524 Inferences About Normal Linear Models∑n
1 μ2
i/σ(^2) , may be computed by replacing eachX
iin the quadratic form by its
meanμi,i=1, 2 ,...,n. This is no fortuitous circumstance; any quadratic form
Q=Q(X 1 ,...,Xn) in normally distributed variables, which is such thatQ/σ^2 is
χ^2 (r, θ), hasθ=Q(μ 1 ,μ 2 ,...,μn)/σ^2 ;andifQ/σ^2 is a chi-square variable (central
or noncentral) for certain real values ofμ 1 ,μ 2 ,...,μn, it is chi-square (central or
noncentral) forallreal values of these means.
We next discuss the noncentralF-distribution. IfUandVare independent and
are, respectively,χ^2 (r 1 )andχ^2 (r 2 ), the random variableF has been defined by
F =r 2 U/r 1 V. Now suppose, in particular, thatUisχ^2 (r 1 ,θ),V isχ^2 (r 2 ), and
U andV are independent. The distribution of the random variabler 2 U/r 1 V is
called anoncentralF-distributionwithr 1 andr 2 degrees of freedom with non-
centrality parameterθ. Note that the noncentrality parameter ofFis precisely the
noncentrality parameter of the random variableU,whichisχ^2 (r 1 ,θ). To obtain the
expectation ofF,usetheE(U) in expression (9.3.3) and the derivation of the ex-
pected value of a centralFgiven in expression (3.6.8). These together immediately
imply that
E(F)=
r 2
r 2 − 2
[
r 1 +θ
r 1
]
, (9.3.4)
provided, of course, thatr 2 >2. Ifθ>0 then the quantity in brackets exceeds one
and, hence, the mean of the noncentralFexceeds the mean of the corresponding
centralF.
We next discuss the noncentralFdistribution for the one-way ANOVA of the
last section.
Example 9.3.1 (Noncentrality Parameter for One-way ANOVA). Consider the
one-way model withblevels, expression (9.2.1), with the hypothesesH 0 :μ 1 =
···=μbversusH 1 :μj =μj′for somej =j′. From expression (9.2.11), theFtest
statistic isF=[Q 4 /(b−1)]/[Q 3 /(n−b)]. In the denominator, the random variable
Q 3 /σ^2 isχ^2 (n−b) under the full model and, hence, in particular, underH 1 .It
follows from Remark 9.8.3 of Section 9.8, though, that the distribution ofQ 4 /σ^2 is
noncentralχ^2 (b− 1 ,θ) under the full model. Recall that
Q 4 /σ^2 =
1
σ^2
∑b
j=1
nj(X·j−X··)^2.
Under the full model,E(X·j)=μjandE(X··)=
∑b
j=1(nj/n)μj. Calling this last
expectationμ, we have from the above discussion that
θ=
1
σ^2
∑b
j=1
nj(μj−μ)^2. (9.3.5)
IfH 0 is true thenμj≡μ,forsomeμ, and, hence,μ=μ. Thus, underH 0 ,θ=0.
UnderH 1 , there are distinctjandj′such thatμj =μj′. In particular, then both
μjandμj′cannot equalμ,soθ>0. Therefore, underH 1 the expectation ofF
exceeds the null expectation.