Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 43 / 3 / ii / Solution

The null model uses one parameter, the day model uses nine, and there are $27$ grouped observations. The four missing entries, in their table order, are thereforeThe test of no day effect is $H_0:{\alpha }_2=\cdots ={\alpha }_9=0$. The nominal binomial likelihood-ratio test statistic is $\Delta D=462.8363$, compared asymptotically with ${\chi }_8^2$. Its nominal $p$-value is about $6.6\times 10^{-95}$, so the common-probability model is decisively inadequate relative to the day model. The fitted probabilities in (i) show substantial differences between days, with days 1 and 7 much higher than days 3 and 8.

However, relative improvement does not establish absolute fit. The day model's deviance goodness-of-fit test compares $D=32.7945$ with ${\chi }_{18}^2$, giving $p\simeq 0.0177$. The counts are large enough for this approximation to be reasonable. Thus at $5\%$ the day-specific binomial model still shows lack of fit. Among the two printed models it is preferred, but it should not be regarded as a fully satisfactory final model.

The residual scale estimate $D/18\simeq 1.822$ suggests overdispersion. Directly computing the Pearson statistic from the day means givesThis corroborates extra dish variation rather than the nominal binomial variance. Dish heterogeneity or dependence among cells in a dish can produce such variation. Residual checks and a quasibinomial regression or an explicit model for dish heterogeneity are sensible refinements. Allowing dispersion of this moderate magnitude would enlarge standard errors but would not plausibly remove the very large day effect. Retain day effects, while accounting for the remaining extra-binomial variation.