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Unformatted text preview: 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 2 = 100 (a) The means are 1 = 175, 2 = 190, 3 = 160, 4 = 200, 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 noncentrality parameter. (b) The means are 1 = 175, 2 = 190, 3 = 160, 4 = 200, 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 = 0 . 01. (c) Suppose that range in the means is max { i }  min { 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 = 0 . 01. Solution: (a) The noncentrality parameter is nc = n r i =1 ( i ) 2 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] > . 9999 . (b) The noncentrality parameter is nc = n r i =1 ( i ) 2 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 (c) The noncentrality parameter is nc = n r i =1 ( i ) 2 2 n 2 2 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 = i + ij where ij are independent N (0 , 2 ) for i = 1 ,...,r and j = 1 ,...,n i ....
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This note was uploaded on 12/22/2011 for the course MAT 3378 taught by Professor G.lamothe during the Spring '11 term at University of Ottawa.
 Spring '11
 G.Lamothe
 Math, Variance

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