= Solution
The <Quasi-Poisson regression> retains the same conditional mean but permits
$$
E(Y_i\mid z_i)=\mu_i,\qquad \log\mu_i=\beta_0+\beta_1z_i,
\qquad \operatorname{Var}(Y_i\mid z_i)=\phi\mu_i.
$$
This is a mean–<variance function> specification through <quasi-likelihood>; it does not assign a full <probability distribution> to each count. For independent observations, the <quasi-score equation> is proportional to $\sum_i x_i(Y_i-\mu_i)=0$, so the mean <statistical parameter> estimates equal those from <Poisson regression>. The output estimates $\widehat\phi=3.515351$ through the <Pearson dispersion estimator>, and inflates the <standard errors> by approximately $\sqrt{\widehat\phi}=1.875$.
The large <residual deviance> relative to 728 residual <statistical degrees of freedom> also signals substantial <overdispersion>. Daily weather, traffic and other omitted conditions may produce greater count variation than a homogeneous <Poisson distribution> allows. The <Quasi-Poisson regression> accounts for that extra marginal <variance>. It still requires a correct conditional mean and an appropriate independence assumption; a common <dispersion parameter> alone does not repair <serial correlation>.
Back to article page
