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409Hw07ans - STAT 409 Fall 2009 Homework#7(Answers 8.1-6 H...

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STAT 409 Fall 2009 Homework #7 ( Answers ) 8.1-6 H 0 : p = 1/6 vs. H 1 : p < 1/6 Left – tailed. or H 0 : p 1/6 vs. H 1 : p < 1/6 n = 8000. a) The test statistic is ( 29 ( 29 8000 6 1 1 6 1 6 1 8000 1 0 0 0 - - = - - = y n n y z p p p . α = 0.05. The critical (rejection) region is z < – z α = – 1.645. b) y = 1265. The observed value of z (test statistic) ( 29 8000 6 1 1 6 1 6 1 8000 1265 - - = z = – 2.05 is less than – 1.645 (the test statistic does fall into the rejection region) so Reject H 0 . OR p-value = P(Z 2.05) = 0.0202 < 0.05 = α , Reject H 0 . c) ( 29 ( - + = - + n p p z p 8000 8000 1265 1 8000 1265 645 . 1 8000 1265 , 0 ˆ 1 ˆ ˆ , 0 α = [ 0 , 0.1648 ]. 1 / 6 0.1667 is not in this interval. This is consistent with the conclusion to reject H 0 .
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8.1-8 The value of the test statistic is z = 390 25 . 0 75 . 0 75 . 0 70 . 0 - = – 2.280. a) Since z = – 2.280 < – 1.645, Reject H 0 . b) Since z = – 2.280 > – 2.326, Do NOT Reject H 0 . c) p-value = P(Z 2.280) = 0.0113. Note that 0.01 < p-value < 0.05. 8.1-10 a) H 0 : p = 0.14 vs. H 1 : p > 0.14. Right – tailed. b) The critical (rejection) region is z > z α = 2.326, where ( 29 86 . 0 14 . 0 14 . 0 1 0 0 0 n n y n n y z p p p - = - - = . c) The observed value of z (test statistic) 590 86 . 0 14 . 0 14 . 0 590 104 - = z = 2.539 is greater than 2.326 (the test statistic does fall into the rejection region) so Reject H 0 .
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