Solution. \(\widehat {\text {HR}} = e^{-0.35} = 0.705\), interval \(\exp (-0.35 \pm 0.235) = (0.557, 0.891)\). Treated patients had about \(30\%\) lower hazard of the event at any time than controls; the interval excludes \(1\), so the effect is significant at the \(5\%\) level (\(z = -2.92\), \(p = 0.004\)).

Solution. \(e^{20 \times 0.025} = e^{0.5} = 1.65\): the \(70\)-year-old’s hazard is \(65\%\) higher.

Solution. \(S_{\text {treated}} = S_{\text {control}}^{\text {HR}} = 0.70^{0.6} = e^{0.6\log 0.70} = e^{-0.214} = 0.807\). One-year mortality falls from \(30\%\) to \(19.3\%\), a relative reduction of \(36\%\), not the \(40\%\) that the hazard ratio suggests: hazard ratios and risk ratios differ unless risks are small.

Solution. At \(6\) the risk set is the subjects with times \(\geq 6\): times \(6, 8, 9, 11, 14\) with \(x = 1, 0, 1, 0, 0\), and the subject who dies has \(x = 1\): factor \(e^\beta /(2e^\beta + 3)\). At \(8\) the risk set is \(8, 9, 11, 14\) with \(x = 0, 1, 0, 0\), and the subject who dies has \(x = 0\): factor \(1/(e^\beta + 3)\).

Solution. Each factor is the conditional probability that the subject who failed was the one who did, given that one failure occurred at that time among those at risk; the common factor \(h_0(t)\) cancels between numerator and denominator. It ignores the information in the exact times of the events (only their order and the risk sets matter), which under the model carries little information about \(\beta \) when \(h_0\) is unknown.

Solution. The likelihood ratio statistic is \(2(-737.2 + 739.9) = 5.4\) on \(1\) degree of freedom, \(p = 0.020\): the covariate improves the model significantly at the \(5\%\) level.