= Solution
Let $\ell_{\rm sat}$ denote the maximized <log-likelihood> in the saturated model, with one fitted mean per observation. With known dispersion, define the <deviance> of a fitted model as $D=2(\ell_{\rm sat}-\ell_{\rm fitted})$; for an exponential-dispersion convention that scales deviance by the dispersion $\phi$, divide that scaled quantity by $\phi$ in the following test.
Let $D_0,D_1$ be the fitted deviances under the null and full models. Their difference is the <likelihood-ratio test statistic>
$$
\boxed{D_0-D_1=2(\ell(\widehat\beta)-\ell(\widehat\beta_0,0)).}
$$
Under the regular identifiable interior null hypothesis, <Wilks theorem> gives the limiting $\chi^2_{p-p_0}$ distribution. Reject for an observed difference above its chosen upper-tail critical value. This concerns the difference of fitted model deviances, not simply the full-model deviance compared with zero.
For a <Poisson regression>, the means are $\mu_i=e^{x_i^T\beta}$ and $\ell=\sum_i(y_i\log\mu_i-\mu_i-\log y_i!)$. Saturation sets $\mu_i=y_i$, taking the limiting value at $y_i=0$. Consequently the <Poisson deviance> is
$$
\boxed{D=2\sum_i\left[y_i\log\frac{y_i}{\widehat\mu_i}-(y_i-\widehat\mu_i)\right],}
$$
with $0\log0=0$. The intercept score implies $\sum_i(y_i-\widehat\mu_i)=0$ at a finite fit, although retaining that term makes the individual deviance contributions clear.
For $y_i=\widehat\mu_i+\delta_i$ with $|\delta_i|/\widehat\mu_i$ small, expand $(\widehat\mu_i+\delta_i)\log(1+\delta_i/\widehat\mu_i)-\delta_i$. The linear term cancels and the result is $\delta_i^2/(2\widehat\mu_i)+O(|\delta_i|^3/\widehat\mu_i^2)$. Therefore
$$
D\simeq\sum_i\frac{(y_i-\widehat\mu_i)^2}{\widehat\mu_i},
$$
the <Pearson chi-squared statistic>. The approximation requires adequately large fitted counts and small relative residuals; it is not an identity for sparse observations.
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