HW6 - Stat 415 HW 6 10.2 The test statistic Y has a...

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Stat 415 HW 6 10.2 The test statistic Y has a binomial distribution with n=20 and p a A type I error occurs if the experimenter concluded that the drug dosage level induces sleep in less that 80% of the people suffering form insomnia when, in fact, drug dosage level does induce sleep in 80% of insomnia. b α = P ( reject H 0 | H 0 true ) = P ( Y 12 | p = 0 . 8) =0.32 - use table 1 Appendix III. c A type II error would occur if the experimenter concluded that the drug dosage level induces sleep in 80% of the people suffering from insomnia when in fact fewer than 80% experience relief. 10.3 a whin n = 20 and p = 0 . 8 , it is necessary to find c such that α = P ( Y c | p = 0 . 8) = 0 . 1 . From table 1, Appendix III, this value is c=11. 10.7 a H 0 : μ = 900 , H a : μ 6 = 900 b The rejection region with α = 0 . 1 is determined by a critical value of z such that P [ | Z | > z 0 ] = 0 . 1 This value is z 0 = 2 . 58 and the rejection region is R = { z : | z | > 2 . 58 } c Z = ¯ y - μ σ/ n
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This note was uploaded on 05/17/2008 for the course MATH 415 taught by Professor Yuzhang during the Spring '08 term at Pennsylvania State University, University Park.

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HW6 - Stat 415 HW 6 10.2 The test statistic Y has a...

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