= Solution
The score equation $X^T(\mu-Y)=0$ is independent of $\gamma$, because $\gamma>0$ is only a common factor. Fixing $\phi=1$ or estimating it therefore gives the same <maximum-likelihood estimator> $\widehat\beta$.
Under the usual full-rank and regularity conditions,
$$
\widehat\beta
\mathrel{\dot\sim}
N_4\!\left(
\beta,\,
\phi(X^TWX)^{-1}
\right),
$$
with $W$ evaluated consistently at the fitted means. Consequently every standard error from "mod2" is $\sqrt{\widehat\phi}$ times the corresponding standard error computed with dispersion one in "mod1". In particular,
$$
\operatorname{SE}_{\rm mod2}(\widehat\beta_{\rm posout})
=\sqrt{0.3103711}\,(0.04556).
$$
Back to article page
