 Solution

Apart from its multinomial coefficient, the likelihood is ${\prod }_{ij}{\theta }_{ij}^{n_{ij}}$. Maximizing its logarithm subject to $\sum {\theta }_{ij}=1$ gives $n_{ij}/{\theta }_{ij}$ equal to a common multiplier, which is $n$. Hence ${\hat{\theta }}_{ij}=n_{ij}/n$.

Under independence the log likelihood separates as ${\sum }_i{n}_{i\cdot }\log {\varphi }_i+{\sum }_j{n}_{\cdot j}\log {\rho }_j$. The two normalization constraints give ${\hat{\varphi }}_i=n_{i\cdot }/n$ and ${\hat{\rho }}_j=n_{\cdot j}/n$, so the fitted probabilities are $n_{i\cdot }{n}_{\cdot j}/n^2$. Therefore twice the difference of the maximized log likelihoods isThe unrestricted model has $rc-1$ free parameters and the independence model has $(r-1)+(c-1)$, leaving $(r-1)(c-1)=4$ degrees of freedom for the table. Under the null, the likelihood-ratio statistic is asymptotically chi-squared with four degrees of freedom. Since $25.192>18.467$, reject independence at the 0.1% significance level. This is evidence of association in the sampled variables; it does not by itself identify a causal direction.