 Solution

For independent Poisson distributions, ignoring the factorial term only when taking likelihood differences,The saturated likelihood maximizes each mean separately at ${\tilde{\mu }}_i=y_i$, including its limiting value at zero. Therefore the Poisson deviance iswhere $0\log (0/{\hat{\mu }}_i)$ is defined to be zero and ${\hat{\mu }}_i=e^{3.24926+0.01898a_i}$. The expression is twice a saturated-model likelihood ratio difference, not twice a likelihood ratio itself.

With sufficiently large fitted counts and regularity, $D$ is approximately ${\chi }_{30-2}^2$ under a correctly specified Poisson regression. Here $D=83.351$ on 28 residual degrees of freedom, with upper-tail probability $2.0935\times 10^{-7}$. The Poisson model fits poorly on this diagnostic. Its deviance is nearly three times its degrees of freedom, consistent with overdispersion, although a wrong mean function, dependence, or influential islands could also cause the discrepancy. The smaller Akaike information criterion than its competitors does not make this an adequate absolute fit.