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hw2mat3378spring2010sol

# hw2mat3378spring2010sol - MAT 3378 3X Spring 2010...

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MAT 3378 3X - Spring 2010 Assignment 2 : Deadline: Friday, June 25 (In Class) Question 1: Suppose that five normal populations have a common variance of σ 2 = 100 (a) The means are μ 1 = 175, μ 2 = 190, μ 3 = 160, μ 4 = 200, μ 5 = 215. Assuming a balanced design of n = 5 observations per population. What is the power of the test for equality of means with a level of significance of 1%? Hint: Compute the non-centrality parameter. (b) The means are μ 1 = 175, μ 2 = 190, μ 3 = 160, μ 4 = 200, μ 5 = 215. Assuming a balanced design how many observations per population must be taken so that the probability of rejecting the hypothesis of equality of means is at least 0.95? Use α = 0 . 01. (c) Suppose that range in the means is max { μ i } - min { μ i } = 40. Assuming a balanced design, how many observations per population must be taken so that the probability of rejecting the hypothesis of equality of means is at least 0.95? Use α = 0 . 01. Solution: (a) The non-centrality parameter is nc = n r i =1 ( μ i - μ · ) 2 σ 2 = 1830 100 n = 91 . 5 . We will reject the null hypothesis of equality of means if F * > F (0 . 99 , 4 , 20) = 4 . 43069 . Power is P [ F (4 , 20 , 91 . 5) > 4 . 43069] > 0 . 9999 . (b) The non-centrality parameter is nc = n r i =1 ( μ i - μ · ) 2 σ 2 = 1830 100 n = 18 . 3 n. Power is Power( n ) = P [ F (4 , 20 , 18 . 3 n ) > F (0 . 99 , 4 , 5 ( n - 1))] . We get Power(2) = 0 . 49582 , Power(3) = 0 . 97114 . Thus, n = 3 is sufficient. 1

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(c) The non-centrality parameter is nc = n r i =1 ( μ i - μ · ) 2 σ 2 n δ 2 2 σ 2 = n (40) 2 2 (100) = 8 n. We will use nc = 8 n . Power( n ) = P [ F (4 , 20 , 18 . 3 n ) > F (0 . 99 , 4 , 5 ( n - 1))] . We get Power(4) = 0 . 89213 , Power(5) = 0 . 97826 . Thus, n = 5 is sufficient. Question 2: Consider the one factor ANOVA model Y ij = μ i + ij where ij are independent N (0 , σ 2 ) for i = 1 , . . . , r and j = 1 , . . . , n i . 1. Using properties of the expectation and variance operators E {} and σ 2 {} operators show that E { Y i · } = μ i and σ 2 { Y i · } = σ 2 /n i . 2. Consider L = r X i =1 c i μ i and ˆ L = r X i =1 c i Y i · .
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