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Unformatted text preview: Section 10.7 Exercise 10.58 . It is given that n 1 = 16, n 2 = 20, s 2 1 = 55 : 7, and s 2 2 = 31 : 4. ( a ) The null and alternative hypotheses for the test are H : & 2 1 = & 2 2 versus H a : & 2 1 6 = & 2 2 : The test statistic is F = s 2 1 s 2 2 = 55 : 7 31 : 4 = 1 : 774 : With ¡ = 0 : 05, df 1 = n & 1 = 15, and df 2 = n & 1 = 19, F &= 2 = F : 025 = 2 : 23. The rejection region is F > 2 : 23. Since the test statistic F = 1 : 774 does not fall in the rejection region, we cannot reject H : & 2 1 = & 2 2 . There is insu&cient evidence to indicate a di¡erence between the population variances under the level of signi¢cance ¡ = 0 : 05. ( b ) Since it is a twotailed test with F = 1 : 774, df 1 = 15, and df 2 = 19, the pvalue is given by pvalue = 2 P f F > 1 : 774 g : Then it is clear that 1 2 ( pvalue) = P f F > 1 : 774 g : Now we read from Table 6 that P f F > 2 : 23 g = 0 : 050 P f F > 1 : 86 g = 0 : 100 P f F > 1 : 774 g = 1 2 ( pvalue) : It follows that 1 2 ( pvalue) > : 100 or pvalue > : 20 : We cannot reject H : & 2 1 = & 2 2 . There is insu&cient evidence to indicate a di¡er ence between the population variances. Exercise 10.59 . It is given that n 1 = 16, n 2 = 20, s 2 1 = 55 : 7, and s 2 2 = 31 : 4. A 95% con¢dence interval for & 2 1 / & 2 2 is s 2 1 s 2 2 !...
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This note was uploaded on 04/20/2008 for the course MTH 1363 taught by Professor Cheng during the Spring '08 term at Creighton.
 Spring '08
 Cheng
 Statistics

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