= Solution
For the Poisson <variance function> $V(\mu)=\mu$, the code computes the <Pearson chi-squared statistic>
$$
X_P^2=\sum_{i=1}^{18}\frac{(Y_i-\widehat\mu_i)^2}{\widehat\mu_i}
$$
and the <Pearson dispersion estimator>
$$
\widehat\phi=\frac{X_P^2}{18-2}=\frac{X_P^2}{16}.
$$
The first quantity measures <goodness of fit>; the second estimates the <dispersion parameter>, which equals one in a correctly specified <Poisson regression>. A standard rough calculation substitutes the residual deviance for the Pearson statistic and gives
$$
\widehat\phi\simeq\frac{75.806}{16}=4.74.
$$
If the reported upper-tail probability $4.908651\times10^{-11}$ is inverted numerically, the actual Pearson statistic used by the code is about $82.93$, giving $\widehat\phi\simeq5.18$. Either calculation reveals severe <overdispersion>.
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