10.6Two-Way Analysis of Variance with Interaction 467
Namely,
reject H 0 int ifSSint/(n−1)(m−1)
SSe/nm(l−1)>F(n−1)(m−1),nm(l−1),α
do not reject H 0 int otherwiseAlternatively, we can compute thep-value. IfFint=v, then thep-value of the test of the
null hypothesis that all interactions equal 0 is
p-value=P{F(n−1)(m−1),nm(l−1)>v}If we want to test the null hypothesis
H 0 r:αi=0,i=1,...,mthen we use the fact that whenH 0 ris true,Xi..is the average ofnlindependent normal
random variables, each with meanμand varianceσ^2. Hence, underH 0 r,
E[Xi..]=μ, Var(Xi..)=σ^2 /nland so
∑mi= 1nl(Xi..−μ)^2
σ^2is chi-square withmdegrees of freedom. Thus, if we let
SSr=∑mi= 1nl(Xi..−ˆμ)^2 =∑mi= 1nl(Xi..−X..)^2then, whenH 0 ris true,
SSr
σ^2is chi-square withm−1 degrees of freedomand so
SSr
m− 1is an unbiased estimator ofσ^2Because it can be shown that, underH 0 r,SSeandSSrare independent, it follows that
whenH 0 ris true
SSr/(m−1)
SSe/nm(l−1)is anFm− 1 ,nm(l−1) random variable