 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 thereforeUse the identity in part (b) to replace $y_i$ by ${\hat{p}}_i$ in its first term:Thus the deviance is twice the sum of the fitted binary information entropies. A deviance goodness-of-fit test against ${\chi }_{n-p}^2$ 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 $\mathrm{logit}{p}_i={\beta }_0$, $\mathrm{logit}{p}_i={\beta }_0+{\beta }_1{\log }_2{T}_i$, and $\mathrm{logit}{p}_i={\beta }_0+{\beta }_1{\log }_2{T}_i+{\beta }_2{S}_i$, with independent Bernoulli random variables conditional on the covariates. Testing the additional severity effect givesUnder $H_0:{\beta }_2=0$, its reference distribution is asymptotically ${\chi }_1^2$. The $p$-value is approximately $1.21\times 10^{-21}$, giving overwhelming evidence that severity improves the fitted model. Holding severity fixed, doubling $T$ increases ${\log }_{2}T$ by one and multiplies the odds byThis is an odds ratio, not a multiplication of the event probability itself.