Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 206 / 2

For independent random variables $Y_i\sim \mathrm{Bernoulli}(p_i)$, the full logistic regression has linear predictorLayout A is the reference level in a regression factor. The smaller model sets ${\beta }_B={\beta }_C=0$, leaving the same logit link and the price predictor. In either caseThe saturated statistical model allows a separate probability per observation; its maximizing probabilities are $p_i=y_i$, with log-likelihood zero under $0\log 0=0$. Hence the binomial deviance for either fitted model isEquivalently, write the summands as $2[y_i\log (y_i/{\hat{p}}_i)+(1-y_i)\log ((1-y_i)/(1-{\hat{p}}_i))]$. The fitted probabilities differ between the two models, but the definition of binomial deviance is the same. A residual deviance for individual binary data need not itself have an accurate chi-squared distribution approximation; the nested-model difference used next has a different asymptotic justification.

There are $120$ observations, since the intercept-only model has $119$ residual degrees of freedom. Adding price introduces one coefficient and decreases the binomial deviance by $148.263-101.920=46.343$. Adding a three-level factor introduces two coefficients and decreases it by $101.920-97.571=4.349$. Thus the missing entries areThe likelihood-ratio test comparing the smaller and full logistic regression tests $H_0:{\beta }_B={\beta }_C=0$ against at least one nonzero layout contrast. Its statistic is $2({\hat{\ell }}_1-{\hat{\ell }}_2)=4.349$. Under ordinary interior-parameter, full-rank and finite-estimate regularity conditions, Wilks theorem gives an asymptotic chi-squared distribution with two degrees of freedom. The reported p-value $0.1136$ exceeds $0.05$, so at that level retain the simpler price-only model.

The table is an analysis of deviance for nested generalized linear models: price is entered before layout. Thus its price row tests the price-only model against an intercept, whereas the layout row tests layout after allowing for price. The latter is the relevant comparison of model1 and model2. Failing to reject a layout effect is not proof that layouts have identical effects; it means these data do not justify the extra two parameters under this test.

Using the smaller logistic regression, approximate pointwise Wald confidence intervals are an estimate plus or minus $1.96$ times its standard error:Both intervals rely on the large-sample normal approximation and correct model assumptions. The negative slope indicates decreasing purchase probability as price rises. A one-unit price increase multiplies the odds by $e^{-0.08207}\simeq 0.9212$; it does not subtract a fixed amount from the probability. There is no significant additional layout effect in the previous likelihood-ratio test, rather than a demonstration of no effect.

In the selected model layout B has the same prediction rule as the other layouts. At price 100 the linear predictor is $5.16258-0.08207(100)=-3.04442$. Applying the inverse logit link givesThe intercept refers to price zero; if that lies outside the observed price range, its direct substantive interpretation would require extrapolation.

The quasibinomial regression retains the logit link and mean function $p_i=(1+e^{-x_i^T\beta }{)}^{-1}$ but uses the working variance function $\mathrm{Var}(Y_i)=\varphi p_i(1-p_i)$. Its quasi-score equation isso multiplying by the common dispersion parameter does not change the coefficient estimates. Iteratively reweighted least squares therefore gives the same fitted means and coefficients as the binomial model. The usual Pearson dispersion estimator isThe estimated coefficient covariance matrix is $\hat{\varphi }(X^{T}WX{)}^{-1}$, with $W_{ii}={\hat{p}}_i(1-{\hat{p}}_i)$. From the intercept standard errors, $(1.08921/1.13031{)}^2\simeq 0.92860$; the rounded slope errors give approximately the same factor. The display uses approximate Student's t-tests with 118 residual degrees of freedom instead of fixed-dispersion normal tests.

For a genuinely individual binary response, the Bernoulli distribution identity $Y_i^2=Y_i$ forces $\mathrm{Var}(Y_i)=p_i(1-p_i)$. Thus $\varphi \ne 1$ is a working quasi-likelihood specification, not a different independent binary distribution with that mean. The mild estimated underdispersion does not establish an improved model, and the mean predictions are unchanged. In particular, a scalar rescaling does not model correlation among users, and usual likelihood-based Akaike information criterion comparisons are unavailable for a family without a specified probability likelihood. The output gives no convincing reason to prefer model3 to model2.