Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 1 / 13I / Solution

The first command fits independent Poisson random variables with means ${\mu }_j$ and a log link. For concentration $x$ and strain indicator $s$, its fitted linear predictor isThe interaction allows different concentration slopes in the two strains. At $x=0$ the second strain's mean is multiplied by $e^{0.33667}$; its slope is $-1.47253-0.12534$. These are multiplicative changes in counts, not changes in probabilities.

The maximum-likelihood estimates use the Poisson regression likelihood. With design matrix $X$, estimated information $X^T\mathrm{diag}({\hat{\mu }}_j)X$ gives the inverse covariance matrix; diagonal square roots give the standard errors. The printed Wald tests use approximately standard-normal coefficient-to-error ratios. Fuel and strain effects are strong, whereas the interaction has $p=0.182$, suggesting a common slope.

$H_2$ removes just the interaction; $H_3$ also removes strain. Their parameter counts are four, three and two. Under the usual large-sample likelihood-ratio test, $D_2-D_1=1.78089$ is compared with ${\chi }_1^2$ and is below $3.841459$, so the interaction is not required at 5%. By contrast, $D_3-D_2=32.61857$ is far above that threshold, so the strain main effect should remain. Comparing $H_3$ directly with $H_1$ uses $D_3-D_1$ and two degrees of freedom.

For an approximate absolute goodness-of-fit test, compare $D_1=84.59557$ with ${\chi }_{70-4}^2$; its upper-tail probability is about $0.061$. For $H_2$, $D_2=86.37646$ on $67$ degrees of freedom has $p\approx 0.056$. The additive model $H_2$ is the preferred parsimonious fit at 5%, subject to the independent Poisson distribution and large-sample assumptions. The printed nine rows are only a subset; the supplied fitted output refers to all 70 observations.