= Solution
In the species-only model, the <log odds> for adult species A is the intercept, so \b[$\widehat\theta_A=0.91191$ with <standard error> 0.07824]. This also follows directly by pooling its vials: there are 570 other-species eggs among 799 eggs eaten, giving $\widehat p_A=570/799$ and
$$
\widehat\theta_A=\log\frac{570}{229}\simeq0.91191.
$$
For a regular <maximum-likelihood estimator>, <asymptotic normality of a maximum likelihood estimator> gives $\widehat\theta_A\overset{\rm approx}{\sim}N(\theta_A,I_A^{-1})$, with information $I_A=N_Ap_A(1-p_A)$. Substitution of $\widehat p_A$ gives $\operatorname{se}(\widehat\theta_A)=\sqrt{1/570+1/229}\simeq0.07824$. An approximate 95% <Wald confidence interval> is
$$
\boxed{0.91191\pm1.96(0.07824)=(0.7586,1.0653).}
$$
This interval treats the selected species-only model as fixed; it does not account separately for uncertainty from choosing that model.
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