= Solution
The <Pearson dispersion estimator> is $\widehat\phi=3.099079$, well above the Poisson value one. A <Quasi-Poisson regression> retains the logarithmic mean model but uses
$$
\mathbb E Y_i=\mu_i,\qquad\operatorname{Var}(Y_i)=\phi\mu_i,\qquad
\log\mu_i=\beta_0+\beta_aa_i.
$$
This is a <quasi-likelihood> specification; it does not by itself identify a count probability distribution. For a common $\phi$, the <quasi-score equation> is the Poisson score divided by $\phi$, so its roots and fitted means are unchanged. Thus
$$
\boxed{\widehat\mu(a)=e^{3.24926+0.01898a},\qquad
\widehat{\operatorname{Var}}(Y\mid a)=3.099079\,\widehat\mu(a).}
$$
Coefficient <standard errors> are multiplied by $\sqrt{\widehat\phi}$. The altitude <standard error> becomes $0.00447\sqrt{3.099079}\approx0.0078691$, giving the requested approximate normal <confidence interval>
$$
\boxed{0.01898\pm1.96(0.0078691)=(0.003557,\,0.034403).}
$$
It excludes zero, so \b[retain altitude on this approximate 5% test]. Its test statistic is about $2.412$ after allowing for <overdispersion>, rather than the Poisson value $4.245$. Rescaling uncertainty cannot repair a systematically misspecified mean; the diagnostics in part (e) motivate reconsidering that mean as well.
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