Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 12J (Statistical Modelling)

The Poisson deviance tests overall fit against the saturated Poisson regression, not whether the single slope is zero. Under fit.1, its large-count approximation is ${\chi }_{60-2}^2={\chi }_{58}^2$. The observed value is 261.72, far above the one-percent upper critical value (about 85.95); its upper-tail probability is about $1.14\times 10^{-27}$. Thus reject fit.1 at the one-percent level. The slope's reported $p$-value 0.421 only compares this inadequate log-linear trend with an intercept-only model; it is not a goodness-of-fit test.

The two design matrices span the same six-dimensional space of arbitrary log means for the six periods. Fit.2 uses the first period as a baseline, whereas fit.3 uses cumulative indicator steps. If $a_j$ is the log mean for period $j$, fit.3 has intercept $a_1$ and jump coefficients $a_{j+1}-a_j$. Thus both fits give identical fitted means and identical maximized likelihood functions. ThereforeDifferent coefficient tests arise because they test different contrasts, not because one full fit explains the observations better.

Use fit.3 if backward selection is intended to combine adjacent periods. Each cumulative indicator is a separate model term; deleting its coefficient enforces equality of the two adjacent period means separated by that boundary. This offers an interpretable reduced stepwise profile. Fit.2 treats the entire factor as one term, so ordinary term-wise backward selection retains all five contrasts or removes period altogether. Removing selected baseline contrasts by hand would instead privilege period one, rather than grouping neighbouring periods. If the scientific objective were simply whether any period effect exists, retaining or deleting the whole factor in fit.2 would be the appropriate different procedure.

Under fit.2, observations in each period are independent Poisson random variables with a common mean, so each sample variance should fluctuate around its sample mean. The plot instead has variances roughly 130–300 for means roughly 80–120, well above the line variance equals mean. This is overdispersion.

Day-to-day traffic variation can produce a Poisson mixture: a random daily intensity increases the marginal variance and can correlate observations from the same day. A useful replacement is a negative binomial regression, with variance $\mu +{\mu }^2/\kappa$, or a Poisson generalized linear mixed model with a day random intercept. A day-factor effect can also be included if inference is restricted to the observed days. A Quasi-Poisson regression estimates a dispersion multiplier for uncertainty, but does not itself explain the shared-day dependence and has no ordinary likelihood-based AIC. The means can retain the period-factor or selected jump structure.