Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

518 Inferences About Normal Linear Models

Since this derivation did not depend onσ, to find the mle ofσ, we substitutex·j

forμjin the logL(Ω). Taking the partial derivative with respect toσwe then get

∂logL(Ω

∂σ

=−(n/2)

2 σ

σ^2

+

1

σ^3

∑b

j=1

∑nj

i=1

(xij−x·j)^2.

Solving this forσ^2 ,weobtain^1 the mle

ˆσΩ^2 =

1

n

∑b

j=1

∑nj

i=1

(xij−x·j)^2. (9.2.7)

Substituting these mles for their respective parameters inL(Ω), after some simpli-
fication, leads to

L(Ω) =ˆ

(

1

2 π

)n/ 2 (

1

ˆσΩ^2

)n/ 2

e−n/^2. (9.2.8)

Hence, the likelihood ratio test rejectsH 0 in favor ofH 1 for small values of the
statisticΛ=ˆ L(ˆω)/L(Ω) or equivalently, for large values ofˆ Λˆ−^2 /n. We can express
this test statistic as a ratio of two quadratic formsQ 3 andQas

Λˆn/^2 = σˆ

2

Ω

σˆ^2 ω

=

∑b

j=1

∑nj

i=1(xij−x·j)

2

∑b

j=1

∑nj

i=1(xij−x··)

2

=dfn

Q 3

Q

. (9.2.9)

In order to rewrite the test statistic in terms of anF-statistic, consider the identity

involvingQ,Q 3 , and another quadratic formQ 4 given by:

Q =

∑b

j=1

∑nj

i=1

(xij−x··)^2 =

∑b

j=1

∑nj

i=1

[(xij−x·j)+(x·j−x··)]^2

=

∑b

j=1

∑nj

i=1

(xij−x·j)^2 +

∑b

j=1

nj(x·j−x··)^2

=dfn Q 3 +Q 4. (9.2.10)

This derivation follows because the cross product term in the second line is 0. Using
this identity, the test statisticΛˆ−^2 /ncan be expressed as

Λˆ−^2 /n=Q^3 +Q^4

Q 3

=1+

Q 4

Q 3

.

As the final version, note that the test rejectsH 0 ifFis too large where

F=

Q 4 /(b−1)

Q 3 /(n−b)

. (9.2.11)

(^1) We are using the fact that the mle ofσ (^2) is the square of the mle ofσ.