Example 66 a study was conducted to determined

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/ £ ¡ ¢ EXAMPLE 6.6 A study was conducted to determined whether persons in suburban district 1 have a different mean income from those in district 2. A random sample of 20 homeowners was taken in district 1. Although 20 homeowners were to be inter- viewed in district 2 also, 1 person refused to provide the information requested, even though the researcher promised to keep the interview confidential. So only 19 observations were obtained from district 2. The data, recorded in thousands of dollars, produced sample means and variances as shown in Table 2. Use these data to construct a 95% confidence interval for ( μ 1 - μ 2 ). SOLUTION Histograms plotted for the two samples suggest that the two populations are mound-shaped (near normal). Also, the sample variances are very similar. The difference in the sample means is ¯ y 1 - ¯ y 2 = 18 . 27 - 16 . 78 = 1 . 49 . 6
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District 1 District 2 Sample size 20 19 Sample mean 18.27 16.78 Sample variance 8.74 6.58 Table 2: The estimate of the common standard deviation σ is s p = s ( n 1 - 1) s 2 1 + ( n 2 - 1) s 2 2 n 1 + n 2 - 2 = r 19(8 . 74) + 18(6 . 58) 20 + 19 - 2 = 2 . 77 . The t -value for a = α/ 2 = . 025 and df = 20 + 19 - 2 = 37 is not listed in Table 4 of the Appendix, but taking the labeled value for the nearest df ( df = 40), we have t = 2 . 021. A 95% confidence interval for the difference in mean incomes for the two districts is of the form ¯ y 1 - ¯ y 2 ± t α/ 2 s p r 1 n 1 + 1 n 2 Substituting into the formula we obtain 1 . 49 ± 2 . 021(2 . 77) r 1 20 + 1 19 or 1 . 49 ± 1 . 79 . Thus, we estimate the difference in mean incomes to lie somewhere in the interval from - . 30 to 3 . 28. If we multiply these limits by $1,000, the confidence interval for the difference in mean incomes is - $300 to $3,280. Since this interval includes both positive and negative values for μ 1 - μ 2 . We are unable to determine whether the mean income for district 1 is larger or smaller than the mean income for district 2. We can also test a hypothesis about the difference between two population means. As with any test procedure, we begin by specifying a research hypothesis for the difference in population means. Thus, we might, for example, specify that the difference μ 1 - μ 2 is greater than some value D 0 . (Note: D 0 will often be 0.) The entire test procedure is summarized here. ¤ § ¥ ƒ A statistical Test for μ 1 - μ 2 , Independent Samples H 0 : μ 1 - μ 2 = D 0 ( D 0 is specified) 7
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H a : 1. μ 1 - μ 2 > D 0 2. μ 1 - μ 2 < D 0 3. μ 1 - μ 2 6 = D 0 T.S. : t = ¯ y 1 - ¯ y 2 - D 0 s p p 1 /n 1 + 1 /n 2 R.R. : For a Type I error α and df = n 1 + n 2 - 2 1. reject H 0 if t > t α 2. reject H 0 if t < - t α 3. reject H 0 if | t | > t α/ 2 / £ ¡ ¢ EXAMPLE 6.7 An experiment was conducted to compare the mean number of tapeworms in the stomachs of sheep that had been treated for worms against the mean number in those that were untreated. A sample of 14 worm-infected lambs was randomly divided into 2 groups. Seven were injected with the drug and the remainder were left untreated. After a 6-month period, the lambs were slaugh- tered and the following worm counts were recorded: Drug-treated sheep 18 43 28 50 16 32 13 Untreated sheep 40 54 26 63 21 37 39 a. Test a hypothesis that there is no difference in the mean number of worms between treated and untreated lambs. Assume that the drug cannot in-
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