stats 363 HW-Section-9-5

# stats 363 HW-Section-9-5 - Section 9.5 Exercise 9.31 a The...

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Unformatted text preview: Section 9.5 Exercise 9.31 . ( a ) The hypotheses are H : p = 0 : 4 versus H a : p 6 = 0 : 4 : This is a two-tailed test. ( b ) With n = 1400 and x = 529, the sample proportion is b p = 529 = 1400 = : 378 and the test statistic is z = : 378 & : 4 q (0 : 4)(0 : 6) 1400 = & 1 : 68 : Hence, the p-value is p-value = P f z < & 1 : 68 g + P f z > 1 : 68 g = 2 (0 : 0465) = 0 : 0930 : Since this p-value is much greater than 0 : 1, we cannot reject H at the 1% level ( p-value > : 01). ( c ) Since the p-value > : 05, H cannot be rejected and there is no su&cient evidence to indicate that p is di¡erent from 0 : 4. Exercise 9.33 . ( a ) The two sets of hypotheses both involve a di¡erent binomial parameter p : H : p = 0 : 6 versus H a : p 6 = 0 : 6 : H : p = 0 : 5 versus H a : p < : 5 : ( b ) For the second set of hypotheses in part ( a ), n = 75 and x = 35, the sample proportion is b p = 35 = 75 = 0 : 4667 and the test statistic is z = : 4667 & : 5 q (0 : 5)(0 : 5) 75 = & : 58 : Hence, the p-value is...
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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-9-5 - Section 9.5 Exercise 9.31 a The...

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