Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

532 Inferences About Normal Linear Models

and

H 0 B: β 1 =···=βb=0versusH 1 B:βj =0,forsomej. (9.5.4)

IfH 0 Ais true, then by (9.5.2) the mean of the (i, j)th cell does not depend on the

level ofA. The second example above is underH 0 B. The cell means remain the

same from column to column for a specified row. We call these hypothesesmain

effecthypotheses.

Remark 9.5.1.The model just described, and others similar to it, are widely used
in statistical applications. Consider a situation in which it is desirable to investigate
the effects of two factors that influence an outcome. Thus the variety of a grain
and the type of fertilizer used influence the yield; or the teacher and the size of the
class may influence the score on a standardized test. LetXijdenote the yield from
the use of varietyiof a grain and typejof fertilizer. A test of the hypothesis that
β 1 =β 2 =···=βb= 0 would then be a test of the hypothesis that the mean yield
of each variety of grain is the same regardless of the type of fertilizer used.

Call the model described around expression (9.5.2) the full model. We want to
determine the mles. If we write out the likelihood function, the summation in the
exponent ofeis

SS=

∑a

i=1

∑b

j=1

(xij−μ−αi−βj)^2.

The mles ofαi,βj,andμminimizeSS. By adding in and subtracting out, we
obtain:

SS=

∑a

i=1

∑b

j=1

{[x··−μ]−[αi−(xi·−x··)]−[βj−(x·j−x··)]+[xij−xi·−x·j+x··]}^2.

(9.5.5)
From expression (9.5.2), we have

∑

iαi=

∑

jβj=0. Further,

∑a

i=1

(xi·−x··)=

∑b

j=1

(x·j−x··)=0

and

∑a

i=1

(xij−xi·−x·j+x··)=

∑b

j=1

(xij−xi·−x·j+x··)=0.

Therefore, in the expansion of the sum of squares, (9.5.5), all cross product terms

are 0. Hence, we have the identity

SS = ab[x··−μ]^2 +b

∑a

i=1

[αi−(xi·−x··)]^2 +a

∑b

j=1

[βj−(x·j−x··)]^2

+

∑a

i=1

∑b

j=1

[xij−xi·−x·j+x··]^2. (9.5.6)