Past exam of the options II course of the University of Oxford / 2011 / O / O4

Ignoring a factor independent of the parameters, the likelihood function isSince every count is positive, a boundary value ${\theta }_{ij}=0$ has zero likelihood. At an interior optimum, a Lagrange multiplier for $\sum {\theta }_{ij}=1$ gives $n_{ij}/{\theta }_{ij}=\lambda$. Summing the resulting ${\theta }_{ij}=n_{ij}/\lambda$ gives $\lambda =N$, soThe log likelihood function is strictly concave on the positive simplex, because its Hessian matrix is diagonal with entries $-n_{ij}/{\theta }_{ij}^2<0$. Hence this critical point is the unique global maximum-likelihood estimate.

Under independence, the log likelihood function separates:Maximizing the two terms over their respective probability simplices, by the same Lagrange multiplier argument as in part (a), givesTherefore the constrained maximum-likelihood estimates areThe associated fitted cell counts are $E_{ij}=N{\hat{\theta }}_{ij}^0=n_{i.}{n}_{.j}/N$.

The log likelihood ratio statistic comparing the unrestricted fit with the independence fit isThe unrestricted model has $rc-1=8$ free parameters. Independence has $(r-1)+(c-1)=4$, so the difference is $(r-1)(c-1)=4$. By Wilks theorem, under the null hypothesis the statistic has asymptotically a chi-squared distribution with $4$ degrees of freedom, rather than $6$, $8$, or $9$.

For the supplied statistic, the approximate p-value is $1-F_4(11.8)=0.02$. Thus reject independence at the 5% level, but not at the 1% level: the data provide evidence of an association. The approximation is reasonable here; the smallest fitted cell count is $27\cdot 99/299\approx 8.94$, so all fitted counts exceed $5$. The statistical hypothesis test concerns association, not a causal effect of personality.