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Unformatted text preview: THE REJECTION REGION METHOD Page 346 11.37 n=20 σ=50 α=.05 X=569 1. Ho: μ = 560 Ha: μ > 560 2. α=.05 3 If the results you get were: what would you do?10 shouldn't come into play because you shouldn’t set it up so it would be this wa 0.1 Do not reject Ho 10 Reject Ho 4. Reject Ho if Z > 1.645 0.45 α=0.05 Z α/2 Didn't divide alpha by 2, This is a ONE sided test. Get the Z from the tables 5. Since Z is NOT greater than 1.645, we do NOT reject Ho. See, does not fall in the rejection region, so w 0.8 Z α Problem 11.39 n=50 σ=6 α=.05 X=(afterbefore)/before= n X Z σ μ = 80 . 20 50 560 569 = = 1. Ho: μ =0 Ha: μ < 0 2. α=.05 3 4. Reject Ho if Z < 1.645 α=0.05 0.45 zZ α=1.45 5. Since Z is not less than 1.645, we cannot reject Ho Problem 11.35 n=100 σ=400 α=.05 X=(afterbefore)/before=5065 1. Ho: μ =5000 Ha: μ > 5000 2. α=.05 3 1.63 4. Reject Ho if Z > 1.645 α=0.05 z z=1.625 Z α=645 5. Since Z is not greater than 1.645, we cannot reject Ho n X Z σ μ = 41 . 1 50 6 2 . 1 = = n X Z σ μ = = = 100 400 5000 5065 Problem 11.33 n=10 σ=0.05 α=.05 X=.493 1. Ho: μ =0.5 Ha: μ ≠.05 2. α=.05 3 4. Reject Ho if Z >Zα = 1.96or Z <Zα =1.96 0.48 α=0.025 α=0.025 0.48 z Z α/2=1.960.44 Z α/2=1.96 5. Since Z.44 and this is not less than Za/2, we cannot reject Ho The PValue method The probability of obtaining A test statistic (Z) as extreme or more extreme than that actually obtained...
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This note was uploaded on 10/21/2008 for the course MBA 803 taught by Professor Cantrell during the Spring '08 term at Clemson.
 Spring '08
 CANTRELL

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