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hw5sol - Stat 511 Due 11am Monday Feb 14 Homework 5...

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Stat 511 Homework 5 Solution Spring 2011 Due: 11am, Monday Feb 14 1. This problem was set to reinforce a point made in lecture about power of various tests in an ANOVA. Consider a study with 4 treatments in a 2 x 2 factorial. The investigators tell you that 1.2 units is an important di±erence and the error variance is 1.0. You consider a design with n = 6 observations for each treatment (i.e. a total of 24 observations). To answer this question, we will still employ the R program ’power.r’ power <- function(n, c, means, sigma2){ xpx <- diag(rep(n,m)) # x’x matrix, m trt levels cb <- c %*% means ncp <- t(cb) %*% solve(c %*% ginv(xpx) %*% t(c)) %*% cb / sigma2; q <- dim(c)[1] # # rows in C df <- m*(n-1) # error df pf(qf(0.95,q,df), q, df, ncp=ncp, lower=F) # P[F(ncp) > crit ] } (a) Calculate the power of the test of the A main e±ect when the population A e±ect is 1.2. cm1 <- t( c(0.5, 0.5, -0.5, -0.5) ) # t() is so have a row vector means1 <- c(101.2,101.2,100,100) # chosen so cb = 1.2 power(6,cm1,means1,1) [1] 0.798381 (b) Calculate the power of the test of the A*B interaction when the population interaction, ( μ 11 μ 12 ) ( μ 21 μ 22 ), is 1.2. cm2 <- t( c(1, -1, 1, -1) ) means2 <- c(1,0.8,1,0) power(6,cm2,means2,1) [1] 0.2879725 The investigators increase factor B to 3 levels, so there are 6 treatments in a 2 x 3 factorial. Again, they consider n = 6 observations for each treatment (i.e. a total of 36 observations now). (c) Calculate the power of the test of the di±erence between level 1 of B and level 2 of B, both averaged over levels of A, when the population di±erence is 1.2. cm1 <- t( c(0.5, 0.5,0, -0.5, -0.5,0) ) means1 <- c(101.2,101.2,100,100,100,100) power(6,cm1,means1,1) [1] 0.8117006 (d) Calculate the power of the test of the A main e±ect when the population A e±ect is 1.2. 1
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Stat 511 Homework 5 Solution Spring 2011 cm2 <- t( 1/3*c(1, -1, 1, -1, 1, -1) ) means2 <-c(2,0.8,2,0.8,2,0.8) power(6,cm2,means2,1) [1] 0.9360686 (e) Explain (i.e. as if to a consulting client) why the test in question 1c) is more powerful than the test in 1a). The ncp for two questions are the same, but d.f associated with F statistics is (1,20) in (a) and (1,30) in 1c) which leads to more powerful test in (c).
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hw5sol - Stat 511 Due 11am Monday Feb 14 Homework 5...

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