Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 218 / 4

Let $T_i=1$ denote Progabide and $T_i=0$ placebo; write $a_i$ for age and $b_i$ for baseline count. The fitted Poisson regression specifies independentIts log-likelihood isThe coefficient table in the original PDF, whose TeX transcription is truncated, gives the maximum-likelihood estimatesFor the treatment interaction contrast in a Poisson regression, the baseline slope of the log mean is $0.0186824$ on placebo and $0.0195506$ on Progabide. At fixed age and baseline $b$, the fitted drug-to-placebo mean ratio is $\exp (-0.3592068+0.0008682b)$. Increasing baseline by one multiplies that ratio by $e^{0.0008682}\approx 1.0008686$. Thus the interaction term describes how the treatment ratio changes with baseline, rather than a uniform additive change in seizure counts. The displayed call makes clear that the source's summary(model1) line is an object-name typo for the epilepsy fit.

The null model has only an intercept:For $\bar{Y}>0$, maximizing its log-likelihood gives $\hat{\mu }=\bar{Y}$ and ${\hat{\beta }}_0=\log \bar{Y}$. The saturated Poisson regression gives each observation its own fitted mean ${\hat{\mu }}_i=Y_i$, allowing zero as a limiting fitted mean. By the definition of Poisson deviance,Since ${\sum }_i(Y_i-\bar{Y})=0$,Use the continuous convention $0\log (0/\bar{Y})=0$. If all counts are zero, both the saturated and null likelihood maxima are attained as limits at zero mean and the deviance is zero.

Under a correctly specified Poisson regression with adequate expected counts, the residual Poisson deviance is approximately chi-squared distribution with $n-p=107-5=102$ residual degrees of freedom. The observed $D=319.86$ givesThus the ordinary Poisson model gives a very poor fit under this goodness-of-fit approximation. This is consistent with overdispersion, although a wrong mean model or dependence can also cause lack of fit; quasi-Poisson variance inflation alone does not establish that the mean model is adequate.

For Quasi-Poisson regression, $\mathrm{Var}(Y_i)=\varphi {\mu }_i$. The Pearson dispersion estimator isIt uses squared Pearson residuals, not the residual deviance divided by 102. The fitted coefficients are unchanged, and their estimated covariance matrix is multiplied by $\hat{\varphi }$. With $\hat{\varphi }=3.24$, standard errors increase by $1.8$, so the main treatment coefficient hasThe main Progabide coefficient is not significant at 5% after this dispersion adjustment. R uses a Student's t-distribution when dispersion is estimated, as described in its GLM summary documentation; the large-sample normal approximation also gives the same decision.

Because an interaction term is present, this tests the treatment contrast at baseline zero. An overall absence of treatment influence requires the joint null ${\beta }_T={\beta }_{TB}=0$. The printed marginal standard errors do not give their covariance, so they are insufficient to calculate that joint Wald test, or the treatment contrast test at an arbitrary nonzero baseline.