treatment groups). Because the sample sizes are both large, we can use the t conﬁdence interval formula mean 1t critical value2 a s b 1n Each sample has df 50 1 49. The closest df value in Appendix Table 3 is for df 40, and the corresponding t critical value for a 95% conﬁdence level is 2.02. The corresponding intervals are 13 b 63 3.71 159.29, 66.712 150 19 b 65 5.43 159.57, 70.432 Brace: 65 2.02 a 150 Cast: 63 2.02 a The chosen conﬁdence level of 95% implies that the method used to construct each of the intervals has a 5% long-run error rate. Assuming that it is reasonable to view these samples as representative of the patient population, we can interpret these intervals as follows: We can be 95% conﬁdent that the mean return-to-work time for those treated with a cast is between 59.29 and 66.71 days, and we can be 95% conﬁdent that the mean return-to-work time for those treated with an ankle brace is between 59.57 and 70.43 days. These intervals are relatively wide, indicating that the values of the treatment means have not been estimated as precisely as we might like. This is not surprising, given the sample sizes and the variability in each sample. Note that the two intervals overlap. This supports the statement that the difference between the two immobilization methods was not signiﬁcant. Formal methods for directly comparing two groups, covered in Chapter 11, could be used to further investigate this issue. ■ A Word to the Wise: Cautions and Limitations .............................................. When working with point and conﬁdence interval estimates, here are a few things you need to keep in mind; 1. In order for an estimate to be useful, we must know something about accuracy. You should beware of point estimates that are not accompanied by a bound on error or some other measure of accuracy. 512 Chapter 9 ■ Estimation Using a Single Sample 2. A conﬁdence interval estimate that is wide indicates that we don’t have very precise information about the population characteristics being estimated. Don’t be fooled by a high conﬁdence level if the resulting interval is wide. High conﬁdence, while desirable, is not the same thing as saying we have precise information about the value of a population characteristic. The width of a conﬁdence interval is affected by the conﬁdence level, the sample size, and the standard deviation of the statistic used (e.g. p or x ) as the basis for constructing the interval. The best strategy for decreasing the width of a conﬁdence interval is to take a larger sample. It is far better to think about this before collecting data and to use the required sample size formulas to determine a sample size that will result in a conﬁdence interval estimate that is narrow enough to provide useful information. 3. The accuracy of estimates depends on the sample size, not the population size. This may be counter to intuition, but as long as the sample size is small relative to the population size (n less than 10% of the population size), the bound on error for estimating a population proportion with 95% conﬁdence is approximately p11 p 2 2 and for estimating a population mean with 95% conﬁdence is apn B s proximately 2 . 1n Note that each of these involves the sample size n, and both bounds decrease as the sample size increases. Neither approximate bound on error depends on the population size. The size of the population does need to be considered if sampling is without replacement and the sample size is more than 10% of the population size. In this Nn case, a ﬁnite population correction factor is used to adjust the bound BN 1 on error (the given bound is multiplied by the correction factor). Since this correction factor is always less than 1, the adjusted bound on error is smaller. 4. Assumptions and “plausibility” conditions are important. The conﬁdence interval procedures of this chapter require certain assumptions. If these assumptions are met, the conﬁdence intervals provide us with a method for using sample data to estimate population characteristics with conﬁdence. When the assumptions associated with a conﬁdence interval procedure are in fact true, the conﬁdence level speciﬁes a correct success rate for the method. However, assumptions (such as the assumption of a normal population distribution) are rarely exactly met in practice. Fortunately, in most cases, as long as the assumptions are approximately met, the conﬁdence interval procedures still work well. In general we can only determine if assumptions are “plausible” or approximately met, and that we are in the situation where we expect the inferential procedure to work reasonably well. This is usually conﬁrmed by knowledge of the data collection process and by using the sample data to check certain “plausibility conditions”. The formal assumptions for the z conﬁdence interval for a population proportion are 1. The sample is a random sample from the population of interest. 2. The sample size is large enough for the sampling distribution of p to be approximately normal. 3. Sampling is without replacement. 9.4 ■ Interpreting and Communicating the Results of Statistical Analyses 513 Whether the random sample assumption is plausible will depend on how the sample was selected and the intended population. Plausibility conditions for the other two assumptions are the following: np 10 and n(1 p) 10 (so the sampling distribution of p is approximately normal), and n is less than 10% of the population size (so sampling with replacement approximates sampling without replacement). The formal assumptions for the t conﬁdence interval for a population mean are 1. The sample is a random sample from the population of interest. 2. The population distribution is normal, so that the distribution of t xm s/ 1n has a t distribution. The plausibility of the random sample assumption, as was the case for proportions, will depend on how the sample was selected and the population of interest. The plausibility conditions for the normal population distribution assumption are the following: A normal probability plot of the data is reasonably straight (indicating that the population distribution is approximately normal), or The data distribution is approximately symmetric and there are no outliers. This may be conﬁrmed by looking at a dotplot, boxplot, stem-and-leaf display, or histogram of the data. (This would indicate that the population is approximately normal.) Alternatively, if n is large (n 30), the sampling distribution of x will be approximately normal even for nonnormal population distributions. This implies that use of the t interval is appropriate even if population normality is not plausible. In the end, you must decide that the assumptions are met or that they are “plausible” and that the inferential method used will provide reasonable results. This is also true for the inferential methods introduced in the chapters that follow. 5. Watch out for the “” when reading published reports. Don’t fall into the trap of thinking conﬁdence interval every time you see a in an expression. As was discussed earlier in this section, published reports are not consistent, and in addition to conﬁdence intervals, it is common to see estimate standard error and estimate sample standard deviation reported. 514 Chapter 9 ■ A c t i v i t y 9.1 Estimation Using a Single Sample Getting a Feel for Conﬁdence Level Explore this applet activity Technology Activity (Applet): Open the applet (available in ThomsonNOW at www.thomsonedu.com/login) called ConﬁdenceIntervals. You should see a screen like the one shown. Getting Started: If the “Method” box does not say “Means,” use the drop-down menu to select Means. In the box just below, select “t” from the drop-down menu. This applet will select a random sample from a speciﬁed normal population distribution and then use the sample to construct a conﬁdence interval for the population mean. The interval is then plotted on the display at the right, and we can see if the resulting interval contains the actual value of the population mean. For purposes of this activity, we will sample from a normal population with mean 100 and standard deviation 5. We will begin with a sample size of n 10. In the applet window, set m 100, s 5 and n 10. Leave the conf-level box set at 95%. Click the “Recalculate” button to rescale the picture on the right. Now click on the sample button. You should see a conﬁdence interval appear on the display on the right hand side. If the interval contains the actual mean of 100, the interval is drawn in green; if 100 is not in the conﬁdence interval, the interval is shown in red. Your screen should look something like the following. Part 1: Click on the “Sample” button several more times, and notice how the conﬁdence interval estimate changes from sample to sample. Also notice that at the bottom of the left-hand side of the display, the applet is keeping track of the proportion of all the intervals calculated so far that include the actual value of m. If we were to construct a large number of intervals, this proportion should closely approximate the capture rate for the conﬁdence interval method. To look at more than 1 interval at a time, change the “Intervals” box from 1 to 100, and then click the sample button. You should see a screen similar to the one at the top of page 515, with 100 intervals is the display on the right-hand side. Again, intervals containing 100 (the value of m in this case) will be green and those that do not contain 100 will be red. Also note that the capture proportion on the left-hand side has also been updated to reﬂect what happened with the 100 newly generated intervals. Activity 9.2 ■ An Alternative Conﬁdence Interval for a Population Proportion 515 Experiment with three other conﬁdence levels of your choice, and then answer the following question: b. In general, is the proportion of computed t conﬁdence intervals that contain m 100 close to the stated conﬁdence level? Continue generating intervals until you have seen at least 1000 intervals, and then answer the following question: a. How does the proportion of intervals constructed that contain m 100 compare to the stated conﬁdence level of 95%? On how many intervals was your proportion based? (Note—if you followed the instructions, this should be at least 1000.) A c t i v i t y 9.2 An Alternative Conﬁdence Interval for a Population Proportion Technology Activity (Applet): This activity presumes that you have already worked through Activity 9.1. Background: In Section 9.2, it was suggested that a conﬁdence interval of the form P mod 1z critical value 2 P mod 11 P mod 2 n B successes 2 is an alternative to the usual n4 large-sample z conﬁdence interval. This alternative interval is preferred by many statisticians because, in repeated sampling, the proportion of intervals constructed that include the actual value of the population proportion, p, tends to be closer to the stated conﬁdence level. In this activity, we will explore how the “capture rates” for the two different interval estimation methods compare. Open the applet (available in ThomsonNOW at www.thomsonedu.com/login) called ConﬁdenceIntervals. You should see a screen like the one shown. where Pmod Part 2: When the population is normal but s is unknown, we construct a conﬁdence interval for a population mean using a t critical value rather than a z critical value. How important is this distinction? Let’s investigate. Use the drop-down menu to change the box just below the method box that’s says “Means” from “t” to “z with s.” The applet will now construct intervals using the sample standard deviation, but will use a z critical value rather than the t critical value. Use the applet to construct at least 1000 95% intervals, and then answer the following question: c. Comment on how the proportion of the computed intervals that include the actual value of the population mean compares to the stated conﬁdence level of 95%. Is this surprising? Explain why or why not. Now experiment with some different samples sizes. What happens when n 20? n 50? n 100? Use what you have learned to write a paragraph explaining what these simulations tell you about the advisability of using a z critical value in the construction of a conﬁdence interval for m when s is unknown. Explore this applet activity 516 Chapter 9 ■ Estimation Using a Single Sample Select “Proportion” from the Method box drop-down menu, and then select “Large Sample z” from the dropdown menu of the second box. We will consider sampling from a population with p .3 using a sample size of 40. In the applet window, enter p .3 and n 40. Note that n 40 is large enough to satisfy np 10 and n(1 p) 10. Set the “Intervals” box to 100, and then use the applet to construct a large number (at least 1000) of 95% conﬁdence intervals. 1. How does the proportion of intervals constructed that include p .3, the population proportion, compare to 95%? Does this surprise you? Explain. Now use the drop-down menu to change “Large Sample z” to “Modiﬁed.” Now the applet will construct A c t i v i t y 9.3 Verifying Signatures on a Recall Petition Background: In 2003, petitions were submitted to the California Secretary of State calling for the recall of Governor Gray Davis. Each of California’s 58 counties then had to report the number of valid signatures on the petitions from that county so that the State could determine whether there were enough valid signatures to certify the recall and set a date for the recall election. The following paragraph appeared in the San Luis Obispo Tribune (July 23, 2003): In the campaign to recall Gov. Gray Davis, the secretary of state is reporting 16,000 veriﬁed signatures from San Luis Obispo County. In all, the County Clerk’s Ofﬁce received 18,866 signatures on recall petitions and was instructed by the state to check a A c t i v i t y 9.4 the alternative conﬁdence interval that is based on Pmod. Use the applet to construct a large number (at least 1000) of 95% conﬁdence intervals. 2. How does the proportion of intervals constructed that include p .3, the population proportion, compare to 95%? Is this proportion closer to 95% than was the case for the large-sample z interval? 3. Experiment with different combinations of values of sample size and population proportion p. Can you ﬁnd a combination for which the large sample z interval has a capture rate that is close to 95%? Can you ﬁnd a combination for which it has a capture rate that is even farther from 95% than it was for n 40 and p .3? How does the modiﬁed interval perform in each of these cases? random sample of 567. Out of those, 84.48% were good. The veriﬁcation process includes checking whether the signer is a registered voter and whether the address and signature on the recall petition match the voter registration. 1. Use the data from the random sample of 567 San Luis Obispo County signatures to construct a 95% conﬁdence interval for the proportion of petition signatures that are valid. 2. How do you think that the reported ﬁgure of 16,000 veriﬁed signature for San Luis Obispo County was obtained? 3. Based on your conﬁdence interval from Step 1, explain why you think that the reported ﬁgure of 16,000 veriﬁed signatures is or is not reasonable. A Meaningful Paragraph Write a meaningful paragraph that includes the following six terms: sample, population, conﬁdence level, estimate, mean, margin of error. A “meaningful paragraph” is a coherent piece writing in an appropriate context that uses all of the listed words. The paragraph should show that you understand the mean- ing of the terms and their relationship to one another. A sequence of sentences that just deﬁne the terms is not a meaningful paragraph. When choosing a context, think carefully about the terms you need to use. Choosing a good context will make writing a meaningful paragraph easier. ■ Summary of Key Concepts and Formulas 517 Summary of Key Concepts and Formulas Term or Formula Comment Point estimate A single number, based on sample data, that represents a plausible value of a population characteristic. Unbiased statistic A statistic that has a sampling distribution with a mean equal to the value of the population characteristic to be estimated. Conﬁdence interval An interval that is computed from sample data and provides a range of plausible values for a population characteristic. Conﬁdence level A number that provides information on how much “conﬁdence” we can have in the method used to construct a conﬁdence interval estimate. The conﬁdence level speciﬁes the percentage of all possible samples that will produce an interval containing the true value of the population characteristic. p 1z critical value2 n p11 p 2 a B p11 p2 n 1.96 2 b B x 1z critical value2 s 1n x 1t critical value2 s 1n n a 1.96s 2 b B A formula used to construct a conﬁdence interval for p when the sample size is large. A formula used to compute the sample size necessary for estimating p to within an amount B with 95% conﬁdence. (For other conﬁdence levels, replace 1.96 with an appropriate z critical value.) A formula used to construct a conﬁdence interval for m when s is known and either the sample size is large or the population distribution is normal. A formula used to construct a conﬁdence interval for m when s is unknown and either the sample size is large or the population distribution is normal. A formula used to compute the sample size necessary for estimating m to within an amount B with 95% conﬁdence. (For other conﬁdence levels, replace 1.96 with an appropriate z critical value.) 518 Chapter 9 ■ Estimation Using a Single Sample Chapter Review Exercises 9.51–9.73 Know exactly what to study! Take a pre-test and receive your Personalized Learning Plan. 9.51 Despite protests from civil libertarians and gay rights activists, many people favor mandatory AIDS testing of certain at-risk groups, and some people even believe that all citizens should be tested. What proportion of the adults in the United States favor mandatory testing for all citizens? To assess public opinion on this issue, researchers conducted a survey of 1014 randomly selected adult U.S. citizens (“Large Majorities Continue to Back AIDS Testing,” Gallup Poll Monthly [1991]: 25–28). The article reported that 466 of the 1014 people surveyed believed that all citizens should be tested. Use this information to estimate p, the true proportion of all U.S. adults who favor AIDS testing of all citizens. 9.54 Seventy-seven students at the University of Virginia were asked to keep a diary of a conversation with their mothers, recording any lies they told during these conversations (San Luis Obispo Telegram-Tribune, August 16, 1995). It was reported that the mean number of lies per conversation was 0.5. Suppose that the standard deviation (which was not reported) was 0.4. a. Suppose that this group of 77 is a random sample from the population of students at this university. Construct a 95% conﬁdence interval for the mean number of lies per conversation for this population. b. The interval in Part (a) does not include 0. Does this imply that all students lie to their mothers? Explain. 9.52 The article “Consumers Show Increased Liking for Diesel Autos” (USA Today, January 29, 2003) reported that 27% of U.S. consumers would opt for a diesel car if it ran as cleanly and performed as well as a car with a gas engine. Suppose that you suspect that the proportion might be different in your area and that you want to conduct a survey to estimate this proportion for the adult residents of your city. What is the required sample size if you want to estimate this proportion to within .05 with 95% conﬁdence? Compute the required sample size ﬁrst using .27 as a preliminary estimate of p and then using the conservative value of .5. How do the two sample sizes compare? What sample size would you recommend for this study? 9.55 The article “Selected Characteristics of High-Risk Students and Their Enrollment Persistence” (Journal of College Student Development [1994]: 54–60) examined factors that affect whether students stay in college. The following summary statistics are based on data from a sample of 44 students who did not return to college after the ﬁrst quarter (the nonpersisters) and a sample of 257 students who did return (the persisters): 9.53 In the article “Fluoridation Brushed Off by Utah” (Associated Press, August 24, 1998), it was reported that a small but vocal minority in Utah has been successful in keeping ﬂuoride out of Utah water supplies despite evidence that ﬂuoridation reduces tooth decay and despite the fact that a clear majority of Utah residents favor ﬂuoridation. To support this statement, the article included the result of a survey of Utah residents that found 65% to be in favor of ﬂuoridation. Suppose that this result was based on a random sample of 150 Utah residents. Construct and interpret a 90% conﬁdence interval for p, the true proportion of Utah residents who favor ﬂuoridation. Is this interval consistent with the statement that ﬂuoridation is favored by a clear majority of residents? Bold exercises answered in back Number of Hours Worked per Week During the First Quarter Nonpersisters Persisters Mean Standard Deviation 25.62 18.10 14.41 15.31 a. Consider the 44 nonpersisters as a random sample from the population of all nonpersisters at the university where the data were collected. Compute a 98% conﬁdence interval for the mean number of hours worked per week for nonpersisters. b. Consider the 257 persisters as a random sample from the population of all persisters at the university where the data were collected. Compute a 98% conﬁdence interval for the mean number of hours worked per week for persisters. ● Data set available online but not required ▼ Video solution available ■ c. The 98% conﬁdence interval for persisters is narrower than the corresponding interval for nonpersisters, even though the standard deviation for persisters is larger than that for nonpersisters. Explain why this happened. d. Based on the interval in Part (a), do you think that the mean number of hours worked per week for nonpersisters is greater than 20? Explain. 9.56 An Associated Press article on potential violent behavior reported the results of a survey of 750 workers who were employed full time (San Luis Obispo Tribune, September 7, 1999). Of those surveyed, 125 indicated that they were so angered by a coworker during the past year that they felt like hitting the coworker (but didn’t). Assuming that it is reasonable to regard this sample of 750 as a random sample from the population of full-time workers, use this information to construct and interpret a 90% conﬁdence interval estimate of p, the true proportion of fulltime workers so angered in the last year that they wanted to hit a colleague. 9.57 The 1991 publication of the book Final Exit, which includes chapters on doctor-assisted suicide, caused a great deal of controversy in the medical community. The Society for the Right to Die and the American Medical Association quoted very different ﬁgures regarding the proportion of primary-care physicians who have participated in some form of doctor-assisted suicide for terminally ill patients (USA Today, July 1991). Suppose that a survey of physicians is to be designed to estimate this proportion to within .05 with 95% conﬁdence. How many primary-care physicians should be included in a random sample? 9.58 Retailers report that the use of cents-off coupons is increasing. The Scripps Howard News Service (July 9, 1991) reported the proportion of all households that use coupons as .77. Suppose that this estimate was based on a random sample of 800 households (i.e., n 800 and p .77). Construct a 95% conﬁdence interval for p, the true proportion of all households that use coupons. 9.59 A manufacturer of small appliances purchases plastic handles for coffeepots from an outside vendor. If a handle is cracked, it is considered defective and must be discarded. A large shipment of plastic handles is received. Bold exercises answered in back Chapter Review Exercises 519 The proportion of defective handles p is of interest. How many handles from the shipment should be inspected to estimate p to within 0.1 with 95% conﬁdence? 9.60 An article in the Chicago Tribune (August 29, 1999) reported that in a poll of residents of the Chicago suburbs, 43% felt that their ﬁnancial situation had improved during the past year. The following statement is from the article: “The ﬁndings of this Tribune poll are based on interviews with 930 randomly selected suburban residents. The sample included suburban Cook County plus DuPage, Kane, Lake, McHenry, and Will Counties. In a sample of this size, one can say with 95% certainty that results will differ by no more than 3 percent from results obtained if all residents had been included in the poll.” Comment on this statement. Give a statistical argument to justify the claim that the estimate of 43% is within 3% of the true proportion of residents who feel that their ﬁnancial situation has improved. 9.61 The McClatchy News Service (San Luis Obispo Telegram-Tribune, June 13, 1991) reported on a study of violence on television during primetime hours. The following table summarizes the information reported for four networks: Network Mean Number of Violent Acts per Hour ABC CBS FOX NBC 15.6 11.9 11.7 11.0 Suppose that each of these sample means was computed on the basis of viewing n 50 randomly selected primetime hours and that the population standard deviation for each of the four networks is known to be s 5. a. Compute a 95% conﬁdence interval for the true mean number of violent acts per prime-time hour for ABC. b. Compute 95% conﬁdence intervals for the mean number of violent acts per prime-time hour for each of the other three networks. c. The National Coalition on Television Violence claims that shows on ABC are more violent than those on the other networks. Based on the conﬁdence intervals from ● Data set available online but not required ▼ Video solution available 520 Chapter 9 ■ Estimation Using a Single Sample Parts (a) and (b), do you agree with this conclusion? Explain. 9.62 The Chronicle of Higher Education (January 13, 1993) reported that 72.1% of those responding to a national survey of college freshmen were attending the college of their ﬁrst choice. Suppose that n 500 students responded to the survey (the actual sample size was much larger). a. Using the sample size n 500, calculate a 99% conﬁdence interval for the proportion of college students who are attending their ﬁrst choice of college. b. Compute and interpret a 95% conﬁdence interval for the proportion of students who are not attending their ﬁrst choice of college. c. The actual sample size for this survey was much larger than 500. Would a conﬁdence interval based on the actual sample size have been narrower or wider than the one computed in Part (a)? 9.63 Increases in worker injuries and disability claims have prompted renewed interest in workplace design and regulation. As one particular aspect of this, employees required to do regular lifting should not have to handle unsafe loads. The article “Anthropometric, Muscle Strength, and Spinal Mobility Characteristics as Predictors of the Rating of Acceptable Loads in Parcel Sorting” (Ergonomics [1992]: 1033–1044) reported on a study involving a random sample of n 18 male postal workers. The sample mean rating of acceptable load attained with a work-simulating test was found to be x 9.7 kg. and the sample standard deviation was s 4.3 kg. Suppose that in the population of all male postal workers, the distribution of rating of acceptable load can be modeled approximately using a normal distribution with mean value m. Construct and interpret a 95% conﬁdence interval for m. 9.64 The Gallup Organization conducted a telephone survey on attitudes toward AIDS (Gallup Monthly, 1991). A total of 1014 individuals were contacted. Each individual was asked whether they agreed with the following statement: “Landlords should have the right to evict a tenant from an apartment because that person has AIDS.” One hundred one individuals in the sample agreed with this statement. Use these data to construct a 90% conﬁdence interval for the proportion who are in agreement with this statement. Give an interpretation of your interval. Bold exercises answered in back 9.65 A manufacturer of college textbooks is interested in estimating the strength of the bindings produced by a particular binding machine. Strength can be measured by recording the force required to pull the pages from the binding. If this force is measured in pounds, how many books should be tested to estimate with 95% conﬁdence to within 0.1 lb, the average force required to break the binding? Assume that s is known to be 0.8 lb. 9.66 Recent high-proﬁle legal cases have many people reevaluating the jury system. Many believe that juries in criminal trials should be able to convict on less than a unanimous vote. To assess support for this idea, investigators asked each individual in a random sample of Californians whether they favored allowing conviction by a 10–2 verdict in criminal cases not involving the death penalty. The Associated Press (San Luis Obispo TelegramTribune, September 13, 1995) reported that 71% supported the 10–2 verdict. Suppose that the sample size for this survey was n 900. Compute and interpret a 99% conﬁdence interval for the proportion of Californians who favor the 10–2 verdict. 9.67 The Center for Urban Transportation Research released a report stating that the average commuting distance in the United States is 10.9 mi (USA Today, August 13, 1991). Suppose that this average is actually the mean of a random sample of 300 commuters and that the sample standard deviation is 6.2 mi. Estimate the true mean commuting distance using a 99% conﬁdence interval. 9.68 In 1991, California imposed a “snack tax” (a sales tax on snack food) in an attempt to help balance the state budget. A proposed alternative tax was a 12¢-per-pack increase in the cigarette tax. In a poll of 602 randomly selected California registered voters, 445 responded that they would have preferred the cigarette tax increase to the snack tax (Reno Gazette-Journal, August 26, 1991). Estimate the true proportion of California registered voters who preferred the cigarette tax increase; use a 95% conﬁdence interval. 9.69 The conﬁdence intervals presented in this chapter give both lower and upper bounds on plausible values for the population characteristic being estimated. In some instances, only an upper bound or only a lower bound is appropriate. Using the same reasoning that gave the large ● Data set available online but not required ▼ Video solution available ■ sample interval in Section 9.3, we can say that when n is large, 99% of all samples have m x 2.33 Use the data of Exercise 9.67 to obtain a 95% conﬁdence interval for the true standard deviation of commuting distance. s 1n (because the area under the z curve to the left of 2.33 is s .99). Thus, x 2.33 is a 99% upper conﬁdence bound 1n for m. Use the data of Example 9.8 to calculate the 99% upper conﬁdence bound for the true wait time for bypass patients in Ontario. 9.70 The Associated Press (December 16, 1991) reported that in a random sample of 507 people, only 142 correctly described the Bill of Rights as the ﬁrst 10 amendments to the U.S. Constitution. Calculate a 95% conﬁdence interval for the proportion of the entire population that could give a correct description. 9.71 When n is large, the statistic s is approximately unbiased for estimating s and has approximately a normal distribution. The standard deviation of this statistic when s the population distribution is normal is ss ⬇ which 12n s can be estimated by . A large-sample conﬁdence 12n interval for the population standard deviation s is then s 1z critical value2 s 12n Bold exercises answered in back Graphing Calculator Explorations 521 9.72 The interval from 2.33 to 1.75 captures an area of .95 under the z curve. This implies that another largesample 95% conﬁdence interval for m has lower limit s s and upper limit x 1.75 . Would you x 2.33 1n 1n recommend using this 95% interval over the 95% interval s discussed in the text? Explain. (Hint: Look x 1.96 1n at the width of each interval.) 9.73 The eating habits of 12 bats were examined in the article “Foraging Behavior of the Indian False Vampire Bat” (Biotropica [1991]: 63–67). These bats consume insects and frogs. For these 12 bats, the mean time to consume a frog was x 21.9 min. Suppose that the standard deviation was s 7.7 min. Construct and interpret a 90% conﬁdence interval for the mean suppertime of a vampire bat whose meal consists of a frog. What assumptions must be reasonable for the one-sample t interval to be appropriate? Do you need a live tutor for homework problems? ● Data set available online but not required Are you ready? Take your exam-prep post-test now. ▼ Video solution available Graphing Calculator Explorations E x p l o r a t i o n 9.1 The Conﬁdence Interval for a Population Proportion Because conﬁdence intervals are widely used, it will come as no surprise to you that your calculator may have a built-in capability to determine a conﬁdence interval for a population proportion. Once again you will need to navigate your calculator’s menu system, looking for key words such as INTR for “interval,” or possibly TESTS. (Conﬁdence intervals are frequently associated with “hypothesis tests,” a topic to come later in Chapter 10.) Once you ﬁnd the right menu, look for words like “1” and “prop” and “z”—together these key words should indicate a one-sample z conﬁdence interval for a proportion. 522 Chapter 9 ■ Estimation Using a Single Sample 1-Prop Zinterval C-Level : x : n : Execute 1-PropZInt x : n : C-Level: Calculate Common screen presentations for conﬁdence interval data entry. Figure 9.10 Once you select the correct choice, you will be presented with a screen for providing the information needed 1-Prop Zinterval to calculate a conﬁdence interval: the number of sucLeft 0.42889 cesses, the sample size, and the conﬁdence level. Two Right0.49024 representative screens are given in Figure 9.10. p̂ 0.46 From Example 9.4, supply the following informan1014 tion: x 466, n 1014, and the C-Level .95. Move your cursor down to Execute or Calculate, and press the Enter, Execute, or Calculate button, depending on your 1-PropZInt calculator. The conﬁdence interval should appear imme(.42889, .49024) diately. Again, two representative screens are given in p̂.459566075 Figure 9.11. n1014 Notice that the formatting in these screens is slightly different. Which of these is the “right” format? Probably F i g u r e 9 . 1 1 Calcuneither one! Recall a previous graphing calculator explolator screens for conration in Chapter 3, where we discussed the differences ﬁdence intervals. between a calculator’s presentation of information and the appropriate way to communicate this information. Check with your instructor to ascertain her or his preferences about what format information is required. You will almost certainly want to report more information than what is presented on these calculator screens. Speciﬁcally, you will notice that the calculator does not verify the appropriateness of the large-sample assumption! Some conﬁdence intervals may stray outside the interval from 0 to 1, which makes sense to the calculator, but not to the statistician. Thus, you may have to modify the interval returned by your calculator. Also, don’t forget that your calculator does not know the context of the problem; you will have to provide that contextual information irrespective of the form of the calculator presentation. Remember: The calculator only does the calculation; you must do the thinking. E x p l o r a t i o n 9.2 A Conﬁdence Interval for a Population Mean Finding the conﬁdence interval for a single mean on your calculator will require you to navigate the menu system much as you did to ﬁnd the conﬁdence interval for a proportion. The conﬁdence interval for the mean will have some added challenges, however: 1. You will need to decide whether to base the interval on the z or t distribution, and 2. You may use previous calculations of the sample mean and standard deviation, or the calculator will evaluate these statistics from data contained in a List. We will use the data from Exercise 9.43 to construct a 99% conﬁdence interval. We have entered the data in List1 and are ready to proceed. Since the population standard deviation is not known, we will use the t conﬁdence interval. For the t conﬁdence interval, the normality of the population becomes an issue. The original data are at hand, and we can assess the plausibility of a normal population via a normal probability plot. Figure 9.12 shows the normal probability plot. After verifying the plausibility of the normality of the population, we can now construct the conﬁdence interval. In this case, we have a choice of entering the sample ■ Graphing Calculator Explorations 523 (a) (a) (b) (b) (a) Edit window for normal probability plot; (b) normal probability plot. Figure 9.12 (a) Conﬁdence interval input screen for raw data; (b) conﬁdence interval input screen for summary statistics. Figure 9.13 calculations or letting the calculator evaluate the sample mean and standard deviation. Based on that choice, we see one of the screens shown in Figure 9.13. Figure 9.14 shows one calculator’s version of the conﬁdence interval. Once again, we caution you that the calculator only calculates—you must still do the thinking and present the solution in the context of the particular problem at hand. Calculated conﬁdence interval. Figure 9.14 This page intentionally left blank