= Solution
For separate binary observations, the <saturated model> fits each outcome exactly and has maximized log-likelihood zero. With $\hat p_i=p_i(\hat\beta)$, the <residual deviance> is therefore
$$
D=-2\ell(\hat\beta)=-2\sum_i\left[y_i\operatorname{logit}\hat p_i+\log(1-\hat p_i)\right].
$$
Use the identity in part (b) to replace $y_i$ by $\hat p_i$ in its first term:
$$
\boxed{D=-2\sum_i\left[\hat p_i\operatorname{logit}\hat p_i+\log(1-\hat p_i)\right]=-2\sum_i\left[\hat p_i\log\hat p_i+(1-\hat p_i)\log(1-\hat p_i)\right].}
$$
Thus the deviance is twice the sum of the fitted binary <information entropies>. A <deviance goodness-of-fit test> against $\chi^2_{n-p}$ is unreliable for individual ungrouped binary observations: there is only one trial per saturated-model parameter, so the usual large-cell-count approximation fails. Deviance differences between fixed-dimensional nested <logistic regression> models can still support a <likelihood-ratio test>, subject to its usual regularity assumptions.
The three tree models are respectively $\operatorname{logit}p_i=\beta_0$, $\operatorname{logit}p_i=\beta_0+\beta_1\log_2T_i$, and $\operatorname{logit}p_i=\beta_0+\beta_1\log_2T_i+\beta_2S_i$, with independent <Bernoulli random variables> conditional on the covariates. Testing the additional severity effect gives
$$
\boxed{2\{\ell(\widehat\beta_{\rm full})-\ell(\widehat\beta_{\rm reduced})\}=655.24-563.90=91.34.}
$$
Under $H_0:\beta_2=0$, its reference distribution is asymptotically $\chi^2_1$. The $p$-value is approximately $1.21\times10^{-21}$, giving overwhelming evidence that severity improves the fitted model. Holding severity fixed, doubling $T$ increases $\log_2T$ by one and multiplies the odds by
$$
\boxed{e^{2.2164}\simeq9.17.}
$$
This is an <odds ratio>, not a multiplication of the event probability itself.
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