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

The logit link makes the probability constant within each group: write $p_i=\{1+\exp [-(\mu +{\alpha }_i)]{\}}^{-1}$, with ${\alpha }_1=0$. Let $S_i={\sum }_{j=1}^k{Y}_{ij}$ and $N=km$. Independence gives $S_i\sim \mathrm{Bin}(N,p_i)$. Up to terms independent of the parameters, the log likelihood isThe likelihood equations areTogether they also imply $S_1-N{\hat{p}}_1=0$. Thus, when all totals satisfy $0<S_i<N$,The group logits are unconstrained apart from their parametrization, and each binomial log likelihood is strictly concave in its logit, so these equations give the unique finite maximum. A total of $0$ or $N$ instead gives a boundary probability and an infinite logit; a finite MLE need not exist in that case.

For $n=2$, keep $0<p_1,p_2<1$ fixed and let $N=km\to \infty$. Since ${\hat{p}}_1=S_1/N$, the central limit theorem gives $\sqrt{N}({\hat{p}}_1-p_1)\stackrel{d}{\to }N(0,p_1(1-p_1))$. The derivative of the logit is $1/[p_1(1-p_1)]$, so the delta method gives the baseline logit estimator in a group-factor binomial model lawThe second group does not improve the baseline estimate because it has its own free offset. Equivalently, the Fisher information matrix for $(\mu ,{\alpha }_2)$ isIts first diagonal entry confirms the same asymptotic variance.

factor(day) treats the nine days as categorical rather than imposing a numerical time trend. rep(400,27) supplies the number of trials for every dish, and prop is the observed surviving proportion. The contrast setting uses treatment coding with day 1 as reference; the ordered-factor polynomial option is unused. Supplying these proportions to glm with the binomial family and weights $400$ gives precisely the grouped-binomial logistic regression likelihood for the counts. The weights represent trial counts, not additional independent dish observations. The binomial family's default link is the logit link and its dispersion is fixed at $1$.

Let $Y_{ij}$ be survivors in dish $j=1,2,3$ on day $i=1,\dots ,9$. Both fitted models assume independent $Y_{ij}\sim \mathrm{Bin}(400,p_i)$. Model 1 specifiesso it has one common probability and one fitted coefficient. Its estimated probability is the total surviving count divided by the total trial count, $\hat{p}=3539/10800\simeq 0.327685$.

Model 2 specifiesIt has nine coefficients and allows a separate probability for each day, while retaining a common probability for the three dishes within that day. The MLEs are the day totals divided by $1200$. In day order they areThe intercept is the first day's logit, and each other coefficient is its day's logit minus that reference. These are additive effects on log odds, not additive changes in survival probability.

The analysis of deviance adds the eight day indicators after an intercept-only model. Its residual deviance compares a fitted model with the saturated statistical model assigning one separate probability to each of the $27$ dishes. For fitted dish probabilities ${\hat{p}}_{ij}$, the binomial deviance isThe binomial scale is one, so this is also scaled deviance. The reported decrease measures the extra fit provided by day. There are $27$ observational units for these residual statistical degrees of freedom, not $10800$ separately fitted Bernoulli responses.

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.