# 11_30ans - STAT 410 Examples for Fall 2009 1 Let X have a...

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STAT 410 Examples for 11/30/2009 Fall 2009 1. Let X have a Binomial distribution with the number of trials n = 8 and with probability of “success” p . We wish to test H 0 : p = 0.60 vs. H 1 : p < 0.60. a) Suppose we observe X = 3. Find the p-value of this test. p-value = P ( value of X as extreme or more extreme than X = x observed | H 0 true ) = P ( X 3 | p = 0.60 ) = 0.1737 . b) Suppose we decided to use the rejection region “Reject H 0 if X 1.” Find the significance level α associated with this rejection region. significance level α = P ( Reject H 0 | H 0 is true ) = P ( X 1 | p = 0.60 ) = 0.0085 . c) Find the “best” rejection region with the significance level α closest to 0.05. Rejection Region for a Left – tailed test: Find a such that P ( X a ) α . Then the Rejection Region is “Reject H 0 if X a .” significance level α = P ( Reject H 0 | H 0 is true ) = P ( X a | p = 0.60 ). P ( X 2 | p = 0.60 ) = 0.0498 0.05. Reject H 0 if X 2 . d) Find the “best” ( randomized ) rejection region with the significance level α = 0.10. 0.0498 = P ( X 2 | p = 0.60 ) < 0.10 < P ( X 3 | p = 0.60 ) = 0.1737. P ( X = 3 | p = 0.60 ) = 0.1239. 0.10 = P ( X 2 | p = 0.60 ) + P ( X = 3 | p = 0.60 ) p p . 0.10 = 0.0498 + 0.1239 p p . p p 0.405. Reject H 0 if X 2, Reject H 0 with probability p p 0.405 if X = 3, Do NOT Reject H 0 if X 4.

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Find the “best” ( randomized ) rejection region with the significance level α = 0.03. 0.0085 = P ( X 1 | p = 0.60 ) < 0.03 < P ( X 2 | p = 0.60 ) = 0.0498. P ( X = 2 | p = 0.60 ) = 0.0413. 0.03 = P ( X 1 | p = 0.60 ) + P ( X = 2 | p = 0.60 ) p p . 0.03 = 0.0085 + 0.0413 p p . p p 0.52. Reject
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11_30ans - STAT 410 Examples for Fall 2009 1 Let X have a...

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