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CHAPTER 11 Hypothesis Tests Two g Sampl n i v l o Inv s or Proporti e n o a ns e M GENDER STEREOTYPES AND ASKING FOR DIRECTIONS Many of us have heard the stereotypical observa- this much lower just by chance. Section 11.6 of tion that men absolutely refuse to stop and ask this chapter will provide some guidance regarding for directions, preferring instead to wander about a statistical test with which you can compare until they “find their own way.” Is this just a these gender results and verify the conclusion for stereotype, or is there really something to it? yourself. There will be no need to stop at Section When Lincoln Mercury was carrying out research 11.5 to ask for directions. prior to the design and introduction of its Source: Jamie LaReau, “Lincoln Uses Gender Stereotypes to Sell Navigation System,” Automotive News, May 30, 2005, p. 32. in-vehicle navigational systems, the company surveyed men and women with regard to their driving characteristics and navigational habits. According to the survey, 61% of the female respondents said they would stop and ask for directions once they figured out they were lost. On the other hand, only 42% of the male respondents said they would stop and ask for directions under the same circumstances. For purposes of our discussion, we’ll assume that Lincoln Mercury surveyed 200 persons of each gender. asking for men—a sample proportion of just 0.42 (84 out of 200) versus 0.61 (122 out of 200)—may have occurred simply by chance variation? Actually, if the population proportions were really the same, there would be only a 0.0001 probability ned way of the direction-asking proportion for men being -fashio les the old two samp g n ri a p Com 364 Reza Estakhrian/Stone/Getty Images Is it possible that the lower rate of direction- 365 • Select and use the appropriate hypothesis test in comparing the means of two independent samples. • Test the difference between sample means when the samples are not independent. • Test the difference between proportions for two independent samples. • Determine whether two independent samples could have come from populations having the same standard deviation. LEARNING OBJECTIVES After reading this chapter, you should be able to: ( ) INTRODUCTION 11.1 One of the most useful applications of business statistics involves comparing two samples to examine whether a difference between them is (1) significant or (2) likely to have been due to chance. This lends itself quite well to the analysis of data from experiments such as the following: Comparing Samples A local YMCA, in the early stages of telephone solicitation to raise funds for expanding its gymnasium, is testing two appeals. Of 100 residents approached with appeal A, 21% pledged a donation. Of 150 presented with appeal B, 28% pledged a donation. SOLUTION Is B really a superior appeal, or could its advantage have been merely the result of chance variation from one sample to the next? Using the techniques in this chapter, we can reach a conclusion on this and similar questions. The approach will be very similar to that in Chapter 10, where we dealt with one sample statistic (either } x or p), its standard error, and the level of significance at which it differed from its hypothesized value. Sections 11.2–11.4 and 11.6 deal with the comparison of two means or two proportions from independent samples. Independent samples are those for which the selection process for one is not related to that for the other. For example, in an experiment, independent samples occur when persons are randomly assigned to the experimental and control groups. In these sections, a hypothesis-testing procedure very similar to that in Chapter 10 will be followed. As before, either one or two critical value(s) will be identified, and then a decision rule will be applied to see if the calculated value of the test statistic falls into a rejection region specified by the rule. When comparing two independent samples, the null and alternative hypotheses can be expressed in terms of either the population parameters (1 and 2, or 1 and 2) or the sampling distribution of the difference between the sample statistics (} x1 and } x2, or p1 and p2). These approaches to describing the null and alternative hypotheses are equivalent and are demonstrated in Table 11.1. EXAMPLEEXAMPLEEXAMPLE EXAMPLE 366 Part 4: Hypothesis Testing TABLE 11.1 When comparing means from independent samples, null and alternative hypotheses can be expressed in terms of the population parameters (on the left) or described by the mean of the sampling distribution of the difference between the sample statistics (on the right). This also applies to testing two sample proportions. For example, H0: 1 5 2 is the same as H0: (p 2p )5 0. 1 2 Null and Alternative Hypotheses Expressed in Terms of the Population Means Hypotheses Expressed in Terms of the Sampling Distribution of the Difference Between the Sample Means Two-tail test: H 0: 1 5 2 H0: ( x} 2 x} ) 5 0 1 2 H1: ( x} 2 x} ) 0 1 2 H0: ( x} 2 x} ) $ 0 H1: ( x} 2 x} ) , 0 H0: ( x} 2 x} ) # 0 1 2 H1: ( x} 2 x} ) . 0 1 2 or H 1: 1 2 Left-tail test: H 0: 1 $ 2 or H 1: 1 , 2 1 1 2 2 Right-tail test: H 0: 1 # 2 or H 1: 1 . 2 Section 11.5 is concerned with the comparison of means for two dependent samples. Samples are dependent when the selection process for one is related to the selection process for the other. A typical example of dependent samples occurs when we have before-and-after measures of the same individuals or objects. In this case we are interested in only one variable: the difference between measurements for each person or object. In this chapter, we present three different methods for comparing the means of two independent samples: the pooled-variances t-test (Section 11.2), the unequal-variances t-test (Section 11.3), and the z-test (Section 11.4). Figure 11.1 summarizes the procedure for selecting which test to use in comparing the sample means. As shown in Figure 11.1, an important factor in choosing between the pooled-variances t-test and the unequal-variances t-test is whether we can assume the population standard deviations (and, hence, the variances) might be equal. Section 11.7 provides a hypothesis-testing procedure by which we can actually test this possibility. However, for the time being, we will use a less rigorous standard—that is, based on the sample standard deviations, whether it appears that the population standard deviations might be equal. ( ) 11.2 THE POOLED-VARIANCES t-TEST FOR COMPARING THE MEANS OF TWO INDEPENDENT SAMPLES Situations can arise where we’d like to examine whether the difference between the means of two independent samples is large enough to warrant rejecting the possibility that their population means are the same. In this type of setting, the alternative conclusion is that the difference between the sample means is small enough to have occurred by chance, and that the population means really could be equal. Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions 367 FIGURE 11.1 Selecting the test statistic for hypothesis tests comparing the means of two independent samples. Hypothesis test m1 – m2 Are the population standard deviations equal? Does s1 = s2? No Yes For any sample sizes Compute the pooled estimate of the common variance as s2p = (n1 – 1)s21 + (n2 – 1)s22 n1 + n2 – 2 and perform a pooled-variances t-test where t= (x1 – x2) – (m1 – m 2)0 √( 1 1 s2p –– + –– n1 n2 ) Perform an unequal-variances t-test where t= (x1 – x2) – (m1 – m 2)0 √ s21 n1 + s22 n2 and [(s21 / n1) + (s22 / n2)]2 df = (s21 / n1)2 (s22 / n2)2 + n1 – 1 n2 – 1 df = n1 + n2 – 2 and (m1 – m2)0 is from H0 and (m1 – m2)0 is from H0 Test assumes samples are from normal populations. When either test can be applied to the same data, the unequal-variances t-test is preferable to the z-test—especially when doing the test with computer assistance. Test assumes samples are from normal populations with equal standard deviations. Section 11.2 See Note 1 Only if n1 and n2 both ≥ 30 A z-test approximation can be performed where z= (x1 – x2) – (m1 – m 2)0 √ s21 s22 + n1 n2 with s21 and s22 as s 21 and s 22 , estimates of and (m1 – m2)0 is from H0 The central limit theorem prevails and there are no limitations on the population distributions. This test may also be more convenient for solutions based on pocket calculators. Section 11.4 See Note 3 Section 11.3 See Note 2 1. Section 11.7 describes a procedure for testing the null hypothesis that 1 5 2. However, when using a computer and statistical software, it may be more convenient to simply bypass this assumption and apply the unequal-variances t-test described in Section 11.3. 2. This test involves a corrected df value that is smaller than if 1 5 2 had been assumed. When using a computer and statistical software, this test can be used routinely instead of the other two tests shown here. The nature of the df expression makes hand calculations somewhat cumbersome. The normality assumption becomes less important for larger sample sizes. 3. When sample sizes are large (each n $ 30), the z-test is a useful alternative to the unequal-variances t-test, and may be more convenient when hand calculations are involved. 4. For each test, Computer Solutions within the chapter describe the Excel and Minitab procedures for carrying out the test. Procedures based on data files and those based on summary statistics are included. 368 Part 4: Hypothesis Testing Our use of the t-test assumes that (1) the (unknown) population standard deviations are equal, and (2) the populations are at least approximately normally distributed. Because of the central limit theorem, the assumption of population normality becomes less important for larger sample sizes. Although it is often associated only with small-sample tests, the t distribution is appropriate when the population standard deviations are unknown, regardless of how large or small the samples happen to be. The t-test used here is known as the pooled-variances t-test because it involves the calculation of an estimated value for the variance that both populations are assumed to share. This pooled estimate is shown in Figure 11.1 as s2p. The number of degrees of freedom associated with the test will be df 5 n1 1 n2 2 2, and the test statistic is calculated as follows: Test statistic for comparing the means of two independent samples, 1 and 2 assumed to be equal: x1 2 } x2) 2 (1 2 2)0 (} t 5 _____________________ wwwwww 1 1 s2p ___ 1 ___ n1 n2 Ï ( with ) x1 and } x2 5 means of samples 1 and 2 where } (1 2 2)0 5 the hypothesized difference between the population means n1 and n2 5 sizes of samples 1 and 2 s1 and s2 5 standard deviations of samples 1 and 2 sp 5 pooled estimate of the common standard deviation (n1 2 1)s21 1 (n2 2 1)s22 s2p 5 _____________________ n1 1 n2 2 2 and df 5 n1 1 n2 2 2 Confidence interval for 1 2 2: (} x1 2 } x2) 6 ty2 Ï ( wwwwww 1 1 2 sp ___ 1 ___ n1 n2 ) with 5 (1 2 confidence coefficient) The numerator of the t-statistic includes (1 2 2)0, the hypothesized value of the difference between the population means. The hypothesized difference is generally zero in tests like those in this chapter. The term in the denominator of the t-statistic is the estimated standard error of the difference between the sample means. It is comparable to the standard error of the sampling distribution for the sample mean discussed in Chapter 10. Also shown is the confidence interval for the difference between the population means. Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions Pooled-Variances t-Test Entrepreneurs developing an accounting review program for persons preparing to take the Certified Public Accountant (CPA) examination are considering two possible formats for conducting the review sessions. A random sample of 10 students are trained using format 1, and then their number of errors is recorded for a prototype examination. Another random sample of 12 individuals are trained according to format 2, and their errors are similarly recorded for the same examination. For the 10 students trained with format 1, the individual performances are 11, 8, 8, 3, 7, 5, 9, 5, 1, and 3 errors. For the 12 students trained with format 2, the individual performances are 10, 11, 9, 7, 2, 11, 12, 3, 6, 7, 8, and 12 errors. These data are in file CX11CPA. SOLUTION Since the study was not conducted with directionality in mind, the appropriate test will be two-tail. The null hypothesis is H0: 1 5 2, and the alternative hypothesis is H1: 1 2. The null and alternative hypotheses may also be expressed as follows: • Null hypothesis H0: • (x}12x}2) 5 0 The two review formats are equally effective. Alternative hypothesis H1: (x}12x}2) 0 The two review formats are not equally effective. In comparing the performances of the two groups, the 0.10 level of significance will be used. Based on these data, the 10 members of group 1 made an average of 6.000 errors, with a sample standard deviation of 3.127. The 12 students trained with format 2 made an average of 8.167 errors, with a standard deviation of 3.326. The sample standard deviations do not appear to be very different, and we will assume that the population standard deviations could be equal. (As noted previously, this rather informal inference can be replaced by the separate hypothesis test in Section 11.7.) In applying the pooled-variances t-test, the pooled estimate of the common variance, s2p, and the test statistic, t, can be calculated as (10 2 1)(3.127)2 1 (12 2 1)(3.326)2 s2p 5 _________________________________ 5 10.484 10 1 12 2 2 and (6.000 2 8.167) 2 0 t 5 ___________________ 5 21.563 wwwwwwww 1 1 10.484 ___ 1 ___ 10 12 Ï ( ) For the 0.10 level of significance, the critical values of the test statistic will be t 5 21.725 and t 5 11.725. These are based on the number of degrees of freedom, df 5 (n1 1 n2 2 2), or (10 1 12 2 2) 5 20, and the specification that EXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXA EXAMPLE 369 370 Part 4: Hypothesis Testing FIGURE 11.2 EXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLE In this two-tail pooled-variances t-test, we are not able to reject the null hypothesis that the two accounting review formats could be equally effective. H0: m(x1 – x2) = 0 The two training formats are equally effective. The two training formats are not equally effective. H1: m(x1 – x2) ≠ 0 x1 and x2 are the mean numbers of errors for group 1 and group 2. Reject H0 Do not reject H0 Reject H0 Area = 0.05 Area = 0.05 m(x1 – x2) = 0 t = –1.725 Test statistic: t = –1.563 t = +1.725 the two tail areas must add up to 0.10. The decision rule is to reject the null hypothesis (i.e., conclude that the population means are not equal for the two review formats) if the calculated test statistic is either less than t 5 21.725 or greater than t 5 11.725. As Figure 11.2 shows, the calculated test statistic, t 5 21.563, falls into the nonrejection region of the test. At the 0.10 level, we must conclude that the review formats are equally effective in training individuals for the CPA examination. For this level of significance, the observed difference between the groups’ mean errors is judged to have been due to chance. Based on the sample data, we will also determine the 90% confidence interval for (1 2 2). For df 5 20 and 5 0.10, this is (} x1 2 } x2) 6 ty2 Ï( wwwww 1 2 1 ) sp ___ 1 ___ 5 (6.000 2 8.167) 6 1.725 n1 n2 Ï10.484( 10 1 12 ) wwwwwwww 1 1 ___ ___ 5 22.167 6 2.392, or from 24.559 to 10.225 The hypothesized difference (zero) is contained within the 90% confidence interval, so we are 90% confident the population means could be the same. As discussed in Chapter 10, a nondirectional test at the level of significance and a 100(1 2 )% confidence interval will lead to the same conclusion. Computer Solutions 11.1 shows Excel and Minitab procedures for the pooled-variances t-test. Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions 371 COMPUTER 11.1 SOLUTIONS Pooled-Variances t-Test for (1 2 2), Population Variances Unknown but Assumed Equal These procedures show how to use the pooled-variances t-test to compare the means of two independent samples. The population variances are unknown, but assumed to be equal. EXCEL Excel pooled-variances t-test for (1 2 2), based on raw data 1. For the sample data (Excel file CX11CPA) on which Figure 11.2 is based, with the label and 10 data values for format 1 in column A, and the label and 12 data values for format 2 in column B: From the Data ribbon, click Data Analysis. Click t-Test: Two-Sample Assuming Equal Variances. Click OK. 2. Enter A1:A11 into the Variable 1 Range box and B1:B13 into the Variable 2 Range box. Enter 0 into the Hypothesized Mean Difference box. Click to select Labels. Specify the significance level for the test by entering 0.10 into the Alpha box. Select Output Range and enter D1 into the box. Click OK. The results will be as shown above. This is a two-tail test, so we refer to the 0.134 p-value. (Excel doesn’t ask whether our test is one-tail or two-tail, but it provides critical values and p-values for both.) 3. To obtain a confidence interval for (1 2 2), it will be necessary to refer to the procedure described below. It is based on summary statistics for the samples. Excel Unstacking Note: When this analysis is based on raw data, Excel requires that the two samples be in two distinctly separate fields (e.g., two separate columns). If the data are in one column and the subscripts identifying group membership are in another column, it will be necessary to “unstack” the data. For example, if the data were in column A and the subscripts (1 and 2, to denote prep formats 1 and 2) were in column B, and each column had a label at the top: Click and drag to select the numbers and labels in columns A and B. From the Data ribbon, click Sort. In the Sort by box, select the category variable in column B (i.e., Format). Click OK. The data for sample 1 will now be listed directly above the data for sample 2—from here, we need only select each grouping of data (e.g., scores) and either move or copy it to its own column. The result will be one column for the scores of group 1 and an adjacent column for the scores of group 2. Excel pooled-variances t-test and confidence interval for (1 2 2), based on summary statistics 1. Using the summary statistics associated with CX11CPA: Open the TEST STATISTICS workbook. 2. Using the arrows at the bottom left, select the t-Test_2 Means (Eq-Var) worksheet. (continued ) 372 Part 4: Hypothesis Testing 3. Enter the sample means, variances, and sizes into the appropriate cells. Enter the hypothesized difference (0, in this case) and the desired alpha level for the test. The calculated t-statistic and a two-tail p-value will be shown at the right. 4. To obtain a confidence interval for (1 2 2), follow steps 1–3, but open the ESTIMATORS workbook and select the t-Estimate_2 Means (Eq-Var) worksheet. Enter the desired confidence level as a decimal fraction (e.g., 0.90). Note: As an alternative, you can use Excel worksheet template TMT2POOL. It simultaneously conducts the test and reports a confidence interval. The steps are described within the template. MINITAB Minitab pooled-variances t-test and confidence interval for (1 2 2), based on raw data 1. For example, using the data (Minitab file CX11CPA) on which Figure 11.2 is based, with the data values for format 1 in column C1 and the data values for format 2 in column C2: Click Stat. Select Basic Statistics. Click 2-Sample t. 2. Select Samples in different columns. Enter C1 into the First box and C2 into the Second box. Click to select Assume equal variances. (Note: If all the data had been in a single column [i.e., “stacked”], it would have been necessary to select the Samples in one column option, then to specify the column containing the data and the column containing the subscripts identifying group membership.) 3. Click Options. Enter the desired confidence level as a percentage (e.g., 90.0) into the Confidence Level box. Enter the hypothesized difference (0) into the Test difference box. Within the Alternative box, select not equal. Click OK. Click OK. Minitab pooled-variances t-test and confidence interval for (1 2 2), based on summary data Follow steps 1 through 3 above, but select Summarized data in step 2 and insert the appropriate summary statistics into the Sample size, Mean, and Standard deviation boxes for each sample. ERCISES X E 11.1 “When comparing two sample means, the t-test should be used only when the sample sizes are less than 30.” Comment. 11.2 An educator is considering two different videotapes for use in a half-day session designed to introduce students to the basics of economics. Students have been randomly assigned to two groups, and they all take the same written examination after viewing the videotape. The scores are summarized here. Assuming normal populations with equal standard deviations, does it appear that the two videotapes could be equally effective? What is the most accurate statement that could be made about the p-value for the test? } 5 77.1 Videotape 1: x 1 } 5 80.0 Videotape 2: x 2 s1 5 7.8 n1 5 25 s2 5 8.1 n2 5 25 11.3 Using independent random samples, a researcher is comparing the number of hours of television viewed last week for high school seniors versus sophomores. The results are shown here. Assuming normal populations Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions with equal standard deviations, does it appear that the average number of television hours per week could be equal for these two populations? What is the most accurate statement that could be made about the p-value for the test? Seniors: Sophomores: } 5 3.9 hours s 5 1.2 hours n 5 32 x 1 1 1 } 5 3.5 hours s 5 1.4 hours n 5 30 x 2 2 2 11.4 An ambulance service located at the edge of town is responsible for serving a large office building in the downtown area. Testing different routes for getting from the ambulance station to the office building, a driver finds that five trips using route A take an average of 5.9 minutes, with a standard deviation of 1.4 minutes; six trips using route B take an average of 4.2 minutes, with a standard deviation of 1.8 minutes. Assuming normal populations with equal standard deviations, and using the 0.10 level, is there a significant difference between the two routes? Construct and interpret the 90% confidence interval for the difference between the population means. 11.5 A maintenance supervisor is comparing the standard version of an instructional booklet with one that has been claimed to be superior. An experiment is conducted in which 26 technicians are divided into two groups, provided with one of the booklets, then given a test a week later. For the 13 using the standard version, the average exam score was 72.0, with a standard deviation of 9.3. For the 13 given the new version, the average score was 80.2, with a standard deviation of 10.1. Assuming normal populations with equal standard deviations, and using the 0.05 level of significance, does the new booklet appear to be better than the standard version? 11.6 A sample of 40 investment customers serviced by an account manager are found to have had an average of $23,000 in transactions during the past year, with a standard deviation of $8500. A sample of 30 customers serviced by another account manager averaged $28,000 in transactions, with a standard deviation of $11,000. Assuming the population standard deviations are equal, use the 0.05 level of significance in testing whether the population means could be equal for customers serviced by the two account managers. Using the appropriate statistical table, what is the most accurate statement we can make about the p-value for this test? Construct and interpret the 95% confidence interval for the difference between the population means. 11.7 Comparing dexterity-test scores of workers on the day shift versus those on the night shift, the production manager of a large electronics plant finds that a sample of 37 workers from the day shift have an average score of 73.1, with a standard deviation of 12.3. For 42 workers from the night shift, the average score was 77.3, with a standard deviation of 8.4. Assuming the population standard deviations are equal, use the 0.05 level of significance in comparing the average scores for the two 373 shifts. Using the appropriate statistical table, what is the most accurate statement we can make about the p-value for this test? Construct and interpret the 95% confidence interval for the difference between the population means. 11.8 Sheila Smith, the manager of a large resort’s main hotel, has been receiving complaints from some guests that they are not being provided with prompt service upon approaching the front desk. In particular, she is concerned that desk staff might be providing female guests with less prompt service than their male counterparts. In observing a sample of 34 male guests, she finds it takes an average of 15.2 seconds, with a standard deviation of 5.9 seconds, for them to be greeted after their arrival at the front desk. For a sample of 39 female guests, the mean and standard deviation are 17.4 seconds and 6.4 seconds, respectively. Assuming the population standard deviations to be equal, use the 0.05 level of significance in examining whether the population mean time for serving female guests might actually be no greater than that for serving male guests. Using the appropriate statistical table, what is the most accurate statement we can make about the p-value for the test? 11.9 Media observers have been examining the number of minutes devoted to business and financial news during the half-hour evening news broadcasts of two local television channels. For each channel, they have randomly selected 10 weekday broadcasts and observed the number of minutes spent on business and financial news during that broadcast. The times measured in these independent samples are shown here. Assuming normal populations with equal standard deviations, use the 0.10 level of significance in testing whether the population means might actually be the same. Using the appropriate statistical table, what is the most accurate statement we can make about the p-value for the test? Construct and interpret the 90% confidence interval for the difference between the population means. Channel 2: 3.8 2.7 4.9 3.4 3.7 4.5 4.2 2.8 3.5 4.6 minutes Channel 4: 3.6 4.0 4.5 5.2 4.8 4.3 5.7 3.5 3.7 5.8 minutes 11.10 In a test of the effectiveness of a new battery design, 16 battery-powered music boxes are randomly provided with either the old design or the new version. Hours of playing time before battery failure were as follows: 8 boxes, new battery type: 3.3, 6.4, 3.9, 5.4, 5.1, 4.6, 4.9, 7.2 hrs 8 boxes, old battery type: 4.2, 2.9, 4.5, 4.9, 5.0, 5.1, 3.2, 4.0 hrs Assuming normal populations with equal standard deviations, use the 0.05 level to determine whether the new battery could be better than the old design. Using the appropriate statistical table, what is the most accurate statement we can make about the p-value for this test? 374 Part 4: Hypothesis Testing 11.11 A nutritionist has noticed a FoodFarm ad stating the company’s peanut butter contains less fat than that produced by a major competitor. She purchases 11 8-ounce jars of each brand and measures the fat content of each. The 11 FoodFarm jars had an average of 31.3 grams of fat, with a standard deviation of 2.1 grams. The 11 jars from the other company had an average of 33.2 grams of fat, with a standard deviation of 1.8 grams. Assuming normal populations with equal standard deviations, use the 0.05 level of significance in examining whether FoodFarm’s ad claim could be valid. What is the most accurate statement that could be made about the p-value for this test? 0.05 level, was the mean return speed for the flattened bat significantly greater than for the regulation bat? Identify and interpret the p-value for the test. Source: Andy Gardiner, ( DATA SET ) Note: Exercises 11.12–11.17 require a computer and statistical software. Source: ”Primates on Facebook,” economist.com, February 26, 2009. 11.12 A study published in the Archives of Internal Medicine examined the prevalence of so-called “difficult” patients who ask for unneeded prescriptions, unnecessarily complain, or otherwise cause extraordinary frustrations for their medical provider. Researchers found the mean age of doctors having the greatest problems with difficult patients was 41, while the mean age of doctors having the least problems with difficult patients was 46. Assume that data file XR11012 contains the ages for doctors in each of these patient-difficulty samples. In a suitable one-tail test at the 0.01 level, was the mean age for doctors having the greatest problems with difficult patients significantly less than that for those having the least problems with difficult patients? Identify and interpret the p-value for the test. Source: Rita Rubin, “‘Difficult Patients Can Test Doctors’ Patience,“ USA Today, February 24, 2009, p. 5D. 11.13 It has been claimed that flattening (“rolling”) the barrel of a graphite baseball bat can stretch the fibers and result in a pitched baseball being returned faster than with a regulation bat. This was a controversial topic during the 2009 baseball College World Series. Assume that data file XR11013 describes the results of a controlled test measuring the return speeds of balls pitched to a flattened bat and a regulation bat, respectively. Based on these sample results, and using a suitable one-tail test at the ( ) 11.3 “NCAA on Guard for ‘Rolled’ Bats,” USA Today, June 17, 2009, p. 8C. 11.14 Comparing the number of Facebook “friends” for men and women, observers have speculated about whether the mean number of friends could be the same for each group. Assume that data file XR11014 lists the number of Facebook friends for independent samples of male and female Facebook users. Based on the results of a two-tail test at the 0.05 level, comment on whether the difference between the sample means could have simply occurred by chance. Identify and interpret the p-value for the test. 11.15 Using the sample results in Exercise 11.14, construct and interpret the 95% confidence interval for the difference between the population means. Is the hypothesized difference (0.00) within the interval? Given the presence or absence of the 0.00 value within the interval, is this consistent with the findings of the hypothesis test conducted in Exercise 11.14? 11.16 An engineer has measured the hardness scores for a sample of conveyor-belt support bearings that have been hardened by two different methods. The first method is used by her company, and the second method is known to be used by a number of other companies in the industry. With the resulting data in file XR11016, use the 0.05 level of significance in comparing the mean hardness scores of the two samples, and comment on the possibility that the difference between the sample means could have occurred by chance. Identify and interpret the p-value for the test. 11.17 Using the sample results in Exercise 11.16, construct and interpret the 95% confidence interval for the difference between the population means. Is the hypothesized difference (0.00) within the interval? Given the presence or absence of the 0.00 value within the interval, is this consistent with the findings of the hypothesis test conducted in Exercise 11.16? THE UNEQUAL-VARIANCES t-TEST FOR COMPARING THE MEANS OF TWO INDEPENDENT SAMPLES When the population standard deviations are unknown and are not assumed to be equal, pooling the sample standard deviations into a single estimate of their common population value is no longer applicable. As a result, s1 and s2 must be used to estimate their respective population standard deviations, 1 and 2. The test assumes the populations to be at least approximately normally distributed, an assumption that becomes less important for larger sample sizes. In the unequal-variances t-test, the t-statistic expression is straightforward, but the df formula is a little more complex—it is a correction formula that provides a Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions 375 df value that is smaller than its counterpart in the preceding section. For accuracy, it’s best to maintain a lot of decimal places if you are computing df with a pocket calculator. If we are using the computer and statistical software, the unequal-variances t-test presents no computational difficulties, and is the preferred method for comparing the means of two independent samples, regardless of the sample sizes. The test statistic, df, and confidence interval expressions for this test are shown here: Unequal-variances t-test for comparing the means of two independent samples, 1 and 2 unknown and not assumed to be equal: x1 2 } x2) 2 (1 2 2)0 (} t 5 _____________________ wwww s22 s21 ___ 1 ___ n1 n2 Ï with where } x1 and } x2 5 means of samples 1 and 2 (1 2 2)0 5 hypothesized difference between the population means n1 and n2 5 sizes of samples 1 and 2 s1 and s2 5 standard deviations of samples 1 and 2 f ( s21yn1 ) 1 ( s22yn2 ) g2 df 5 __________________ ( s21yn1 )2 _______ ( s22yn2 )2 _______ 1 n1 2 1 n2 2 1 Confidence interval for 1 2 2: with } 2} (x x2) 6 t/2 1 Ï wwww s2 s2 1 2 ___ 1 ___ n1 n2 5 (1 2 confidence coefficient) Unequal-Variances t-Test The makers of Graphlex, a graphite additive for engine oil, have conducted an experimental study to determine the effectiveness of their product in improving the fuel efficiency of automobiles. In cooperation with the Metropolitan Cab Company, they’ve randomly divided the company’s cabs into two groups of equal size. Graphlex was added to the engine oil of the 45 cabs in the experimental group, while the 45 cabs in the control group continued to operate with the usual lubricant. Drivers were not informed of the experiment. After 1 month, fuel efficiency records were examined. For the 45 cabs using Graphlex, the average cab achieved 18.94 miles per gallon (mpg), with a standard deviation of 3.90 mpg. For the 45 cabs not using Graphlex, the average mpg was 17.51, with a standard deviation of 2.87 mpg. The underlying data are in file CX11MPG. Graphlex is preparing a national advertising campaign to promote its ability to improve fuel efficiency. EXAMPLEEXAMPLEEXAMPLEEXAM EXAMPLE 376 Part 4: Hypothesis Testing FIGURE 11.3 The makers of Graphlex claim their graphite oil additive improves the fuel efficiency of automobiles. In this right-tail unequal-variances test at the 0.05 level, the results indicate they may be correct. H0: m(x1 – x2) ≤ 0 mpg with Graphlex is no higher than for ordinary oil. Graphlex improves fuel economy. H1: m(x1 – x2) > 0 x1 and x2 are mpg averages for cabs with and without Graphlex. Do not reject H0 Reject H0 Area = 0.05 m(x1 – x2) = 0 LEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLE t = +1.664 Test statistic: t = +1.98 SOLUTION The results will be evaluated at the 0.05 level of significance. Since the purpose of the study was to determine whether cabs using Graphlex (group 1) get better fuel economy than cabs without it (group 2), the null and alternative hypotheses are H0: 1 # 2, and H1: 1 . 2. Expressed in terms of the sampling distribution of the difference between sample means, the null and alternative hypotheses can be stated as follows: • • Null hypothesis H0: (}x12}x2) # 0 mpg with Graphlex is no higher than with conventional oil. Alternative hypothesis H1: (}x12}x2) . 0 mpg with Graphlex is higher. For these data, the values for the test statistic (t) and the number of degrees of freedom (df ) are calculated as x1 2 } x2) 2 (1 2 2)0 (} (18.94 2 17.51) 2 0 t 5 _____________________ 5 ___________________ 5 1.98 2 2 wwww wwwwwww s1 s2 3.902 _____ 2.872 _____ ___ 1 1 ___ n1 n2 45 45 and [(3.902Y45) 1 (2.872Y45)]2 3 ( s21Yn1 ) 1 ( s22Yn2 ) 42 df 5 _________________ 5 ________________________ 5 80.85, rounded to 81 (3.902Y45)2 (2.872Y45)2 ( s21Yn1 )2 _______ ( s22Yn2 )2 ___________ ___________ _______ 1 1 45 2 1 45 2 1 n1 2 1 n2 2 1 Ï Ï For the 0.05 level of significance, df 5 81, and a right-tail test, the critical value of the test statistic is t 5 11.664. The decision rule is, “Reject H0 if the calculated test statistic is greater than t 5 11.664, otherwise do not reject.” As Figure 11.3 shows, the calculated test statistic exceeds the critical value, the null hypothesis is rejected, and we conclude that Graphlex really works. Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions 377 NOTE As we saw in Chapter 10, different levels of significance can lead to different conclusions. For example, at the 0.01 level (critical t 5 12.373), the null hypothesis would not have been rejected. While the Graphlex Company would likely stress the additive’s effectiveness by relying on the 0.05 level of significance, the manufacturer of a competing brand would tend to prefer a more demanding test (e.g., 5 0.01) in order to boast that Graphlex has no effect. You may have noticed that the degrees of freedom value for this test, df 5 81, was near the upper part of the df range of the t distribution table. Should you encounter a test in which df happens to be over 100, just use the “infinity” row of the table to determine the critical value(s) of the test statistic. This row represents the normal distribution, toward which the t distribution converges as df becomes larger. Based on the sample data, we will also determine the 90% confidence interval for (1 2 2). For df 5 81 and 5 0.10, this will be x1 2 } x2) 6 ty2 (} Ï wwww s2 s2 Ï 1 2 ___ 1 ___ 5 (18.94 2 17.51) 6 1.664 n1 n2 wwwwwww 3.902 2.872 _____ 1 _____ 45 45 5 1.43 6 1.20, or from 0.23 to 2.63 Computer Solutions 11.2 shows Excel and Minitab procedures for the unequal-variances t-test. COMPUTER 11.2 SOLUTIONS Unequal-Variances t-Test for (1 2 2), Population Variances Unknown and Not Equal These procedures show how to use the unequal-variances t-test to compare the means of two independent samples. The population variances are unknown and not assumed to be equal. EXCEL Excel unequal-variances t-test for (1 2 2), based on raw data 1. For the sample data (Excel file CX11MPG) on which Figure 11.3 is based, with the label and 45 mpg values for the Graphlex cabs in column A, and the label and 45 data values for the standard cabs in column B: From the Data ribbon, click Data Analysis. Click t-Test: Two-Sample Assuming Unequal Variances. Click OK. (continued ) 378 Part 4: Hypothesis Testing 2. Enter A1:A46 into the Variable 1 Range box and B1:B46 into the Variable 2 Range box. Enter 0 into the Hypothesized Mean Difference box. Click to select Labels. Specify the significance level by entering 0.05 into the Alpha box. Select Output Range and enter D1 into the box. Click OK. This is a one-tail test, so we refer to the 0.026 p-value. 3. To obtain a confidence interval for (1 2 2), it will be necessary to refer to the procedure described next. It is based on summary statistics for the samples. Note: For an analysis based on raw data, Excel requires the two samples to be in two distinctly separate fields (e.g., two separate columns). If the data are in one column and the subscripts identifying group membership are in another column, see the Excel Unstacking Note in Computer Solutions 11.1. Excel unequal-variances t-test and confidence interval for (1 2 2), based on summary statistics 1. Using the summary statistics associated with CX11MPG (the test is summarized in Figure 11.3): Open the TEST STATISTICS workbook. 2. Using the arrows at the bottom left, select the t-Test_2 Means (Uneq-Var) worksheet. 3. Enter the sample means, variances, and sizes into the appropriate cells. Enter the hypothesized difference (0, in this case) and the desired alpha level for the test. The calculated t-statistic and a one-tail p-value will be shown at the right. 4. To get a confidence interval for (1 2 2), follow steps 1–3, above, but open the ESTIMATORS workbook and select the t-Estimate_2 Means (Uneq-Var) worksheet. Enter the desired confidence level as a decimal fraction (e.g., 0.90). Note: As an alternative, you can use Excel worksheet template TMT2UNEQ. It simultaneously conducts the test and reports a confidence interval. The steps are described within the template. MINITAB Minitab unequal-variances t-test and confidence interval for (1 2 2), based on raw data 1. For example, using the data (Minitab file CX11MPG) on which Figure 11.3 is based, with the data values for the Graphlex cabs in column C1 and the data values for the standard cabs in column C2: Click Stat. Select Basic Statistics. Click 2-Sample t. 2. Select Samples in different columns. Enter C1 into the First box and C2 into the Second box. Do NOT select the “Assume equal variances” option. (Note: If all the data had been in a single column [i.e., “stacked”], it would have been necessary to select the Samples in one column option, then to specify the column containing the data and the column containing the subscripts identifying group membership.) 3. Click Options. Enter the desired confidence level as a percentage (e.g., 90.0) into the Confidence Level box. Enter the hypothesized difference (0) into the Test difference box. Within the Alternative box, select greater than. Click OK. Click OK. Minitab unequal-variances t-test and confidence interval for (1 2 2), based on summary data Follow steps 1 through 3 above, but select Summarized data in step 2 and insert the appropriate summary statistics into the Sample size, Mean, and Standard deviation boxes for each sample. Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions 379 ERCISES X E 11.18 In two independent samples from populations that are normally distributed, 2 x1 5 35.0, s1 5 5.8, n1 5 12 and 2 x2 5 42.5, s2 5 9.3, n2 5 14. Using the 0.05 level of significance, test H0: 1 5 2 versus H1: 1 ? 2. 11.19 In two independent samples, 2 x1 5 165.0, s1 5 21.5, n1 5 40 and 2 x2 5 172.9, s2 5 31.3, n2 5 32. Using the 0.10 level of significance, test H0: 1 $ 2 versus H1: 1 , 2. 11.20 In two independent samples, 2 x1 5 125.0, s1 5 21.5, n1 5 40, and 2 x2 5 116.4, s2 5 10.8, n2 5 35. Using the 0.025 level of significance, test H0: 1 # 2 versus H1: 1 . 2. 11.21 Fred and Martina, senior agents at an airline secu- rity checkpoint, carry out advanced screening procedures for hundreds of randomly selected passengers per day. For a random sample of 30 passengers recently processed by Fred, the mean processing time was 124.5 seconds, with a standard deviation of 20.4 seconds. For a random sample of 36 passengers recently processed by Martina, the corresponding mean and standard deviation were 133.0 seconds and 38.7 seconds, respectively. Using the 0.05 level of significance, can we conclude that the population mean processing times for Fred and Martina could be the same? Using the appropriate statistical table, what is the most accurate statement we can make about the p-value for the test? Construct and interpret the 95% confidence interval for the difference between the population means. 11.22 A tire company is considering switching to a new type of adhesive designed to improve tire reliability in high-temperature and overload conditions. In laboratory “torture” tests with temperatures and loads 90% higher than the maximum normally encountered in the field, 15 tires constructed with the new adhesive run an average of 65 miles before failure, with a standard deviation of 14 miles. For 18 tires constructed with the conventional adhesive, the mean mileage before failure was 53 miles, with a standard deviation of 22 miles. Assuming normal populations and using the 0.05 level of significance, can we conclude that the new adhesive is superior to the old under such test conditions? What is the most accurate statement that could be made about the p-value for this test? 11.23 The credit manager for Braxton’s Department Store, in examining the accounts for various types of customers served by the establishment, has noticed that the mean outstanding balance for a sample of 20 customers from the local ZIP code is $375, with a standard deviation of $75. For a sample of 25 customers from a nearby ZIP code, the mean and standard deviation were $425 and $143, respectively. Assuming normal populations and using the 0.05 level of significance, examine whether credit customers from the two ZIP codes might have the same mean outstanding balance. What is the most accurate statement that could be made about the p-value for this test? Construct and interpret the 95% confidence interval for the difference between the population means. Is the hypothesized difference (0.00) within the interval? Given the presence or absence of the 0.00 value within the interval, is this consistent with the findings of the hypothesis test? 11.24 Safety engineers at a manufacturing plant are evaluating two brands of 50-ampere electrical fuses for possible purchase and use in the plant. One of the performance characteristics they are considering is how long the fuse will carry 50 amperes before it blows. In a sample of 35 fuses from the Shockley Fuse Company, the mean time was found to be 240 milliseconds, with a standard deviation of 50 milliseconds. In comparable tests of a sample of 30 fuses from the Fusemaster Corporation, the mean time was 221 milliseconds, with a standard deviation of 28 milliseconds. Using the 0.10 level of significance, examine whether the population mean times for fuses from the two companies might be the same. What is the most accurate statement that could be made about the p-value for this test? Construct and interpret the 90% confidence interval for the difference between the population means. Is the hypothesized difference (0.00) within the interval? Given the presence or absence of the 0.00 value within the interval, is this consistent with the findings of the hypothesis test? 11.25 According to a national Gallup poll, men visit the doctor’s office an average of 3.8 times per year, while women visit an average of 5.8 times per year. In a similar poll conducted in a Midwest county, a sample of 50 men visited the doctor an average of 2.2 times in the past year, with a standard deviation of 0.6 visits. For a sample of 40 women from the same county, the mean and standard deviation were 3.9 and 0.9, respectively. In a twotail test at the 0.05 level, test whether the observed difference between x}1 and x}2 is significantly different from the (3.8 2 5.8)5 22.0 visits per year that was found for the nation as a whole. Construct and interpret the 95% confidence interval for (1 2 2) for the county. Is the hypothesized difference (22.0) within the interval? Given the presence or absence of the 22.0 value within the interval, is this consistent with the findings of the hypothesis test? Source: Cindy Hall and Marcy E. Mullins, “Doctors See Women More Often,” USA Today, May 2, 2000, p. 7D. 380 Part 4: Hypothesis Testing 11.26 One of the measures of the effectiveness of a stimulus is how much the viewer’s pulse rate increases on exposure to it. In testing a lively new music theme for its television commercials, an advertising agency shows ads with the new music to a sample of 25 viewers. Their mean pulse rate increase is 20.5 beats per minute, with a standard deviation of 7.4. For a comparable sample of 25 viewers seeing the same ads with the previous music theme, the mean pulse rate increase is 16.4 beats per minute, with a standard deviation of 4.9. Assuming normal populations and using the 0.025 level of significance, can we conclude that the new music theme is better than the old in terms of increasing the pulse rate of viewers? What is the most accurate statement that could be made about the p-value for this test? ( DATA SET ) Note: Exercises 11.27–11.30 require a computer and statistical software. 11.27 According to a Yankelovich poll, women spend an average of 19.1 hours shopping during the month of December, compared to 12.7 hours for men. Assuming that file XR11027 contains the survey data underlying these results, use the 0.01 level in examining whether the sample mean for women is significantly higher than that for men. Identify and interpret the p-value for the test. Source: Anne R. Carey and Gary Visgaitis, “‘Tis the Season, Health,” USA Today 1999 Snapshot Calendar, December 8. 11.28 When companies are designing a new product, one of the steps typically taken is to see how potential buyers react to a picture or prototype of the proposed product. The product-development team for a notebook computer company has shown picture A to a large sample of potential buyers and picture B to another, asking each person to indicate what they “would expect to pay” for such a product. The data resulting from the two pictures are provided in ( ) 11.4 file XR11028. Using the 0.05 level of significance, determine whether the prototypes might not really differ in terms of the price that potential buyers would expect to pay. Identify and interpret the p-value for the test. Construct and interpret the 95% confidence interval for the difference between the population means. Is the hypothesized difference (0.00) within the interval? Given the presence or absence of the 0.00 value within the interval, is this consistent with the findings of the hypothesis test? 11.29 It has been reported that the average visitor from Japan spent $3120 during a trip to the United States, while the average for a visitor from the United Kingdom was $2654. Assuming that file XR11029 contains the survey data underlying these results, use the 0.05 level in examining whether the sample mean for Japanese visitors is significantly higher than that for visitors from the United Kingdom. Identify and interpret the p-value for the test. Source: The World Almanac and Book of Facts 2009, p. 127. 11.30 During May 2009, visitors to usatoday.com spent an average of 12.2 minutes per visit, compared to 11.0 minutes for visitors to washingtonpost.com. Assuming that file XR11030 contains the sample data underlying these results, use the 0.01 level of significance in examining whether the population mean visiting times for the two sites might really be the same. Identify and interpret the p-value for the test. Construct and interpret the 99% confidence interval for the difference between the population means. Is the hypothesized difference (0.00) within the interval? Given the presence or absence of the 0.00 value within the interval, is this consistent with the findings of the hypothesis test? Source: Jennifer Saba, “Average Time Spent on Top 30 Newspaper Web Sites Declines,” editorandpublisher.com, June 22, 2009. THE z-TEST FOR COMPARING THE MEANS OF TWO INDEPENDENT SAMPLES The z-test approximation is included in Figure 11.1 and presented here as an alternative to the unequal-variances t-test whenever both n1 and n2 are $30. Besides requiring no assumptions about the shape of the population distributions, it offers the advantages of slightly greater simplicity and avoidance of the cumbersome df correction formula used in the unequal-variances t-test; thus, it can be useful to those who are not relying on a computer and statistical software. This test has been popular for many years as a method for comparing the means of two large, independent samples when 1 and 2 are unknown, and of two independent samples of any size when 1 and 2 are known and the two populations are normally distributed. Like the unequal-variances t-test, the z-test approximation does not assume the population standard deviations are equal, and s1 and s2 are used to estimate their respective population standard deviations, 1 and 2. Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions 381 z-test approximation for comparing the means of two independent samples, 1 and 2 unknown, and each n 30: } 2x } ) 2 ( 2 ) (x 1 2 1 2 0 } and x } 5 means of samples 1 and 2 where x z 5 _____________________ 1 2 wwww s22 s21 ( 2 ) 5 hypothesized difference 1 2 0 ___ ___ between the population means n1 n2 n1 and n2 5 sizes of samples 1 and 2 s1 and s2 5 standard deviations of samples 1 and 2 Confidence interval for 1 2 2: wwww s22 s21 } 2x } )6z ___ ___ (x 1 1 2 y2 n n2 1 with 5 (1 2 confidence coefficient) Ï Ï z-Test A university’s placement center has collected data comparing the starting salaries of graduating students with surnames beginning with the letters A through M with those whose surnames begin with N through Z. For a sample of 30 students in the A–M category, the average starting salary was $37,233.33, with a standard deviation of $3475.54. For a sample of 36 students with surnames beginning with N–Z, the average starting salary was $35,855.81, with a standard deviation of $2580.02. The underlying data are in file CX11GRAD. SOLUTION For this study, the null hypothesis is that there is no difference between the population means, or H0: 1 5 2. Because the intent of the test is nondirectional, the null hypothesis can be rejected by an extreme difference in either direction, and the alternative hypothesis is H1: 1 ? 2. For testing the null hypothesis, we’ll use the 0.02 level of significance. As described in Table 11.1, the null and alternative hypotheses can also be stated as follows: • Null hypothesis H0: (2x 22x ) 5 0 1 • 2 Alternative hypothesis H1: (2x 22x ) Þ 0 1 2 The starting salaries are the same for both populations. The starting salaries are not the same. For these data, the calculated value of the test statistic, z, can be computed as } 2x } ) 2 0 (37,233.33 2 35,855.81) 2 0 (x 1 2 z 5 ____________ 5 __________________________ 5 1.80 2 wwww wwwwwwwwww s1 s22 3475.542 _________ 2580.022 _________ 1 ___ 1 ___ 30 36 n1 n2 Ï Ï For the 0.02 level of significance, the critical values will be z 5 22.33 and z 5 12.33. The decision rule will be to reject the null hypothesis of equal population means if the calculated z is either less than 22.33 or greater than 12.33, as shown in Figure 11.4. Because the calculated test statistic, z 5 1.80, falls into the nonrejection region of the diagram, the null hypothesis cannot be rejected at EXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAM EXAMPLE 382 Part 4: Hypothesis Testing FIGURE 11.4 This is an application of the z-test in comparing two sample means. From the results, we are not able to reject the possibility that graduates with surnames beginning with A–M receive the same starting salaries as graduates whose names begin with N–Z. H0: m(x1 – x2) = 0 Starting salaries are the same for the two groups. Starting salaries are not the same. H1: m(x1 – x2) ≠ 0 x1 and x2 are the mean starting salaries for the A–M and N–Z samples. Reject H0 Do not reject H0 Reject H0 Area = 0.01 Area = 0.01 m(x1 – x2) = 0 z = –2.33 z = +2.33 LEEXAMPLE Test statistic: z = +1.80 the 0.02 level of significance. From this analysis, we cannot conclude that people with surnames in the first part of the alphabet receive different starting salaries than persons whose names are in the latter portion. The approximate p-value for this test can be determined by finding the area to the right of the calculated test statistic, z 5 1.80, then (because it is a nondirectional test) multiplying this quantity by 2. Referring to the normal distribution table, we find the cumulative area to z 5 1.80 is 0.9641. Subtracting 0.9641 from 1.0000 gives the right-tail area as 0.0359. The approximate p-value for this two-tail test will be 2(0.0359), or 0.0718. Based on the sample data, we will also determine the 98% confidence interval for (1 2 2). This corresponds to 5 0.02 and, to the best accuracy possible using the normal table with z 5 2.33, the interval will be wwww s21 s22 ___ } 2x } )6z (x 1 ___ 1 2 /2 n1 n2 Ï 5 (37,233.33 2 35,855.81) 6 2.33 Ï wwwwwwwwww 3475.542 2580.022 _________ 1 _________ 30 36 5 1377.52 6 1785.99 or from 2408.47 to 13163.51 The hypothesized difference (zero) is contained within the 98% confidence interval, so we are 98% confident the population means could be the same. As we discussed in Chapter 10, a nondirectional test at the level of significance and a 100(1 2 )% confidence interval will lead to the same conclusion. Computer Solutions 11.3 describes Excel procedures for the z-test and confidence interval for the difference between population means. The computer-generated confidence interval differs very slightly from ours because our z-value (2.33) contained only the standard two decimal places from the normal distribution table. Seeing Statistics Applet 14, at the end of the chapter, allows you to visually examine the sampling distribution of the difference between sample means and how it responds to the changes you select. Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions 383 COMPUTER 11.3 SOLUTIONS The z-Test for (1 2 2) These procedures show how to use the z-test to compare the means of two independent samples. Each sample size should be 30. Excel offers the z-test; Minitab does not. EXCEL Excel z-test for (1 2 2), based on raw data 1. Open Excel data file CX11GRAD. It corresponds to Figure 11.4 and has the label and 30 salary values for the A–M group in column A, and the label and 36 salary values for the N–Z group in column B. 2. Excel does this test assuming the population variances are known. To estimate these using the sample variances, first compute the variances of the two samples: Select any available cell and enter 5VAR(A2:A31) and hit return. Select another available cell, enter 5VAR(B2:B37), and hit return. From the Data ribbon, click Data Analysis. Click z-Test: Two Sample for Means. Click OK. 3. Enter A1:A31 into the Variable 1 Range box and B1:B37 into the Variable 2 Range box. Enter 0 into the Hypothesized Mean Difference box. Click to select Labels. Specify the significance level by entering 0.02 into the Alpha box. 4. Enter the variances for columns A and B, obtained in step 2, into the Variable 1 Variance (known) and Variable 2 Variance (known) boxes, respectively. Select Output Range and enter D1 into the box. Click OK. This is a two-tail test, so we refer to the 0.072 p-value. 5. To get a confidence interval for (1 2 2), it will be necessary to refer to the Excel worksheet template and procedure described below. It is based on summary statistics for the samples. Note: For an analysis based on raw data, Excel requires the two samples to be in two distinctly separate fields (e.g., two separate columns). If the data are in one column and the subscripts identifying group membership are in another column, see the Excel Unstacking Note in Computer Solutions 11.1. Excel z-test and confidence interval for (1 2 2), based on summary data (continued ) 384 Part 4: Hypothesis Testing 1. Open Excel worksheet TMZ2TEST. Enter the hypothesized difference between the population means (0) into cell B3. Enter the summary statistics for the example on which Figure 11.4 is based into cells B6:B11. Enter the desired confidence level as a decimal fraction (e.g., 0.98) into cell B13. 2. Refer to the p-value that corresponds to the type of test being conducted. The test in Figure 11.4 is a two-tail test, so the p-value is in cell D11, or 0.072. ERCISES X E 11.31 Under what conditions is it appropriate to use the z-test as an approximation to the unequal-variances t-test when comparing two sample means? 11.32 For the following independent random samples, use the z-test and the 0.01 level of significance in testing H0: 1 5 2 versus H1: 1 ? 2. } 5 33.5 x 1 } 5 27.6 x 2 s1 5 6.4 s2 5 2.7 n1 5 31 n2 5 30 11.33 For the following independent random samples, use the z-test and the 0.05 level of significance in testing H0: 1 # 2 versus H1: 1 . 2. } 5 85.2 x 1 } 5 81.7 x 2 s1 5 9.6 s2 5 4.1 n1 5 40 n2 5 32 11.34 Repeat Exercise 11.20 using the z-test approximation to the unequal-variances t-test. of significance in determining whether the population means might actually be the same. Determine and interpret the p-value for the test; then construct and interpret the 95% confidence interval for the difference between the population means. 11.38 A study of 102 patients who had digestive-tract surgery found that those who chewed a stick of gum for 15 minutes at mealtimes ended up being discharged after an average of 4.4 days in the hospital, compared to an average of 5.2 days for their counterparts who were not provided with gum. Assuming (1) a sample size of 51 for each group and (2) sample standard deviations of 1.3 and 1.9 days, respectively, use a z-test and the 0.01 level of significance in examining whether the mean hospital stay for the gum-chewing patients could have been this much lower simply by chance. Determine and interpret the p-value for the test. Source: Jennifer Bails, “Chewing Gum Might Help Digestive-Tract Patients Get Home Sooner,” Pittsburgh Tribune-Review, October 15, 2005, p. A4. 11.35 Repeat Exercise 11.21 using the z-test approxima- tion to the unequal-variances t-test. 11.36 According to the U.S. Bureau of Labor Statistics, personal expenditures for entertainment fees and admissions averaged $349 per person in the Northeast and $420 in the West. Assuming that these data involved (1) sample sizes of 800 and 600 and (2) standard deviations of $215 and $325, respectively, use a z-test and the 0.01 level of significance in testing the difference between these means. Determine and interpret the p-value for the test. Construct and interpret the 99% confidence interval for the test. Source: U.S. Bureau of Labor Statistics, Consumer Expenditure Survey, Interview Survey, annual. 11.37 Surveys of dog owners in two adjacent towns found the average veterinary-related expense total for dog owners in town A during the preceding year was $240, with a standard deviation of $55, while the comparable results in nearby town B were $225 and $32, respectively. Assuming a sample size of 30 dog owners surveyed in each town, use a z-test and the 0.05 level 11.39 A study found that women of normal weight missed an average of 3.4 days of work due to illness during the preceding year, compared to an average of 5.2 days for women who were considered overweight. Assuming (1) a sample size of 30 for each group and (2) sample standard deviations of 2.5 and 3.6 days, respectively, use a z-test and the 0.05 level of significance in examining whether the mean number of absence days for the overweight group could have been this much higher simply by chance. Determine and interpret the p-value for the test. Source: Nanci Hellmich, “Heavy Workers, Hefty Price,” USA Today, September 9, 2005, p. 7D. 11.40 According to a psychologist at the Medical College of Virginia, soccer players who “head” the ball 10 or more times a game are at risk for brain damage that could lower their intellectual abilities. In tests involving young male soccer players, those who typically headed the ball 10 or more times a game recorded a mean IQ score of 103, compared to a mean of 112 for those who usually headed the ball once or less a game. Assuming 30 players in each sample, with sample Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions standard deviations of 17.4 and 14.5 IQ points for the “headers” and “nonheaders,” respectively, use an appropriate one-tail z-test and the 0.01 level of significance in reaching a conclusion as to whether frequent “heading” of the ball by soccer players might lower intellectual performance. Determine and interpret the p-value for the test. Source: Marilyn Elias, “Heading Soccer Ball Can Lower IQ,” USA Today, August 14, 1995, p. 1D. ( DATA SET ) Note: Exercises 11.41–11.43 require a computer and statistical software. 11.41 In their 2003 and 2008 studies on how long wireless customers have to spend on hold before speaking with a customer service representative, J.D. Power and Associates found the mean times to be 3.3 minutes and 4.4 minutes, respectively. Assume that data file XR11041 contains the data, in minutes, for times on hold during the 2003 and 2008 studies. Use a one-tail test and the 0.05 level of significance in concluding whether the population mean time for 2008 could be greater than that for 2003. Determine and interpret the p-value for the test. Source: J.D. Power and Associates press release, “Customer Hold Times for Wireless Phone Customers Reach an All-Time High,” jdpower.com, August 14, 2008. 11.42 In planning the processes to be incorporated into a new manufacturing facility, engineers have proposed two possible assembly procedures for one of the phases in the production sequence. Sixty of the eventual production line workers have taken part in preliminary tests, with data representing productivity (units produced in 1 hour) for 30 workers using procedure A and 30 using procedure B. Given the data in file XR11042, use the 0.10 level of significance in determining whether the two procedures might be equally efficient. Identify and interpret the p-value for the test, then construct and interpret the 90% confidence interval for the difference between the population means. 11.43 A study has been conducted to examine the effectiveness of a new experimental program for preparing high school students for the Scholastic Aptitude Test. Eighty students have been randomly divided into two groups of 40. The eventual SAT scores for those exposed to the new program and the conventional program are listed in data file XR11043. Use a one-tail test and the 0.025 level of significance in concluding whether the experimental program might be better than the conventional method in preparing students for the SAT. Determine and interpret the p-value for the test. ( ) COMPARING TWO MEANS WHEN THE SAMPLES ARE DEPENDENT 11.5 In previous comparisons of two sample means, the samples were independent from each other. That is, the selection process for one sample was not related to the selection process for the other. However, there may be times when we wish to test hypotheses involving samples that are not independent. For example, we may wish to examine the before-and-after productivity of individual employees after a change in their workstation layout, or compare the before-and-after reading speeds of individual participants in a speed-reading course. In such cases, we do not really have two different samples of persons, but rather before and after measurements for the same individuals. As a result, there will be just one variable: the difference recorded for each individual. Tests in which the samples are not independent are also referred to as paired observations, or matched pairs, and they are essentially the same as those discussed in Chapter 10 for the mean of a single sample. The variable under consideration in this case is d 5 (x1 2 x2), where x1 and x2 are the before and after measurements, respectively. As in the tests for one sample mean, the null and alternative hypotheses will be one of the following, with the test statistic calculated as shown here: Null Hypothesis H0: H0: H0: d 5 0 d $ 0 d # 0 Alternative Hypothesis H1: H1: H1: d ± 0 d , 0 d . 0 385 Type of Test Two-tail Left-tail Right-tail 386 Part 4: Hypothesis Testing Test statistic for comparing the means for paired observations: } d where d 5 for each individual or test unit, (x1 2 x2), t 5 _______ sd Ïw n the difference between the two measurements } d 5 the average difference, 5 odi yn n 5 number of pairs of observations wwwwww o d i2 2 nd} 2 sd 5 the standard deviation of d, or ___________ n21 df 5 n 2 1 Confidence interval for d: sd } d 6 ty2 ____ n Ïw Ï EXAMPLEEXAMPLEEXA EXAMPLE Dependent Samples Exploring ways to increase office productivity, a company vice president has ordered 12 ergonomic keyboards for distribution to a sample of secretarial employees. If the keyboards substantially increase productivity, she plans to replace all of the firm’s current keyboards with the new models. Prior to delivery of the keyboards, each of the 12 sample members types a standard document on his or her old keyboard, and the number of words per minute is measured. After receiving the new keyboards and spending a few weeks becoming familiar with their operation, each employee then types the same document using the ergonomic model. Table 11.2 shows the number of words per minute each of the 12 persons typed in each test. The data are also in file CX11TYPE. TABLE 11.2 For dependent samples, only one variable is tested: the difference between measurements. For each of 12 individuals, the typing speed for generating a standard document is shown before and after learning to use an ergonomic keyboard. Person x1, Words/Minute with Old Keyboard x2, Words/Minute with New Keyboard Difference d 5 (x1 2 x2) d2 1 2 3 4 5 6 7 8 9 10 11 12 25.5 59.2 38.4 66.8 44.9 47.4 41.6 48.9 60.7 41.0 36.1 34.4 43.6 69.9 39.8 73.4 50.2 53.9 40.3 58.0 66.9 66.5 27.4 33.7 218.1 210.7 21.4 26.6 25.3 26.5 1.3 29.1 26.2 225.5 8.7 0.7 327.61 114.49 1.96 43.56 28.09 42.25 1.69 82.81 38.44 650.25 75.69 0.49 278.7 5 ^d 1407.33 5 ^d 2 SOLUTION Because the vice president doesn’t want to replace the current stock of keyboards unless the ergonomic model is clearly superior, the burden of proof is on the new model, and a one-tail test is appropriate. The 0.025 level will be used to examine whether the new keyboard has significantly increased typing speeds. For each person in the sample, the difference in typing speed between the first and second measurements is d 5 (x1 2 x2) words per minute. The null and alternative hypotheses will be as follows: • Null hypothesis H0: • d $ 0 Typing with the ergonomic keyboard is no faster than with the current keyboard. Alternative hypothesis H1: d , 0 The ergonomic keyboard is faster. The sample mean and standard deviation for d are calculated as in Chapter 3, and can be expressed as 2 10.7 2 1.4 2 6.6 2 5.3 2 6.5 1 1.3 2 9.1 2 6.2 2 25.5 1 8.7 1 0.7 2 218.1 d 5 _______________________________________________________________ _____ 12 5 26.558 sd 5 Ï }2 wwwww 2 o di 2 nd __________ n21 5 Ï wwwwwwwwwwww 1407.33 2 12(26.558)2 ______________________ 5 9.001 12 2 1 and the test statistic is calculated as } 26.558 d t 5 ______ 5 ___________ 5 22.524 sdYÏw n 9.001YÏww 12 The number of degrees of freedom for the test is df 5 (n 2 1), or (12 2 1) 5 11. For the 0.025 level of significance in a left-tail test, the critical value for the test statistic will be t 5 22.201. This is obtained by referring to the 5 0.025 column and df 5 11 row of the table. The decision rule is, “Reject the null hypothesis if the calculated test statistic is less than t 5 22.201, otherwise do not reject.” As Figure 11.5 shows, the calculated test statistic is less than the critical value and falls into the rejection region for the test. As a result, the null hypothesis is rejected, and we conclude that the ergonomic keyboard does increase typing speeds. Following through with the intent of her test, the vice president should order them for all secretarial personnel. Based on the sample data, we will also determine the 95% confidence interval for d. This corresponds to 5 0.05. With df 5 12 2 1 5 11 and t 5 2.201, the interval will be sd } 9.001 d 6 ty2 ____ 5 26.558 6 2.201 ______ n Ïw 12 Ïww 5 26.558 6 5.719, or from 212.277 to 20.839 PLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPL Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions 387 388 Part 4: Hypothesis Testing FIGURE 11.5 A summary of the hypothesis test for the paired observations in Table 11.2. At the 0.025 level in a one-tail test, we conclude that the ergonomic keyboard increases typing speeds. H0: md ≥ 0 The ergonomic keyboard is no faster than the current keyboard. The ergonomic keyboard increases typing speed. H1: md < 0 For each person, d = x1 – x2; x1 = words/minute with current keyboard. x2 = words/minute with ergonomic keyboard. Reject H0 Do not reject H0 Area = 0.025 md = 0 t = –2.201 Test statistic: t = –2.524 Computer Solutions 11.4 describes Excel and Minitab procedures for the t-test and confidence interval when comparing the means of dependent samples. In a one-tail test, Minitab will provide either an upper or a lower confidence limit for d , depending on the directionality of the test. COMPUTER 11.4 SOLUTIONS Comparing the Means of Dependent Samples These procedures show how to use a t-test to compare sample means when the samples are not independent. EXCEL Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions 389 Excel t-test for comparing the means of dependent samples, based on raw data 1. For the sample data (Excel file CX11TYPE) on which Figure 11.5 is based, with the label and 12 “old keyboard” data values in column A, and the label and 12 “new keyboard” data values in column B: From the Data ribbon, click Data Analysis. Click t-Test: Paired Two-Sample For Means. Click OK. 2. Enter A1:A13 into the Variable 1 Range box and B1:B13 into the Variable 2 Range box. Enter 0 into the Hypothesized Mean Difference box. Click to select Labels. Specify the significance level by entering 0.025 into the Alpha box. Select Output Range and enter D1 into the box. Click OK. This is a one-tail test, so we refer to the 0.014 p-value. 3. To obtain a confidence interval for (1 2), it will be necessary to refer to the procedure described below. It is based on summary statistics for the samples. Excel t-test for comparing the means of dependent samples, based on summary statistics 1. Using the summary statistics associated with CX11TYPE (the test is summarized in Figure 11.5): Open the TEST STATISTICS workbook. 2. Using the arrows at the bottom left, select the t-Test_Mean worksheet. 3. For d 5 x1 2 x2, enter the mean of d (26.558), the standard deviation of d (9.001), and the number of pairs (12) into the appropriate cells. Enter the hypothesized difference (0) and the desired alpha level for the test (0.025). The calculated t-statistic and a one-tail p-value will be shown at the right. 4. To get a confidence interval for d , follow steps 1–3, but open the ESTIMATORS workbook and select the t-Estimate_Mean worksheet. Enter the desired confidence level as a decimal fraction (e.g., 0.95). Note: As an alternative, you can use Excel worksheet template TMTTEST. The steps are described within the template. MINITAB Minitab t-test for comparing the means of dependent samples, based on raw data 1. For example, using the data (Minitab file CX11TYPE) on which Figure 11.5 is based, with the “old keyboard” data values in column C1 and the “new keyboard” data values in column C2: Click Stat. Select Basic Statistics. Click Paired t. 2. Select Samples in columns and enter C1 into the First sample box and C2 into the Second sample box. 3. Click Options. Enter the desired confidence level as a percentage (e.g., 95.0) into the Confidence Level box. Enter the hypothesized difference (0) into the Test mean box. Within the Alternative box, select less than. Click OK. Click OK. Minitab t-test for comparing the means of dependent samples, based on summary data Follow steps 1 through 3 above, but select Summarized data (differences) in step 2 and insert the appropriate summary statistics (number of pairs, the mean of d, and the standard deviation of d) into the Sample size, Mean, and Standard deviation boxes. 390 Part 4: Hypothesis Testing ERCISES X E 11.44 A pharmaceutical firm has checked the cholesterol levels for each of 30 male patients, then provided them with fish-oil capsules to take on a daily basis. The cholesterol levels are rechecked after a 1-month period. Does this study involve independent samples or dependent samples? 11.45 Each of 20 consumers is provided with a pack- age containing two different brands of instant coffee. A week later, they are asked to rate the taste of each coffee on a scale of 1 (poor taste) to 10 (excellent taste). Is this an example of independent samples or dependent samples? 11.46 A university president randomly selects 10 tenured faculty from the College of Arts and Sciences and 10 tenured faculty from the College of Business. Each faculty member is then asked to rate his or her job satisfaction on a scale of 1 (very dissatisfied) to 10 (very satisfied). Would this be an example of independent samples or dependent samples? 11.47 A trucking firm is considering the installation of a new, low-restriction engine air filter for its long-haul trucks, but doesn’t want to make the switch unless the new filter can be shown to improve the fuel economy of these vehicles. A test is set up in which each of 10 trucks makes the same run twice—once with the old filtration system and once with the new version. Given the sample results shown below, use the 0.05 level of significance in determining whether the new filtration system could be superior. Truck Number 1 2 3 4 5 6 7 8 9 10 Current Filter New Filter 7.6 mpg 5.1 10.4 6.9 5.6 7.9 5.4 5.7 5.5 5.3 7.3 mpg 7.2 6.8 10.6 8.8 8.7 5.7 8.7 8.9 7.1 11.48 In an attempt to measure the emotional effect of a proposed billboard ad, an advertising agency checks the pulse rate of 10 persons before and after they are shown a photograph of the billboard. The agency believes that an effective billboard will increase the pulse rate of those who view it. In its test, the agency found the mean change in pulse rate was 15.7 beats per minute, with a standard deviation of 1.6. Using the 0.01 level of significance, examine whether the billboard stimulus could meet the agency’s criterion for effectiveness. 11.49 The students in an aerobics class have been weighed both before and after the 5-week class, with the following results: Person Number Weight Before Weight After 1 2 3 4 5 6 7 8 198 lb 154 124 110 127 162 141 180 194 lb 151 126 104 123 155 129 165 Using the 0.05 level of significance, evaluate the effectiveness of the program. Using the appropriate statistical table, what is the most accurate statement we can make about the p-value for this test? ( DATA SET ) Note: Exercises 11.50 and 11.51 require a computer and statistical software. 11.50 For a special pre–New Year’s Eve show, a radio station personality has invited a small panel of prominent local citizens to help demonstrate to listeners the adverse effect of alcohol on reaction time. The reaction times (in seconds) before and after consuming four drinks are in data file XR11050. At the 0.005 level, has the program host made his point? Identify and interpret the p-value for the test. 11.51 A plant manager has collected productivity data for a sample of workers, intending to see whether there is a difference in the number of units they produce on Monday versus Thursday. The results (units produced) are in data file XR11051. Using the 0.01 level of significance, evaluate the null hypothesis that there is no difference in worker productivity between the two days. Identify and interpret the p-value for the test, then construct and interpret the 99% confidence interval for the mean difference in productivity between the days. Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions 391 ( ) COMPARING TWO SAMPLE PROPORTIONS 11.6 The comparison of sample proportions from two independent samples is a frequent subject for statistical analysis. The following are but a few of the possibilities: • • • Comparing the percentage of defective parts between shipments provided by two different suppliers. Determining whether the proportion of headache sufferers getting relief from a new medication is significantly greater than for those using aspirin. Comparing the enlistment percentage of high school seniors who have viewed version A of a recruiting film versus those seeing version B. In this section, tests assume that both sample sizes are large (each n $ 30). In addition, n1p1, n1(1 2 p1), n2p2, and n2(1 2 p2) should all be $5. (These requirements are necessary in order that the normal distribution used here will be a close approximation to the binomial distribution.) As in the comparison of means from independent samples, tests involving proportions can be either nondirectional or directional. Possible null and alternative hypotheses are similar to those summarized in Table 11.1. Unlike the previous sections, our choice of test statistic will depend on the hypothesized difference between the population proportions, (1 2 2)0. In the vast majority of practical applications, the hypothesized difference will be zero and the appropriate test statistic will be the first of the two alternatives shown below. Accordingly, that will be our emphasis in this section. The confidence interval for (1 2 2) is not affected. Test statistic for comparing proportions of two independent samples: 1. When the hypothesized difference is zero (the usual case): (p1 2 p2) z 5 ___________________ where p1 and p2 5 the sample proportions wwwwwwwww 1 1 } } ___ ___ n1 and n2 5 the sample sizes p(1 2 p ) 1 n1 n2 } 5 pooled estimate of the p population proportion n1p1 1 n2p2 } 5 ____________ with p n1 1 n2 Ï 2. ( ) When the hypothesized difference is (1 2 2)0 Þ 0: (p1 2 p2) 2 (1 2 2)0 z 5 _________________________ wwwwwwwwwwww p1(1 2 p1) p2(1 2 p2) __________ 1 __________ n1 n2 Ï In either case, confidence interval for (1 2 2): wwwwwwwwwwww p (1 2 p ) p (1 2 p ) Ï (p1 2 p2) 6 z/2 1 1 2 2 __________ 1 __________ n1 n2 Sample Proportions In a 10-year study sponsored by the National Heart, Lung and Blood Institute, 3806 middle-age men with high cholesterol levels but no known heart problems were divided into two groups. Members of the first group received a new drug designed to lower cholesterol levels, while the second group received daily EXAMPLEEXAMPL EXAMPLE XAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEX 392 Part 4: Hypothesis Testing dosages of a placebo. Besides lowering cholesterol levels, the drug appeared to be effective in reducing the incidence of heart attacks. During the 10 years, 155 of those in the first group suffered a heart attack, compared to 187 in the placebo group.1 Assume the underlying data are in file CX11HRT, coded as 1 5 did not have a heart attack, and 2 5 had a heart attack. SOLUTION If we assume the 3806 participants were randomly divided into two groups, there would have been 1903 men in each group. Under this assumption, the sample proportions for heart attacks within the two groups are p1 5 155y1903, or p1 5 0.0815, and p2 5 187y1903, or p2 5 0.0983. Since the intent of the study was to evaluate the effectiveness of the new drug, the hypothesis test will be directional. In terms of the population proportions, the null and alternative hypotheses are H0: 1 $ 2 and H1: 1 , 2. The hypotheses can also be expressed as • Null hypothesis H0: • (p 2p ) $ 0 1 Users of the new drug are at least as likely to experience a coronary. 2 Alternative hypothesis H1: (p 2p ) , 0 1 Users of the new drug are less likely to experience a coronary. 2 In testing the null hypothesis, we will use the 0.05 level of significance. The pooled estimate of the (assumed equal) population proportions is calculated as n1p1 1 n2p2 (1903)(0.0815) 1 (1903)(0.0983) } 5 ___________ p 5 ______________________________ 5 0.0899 n 1 1 n2 1903 1 1903 The calculated value of the test statistic, z, is p 1 2 p2 0.0815 2 0.0983 z 5 ___________________ 5 __________________________________ 5 21.81 wwwwwwww wwwwwwwwwwwwwwww 1 1 1 1 }(1 2 p }) ___ p 1 ___ 0.0899(1 2 0.0899) _____ 1 _____ n1 n2 1903 1903 Ï ( ) Ï ( ) For the 0.05 level in this left-tail test, the critical value of z will be z 5 21.645. The decision rule is, “Reject H0 if the calculated test statistic is , 21.645, otherwise do not reject.” As Figure 11.6 shows, the calculated test statistic, z 5 21.81, is less than the critical value and falls into the rejection region. At the 0.05 level of significance, the null hypothesis is rejected, and we conclude that the new medication is effective. Using the normal distribution table, we find the cumulative area to z 5 21.81 to be 0.0351. This is the approximate p-value for the test. Based on the sample data, we will also determine the 90% confidence interval for (1 2 2). With z 5 1.645, this will be (p1 2 p2) 6 zy2 Ï 5 (0.0815 2 0.0983) 6 1.645 Ï wwwwwwwwwww p (1 2 p ) p (1 2 p ) 1 1 2 2 __________ 1 __________ n1 wwwwwwwwwwwwwwwwwwww 0.0815(1 2 0.0815) 0.0983(1 2 0.0983) __________________ 1 __________________ 1903 5 20.0168 6 0.0152, or from 20.0320 to 20.0016 1Source: n2 1903 “News from the World of Medicine,” Reader’s Digest, May 1984, p. 222. Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions 393 FIGURE 11.6 H0: m(p1 – p2) ≥ 0 The medication is no more effective than the placebo. H1: m(p1 – p2) < 0 The medication reduces the likelihood of a coronary. p1 and p2 are the proportions of each sample who had a coronary during the 10 years of the study. Reject H0 At the 0.05 level of significance, the medication that was the subject of this study appears to have been effective in reducing the incidence of heart attacks in middle-age men. Do not reject H0 Area = 0.05 m(p1 – p2) = 0 z = –1.645 Test statistic: z = –1.81 Computer Solutions 11.5 describes Excel and Minitab procedures for the z-test and confidence interval when comparing the proportions from two independent samples. In a one-tail test, Minitab will provide either an upper or a lower confidence limit for (1 2 2), depending on the directionality of the test. COMPUTER 11.5 SOLUTIONS The z-Test for Comparing Two Sample Proportions These procedures show how to use the z-test to compare proportions from two independent samples. EXCEL—USING SUMMARY DATA (continued ) 394 Part 4: Hypothesis Testing Excel comparison of p1 and p2 when the hypothesized value of (1 2 2) is 0, for summary data 1. For the summary data described in Figure 11.6: Open the TEST STATISTICS workbook. 2. Using the arrows at the bottom left, select the z-Test_2 Proportions (Case 1). For each sample, enter the sample proportion and the sample size, as shown above. You can also enter the alpha level for the test (0.05). Along with the z statistic, the printout lists both one-tail and two-tail p-values. Because this is a one-tail test, the p-value is 0.035. Note: As an alternative, you can use Excel worksheet template TM2PTEST. The steps are described within the template. Excel comparison of p1 and p2 when the hypothesized value of (1 2 2) is not 0, for summary data Follow the procedure above, but select z-Test_2 Proportions (Case 2) in step 2, then specify the value of (1 2 2) associated with the null hypothesis. Excel confidence interval for (1 2 2), based on summary data Follow the first procedure, but open the ESTIMATORS workbook and select the z-Estimate_2 Proportions worksheet, then specify the desired confidence level as a decimal fraction (e.g., 0.90). Note: As an alternative, you can use Excel worksheet template TM2PTEST. The steps are described within the template. EXCEL—USING RAW DATA Excel comparison of p1 and p2 for raw data, regardless of the hypothesized value of (1 2 2) 1. For the sample data (Excel file CX11HRT) on which Figure 11.6 is based, with the labels and data values for the drug and placebo groups in columns A and B, and coded as 1 5 did not have a heart attack and 2 5 had a heart attack: Click on any cell within the data field. From the Add-Ins ribbon, click Data Analysis Plus. Click Z-Test: Two Proportions. Click OK. 2. Enter A1:A1904 into the Variable 1 Range box. Enter B1:B1904 into the Variable 2 Range box. Enter 2 into the Code for Success box. Enter the hypothesized difference (in this case, 0) into the Hypothesized Difference box. Click Labels. Enter 0.05 into the Alpha box. Click OK. Excel confidence interval for (1 2 2), based on raw data Follow the preceding procedure, but select Z-Estimate: Two Proportions from the Data Analysis Plus menu. Specify the desired confidence interval by entering the appropriate value for alpha. For example, to get a 95% confidence interval, input 0.05 into the Alpha box in step 2. MINITAB—USING SUMMARY DATA Minitab comparison of p1 and p2 when the hypothesized value of (1 2 2) is 0, for summary data 1. Using the summary statistics associated with Figure 11.6: Click Stat. Select Basic Statistics. Click 2 Proportions. Select Summarized Data. For sample 1, enter the sample size (1903) into the Trials box. Multiply the sample Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions 395 proportion (0.0815) times the sample size (1903) to get the number of “successes” (0.0815)(1903) 5 155.1, rounded to 155, and enter this into the Events box for sample 1. Repeat this for sample 2, entering 1903 into the Trials box and (0.0983)(1903) 5 187.1, rounded to 187 into the Events box. 2. Click Options. Enter the desired confidence level as a percentage (90.0) into the Confidence Level box. Enter the hypothesized difference between the population proportions (0) into the Test difference box. Within the Alternative box, select less than. Click to select Use pooled estimate of p for test. Click OK. Click OK. Minitab comparison of p1 and p2 when the hypothesized value of (1 2 2) is not 0, for summary data Follow the previous procedure, but in step 2 do NOT select “Use pooled estimate of p for test.” Specify the hypothesized nonzero difference between the population proportions. MINITAB —USING RAW DATA Minitab comparison of p1 and p2 when the hypothesized value of (1 2 2) is 0, for raw data 1. For the sample data (Minitab file CX11HRT) on which Figure 11.6 is based, with the data for the drug group in column C1, the data for the placebo group in column C2, and data coded as 1 5 did not have a heart attack and 2 5 had a heart attack: Click Stat. Select Basic Statistics. Click 2 Proportions. Select Samples in different columns. Enter C1 into the First box and C2 into the Second box. Note that Minitab will select the larger of the two codes (i.e., 2 5 had heart attack) as the “success” or “event.” 2. Click Options. Enter the desired confidence level as a percentage (90.0) into the Confidence Level box. Enter the hypothesized difference between the population proportions (0) into the Test difference box. Within the Alternative box, select less than. Click to select Use pooled estimate of p for test. Click OK. Click OK. Minitab comparison of p1 and p2 when the hypothesized value of (1 2 2) is not 0, for raw data Follow the previous procedure, but in step 2 do NOT select “Use pooled estimate of p for test.” Specify the hypothesized nonzero difference between the population proportions. ERCISES X E 11.52 Summary data for two independent samples are p1 5 0.36, n1 5 150, and p2 5 0.29, n2 5 100. Use the 0.025 level of significance in testing H0: 1 # 2 versus H1: 1 . 2. 11.53 Summary data for two independent samples are p1 5 0.31, n1 5 400, and p2 5 0.38, n2 5 500. Use the 0.05 level of significance in testing H0: 1 5 2 versus H1: 1 Þ 2. 11.54 A bank manager has been presented with a new brochure that was designed to be more effective in attracting current customers to a personal financial counseling session that would include an analysis of additional banking services that could be advantageous to both the bank and the customer. The manager’s assistant, who created the new brochure, randomly selects 400 current customers, then randomly chooses 200 to receive the standard brochure that has been used in the past, with the other 200 receiving the promising new brochure that he has developed. Of those receiving the standard brochure, 35% call for more information about the counseling session, while 42% of those receiving the new brochure call for more information. Using the 0.10 level of significance, is it possible that the superior performance of the new brochure was just due to chance and that the new brochure might really be no better than the old one? 11.55 In examining the ability of users to complete a variety of information-seeking tasks on their mobile devices, Nielsen Norman Group assigned sample members to get the answers to a variety of questions like, “How many calories are there typically in a slice of thincrust pizza?” The success rate for achieving such tasks was 75% for those using touch-screen phones like the iPhone to access the mobile Internet, compared to 80% for persons accessing websites on a conventional personal computer. Assuming these percentages to be based 396 on independent samples of 500 persons each, and using the 0.01 level of significance, can we conclude that the population success rate with a conventional PC is greater than that when using the mobile Internet and touchscreen phone? Source: “Study: Mobile Web a Throwback to ‘90s,” Part 4: Hypothesis Testing and the 0.10 level of significance in examining the difference between these two rates. Determine and interpret the p-value for the test. Source: John Diamond, “Some Veterans Suffer Bone, Muscle Ailments,” The Indiana Gazette, January 22, 1997, p. 5. United Dairy Industry Association, Cottage Cheese: Attitudes & Usage, 11.62 According to the ICR Research Group, 63% of Americans in the 18–34 age group say they are comfortable filing income tax returns electronically, compared to just 49% of those who are 55–64. Using the 0.025 level of significance, and assuming there were 200 persons surveyed from each age group, examine whether Americans in the 18–34 age group might be more comfortable with electronic filing than their counterparts in the 55–64 group. Determine and interpret the p-value for the test. p. 21. Source: Anne R. Carey and Genevieve Lynn, “E-Filing: No Problem,” USA USA Today, July 20, 2009, p. 3B. 11.56 In a study by the United Dairy Industry Association, 42% of 655 persons age 35–44 said they seldom or never ate cottage cheese. For 455 individuals age 45–54, the corresponding percentage was 34%. At the 0.01 level, can we reject the possibility that the population percentages could be equal for these two groups? Source: 11.57 During the course of a year, 38.0% of the 213 merchant ships lost were general cargo carriers, while 43.8% of the 219 ships lost during the following year were in this category. Using a two-tail test at the 0.05 level, examine whether this difference could have been the result of chance variation from one year to the next. Source: Bureau of the Census, Statistical Abstract of the United States 1996, p. 657. 11.58 Of 200 subjects approached by interviewer A, 45 refused to be interviewed. Of 120 approached by interviewer B, 42 refused an interview. At the 0.05 level of significance, can we reject the possibility that the interviewers are equally capable of obtaining interviews? 11.59 Community National Bank is using an observational study to examine the utilization of its new 24-hour banking machine. Of 300 males who used the machine last week, 42% made two or more transactions before leaving. Of 250 female users during the same period, 50% made at least two transactions while at the machine. At the 0.10 level, do males and females differ significantly in terms of making multiple transactions? Determine and interpret the p-value for this test, then construct and interpret the 90% confidence interval for 1 2 2. 11.60 A telephone sales solicitor, trying to decide between two alternative sales pitches, randomly alternated between them during a day of calls. Using approach A, 20% of 100 calls led to requests for the mailing of additional product information. For approach B in another 100 calls, only 14% led to requests for the product information mailing. At the 0.05 level, can we conclude that the difference in results was due to chance? Determine and interpret the p-value for this test, then construct and interpret the 95% confidence interval for 1 2 2. 11.61 In a preliminary study, the U.S. Veterans Affairs Department found that 30.9% of the 81 soldiers who were near an accidental nerve gas release just after the 1991 Persian Gulf War had muscle and bone ailments, compared to a rate of 23.5% for the 52,000 Gulf veterans who were not near that area. Use an appropriate one-tail test Today, March 22, 2000, p. 1B. ( DATA SET ) Note: Exercises 11.63–11.65 require a computer and statistical software. 11.63 An American Express survey of small-business owners found that 71% of female owners feel stressed by their work/life balance, compared to 62% of their male counterparts. Assume that file XR11063 contains the underlying sample data for female and male smallbusiness owners, respectively, with the data coded so that 1 5 feels no stress in work/life balance and 2 5 feels stress in work/life balance. Using the 0.01 level of significance, evaluate the null hypothesis that female and male small-business owners might really be equally likely to feel stress in their work/life balance. Identify and interpret the p-value for the test, then construct and interpret the 99% confidence interval for the difference between the population proportions. Source: Jeff Cornwall, “Small Business Owners Still Struggle with Balance,” smartbiz.com, May 30, 2007. 11.64 Attempting to improve the quality of services pro- vided to customers, the owner of a chain of high-fashion department stores randomly selected a number of clerks for special training in customer relations. Of this group, only 10% were the subject of complaints to the store manager during the 3 months following the training. On the other hand, 15% of a sample of untrained clerks were mentioned in customer complaints to the manager during this same period. The data are in file XR11064, with data for each group coded as 1 5 not mentioned in a complaint and 2 5 mentioned in a complaint. Using the 0.05 level of significance, does the training appear to be effective in reducing the incidence of customer dissatisfaction with sales personnel? Identify and interpret the p-value for the test. 11.65 A study by the National Marine Manufacturers Association found that 12.2% of those who participated in sailing were females age 25–34. Of those who participated in horseback riding during the same period, 14.7% were females in this age group. Assume that file XR11065 contains the underlying data for each activity group, Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions coded as 1 5 not a female in this age group and 2 5 a female in this age group. Using the 0.10 level of significance, test whether the population proportion could be equal for females age 25–34 participating in each of these activities. Identify and interpret the p-value for the test, then construct and interpret the 90% confidence interval for the difference between the population proportions. Source: National Marine Manufacturers Association, The Boating Market: A Sports Participation Study, p. 40. COMPARING THE VARIANCES OF TWO INDEPENDENT SAMPLES There are occasions when it is useful to compare the variances of two independent samples. For example, we might be interested in whether one manufacturing process differs from another in terms of the amount of variation among the units produced. We can examine two different portfolio strategies to determine whether there is significantly more variation in the performances of the investments in one of the portfolios than in the other. We can also compare the variances of two independent samples to determine the permissibility of using the pooled-variances t-test of Section 11.2, which assumes that the standard deviations (and, thus, the variances as well) of the respective populations are equal. The test in this section involves the F distribution. Like the t distribution, it is a family of distributions and is continuous. Unlike the t distribution, however, its exact shape is determined by two different degrees of freedom instead of just a single value. From a theoretical standpoint, the F distribution is the sampling distribution of s21ys22 that would result if two samples were repeatedly drawn from the same, normally distributed population. In terms of the hypothesis-testing procedure introduced in Chapter 10, the test can be described as follows: 1. Formulate null and alternative hypotheses. The null and alternative hypoth2 2 2 2 eses are H0: 1 5 2 and H1: 1 2. 2. Select the significance level, . There are three F distribution tables in Appendix A (Table A.6, Parts A–C). They represent upper-tail areas of 0.05, 0.025, and 0.01, respectively. Since these are one-tail areas, they represent 5 0.10, 5 0.05, and 5 0.02 for our two-tail test. 3. Calculate the test statistic. The calculated test statistic is s21 s22 __ __ F5 or , whichever is larger s22 s21 4. Identify the critical value of the test statistic and state the decision rule. Although 2 2 the test is nondirectional (i.e., H1: 1 2), there will be just one critical value of F. This is because we have selected the larger of the two ratios in step 3. The critical value of F will be F(y2, n1, n2) 397 where 5 specified level of significance: 0.10, 0.05, or 0.02 n1 5 (n 2 1), where n is the size of the sample that had the larger variance n2 5 (n 2 1), where n is the size of the sample that had the smaller variance The critical value is found by consulting the F table that corresponds to y2 (0.05, 0.025, or 0.01), with n1 5 the number of degrees of freedom associated with the numerator of the F ratio, and n2 5 the number of degrees of freedom for the denominator. If n1 or n2 happens to be one of the larger values not ( ) 11.7 398 Part 4: Hypothesis Testing included in the table, interpolate between the listed entries. The decision rule is, “Reject H0 if calculated F . critical F, otherwise do not reject.” 5. Compare the calculated and critical values and reach a conclusion. If the calculated F exceeds the critical F, we are not able to assume that the population variances are equal. 6. Make the related decision. This will depend on the purpose for the test. For example, if the variance for one investment portfolio strategy differs significantly from the variance for another, we may wish to pursue the one for which the variance is lower. The ability to assume equal variances would also allow us to compare two sample means with the pooled-variances t-test of Section 11.2. However, keep in mind that the unequal-variances t-test of Section 11.3 can routinely be applied without having to go through the inconvenience of the test for variance equality. EXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEE EXAMPLE Comparing Variances A sample of 9 technicians exposed to the standard version of a training film required an average of 31.4 minutes to service a compressor system, with a standard deviation of 14.5 minutes. For 7 technicians viewing an alternative version of the film, the average time required was 22.3 minutes, with a standard deviation of 10.2 minutes. If the sampled populations are approximately normally distributed, 2 2 can 1 and 2 be assumed to be equal? The underlying data are in file CX11TECH. SOLUTION Formulate Null and Alternative Hypotheses The null and alternative hypotheses are H0: 1 5 2 and H1: 1 2. 2 2 2 2 Select the Significance Level, For this test, the level of significance will be 5 0.02. Calculate the Test Statistic The calculated F statistic will be s22 10.22 s2 14.52 __1 5 _____ __ or whichever is larger 5 _____, s22 10.22 s21 14.52 Since the first ratio is larger, the calculated F is 14.52 _____ 5 2.02 10.22 Identify the Critical Value of the Test Statistic and State the Decision Rule The sample associated with the numerator of the F statistic is the one having the larger variance. It had a sample size of 9, so n1 is 9 2 1, or n1 5 8. The sample associated with the denominator of the F statistic is the one having the smaller variance. It had a sample size of 7, so n2 5 7 2 1, or n2 5 6. The specified level of significance was 5 0.02, and the critical value of F will be F(y2, n1, n2) 5 F(0.01, 8, 6). Referring to the F table with an upper-tail area of 0.01, with n1 5 8 and n2 5 6, we find the critical value to be 8.10. The decision rule is, “Reject H0 if calculated F . 8.10, otherwise do not reject.” Compare the Calculated and Critical Values and Reach a Conclusion As the test summary in Figure 11.7 shows, the calculated F (2.02) does not exceed the critical value (8.10), so the null hypothesis of equal population variances is not rejected. Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions 399 FIGURE 11.7 H0: s1 = s2 H1: s1 ≠ s2 (or H0: (or H1: s 21 s 21 Do not reject H0 = s 22 ) ≠ s 22 ) The F distribution can be used to test the null hypothesis that the population variances are equal. In this test at the 5 0.02 level, the calculated F was less than the critical F, so the population variances can be assumed to be equal. Reject H0 Area = a/2 = 0.01 F = 8.10 Make the Related Decision At the 0.02 level of significance, we conclude that the variances in service times associated with the two different technician-training films could be equal. If we were to compare the means of the two samples, the population standard deviations could be assumed to be equal and it would be permissible to apply the pooledvariances t-test of Section 11.2. Computer Solutions 11.6 describes Excel and Minitab procedures for testing whether two population variances could be equal. EXAMPLEEXAMPLE Test statistic: F = 2.02 COMPUTER 11.6 SOLUTIONS Testing for the Equality of Population Variances These procedures show how to use the F-test to compare the variances of two independent samples. EXCEL (continued ) 400 Part 4: Hypothesis Testing Excel F-test comparing sample variances, based on raw data 1. For Excel data file CX11TECH on which Figure 11.7 is based, the label and 9 data values for method 1 are in column A, and the label and 7 data values for method 2 are in column B. From the Data ribbon, click Data Analysis. Click F-Test Two-Sample for Variances. Click OK. 2. Enter A1:A10 into the Variable 1 Range box. Enter B1:B8 into the Variable 2 Range box. (The sample with the greater variance should always be specified as “Variable 1.”) 3. Select Labels. Specify the significance level by entering 0.01 into the Alpha box. (We are doing a two-tail test at the 0.02 level, but Excel performs only a one-tail test, so we must tell Excel to do its one-tail test at the 0.01 level to get comparable results.) 4. Select Output Range and enter D1 into the box. Click OK. The p-value for our two-tail test is 0.407—that is, double the 0.2035 value that Excel shows for its one-tail test. Excel F-test comparing sample variances, based on summary statistics 1. Using the summary statistics associated with CX11TECH (the test is summarized in Figure 11.7): Open the TEST STATISTICS workbook. 2. Using the arrows at the bottom left, select the F-Test_2 Variances worksheet. 3. Enter the sample sizes and variances into the appropriate cells along with an alpha level for the test. The calculated F-statistic and a two-tail p-value will be shown at the right. MINITAB Minitab F-test comparing sample variances, based on raw data 1. Using Minitab data file CX11TECH on which Figure 11.7 is based, with the data for method 1 in column C1 and the data for method 2 in column C2: Click Stat. Select Basic Statistics. Click 2 Variances. 2. Select Samples in different columns. Enter C1 into the First box and C2 into the Second box. Click OK. A portion of the printout, including the two-tail p-value, is shown here. Minitab F-test for comparing sample variances, based on summary data Follow steps 1 and 2 above, but select Sample variances in step 2 and insert the appropriate summary statistics for each sample into the Sample size and Variance boxes. ERCISES X E 11.66 For two samples from what are assumed to be normally distributed populations, the sample sizes and standard deviations are n1 5 10, s1 5 23.5, n2 5 9, and s2 5 10.4. At the 0.10 level of significance, test the null hypothesis that the population variances are equal. Would your conclusion be different if the test had been conducted at the 0.05 level? At the 0.02 level? 11.67 For two samples from what are assumed to be normally distributed populations, the sample sizes and standard deviations are n1 5 28, s1 5 103.1, n2 5 41, and s2 5 133.5. At the 0.10 level of significance, test the null hypothesis that the population variances are equal. Would your conclusion be different if the test had been conducted at the 0.05 level? At the 0.02 level? Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions 401 11.68 Given the information in Exercise 11.10, and using the 0.02 level of significance in comparing the sample standard deviations, were we justified in assuming that the population standard deviations are equal? Would your conclusion change if the standard deviation of playing times for music boxes equipped with the new battery design had been 20% larger than the value calculated in Exercise 11.10? average time for shipments to be received is about the same, regardless of supplier, but the purchasing agent is concerned about company 2’s higher variability in shipping time. Using the 0.025 level of significance in a one-tail test, should the purchasing agent conclude that company 2’s higher standard deviation in shipping times is due to something other than chance? 11.69 Given the information in Exercise 11.11, and using the 0.05 level of significance in comparing the sample standard deviations, were we justified in assuming that the population standard deviations were equal? Would your conclusion change if the standard deviation for the FoodFarm jars had been 3.0 grams? ( DATA SET ) Note: Exercises 11.72–11.74 require a computer and statistical software. 11.70 According to the National Association of Homebuilders, the average life expectancies of a dishwasher and a microwave oven are about the same: 9 years. Assume that their finding was based on a sample of n1 5 60 dishwashers and n2 5 40 microwave ovens, and that the corresponding sample standard deviations were s1 5 3.0 years and s2 5 3.7 years. Using the 0.02 level of significance, examine whether the population standard deviations for the lifetimes of these two types of appliances could be the same. Source: National Association of Homebuilders, Study of Life Expectancy of Home Components, February 2007, p. 5. 11.71 In conducting her annual review of suppliers, a purchasing agent has collected data on a sample of orders from two of her company’s leading vendors. On average, the 24 shipments from company 1 have arrived 3.4 days after the order was placed, with a standard deviation of 0.4 days. The 30 shipments from company 2 arrived an average of 3.6 days after the order was placed, with a standard deviation of 0.7 days. The 11.72 A researcher has observed independent samples of females and males, recording how long each person took to complete his or her shopping at a local mall. The respective times, in minutes, are listed in file XR11072. Using the 0.025 level of significance in a onetail test, would females appear to exhibit more variability than males in the length of time shopping in this mall? 11.73 For independent samples of customers of Internet providers A and B, the ages are as listed in file XR11073. Using the 0.05 level of significance in a nondirectional test, examine whether the variation in ages could be the same for customers of the two providers. 11.74 An investment analyst has data for the past 8 years on each of two mutual funds, with the annual rates of return for each listed in file XR11074. On average, each fund has had an excellent annual rate of return over the time period for his data, but the analyst is concerned that a mutual fund with greater variation in rate of return tends to involve greater risk for his investment clients. Using the 0.05 level in a nondirectional test, examine whether the two mutual funds have differed significantly in their performance variability. SUMMARY • Comparing sample means or proportions One of the most useful applications of business statistics involves comparing two samples to examine whether a difference between them is (1) significant or (2) more likely due to chance variation from one sample to the next. As with testing hypotheses for just a single sample, comparing the means or proportions from two samples requires comparing the calculated value of a test statistic with its critical value(s), then deciding whether the null hypothesis should be rejected. • Independent versus dependent samples Independent samples are those for which the selection process for one is not related to the selection process for the other. An example of independent samples occurs when subjects are randomly assigned to the experimental and control groups of an experiment. Samples are dependent when the selection process for one is related to the selection process for the other. A typical example of dependent samples occurs with before-and-after measurements for the same individuals or test units. In this case, there is really only one variable: the difference between the two measurements recorded for each individual or test unit. ( ) 11.8 402 Part 4: Hypothesis Testing • Tests for comparing means of independent samples When comparing the means of independent samples, the t-test is applicable whenever the population standard deviations are unknown. The pooled-variances t-test is used whenever the population standard deviations are assumed to be equal, regardless of the sample size. The population standard deviations are typically unknown. The chapter describes the unequal-variances t-test to be used if the population standard deviations are unknown and cannot be assumed to be equal. The unequal-variances t-test involves a special correction formula for the number of degrees of freedom (df ). The z-test can be used as a close approximation to the unequal-variances t-test when the population standard deviations are not assumed to be equal, but samples are large (each n $ 30). • Comparing proportions from independent samples Comparing proportions from two independent samples involves a z-test. When samples are large and other conditions described in the chapter are met, the normal distribution is a close approximation to the binomial. • Comparing variances for independent samples A special hypothesis test can be applied to test the null hypothesis that the population variances are equal for two independent samples. Based on the F distribution, it has many possible applications, including determining whether the pooled-variances t-test is appropriate for a given set of data. A T I O N S E Q U Pooled-variances t-test for comparing the means of two independent samples, 1 and 2 unknown and assumed to be equal } 2} x2) 2 (1 2 2)0 (x 1 • t 5 _____________________ where } x1 and } x2 5 means of samples 1 and 2 wwwwww 1 1 2 ___ ___ ( 2 sp 1 1 2)0 5 hypothesized difference between n1 n2 the population means Ï( ) n1 and n2 5 sizes of samples 1 and 2 s1 and s2 5 standard deviations of samples 1 and 2 (n1 2 1)s21 1 (n2 2 1)s22 with s2p 5 _____________________ n1 1 n2 2 2 • and df 5 n1 1 n2 2 2 Confidence interval for 1 2 2: x1 2 } x2) 6 ty2 (} Ï ( wwwwww 1 1 2 sp __ 1 __ n1 n2 ) Unequal-variances t-test for comparing the means of two independent samples, 1 and 2 unknown and not assumed to be equal } 2x } ) 2 ( 2 ) (x 1 2 1 2 0 • t 5 _____________________ where 2 x1 and 2 x2 5 means of samples 1 and 2 2 2 wwww s1 s2 ( 2 1 2)0 5 hypothesized difference between ___ 1 ___ n1 n2 the population means n1 and n2 5 sizes of samples 1 and 2 s1 and s2 5 standard deviations of samples 1 and 2 3 ( s21Yn1 ) 1 ( s22Yn2 ) 42 with df 5 _________________ ( s21Yn1 )2 _______ ( s22Yn2 )2 _______ 1 n1 2 1 n2 2 1 Ï Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions • Confidence interval for 1 2 2: Ï } 2x })6t (x 1 2 ay2 wwww s2 s2 1 2 __ 1 __ n1 n2 z -test approximation for comparing the means of two independent samples, 1 and 2 unknown, and each n 30 } 2x } ) 2 ( 2 ) (x 1 2 1 2 0 } and x } 5 means of samples 1 and 2 • z 5 _____________________ where x 1 2 wwww s21 s22 (1 2 2)0 5 hypothesized difference between ___ 1 ___ the population means n1 n2 n1 and n2 5 sizes of samples 1 and 2 s1 and s2 5 standard deviations of samples 1 and 2 • Confidence interval for 1 2 2: Ï } 2x })6z (x 1 2 ay2 Ï wwww s2 s2 1 2 __ 1 __ n1 n2 Comparing proportions from two independent samples • z-test, with test statistic 1. When the hypothesized difference is zero (the usual case): (p1 2 p2) z 5 ___________________ where p1 5 observed proportion, sample 1 wwwwwwwww 1 1 }(1 2 p }) ___ p2 5 observed proportion, sample 2 p 1 ___ n1 n2 n1 5 sample size, sample 1 n2 5 sample size, sample 2 } 5 pooled estimate of the population p proportion Ï ( ) 1p1 1 n2p2 }5 n ___________ with p n1 1 n2 2. When the hypothesized difference is (1 2 2)0 Þ 0: (p1 2 p2) 2 (1 2 2)0 z 5 ________________________ wwwwwwwwwww p1(1 2 p1) p2(1 2 p2) __________ 1 __________ n1 n2 Ï • Confidence interval for (1 2 2): wwwwwwwwwww p (1 2 p ) p (1 2 p ) Ï (p1 2 p2) 6 zy2 1 1 2 2 __________ 1 __________ n1 n2 Comparing the means when samples are dependent • t-test, with test statistic: } d t 5 ______ sd yÏw n where d 5 for each individual or test unit, (x1 2 x2), the difference between the two measurements } d 5 the average difference, 5 odi yn n 5 number of pairs of observations sd 5 the standard deviation of d, or df 5 n 2 1 • Confidence interval for d: sd } d 6 ty2 ____ n Ïw wwwww od 2 2 n}d2 Ï i _________ n21 403 404 Part 4: Hypothesis Testing Comparing variances from two independent samples • • F-test, with test statistic: Critical value of F: F(ay2, n1, n2) s21 F 5 __ s22 s22 or __, s21 whichever is larger where a 5 specified level of significance for a nondirectional test: 0.10, 0.05, or 0.02 n1 5 (n 2 1), where n is the size of the sample that had the larger variance n2 5 (n 2 1), where n is the size of the sample that had the smaller variance TER EXERCISES P A H C Reminder: Be sure to refer to Figure 11.1 in selecting the testing procedure for an exercise. When either test can be applied to the same data, the unequal-variances t-test is preferable to the z-test— especially when doing the test with computer assistance. 11.75 Suspecting that television repair shops tend to charge women more than they do men, Emily disconnected the speaker wire on her portable television and took it to a sample of 12 shops. She was given repair estimates that averaged $85, with a standard deviation of $28. Her friend John, taking the same set to another sample of 9 shops, was provided with an average estimate of $65, with a standard deviation of $21. Assuming normal populations with equal standard deviations, use the 0.05 level in evaluating Emily’s suspicion. Using the appropriate statistical table, what is the approximate p-value for this test? 11.76 The manufacturer of a small utility trailer is interested in comparing the amount of fuel used in towing the trailer with that required to overcome the weight and wind resistance of a rooftop carrier. Over a standard test route at highway speeds, 8 trips with the trailer resulted in an average fuel consumption of 0.28 gallons, with a standard deviation of 0.03. Under similar conditions, 10 trips with the rooftop carrier led to an average consumption of 0.35 gallons, with a standard deviation of 0.05 gallons. Assuming normal populations with equal standard deviations, use the 0.01 level in examining whether pulling the trailer uses significantly less fuel than driving with the rooftop carrier. 11.77 A compressor manufacturer is testing two differ- ent designs for an air tank. Testing involves pumping air into a tank until it bursts, then noting the air pressure just prior to tank failure. Four tanks of design A are found to fail at an average of 1400 pounds per square inch (psi), with a standard deviation of 250 psi. Six tanks of design B fail at an average of 1620 psi, with a standard deviation of 230 psi. Assuming normal populations with equal standard deviations, use the 0.10 level of significance in comparing the two designs. Construct and interpret the 90% confidence interval for the difference between the population means. 11.78 An industrial engineer has designed a new workstation layout that she claims will increase production efficiency. For a sample of 8 workers using the new workstation layout, the average number of units produced per hour is 36.4, with a standard deviation of 4.9. For a sample of 6 workers using the standard layout, the average number of units produced per hour is 30.2, with a standard deviation of 6.2. Assuming normal populations with equal standard deviations, use the 0.025 level of significance in testing the engineer’s claim. 11.79 Bob Techster is very serious about winning this year’s All-American Soap Box Derby competition. Since there is no wind-tunnel time available at the aerospace firm where his father works, Bob is using a local hill to test the effectiveness of an aerodynamic design change. In 10 runs with the standard car, the average time was 22.5 seconds, with a standard deviation of 1.3 seconds. With the design change, 12 runs led to an average time of 21.1 seconds, with a standard deviation of 0.9 seconds. Assuming normal populations with equal standard deviations, use the 0.01 level in helping Bob determine whether the modifications are really effective in improving his racer’s speed. Using the appropriate statistical table, what is the approximate p-value for this test? 11.80 The developer of a new welding rod claims that spot welds using his product will have greater strength Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions than conventional welds. For 45 welds using the new rod, the average tensile strength is 23,500 pounds per square inch, with a standard deviation of 600 pounds. For 40 conventional welds on the same materials, the average tensile strength is 23,140 pounds per square inch, with a standard deviation of 750 pounds. Use the 0.01 level in testing the claim of superiority for the new rod. Using the appropriate statistical table, what is the approximate p-value for this test? 11.81 Independent random samples of vehicles traveling past a given point on an interstate highway have been observed on Monday versus Wednesday. For 16 cars observed on Monday, the average speed was 59.4 mph, with a standard deviation of 3.7 mph. For 20 cars observed on Wednesday, the average speed was 56.3 mph, with a standard deviation of 4.4 mph. At the 0.05 level, and assuming normal populations, can we conclude that the average speed for all vehicles was higher on Monday than on Wednesday? What is the most accurate statement that can be made about the p-value for this test? 11.82 A sample of 25 production employees has been tested twice on a standard test of manual dexterity. The average change in the time required to finish the test was a decrease of 1.5 minutes, with a standard deviation of 0.3 minutes. At the 0.05 level, can we conclude that the average production employee will complete the test more quickly the second time he or she takes it? 11.83 For a sample of 48 finance majors, the average time spent reading each issue of the campus newspaper is 19.7 minutes, with a standard deviation of 7.3 minutes. The corresponding figures for a sample of 40 management information systems majors are 16.3 and 4.1 minutes. Using the 0.01 level of significance, test the null hypothesis that the population means are equal. Using the appropriate statistical table, what is the approximate p-value for this test? Construct and interpret the 99% confidence interval for the difference between the population means. 11.84 Two machines are supposed to be producing steel bars of approximately the same length. A sample of 35 bars from one machine has an average length of 37.013 inches, with a standard deviation of 0.095 inches. For 38 bars produced by the other machine, the corresponding figures are 36.974 inches and 0.032 inches. Using the 0.05 level of significance, test the null hypothesis that the population means are equal. Using the appropriate statistical table, what is the approximate p-value for this test? Construct and interpret the 95% confidence interval for the difference between the population means. 11.85 Observing a sample of 35 morning customers at a convenience store, a researcher finds their average stay in the store is 3.0 minutes, with a standard deviation of 0.9 minutes. For a sample of 43 afternoon customers, the mean and standard deviation were 3.5 and 1.8 minutes, respectively. Using the 0.05 level of significance, determine 405 whether the population means could be equal. Construct and interpret the 95% confidence interval for the difference between the population means. 11.86 In an experiment comparing a new long-distance golf ball with the conventional design, each of 20 golfers hits one drive with each ball. The average golfer in the sample hit the new ball 9.3 yards farther, with a standard deviation of 10.5 yards. At the 0.005 level, evaluate the effectiveness of the new ball in increasing distance. Using the appropriate statistical table, what is the approximate p-value for this test? 11.87 The idling speed of 14 gasoline-powered generators is measured with and without an oil additive that is designed to lower friction. With the additive installed, the mean change in speed was 123 revolutions per minute (rpm), with a standard deviation of 3.5 rpm. At the 0.05 level of significance, is the additive effective in increasing engine rpm? 11.88 A motel manager, concerned with customer theft of towels, decided that the theft rate might be reduced by changing from white, imprinted towels to a drab green version. Of the 120 guests provided with the white towels, 35% took at least one towel with them when they checked out. Of the 160 guests given the drab green towels, only 25% checked out with one or more towels in their possession. At the 0.01 level of significance, can we conclude that the manager’s idea is effective in reducing the rate of towel theft? 11.89 An Internal Revenue Service manager is comparing the results of recent taxpayer interviews conducted by two auditors. Of 200 taxpayers audited by Ms. Smith, 30% had to pay additional taxes. Of 100 audited by Mr. Burke, only 19% paid additional taxes. At the 0.01 level of significance, can the manager conclude that Ms. Smith is a more effective auditor than Mr. Burke? 11.90 A pharmaceutical manufacturer has come up with a new drug intended to provide greater headache relief than the old formula. Of 250 patients treated with the previous medication, 130 reported “fast relief from headache pain.” Of 200 individuals treated with the new formula, 128 said they got “fast relief.” At the 0.05 level, can we conclude that the new formula is better than the old? Using the appropriate statistical table, what is the approximate p-value for this test? 11.91 In tests of a speed-reading course, it is found that 10 subjects increased their reading speed by an average of 200 words per minute, with a standard deviation of 75 words. At the 0.10 level, does this sample information suggest that the course is really effective in improving reading speed? Using the appropriate statistical table, what is the approximate p-value for this test? 11.92 A maintenance engineer has been approached by a supplier who promises less downtime for machines 406 using its new molybdenum-based super lubricant. Of 120 machines maintained with the new lubricant, only 20% were out of service for more than an hour during the month of the test, while 30% of the 100 machines using the previous lubricant were down for more than an hour. Using the 0.05 level in an appropriate test, evaluate the supplier’s contention. Using the appropriate statistical table, what is the approximate p-value for this test? 11.93 According to the National Center for Education Statistics, 88% of elementary, middle-school, and secondaryschool students from families earning $75,000 or more per year use a computer at school, compared to 81% from families earning less than $20,000. Using a one-tail test at the 0.01 level, and assuming that the percentages are from independent samples of 500 families each, is the percentage of more-affluent students using computers at school significantly greater than for those from lower-income families? Determine and interpret the p-value for the test. Source: Bureau of the Census, Statistical Abstract of the United States 2009, p. 163. 11.94 A study by Experian found that 20% of consum- ers in New Jersey have 10 or more credit cards, compared to 11% in Tennessee. In a two-tail test at the 0.05 level of significance, and assuming the percentages are from independent samples of 100 consumers each, can we conclude that New Jersey and Tennessee consumers might not differ in terms of the percentage who have 10 or more credit cards? Determine and interpret the p-value for the test, then construct and interpret the 95% confidence interval for the difference between the population proportions. Source: “1 in 7 Americans Carry 10 or More Credit Cards,” moneycentral.msn.com, July 25, 2009. 11.95 Given the information in Exercise 11.75, and using the 0.05 level of significance in comparing the sample standard deviations, were we justified in assuming that the population standard deviations were equal? Would your conclusion change if the standard deviation of the estimates received by Emily had been $35? 11.96 Given the information in Exercise 11.76, and using the 0.01 level of significance in comparing the sample standard deviations, were we justified in assuming that the population standard deviations were equal? Would your conclusion change if the standard deviation of the data values with the rooftop carrier had been 0.07 gallons? ( DATA SET ) Note: Exercises 11.97–11.101 require a computer and statistical software. 11.97 Three years after receiving their degrees, graduates of a university’s MBA program have reported their annual salary rates, with a portion of the data listed in file XR11097. Graduates in one of the groups represented in the data are employed by consulting firms, while another group consists of graduates who are with national-level corporations. Considering these groups as data from independent samples, use the 0.05 level in determining whether the sample means differ Part 4: Hypothesis Testing significantly from each other. Identify and interpret the p-value for the test, then generate and interpret the 95% confidence interval for the difference between the population means. 11.98 A supermarket manager is examining whether a new plastic bagging configuration in the produce area might make it more efficient for customers who are bagging their own fruits and vegetables. She observes a sample of customers using the new system and a sample of customers using the standard system, and she notes how many seconds it took each of them to unroll and open the plastic bag in preparation for bagging the veggies. The data are in file XR11098. Considering these as data from independent samples, use the 0.025 level of significance in examining whether the population mean for the new bagging system might be less than that for the conventional system. Identify and interpret the p-value for the test. 11.99 A company that makes an athletic shoe designed for basketball has stated in its advertisements that the shoe increases the jumping ability of players who wear it. The general manager of a professional team has conducted a test in which each player’s vertical leap is measured with the current shoe versus the new model, with the data as listed in file XR11099. At the 0.05 level of significance, evaluate the claim that has been made by the shoe company. Identify and interpret the p-value for the test. 11.100 The data in file XR11100 are the weights (in grams) for random samples of grain packages filled by two different filling machines. The machines have a fine adjustment for the mean amount of fill, but the standard deviations are inherent in the design of the grain delivery mechanism. Based on these data, and using the 0.01 level of significance, is there reason to conclude that the population standard deviations might not be equal for the quantities being delivered from the two machines? Identify and interpret the p-value for the test. 11.101 An anthropologist studying personal advertisements in a Utica, New York, newspaper has observed whether the advertiser included a mention of an interest in the outdoors in his or her ads. According to the anthropologist, citing the outdoors “may be taken to imply not only good life habits, but also sound character.” Overall, 58% of men and 62% of women mentioned the outdoors in their ad. Assuming that the data are in file XR11101, coded as 1 5 did not mention the outdoors and 2 5 mentioned the outdoors, use the 0.10 level of significance in examining whether the population percentages for men and women who mention the outdoors might be the same. Identify and interpret the p-value for the test, then construct and interpret the 90% confidence interval for the difference between the population proportions. Source: Karen S. Peterson, “Personal Ads Get Back To Nature,” USA Today, November 23, 1999, p. 1D. Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions G INTE 407 R AT E D C A S E S Thorndike Sports Equipment The Thorndikes have submitted a bid to be the sole supplier of swimming goggles for the U.S. Olympic team. OptiView, Inc. has been supplying the goggles for many years, and the Olympic committee has said it will switch to Thorndike only if the Thorndike goggles are found to be significantly better in a standard leakage test. For purposes of fairness, the committee has purchased 16 examples from each manufacturer in the retail marketplace. This is to avoid the possibility that either manufacturer might supply goggles that have been specially modified for the test. Testing involves installing the goggles on a surface that simulates the face of a swimmer, then submitting them to increasing water pressure (expressed in meters of water depth) until the goggles leak. The greater the number of meters before leakage, the better the quality of the goggles. Both companies have received copies of the test results and have an opportunity to offer their respective comments before the final decision is made. Ted Thorndike has just received his company’s copy of the results, rounded to the nearest meter of water depth. The data are also listed in file THORN11. Thorndike Goggles (meters) 82 106 117 114 91 106 95 95 110 101 81 92 101 94 108 108 86 107 100 73 OptiView Goggles (meters) 73 92 95 108 83 94 106 77 70 109 103 90 1. Based on analysis of these data, formulate a commentary that Ted Thorndike might wish to make to the committee. 2. Based on analysis of these data, formulate a commentary that OptiView might wish to make to the committee. 3. What would be your recommendation to the committee? Springdale Shopping Survey Item C of the Springdale Shopping Survey, introduced at the end of Chapter 2, describes variables 7–9 of the survey. These variables represent the general attitude respondents have toward each of the three shopping areas, and range from 5 (like very much) to 1 (dislike very much). Samples involving consumer groups often differ in their results, and managers find it useful to determine whether the differences could be due to sampling variation or whether there is “something going on” regarding the attitudes, perceptions, and behaviors of one consumer group versus another. Variable 7 5 Attitude toward Springdale Mall Variable 8 5 Attitude toward Downtown Variable 9 5 Attitude toward West Mall Part I: Attitude Comparisons Based on Marital Status of Respondents 1. For variable 7 (attitude toward Springdale Mall), carry out an appropriate hypothesis test to determine whether married persons (code 5 1 on variable 28, marital status) have a different mean attitude than unmarried persons (code 5 2 on variable 28). Interpret the resulting computer printout, including the p-value for the test. 2. Repeat step 1 for variable 8 (attitude toward Downtown). 3. Repeat step 1 for variable 9 (attitude toward West Mall). 4. Comment on the extent to which attitudes toward each of the shopping areas differ between married and unmarried respondents. Part II: Attitude Comparisons Based on Gender of Respondent Repeat Part I, using variable number 26 (gender of respondent) instead of variable 28 (marital status of respondent) as the basis on which the groups are identified. 408 Part 4: Hypothesis Testing INESS S U B CASE Circuit Systems, Inc., located in northern California, is a company that produces integrated circuit boards for the microcomputer industry. In addition to salaried management and office staff personnel, Circuit Systems currently employs approximately 250 hourly production workers involved in the actual assembly of the circuit boards. These hourly employees earn an average of $11.00 per hour. Thomas Nelson, the Director of Human Resources at Circuit Systems, has been concerned with hourly employee absenteeism within the company. Presently, each hourly employee earns 18 days of paid sick leave per year. Thomas has found that many of these employees use most or all of their sick leave well before the year is over. After an informal survey of employee records, Thomas is convinced that while most hourly employees make legitimate use of their sick leave, there are many who view paid sick leave as “extra” vacation time and “call in sick” when they want to take off from work. This has been a source of conflict between the hourly production workers and management. The problem is due in part to a restrictive vacation policy at Circuit Systems in which hourly employees receive only one week of paid vacation per year in addition to a few paid holidays. With only one week of paid vacation and a few paid holidays, the hourly production employees work a 50-week year, not counting paid sick leave. In an effort to save money and increase productivity, Thomas has developed a two-point plan that was recently approved by the president of Circuit Systems. To combat the abuse of paid sick leave, hourly workers will now be allowed to convert unused paid sick leave to cash on a “three-for-one” basis, i.e., each unused day of sick leave can be converted into an additional one-third of a day’s pay. An hourly employee could earn up to an additional six days of pay each year if he or she does not take any paid sick leave during the year. Even though a worker could gain more time off by dishonestly “phoning in sick,” Thomas hopes that the majority of hourly employees will view this approved conversion of sick leave into extra pay as a more acceptable alternative. In the second part of his plan, Thomas is instituting a voluntary exercise program for hourly employees to improve their overall health. At an annual company expense of $200 for each hourly employee who participates, Circuit Systems will subsidize membership in a local health club. In return, the participating employee is required to exercise at least three times per week outside of regular working hours to maintain his or her free membership. Circuit Systems believes that, in the long term, an investment in employees’ physical well-being may increase their productivity as well as reduce the company’s future health insurance premiums. In discussions with hourly employees, Thomas has found that many of them approve of the exercise program and are willing to participate. Many of the supervisors that Thomas has spoken with believe that the paid sick leave conversion and the exercise program may help in curbing the absenteeism problem, but others do not give it much hope for succeeding and think the cost would outweigh any benefits. The president of Circuit Systems agreed to give the proposal a one-year trial period. At the end of the trial period, Thomas must evaluate the new anti-absenteeism plan, present the results, and make a recommendation to either continue or discontinue the plan. Assignment Over the next year, during which time the sick leave conversion and exercise program are in place, Thomas Nelson has maintained data on employee absences, use of the sick leave conversion privilege, participation in the exercise program, and other pertinent information. He has also gone back to collect data from the year prior to starting the new program in order to better evaluate the new program. His complete data are in the file CIRCUIT. A description of this data set is given in the Data Description section. Using this data set and other information given in the case, help Thomas Nelson evaluate the new program to determine whether it is effective in reducing the average cost of absenteeism by hourly employees, thereby increasing worker productivity. In particular, you need to compare this year’s data to last year’s data to determine whether there has been a reduction in the average cost of absenteeism per hourly production worker by going to the new program. The case questions will assist you in your analysis of the data. Use important details from your analysis to support your recommendation. Data Description The CIRCUIT file contains data for the past two years on the 233 hourly production employees in the company who were with the company for that entire period of time. A partial listing of the data is shown here. Photodisc/Getty Images Circuit Systems, Inc. (A) Chapter 11: Hypothesis Tests Involving Two Sample Means or Proportions Employee 6631 7179 2304 9819 4479 1484 A 409 Hourly Pay Sick Leave Last Year Sick Leave This Year Exercise Program $10.97 11.35 10.75 10.96 10.59 11.41 A 3.50 24.00 18.00 21.25 16.50 16.50 A 2.00 12.50 12.75 14.00 11.75 9.75 A 0 0 0 0 0 1 A These data are coded as follows: Employee: Employee ID number. Hourly Pay: Hourly pay of the employee in both years. Unfortunately, due to economic conditions, there were no pay raises last year. Sick Leave Last Year: Actual number of days of sick leave taken by the employee last year before the new program started. Sick Leave This Year: Actual number of days of sick leave taken by the employee this year under the new program. Exercise Program: 1, if participating in the exercise program. 0, if not participating. 1. Using the method presented in the chapter for comparing the means of paired samples, compare the two years in terms of days missed before and after the new program was implemented. On this basis alone, does it appear that the program has been effective in reducing the number of days missed? 2. Keeping in mind that the goal of the program is to reduce the cost of absenteeism, you will need to create two new variables for each employee: (1) the cost of paid absences last year, and (2) the cost associated with absences this year. A few hints on creating these variables for each person: For (1), assume an 8-hour workday and consider both the person’s daily pay and his or her number of absence days. For (2), assume an 8-hour workday and keep in mind that the total cost associated with absenteeism for each person must include the cost of paid absences, the extra pay for unused sick leave (if any), and health club membership (if applicable). You might call these new variables Cost_Before and Cost_After. Use these new variables in repeating the procedure you followed in Question 1, then discuss the results and make a recommendation to Mr. Nelson regarding the effectiveness and possible continuation of the new program. 3. Using the new variables you created in Question 2, use this year’s costs and an appropriate statistical test in evaluating the effectiveness of the exercise program. Discuss the results and make a recommendation to Mr. Nelson regarding the effectiveness and possible continuation of the company-paid health club memberships. TATISTICS: APPLET 1 S G N I 4 SEE Distribution of Difference Between Sample Means This applet has three parts: 1. The upper portion shows two normally distributed populations. For purposes of simplicity in the applet, their standard deviations are the same. By moving the slider at the top of the applet, we can shift one of the population curves back and forth, thus changing the difference between the population means. By moving the upper of the two sliders at the right, we can change the standard deviations of the populations. 2. The center portion of the applet shows the sampling distribution of the means for simple random samples taken from each of the populations at the top. To reduce complexity within the applet, the sample sizes are the same. By moving the lower of the two sliders at the right, we can change the sizes of the samples. 3. The bottom portion of the applet shows the sampling distribution of (}x1 2 x}2), the difference between the sample means. The standard error of this sampling distribution is shown at the lower right as “sigma,” and the applet calculates it using the formula provided in text Section 11.4, but with 1 and 2 replacing s1 and s2, respectively. Despite the small sample sizes, this is appropriate because the populations are normal and the population standard deviations are known. In real-world applications, we almost never know both population standard deviations, but don’t let this detract from your use of this very interesting applet. Applet Exercises 14.1 Set the top slider so that the difference between the population means is 23.0. Set the slider at the upper right so that the standard deviation of each population is 2.5. Set the slider at the lower right so that n1 5 n2 5 20. a. Is there very much overlap between the two population curves at the top of the applet? b. Is there very much overlap between the two sampling distribution curves in the center part of the applet? c. Viewing the bottom portion of the applet, and assuming that a sample is going to be taken from each of the two populations, does it seem very likely that (x}1 2 x}2) will be greater than zero? 14.2 Repeat Applet Exercise 14.1, but with the top slider set so the difference between the sample means is 10.5. 14.3 Repeat Applet Exercise 14.1, but with the top slider set so the difference between the sample means is 13.0. 14.4 Use the top slider to gradually change the difference between the population means from 20.5 to 4.5. Describe how this affects the graphs in the three portions of the applet. 14.5 Use the upper right slider to gradually increase the population standard deviations from 1.1 to 3.0. Describe how this affects the graphs in the three portions of the applet. 14.6 Use the lower right slider to gradually increase the sample sizes from 2 to 20. Describe how this affects the graphs in the three portions of the applet. Try out the Applets: http://www.cengage.com/bstatistics/weiers