Introduction to Probability and Statistics for Engineers and Scientists

10.6Two-Way Analysis of Variance with Interaction 465

Therefore, with a “hat” over a parameter denoting the estimator of that parameter, we
obtain from the preceding that unbiased estimators are given by

μˆ=X...

βˆj=X.j.−X...

αˆi=Xi..−X...

γˆij=Xij.−ˆμ−βˆj−ˆαi=Xij.−Xi..−X.j.+X...

To develop tests for the null hypothesesH 0 int,H 0 r, andH 0 c, start with the fact that

∑l

k= 1

∑n

j= 1

∑m

i= 1

(Xijk−μ−αi−βj−γij)^2

σ^2

is a chi-square random variable withnmldegrees of freedom. Therefore,

∑l

k= 1

∑n

j= 1

∑m

i= 1

(Xijk−ˆμ−ˆαi−βˆj−ˆγij)^2

σ^2

will also be chi-square, but with 1 degree of freedom lost for each parameter that is
estimated. Now, since

∑
iαi=0, it follows thatm−1oftheαineed to be estimated;
similarly,n−1oftheβjneed to be estimated. Also, since

∑

iγij=

∑
jγij=0, it follows
that if we arrange all theγijin a rectangular array havingmrows andncolumns, then all
the row and column sums will equal 0, and so the values of the quantities in the last row
and last column will be determined by the values of all the others; hence we need only
estimate (m−1)(n−1) of these quantities. Because we also need to estimateμ, it follows
that a total of

n− 1 +m− 1 +(n−1)(m−1)+ 1 =nm
parameters needs to be estimated. Since

μˆ+ˆαi+βˆj+ˆγij=Xij.

it thus follows from the preceding that if we let

SSe=

∑l

k= 1

∑n

j= 1

∑m

i= 1

(Xijk−Xij.)^2

then
SSe
σ^2

is chi-square withnm(l−1) degrees of freedom

Therefore,

SSe

nm(l−1)

is an unbiased estimator ofσ^2