Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 5J / Solution

These are nested Poisson regression models with independent counts and the default log link function. If $x_i$ is biomass and $g_i$ is the three-level pH group, model 1 specifies $\log {\mu }_i=\alpha +\beta x_i$: one common intercept and slope. Model 2 specifies $\log {\mu }_i={\alpha }_{g_i}+\beta x_i$: three group-specific intercepts and a common slope. Model 3 specifies $\log {\mu }_i={\alpha }_{g_i}+{\beta }_{g_i}{x}_i$: three intercepts and three slopes, because the multiplication syntax includes both main effects and their interaction. The parameter counts are therefore $2,4,6$.

The analysis of deviance compares model 1 against model 2 by testing equality of the three intercepts, conditional on a common biomass slope. The likelihood-ratio test statistic is $407.67-99.24=308.43$ with two additional parameters, giving the reported tiny p-value. The next test compares model 2 against model 3 by testing equality of their three slopes. Its statistic is $99.24-83.20=16.04$, again with two degrees of freedom and reported $p=3.288\times 10^{-4}$. Both restrictions are rejected. Thus model 3 is the supported choice: both baseline richness and its biomass dependence vary with pH. Under the Poisson assumptions, residual deviance $83.20$ on $84$ residual degrees of freedom is also consistent with a satisfactory fit, although these nested tests themselves establish improvement rather than every possible aspect of model adequacy.

The prediction from model 2 is on its linear-predictor scale. Exponentiating gives its fitted means. The final expression is twice the saturated log likelihood minus the model-2 log likelihood, i.e. the Poisson deviance of model 2. ConsequentlyIn the saturated model each mean equals the corresponding observed count, with the usual continuous convention for a zero count.