9.2. One-Way ANOVA 517

analysis of variance(ANOVA). In short, we say that this example is a one-way

ANOVA problem.

Here thefull modelparameter space is

Ω={(μ 1 ,μ 2 ,...,μb,σ^2 ):−∞<μj<∞, 0 <σ^2 <∞},

while thereduced model(full model underH 0 ) parameter space is

ω={(μ 1 ,μ 2 ,...,μb,σ^2 ):−∞<μ 1 =μ 2 =···=μb=μ<∞, 0 <σ^2 <∞}.

The likelihood functions, denoted byL(Ω) andL(ω) are, respectively,

L(Ω) =

(

1

2 πσ^2

)ab/ 2

exp

⎡

⎣−^1

2 σ^2

∑b

j=1

∑nj

i=1

(xij−μj)^2

⎤

⎦.

and

L(ω)=

(

1

2 πσ^2

)ab/ 2

exp

⎡

⎣−^1

2 σ^2

∑b

j=1

∑nj

i=1

(xij−μ)^2

⎤

⎦

We first consider the reduced model. Notice that it is just a one sample model
with sample size nfrom aN(μ, σ^2 ) distribution. We have derived the mles in
Example 4.1.3 of Chapter 4, which, in this notation, are given by

μˆω=^1 n

∑b

j=1

∑nj

i=1xij=x·· and ˆσ

2

ω=

1

n

∑b

j=1

∑nj

i=1(xij−x··)

(^2). (9.2.3)
The notationx··denotes that the mean is taken over both subscripts. This is often
called thegrand mean. EvaluatingL(ω) at the mles, we obtain after simplification:
L(ˆω)=
(
1
2 π
)n/ 2 (
1
σˆ^2 ω
)n/ 2
e−n/^2. (9.2.4)
Next, we consider the full model. The log of its likelihood is
logL(Ω) =−(n/2) log(2π)−(n/2) log(σ^2 )−
1
2 σ^2
∑b
j=1
∑nj
i=1
(xij−μj)^2. (9.2.5)
Forj=1,...,b, the partial of the log ofL(Ω) with respect toμjresults in
∂logL(Ω
∂μj

1
σ^2
∑nj
i=1
(xij−μj).
Setting this partial to 0 and solving forμj, we obtain the mle ofμjwhich we denote
by
μˆj=
1
nj
∑nj
i=1
xij=x·j,j=1,...,b. (9.2.6)