= Solution
The generalized <likelihood ratio> compares the constrained maximum from part (b) with the unrestricted maximum from part (a). Taking minus twice its logarithm gives
$$
\boxed{\Lambda=2\sum_{i,j}n_{ij}\log\frac{n_{ij}N}{n_{i.}n_{.j}}.}
$$
The unrestricted $3\times3$ model has $9-1=8$ free parameters; the independence model has $(3-1)+(3-1)=4$. Thus the asymptotic null distribution is $\chi^2_4$, with four degrees of freedom. The supplied value gives
$$
p\text{-value}\approx1-F_4(11.8)=0.02.
$$
Therefore \b[reject independence at the 5% level], but not at the 1% level. There is evidence of an association between the two classifications. The fitted expected counts are all about nine or larger, so the usual chi-squared approximation has no very small expected cells here.
Back to article page
