Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 37 / 3

Index independent experimental vials by $r$, write $N_r$ for the observed total eaten and $E_r$ for the number of other-species eggs eaten. The grouped-binomial logistic regression usesHere $D_r\in \{1,2,3\}$ and $S_r\in \{A,B,C\}$ identify day and adult species, and the free effects are ${\beta }_0,d_2,d_3,s_B,s_C$. There is no day-species interaction. The weights in the binomial generalized linear model are the denominators $N_r$, so fitting the proportions $E_r/N_r$ gives this binomial likelihood, not 23 Bernoulli observations. The dispersion is fixed at one. This is a model for the composition of eggs eaten conditional on the total, rather than a model for the total appetite.

There are 23 independent vial responses. The intercept-only fit has 22 residual statistical degrees of freedom and deviance 33.631. Adding two day indicators removes two statistical degrees of freedom and reduces the residual deviance by $33.631-33.358=0.273$. Adding the two species indicators then reduces it by $33.358-14.646=18.712$. Thus the six replacements, in their printed order, are $22,33.631,2,0.273,18.712,14.646$.

The residual deviance statistical degrees of freedom are $23-1=22$, $23-3=20$, and $23-5=18$. These sequential analysis-of-deviance rows compare nested generalized linear models in the displayed order. In an unbalanced design, the first day row does not test day after controlling for species.

Compare the species-only model against the additive day-and-species model. The null hypothesis is $H_0:d_2=d_3=0$, with the alternative that at least one is nonzero. The likelihood-ratio statistic is the reduction in residual deviance,Under the null hypothesis, regular binomial likelihood asymptotics give $G^2\stackrel{\mathrm{approx}}{\sim }{\chi }_2^2$, because the larger model has two additional parameters. For this chi-squared distribution the upper-tail probability is $\exp (-0.341/2)\simeq 0.843$. There is no evidence of a day effect after controlling for species, and the species-only model is a reasonable simpler choice. The sequential day-only statistic 0.273 from the preceding table answers a different question and must not replace this comparison.

In the species-only model, the log odds for adult species A is the intercept, so ${\hat{\theta }}_A=0.91191$ with standard error 0.07824. This also follows directly by pooling its vials: there are 570 other-species eggs among 799 eggs eaten, giving ${\hat{p}}_A=570/799$ andFor a regular maximum-likelihood estimator, asymptotic normality of a maximum likelihood estimator gives ${\hat{\theta }}_A\stackrel{\mathrm{approx}}{\sim }N({\theta }_A,I_A^{-1})$, with information $I_A=N_A{p}_A(1-p_A)$. Substitution of ${\hat{p}}_A$ gives $\mathrm{se}({\hat{\theta }}_A)=\sqrt{1/570+1/229}\simeq 0.07824$. An approximate 95% Wald confidence interval isThis interval treats the selected species-only model as fixed; it does not account separately for uncertainty from choosing that model.

The relevant no-preference benchmark is $p_A=2/3$, because two of the three equally supplied egg types count as other-species eggs. Its log odds is $\log 2\simeq 0.6931$, which is below the entire confidence interval. Under the fitted binomial model, species A preferentially consumes other-species eggs. The estimate is ${\hat{p}}_A\simeq 0.7134$, and transforming the log-odds confidence interval by $e^{\theta }/(1+e^{\theta })$ gives approximately $(0.6810,0.7437)$.

Comparing the interval with zero would test equal numbers of own- and other-species eggs eaten, which is not the no-preference hypothesis for this supply. The conclusion remains model-dependent: eggs are taken from a finite supply without replacement, and depletion or dependence within a vial can affect the binomial approximation. This fitted analysis supports an association with egg type; it does not by itself establish the mechanism of recognition or avoidance.