= Solution
For <independent> <binomial distributions>, the <log-likelihood>, up to known binomial coefficients, is
$$
\ell(p)=\sum_{i=1}^m\{y_i\log p_i+(n_i-y_i)\log(1-p_i)\}.
$$
The <saturated statistical model> maximizes each term at $\widetilde p_i=y_i/n_i$. Under the <logit link>, the fitted probabilities are $\widehat p_i=(1+e^{-x_i^T\widehat\beta})^{-1}$. Subtracting the fitted <log-likelihood> from the saturated <log-likelihood> gives the <binomial deviance>
$$
\boxed{D_1=2\sum_{i=1}^m\left[y_i\log\frac{y_i}{n_i\widehat p_i}
+(n_i-y_i)\log\frac{n_i-y_i}{n_i(1-\widehat p_i)}\right].}
$$
A zero-count summand uses the continuous convention $0\log0=0$. For the restricted model, replace $\widehat p_i$ by $\widehat p_i^{(0)}=(1+e^{-\widetilde x_i^T\widehat{\widetilde\beta}})^{-1}$ to obtain $D_0$. Since the saturated likelihood cancels, $D_0-D_1=2(\widehat\ell_1-\widehat\ell_0)$ is the <likelihood-ratio test statistic>. For the nested restriction setting $p-k$ coefficients to zero, <Wilks theorem> gives the asymptotic <chi-squared distribution> $\chi^2_{p-k}$ under regularity and an interior true parameter; reject for a large difference.
For the six anther cells, $\omega_0$ gives a common success probability, $\omega_1$ gives a separate probability for each storage condition, $\omega_2$ gives a common intercept and a log-force slope, and $\omega_3$ gives storage-specific intercepts with a common log-force slope. Their parameter counts are $1,2,2,3$, so their residual <statistical degrees of freedom> are $5,4,4,3$. In $\omega_3$, storage has a constant effect on <log odds>; multiplying force by $r$ multiplies the <odds> by $r^\beta$ in either storage group.
The relevant nested <likelihood-ratio tests> at the $5\%$ level are:
* Adding storage to $\omega_0$ gives $D_0-D_1=5.279>3.84$ on one <statistical degree of freedom>, with $p\simeq0.0216$.
* Adding log-force to $\omega_0$ gives $D_0-D_2=2.360<3.84$ on one <statistical degree of freedom>, with $p\simeq0.1245$.
* Adding log-force to the storage model gives $D_1-D_3=2.554<3.84$ on one <statistical degree of freedom>, with $p\simeq0.1100$.
* Adding storage to the log-force model gives $D_2-D_3=5.473>3.84$ on one <statistical degree of freedom>, with $p\simeq0.0193$.
* Adding both terms to $\omega_0$ gives $D_0-D_3=7.833>5.99$ on two <statistical degrees of freedom>, with $p\simeq0.0199$.
The two two-parameter models $\omega_1$ and $\omega_2$ are not nested, so their <deviance> difference has no ordinary nested <chi-squared distribution> calibration. \b[The preferred parsimonious model is $\omega_1$, with a storage effect and no log-force effect.] Its <deviance goodness-of-fit test> is acceptable: $D_1=5.173<9.49$ on four <statistical degrees of freedom>. The common-probability model also passes its separate goodness-of-fit comparison, $10.452<11.07$, but this does not contradict the relative evidence for adding storage: the two tests assess different null hypotheses. These are large-sample approximations; sufficiently large expected successes and failures, not merely the number of anthers, underpin their calibration.
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