Solution. Every quantile is multiplied by \(1.5\): the median is \(15\) months and the upper quartile \(27\) months.

Solution. With \(h_0(t) = \lambda \) constant, \(h(t \mid x) = e^{-\beta ^{\mathsf T}x}h_0(te^{-\beta ^{\mathsf T}x}) = \lambda e^{-\beta ^{\mathsf T}x}\), so a unit increase in \(x_j\) multiplies the hazard by \(e^{-\beta _j} = 1/e^{\beta _j}\). This is Theorem 10.2.1 with \(\sigma = 1\).

Solution. Time ratio \(e^{-0.4} = 0.67\): smokers’ survival times are two-thirds of non-smokers’. Hazard ratio \(e^{0.4/0.8} = e^{0.5} = 1.65\): at any time, smokers’ hazard is \(65\%\) higher. The shape is \(k = 1/0.8 = 1.25\).

Solution. By \(e^{-\alpha \beta \times 10} = e^{-1.72 \times 0.360} = e^{-0.619} = 0.54\): ten points almost halve the odds of having died by any time, the proportional odds reading of the same model.

Solution. Medians scale with the time ratios. The adenocarcinoma patient: \(107.2 \times 0.4605 = 49\) days. The squamous patient with \(10\) more points: \(107.2 \times 1.4325 = 154\) days.

Solution. \(S(T \mid x)\) is uniform on \((0, 1)\), so \(H(T \mid x) = -\log S(T \mid x)\) is unit exponential; with estimated parameters this holds approximately, and censored \(t_i\) give censored residuals. Few residuals are large, so the estimated cumulative hazard at large \(r\) rests on a handful of subjects and wanders from the line even for a correct model.

Solution. The Weibull hazard ratio is \(e^{-\log (1.32)/0.78} = e^{-0.356} = 0.70\), the Cox value, so they agree. For a clinician: patients on the treatment have, at any moment, \(30\%\) lower risk of dying; equivalently, their survival times are about a third longer, so a median of \(12\) months becomes about \(16\).