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Six Independent Samples of 100 Values Each Are Randomly Drawn μ1=μ2=μ3=μ4=μ5=μ6\mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 } = \mu _ { 4 } = \mu _ { 5 } = \mu _ { 6 }

Question 44

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Six independent samples of 100 values each are randomly drawn from populations that are normally distributed with equal variances. You wish to test the claim that μ1=μ2=μ3=μ4=μ5=μ6\mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 } = \mu _ { 4 } = \mu _ { 5 } = \mu _ { 6 } .
If you test the individual claims μ1=μ2,μ1=μ3,μ1=μ4,,μ5=μ6\mu _ { 1 } = \mu _ { 2 } , \mu _ { 1 } = \mu _ { 3 } , \mu _ { 1 } = \mu _ { 4 } , \ldots , \mu _ { 5 } = \mu _ { 6 } , how many ways can you pair off the means?
i) Assume that the tests are independent and that for each test of equality between two means, there is a 0.900.90 probability of not making a type I error. If all possible pairs of means are tested for equality, what is the probabi making no type I errors?
ii) If you use analysis of variance to test the claim that μ1=μ2=μ3=μ4=μ5=μ6\mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 } = \mu _ { 4 } = \mu _ { 5 } = \mu _ { 6 } at the 0.100.10 level of significan what is the probability of not making a type I error?

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i)15
ii) ...

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