Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 218 / 1

The first fit is a Poisson regression with independent responsesThe fitted intercept gives $e^{3.8879}$ expected visitors when spending on both services is zero. Holding advert2 fixed, increasing advert1 by one unit, namely one hundred pounds, multiplies the expected visitor count by $e^{0.1923}\simeq 1.212$.

The second fit is a Quasi-Poisson regression. It retains the logarithmic link function and mean modelbut assumes only $\mathrm{Var}(Y_i)=\varphi {\mu }_i$ for an unknown dispersion parameter $\varphi$, rather than a complete Poisson distribution.

The quasi-Poisson standard errors are the Poisson standard errors multiplied by $\sqrt{\hat{\varphi }}$, and the Pearson estimator divides the Pearson chi-squared statistic by the residual degrees of freedom $40-3=37$. Therefore the requested sum of squared Pearson residuals isup to the rounding in the printed standard errors.

The third fit is a negative binomial regression. The Poisson model is obtained at the boundary where the negative-binomial overdispersion tends to zero, so the likelihood-ratio test has the asymptotic null distribution $\frac{1}{2}{\chi }_0^2+\frac{1}{2}{\chi }_1^2$. Since model3 has one additional parameter,The supplied output gives $\mathbb{P}({\chi }_1^2\ge D)=0.04608555$, and hence the boundary-corrected p-value is

The stated mean and variance identify ${\lambda }_i$ as a gamma distribution with shape ${\nu }_i$ and scale ${\alpha }_i$. A Poisson-gamma mixture is negative binomial, withWhen ${\nu }_i=\nu$, this is exactly the constant-shape variance model fitted by the negative binomial regression, so the most appropriate printed p-value is $0.0798$ from model3.

When ${\alpha }_i=\alpha$, the law of total variance givesThis is the quasi-Poisson mean-variance relation with constant dispersion, so the most appropriate printed p-value is $0.0982$ from model2.

Among all depth-one regression tree splits, the best split separates the second observation from the first and third by cutting advert1 between $0.89$ and $1.13$. The test point with advert1 equal to zero reaches the leaf containing responses $60$ and $53$, so the output is their arithmetic mean,This decision stump makes a piecewise-constant prediction far outside the observed predictor range and cannot extrapolate the spending trend towards the origin. Its shallow structure and leaf averaging keep its variance of an estimator modest, while that extrapolation failure can produce substantial bias of an estimator.