stats 363 HW-Section-10-7

stats 363 HW-Section-10-7 - Section 10.7 Exercise 10.58 It...

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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 0 : ° 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 0 : 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 0 : ° 2 1 = ° 2 2 . There is insu°cient evidence to indicate a di±erence between the population variances under the level of signi²cance ± = 0 : 05. ( b ) Since it is a two-tailed test with F = 1 : 774, df 1 = 15, and df 2 = 19, the p -value is given by p -value = 2 P f F > 1 : 774 g : Then it is clear that 1 2 ( p -value) = 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 ( p -value) : It follows that 1 2 ( p -value) > 0 : 100 or p -value > 0 : 20 : We cannot reject H 0 : ° 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% con²dence interval for ° 2 1 / ° 2 2 is s 2 1 s 2 2 ! 1 F °= 2; df 1 ;df 2 < ° 2 1 ° 2 2 < s 2 1 s 2 2 ! F °= 2; df 2 ;df 1 with df 1 = 15, df 2 = 19, and ± = 0 : 05. We read from Table 6 that F °= 2; df 1 ;df 2 = F 0 : 025;15 ; 19 = 2 : 62 and F °= 2; df 2 ;df 1 = F 0 : 025;19 ; 15 ± 2 : 76 Hence, the desired 95% con²dence interval for °
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