stats 363 HW-Section-10-5

stats 363 HW-Section-10-5 - Section 10.5 Exercise 10.35. It...

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Section 10.5 Exercise 10.35 . It is given that n = 10, d = 0 : 3, s 2 d = 0 : 16 ( s d = 0 : 4). ( a ) The null and alternative hypotheses for the test are H 0 : 1 2 = 0 or H 0 : d = 0 : versus H a : 1 2 6 = 0 or H a : d 6 = 0 : The test statistic is t = d d s d = p n = 0 : 3 & 0 0 : 4 = p 10 = 2 : 372 : With ± = 0 : 05 and df = n 1 = 9, t 2 = t 0 : 025 = 2 : 262. The rejection region is j t j > 2 : 262. Since the test statistic t = 2 : 372 falls inside the rejection region, we reject H 0 : 1 2 between the population means under the level of signi²cance ± = 0 : 05. This is a two-tailed test with t = 2 : 372 and df = 9: The p -value is given by p -value = P f t < 2 : 372 g + P f t > 2 : 372 g = 2 P f t > 2 : 372 g : Then it is clear that 1 2 ( p -value) = P f t > 2 : 372 g : Now we read from Table 4 that P f t > 2 : 262 g = 0 : 025 P f t > 2 : 372 g = 1 2 ( p -value) P f t > 2 : 821 g = 0 : 010 : It follows that 0 : 010 < 1 2 ( p -value) < 0 : 025 or 0 : 01 < p -value < 0 : 05 : ( b ) With ± = 0 : 05 and df = n 1 = 9, t 2 = t 0 : 025 = 2 : 262. A 95% con²dence interval for 1 2 = d is d ± t 2 s d p n ! = 0 : 3 ± 2 : 262 0 : 4 p 10 ! = 0 : 3 ± 0 : 286 or 0 : 014 1 2 < 0 : 586 :
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c ) Using s 2 d = 0 : 16 and B = 0 : 1, the inequality to be solved is approximately 1 : 96 s s p n ! & 0 : 1 or 1 : 96 0 : 04 p n ! & 0 : 1 : Thus, p n ± (1 : 96) (0 : 4) 0 : 1 = 7 : 84 Hence, n ± (7 : 84) 2 = 61 : 4656 or n = 62 : Exercise 10.37 . n = 5), along with d and s 2 d , is given as follows: i 1 2 3 4 5 x 1 i 1 : 3 1 : 6 1 : 1 1 : 4 1 : 7 x 2 i 1 : 2 1 : 5 1 : 1 1 : 2 1 : 8 d i = x 1 i ² x 2 i 0 : 1 0 : 1 0 : 0 0 : 2 ² 0 : 1 n X i =1 d i = 0 : 3 d 2 i 0 : 01 0 : 01 0 : 00 0 : 04 0 : 01 n X i =1 d 2 i = 0 : 07 d = 1 n n X i =1 d i = 0 : 3 5 = 0 : 06 s 2 d = 1 n ² 1 n X i =1 d i ² d ± 2 = 1 n ² 1 2 4 n X i =1 d 2 i ² 1 n n X i =1 d i ! 2 3 5 = 1 5 ² 1 ² 0 : 07 ² 1 5 (0 : 3) 2 ³ = 0 : 013 The hypotheses to be tested are H 0 : 1 ² 2 = 0 or H 0 : d = 0 : versus H a : 1 ² 2 6 = 0 or H a : d 6 = 0 : The test statistic is t = d ² d s d = p n = 0 : 06 ² 0 q 0 : 013 = 5 = 1 : 177 : ( a ) With ± = 0 : 05 and df = n ² 1 = 4, t 2 = t 0 : 025 = 2 : 776. The rejection region is j t j > 2 : 776. Since the test statistic t = 1 : 177 does not fall in the rejection region, we cannot reject H 0 : 1 ² 2 = 0. There is insu±cient evidence to indicate ± = 0 : 05. 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.

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stats 363 HW-Section-10-5 - Section 10.5 Exercise 10.35. It...

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