460 CHAPTER 9 Inferences from Two Samples Alternative Methods: Resampling Methods of Bootstrapping and Randomization The claim that the mean height of U.S. Army male personnel in 1988 is less than the mean height of this population in 2012 can be tested by using the resampling methods of bootstrapping and randomization. See Section 9-5 “Resampling: Using Technology for Inferences.” Using E = 41.049035, x1 = 1739.417, and x2 = 1777.8, we can now use simple substitution to find the confidence interval as shown here: 1x1 - x22 - E 6 m1 - m2 6 1x1 - x22 + E -79.4 mm 6 m1 - m2 6 2.7 mm The confidence interval limits of (-77.9 mm, 1.1 mm) found from technology are more accurate, but we can see that the manual calculations give us a result that is reasonably good, even though we used a simplified method for finding the number of degrees of freedom (instead of getting more accurate results by using Formula 9-1 to compute the number of degrees of freedom). YOUR TURN. Do Part (b) of Exercise 5 “Better Tips by Giving Candy.” INTERPRETATION We are 90% confident that the confidence interval actually does contain the difference between the mean height in 1988 and the mean height in 2012. Because the confidence interval does contain 0, it suggests that there is not a significant difference between the mean height in 1988 and the mean height in 2012. PART 2 Alternative Methods Part 1 of this section dealt with situations in which the two population standard deviations are unknown and are not assumed to be equal. In Part 2 we address two other situations: 1. The two population standard deviations are unknown but are assumed to be equal. 2. The two population standard deviations are both known. Alternative Method: Assume That S1 = S2 and Pool the Sample Variances Even when the specific values of s1 and s2 are not known, if it can be assumed that they have the same value, the sample variances s2 1 and s 2 2 can be pooled to obtain an estimate of the common population variance s 2. The pooled estimate of S 2 is denoted by s2 p and is a weighted average of s 2 1 and s 2 2, which is used in the test statistic for this case: Test Statistic t = 1x1 - x22 - 1m1 - m22 Bs2 p n1 + s2 p n2 where s2 p = 1n1 - 12s 2 1 + 1n2 - 12s 2 2 1n1 - 12 + 1n2 - 12 (pooled sample variance) and the number of degrees of freedom is df = n1 + n2 - 2.
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