stats 363 HW-Section-10-4

# stats 363 HW-Section-10-4 - Section 10.4 Exercise 10.19(a...

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Section 10.4 Exercise 10.19 . ( a ) n 1 = 10, n 2 = 4, s 2 1 = 3 : 4, s 2 2 = 4 : 9: s 2 = ( n 1 1) s 2 1 + ( n 2 1) s 2 2 n 1 + n 2 2 = (10 1) (3 : 4) + (4 1) (4 : 9) 10 + 4 2 = 3 : 775 : ( b ) n 1 = 12, n 2 = 21, s 2 1 = 18, s 2 2 = 23: s 2 = ( n 1 1) s 2 1 + ( n 2 1) s 2 2 n 1 + n 2 2 = (12 1) (18) + (21 1) (23) 12 + 21 2 = 21 : 2258 : Exercise 10.21 . It is given that n 1 = 16, n 2 = 13, x 1 = 34 : 6, x 2 = 32 : 2, s 2 1 = 4 : 8, s 2 2 = 5 : 9. ( a ) The null and alternative hypotheses for the test are H 0 : 1 2 = 0 versus H a : 1 2 6 = 0 : ( b ) With ± = 0 : 01 and df = n 1 + n 2 2 = 27, t 2 = t 0 : 005 = 2 : 771. The rejection region is j t j > 2 : 771. ( c ) The pooled sample variance is s 2 = ( n 1 1) s 2 1 + ( n 2 1) s 2 2 n 1 + n 2 2 = (16 1) (4 : 8) + (13 1) (5 : 9) 16 + 13 2 = 5 : 2889 and the test statistic is t = x 1 x 2 r s 2 1 n 1 + 1 n 2 ± = 34 : 6 32 : 2 r 5 : 2889 1 16 + 1 13 ± = 2 : 795 : ( d ) This is a two-tailed test with t = 2 : 795 and 27 df : The p -value is given by p -value = P f t < 2 : 795 g + P f t > 2 : 795 g = 2 P f t > 2 : 795 g : Then it is clear that 1 2 ( p -value) = P f t > 2 : 795 g : Now we read from Table 4 that P f t > 2 : 473 g = 0 : 010 P f t > 2 : 771 g = 0 : 005 P f t > 2 : 795 g = 1 2 ( p -value) :

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It follows that 1 2 ( p -value) < 0 : 005 or p -value < 0 : 01 : ( e ) Since p -value < 0 : 01, we reject H 0 : 1 2 = 0. There is su&cient evi- dence to indicate a di±erence between the population means. Exercise 10.23
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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.

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stats 363 HW-Section-10-4 - Section 10.4 Exercise 10.19(a...

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