Solution. \(2(-612.4 + 618.9) = 13.0\) on \(3\) degrees of freedom; the \(5\%\) critical value is \(7.81\) and \(p = 0.0046\). The factor is needed, even if some of the individual indicators are not significant.

Solution. \(3.17 \times 0.969^{-30} = 3.17 \times e^{30 \times 0.0315} = 3.17 \times 2.57 = 8.2\), using \(\log 0.969 = -0.0315\). The combination of tumour type and poorer performance gives about eight times the hazard.

Solution. The order affects the risk sets: whichever tied subject is placed first remains in the risk set for the others. Different arbitrary orders give different partial likelihoods. The Breslow method avoids choosing by keeping all tied subjects in every denominator; Efron’s and the exact method average over the possible orders.

Solution. At \(10\) to \(20\) events per coefficient, \(3\) to \(6\) coefficients. More risks over-fitting: coefficients that are too large, unstable selection, and optimistic predictions.

Solution. Adjusting includes the covariate in \(e^{\beta ^{\mathsf T}x}\), assuming that its effect multiplies the hazard by a constant factor and estimating that factor. Stratifying gives each level its own baseline hazard, making no assumption about the shape of its effect over time, but estimating no hazard ratio for it. Prefer stratification when the covariate violates proportional hazards or is a nuisance factor with many levels, such as hospital.

Solution. \(\hat S_2 = \hat S_1^{\exp (1.1 - 0.2)} = 0.6^{e^{0.9}} = 0.6^{2.46} = 0.285\). The second patient, with the higher linear predictor, has the higher hazard at all times and the lower survival, and a concordant model predicts that this patient dies first.