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*(n1)
# 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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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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 Statistics, Normal Distribution, Statistical hypothesis testing

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