= Solution
The null model uses one parameter, the day model uses nine, and there are $27$ grouped observations. The four missing entries, in their table order, are therefore
$$
\boxed{26,\qquad8,\qquad495.6308-32.7945=462.8363,\qquad18.}
$$
The test of no day effect is $H_0:\alpha_2=\cdots=\alpha_9=0$. The nominal binomial <likelihood-ratio test> statistic is $\Delta D=462.8363$, compared asymptotically with $\chi_8^2$. Its nominal $p$-value is about $6.6\times10^{-95}$, so the common-probability model is decisively inadequate relative to the day model. The fitted probabilities in (i) show substantial differences between days, with days 1 and 7 much higher than days 3 and 8.
However, relative improvement does not establish absolute fit. The day model's <deviance goodness-of-fit test> compares $D=32.7945$ with $\chi_{18}^2$, giving $p\simeq0.0177$. The counts are large enough for this approximation to be reasonable. Thus at $5\%$ the day-specific binomial model still shows lack of fit. Among the two printed models it is preferred, but it should not be regarded as a fully satisfactory final model.
The residual scale estimate $D/18\simeq1.822$ suggests <overdispersion>. Directly computing the Pearson statistic from the day means gives
$$
X^2=\sum_{i,j}\frac{(Y_{ij}-400\widehat p_i)^2}{400\widehat p_i(1-\widehat p_i)}\simeq32.70675,\qquad X^2/18\simeq1.81704.
$$
This corroborates extra dish variation rather than the nominal binomial <variance>. Dish heterogeneity or dependence among cells in a dish can produce such variation. Residual checks and a <quasibinomial regression> or an explicit model for dish heterogeneity are sensible refinements. Allowing dispersion of this moderate magnitude would enlarge <standard errors> but would not plausibly remove the very large day effect. \b[Retain day effects, while accounting for the remaining extra-binomial variation.]
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