POWER OF THE F-TEST
Suppose C is a q × p matrix such that Cβ is testable.
We have established that
−1
(Cβ̂ − d)/q
−1
(Cβ − d)
(Cβ̂ − d)0 [C(X0 X)− C0 ]
F =
σ̂ 2
∼ Fq,n−r (δ 2 ) where
δ2 =
c
Copyright 2012
(Iowa State University)
(Cβ − d)0 [C(X0 X)− C0 ]
σ2
.
Statistics 511
1 / 10
Let Fq,n−r,1−α denote the 1 − α quantile of the central F
distribution with q and n − r d.f.
The power of the significance level α test of
H0 : Cβ = d
is given by
P(F ≥ Fq,n−r,1−α ).
The power is an increasing function of the noncentrality
parameter δ 2 .
c
Copyright 2012
(Iowa State University)
Statistics 511
2 / 10
x=seq(0,15,by=.01)
y1=df(x,3,26)
y2=df(x,3,26,ncp=10)
plot(c(x,x),c(y1,y2),pch=" ",
xlab="x",ylab="Density",
main="A Comparison of Central and Non-Central
F Densities")
lines(x,y1,col="blue",lwd=2)
lines(x,y2,lty=2,col="red",lwd=2)
legend(6,.6,
c(expression(paste(F[list(3,26)]," density")),
expression(paste(F[list(3,26)](10)," density"))),
lty=1:2, col=c("blue","red"), lwd=2)
qf(.95,3,26)
2.975154
1-pf(qf(.95,3,26),3,26,10)
0.6895487
plot(c(x,x),c(y1,y2),pch=" ",
xlab="x",ylab="Density",
main="A Comparison of Central and Non-Central F
Densities")
lines(x,y1,col="blue",lwd=2)
lines(x,y2,lty=2, col="red",lwd=2)
lines(x,df(x,3,36,15),lty=3, col="black", lwd=2)
legend(6,.6,
c(expression(paste(F[list(3,26)]," density")),
expression(paste(F[list(3,26)](10)," density")),
expression(paste(F[list(3,26)](15)," density"))),
lty=1:3, col=c("blue","red","black"), lwd=2)
1-pf(qf(.95,3,26),3,26,15)
0.8675232
Interpretation of the Noncentrality Parameter
The noncentrality parameter
δ2 =
(Cβ − d)0 [C(X0 X)− C0 ]
σ2
−1
(Cβ − d)
quantifies the discrepancy between Cβ and d with respect to
Var(Cβ̂) = σ 2 C(X0 X)− C0 .
c
Copyright 2012
(Iowa State University)
Statistics 511
3 / 10
All else being equal, the noncentrality parameter
δ2 =
(Cβ − d)0 [C(X0 X)− C0 ]
σ2
−1
(Cβ − d)
increases as . . .
the distance between Cβ and d increases,
σ 2 decreases,
the design, as defined by X, improves (e.g., sample size
increases).
c
Copyright 2012
(Iowa State University)
Statistics 511
4 / 10
Consider a completely randomized design (CRD) with three
treatments.
Suppose n = 30 experimental units are available.
Let ni denote the number of experimental units assigned to
treatment i for i = 1, 2, 3.
Suppose we consider the model
Yij = µi + ij
c
Copyright 2012
(Iowa State University)
i = 1, 2, 3;
j = 1, . . . , ni ;
∼ N(0, σ 2 I)
Statistics 511
5 / 10
Consider the test of H0 : µ1 = µ2 = µ3 .
This null hypothesis is equivalent to H0 : Cβ = 0, where


µ1
1 −1 0
C=
and β =  µ2  .
0 1 −1
µ3
(Actually, C could be any 2 × 3 matrix with row space equal
to the row space of the C matrix above.)
c
Copyright 2012
(Iowa State University)
Statistics 511
6 / 10



n1 0 0
X 0 X =  0 n2 0 
0 0 n3
−1 0
C =
C(X X)
0
−1 0 −1
C(X X)
C
c
Copyright 2012
(Iowa State University)
"
=
1
n2
+
1
n2
0

(X0 X)−1 =  0
0
"
0
1
n1
1
n3
1
n1
+ n12
− n12
1
n1
1
n2
+
− n12
1
1
n2 + n3
1
n2
0
1
n3
#
#
1
n2

0

0 
1
1
n1 n2
+
1
n1 n3
+
1
n2 n3
Statistics 511
7 / 10
"
=
1
n2
+
1
n2
1
n3
1
n1
1
n2
+
#
1
n2
n1 n2 n3
n
Suppose µ1 = 3, µ2 = 2, µ3 = 1, and σ 2 = 1
Then Cβ − d =
δ
2
1
1
and
= [1, 1]
=
c
Copyright 2012
(Iowa State University)
1
n2
+
1
n2
1
n3
1
n1
1
n2
+
1
n2
1
1
n1 n2 n3
n
4n1 n3 + n1 n2 + n2 n3
.
n
Statistics 511
8 / 10
Now suppose µ1 = 3, µ2 = µ3 = 1, σ 2 = 1
Then Cβ − d =
δ
2
2
0
and
= [2, 0]
=
c
Copyright 2012
(Iowa State University)
1
n2
+
1
n2
1
n3
1
n1
1
n2
+
1
n2
2
0
n1 n2 n3
n
4n1 (n2 + n3 )
.
n
Statistics 511
9 / 10
µ1 µ2 µ3
3 2 1
3 2 1
3 2 1
3 1 1
3 1 1
3 1 1
c
Copyright 2012
(Iowa State University)
n1
5
10
12
5
10
12
n2
20
10
6
20
10
6
n3
5
10
12
5
10
12
δ2
10.0
20.0
24.0
16.6̄
26.6̄
28.8
Statistics 511
10 / 10