Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 4 / 5I / Solution

The first two commands create vectors of observed successes and corresponding numbers of trials. The next two create the class labels and convert them into an R factor, so the labels are treated categorically. The final command fits a grouped-binomial logistic regression, using the proportions as responses and trial totals as binomial weights, and prints its summary. The model is $\log [{\pi }_i/(1-{\pi }_i)]={\beta }_0+{\beta }_b{1}_{\{S_i=b\}}$, with class a as reference.

The intercept estimates the log odds in class a. Its value corresponds to probability $39/287\approx 0.1359$. The class-b coefficient is the estimated increase in log odds, about $0.4999$, so the fitted probability in class b is $35/170\approx 0.2059$ and the estimated odds ratio is $e^{0.4999}\approx 1.6485$. The printed standard errors measure uncertainty in these estimated coefficients. The z values are estimate divided by standard error: the intercept tests whether the class-a probability is $1/2$, while the class-b z value tests equal class probabilities. The two-sided normal-reference P-value for the latter is about $0.051$, so it is just above a conventional $5\%$ threshold; this is a borderline result rather than strong evidence of a class effect.

The residual deviance compares the fitted two-parameter model with the four-probability saturated model. Its value $1.9369$ on $4-2=2$ residual degrees of freedom is small relative to a ${\chi }_2^2$ reference, with approximate tail probability $e^{-1.9369/2}\approx 0.380$. There is no evidence of lack of fit by that check. Four Fisher scoring iterations were needed for numerical fitting. These interpretations presume independent binomial counts and the usual large-sample approximations; weights here encode trial counts, not extra independent observations.