Solution. At \(2\): \(\frac 17 = 0.143\); at \(4\): \(+\frac 26\), \(\hat H = 0.476\); at \(7\): \(+\frac 13\), \(\hat H = 0.810\); at \(11\): \(+1\), \(\hat H = 1.810\). Then \(\exp (-\hat H) = 0.867\), \(0.621\), \(0.445\), \(0.164\), against Kaplan–Meier \(0.857\), \(0.571\), \(0.381\), \(0\). The last difference is large because the final factor \(1 - \frac 11 = 0\) of the Kaplan–Meier product has no counterpart: \(e^{-1} = 0.37\).

Solution. A constant hazard, the exponential model with \(\lambda = 0.05\) per month. Mean \(1/\lambda = 20\) months, median \(\log 2/\lambda = 13.9\) months.

Solution. A censored subject was observed for part of the interval; if censoring times are spread evenly, the average is half the interval, so each counts as half a subject at risk. If censoring happens mostly at the start, the censored subjects were at risk for less than half the interval, \(n_i'\) is overstated, \(\hat q_i\) too small, and survival overestimated; then \(n_i - c_i\) would be more accurate.

Solution. Year 1: \(n = 100\), \(n' = 98\), \(q = 0.102\), \(p = 0.898\). Year 2: \(n = 86\), \(n' = 83\), \(q = 0.181\), \(p = 0.819\). Year 3: \(n = 65\), \(n' = 62.5\), \(q = 0.320\), \(p = 0.680\). \(\hat S(3) = 0.898 \times 0.819 \times 0.680 = 0.500\): half the machines last three years.

Solution. \(H_2(t) = \int _0^tch_1 = cH_1(t)\), so \(\log H_2 - \log H_1 = \log c\), a constant. Plotting \(\log \hat H\) (equivalently \(\log (-\log \hat S)\)) for both groups against \(\log t\) should therefore give parallel curves; curves that converge, diverge or cross suggest that the hazards are not proportional, which matters for the Cox model.

Solution. \(P(\text {survive to } 63) = 0.990 \times 0.989 \times 0.988 = 0.967\). The probability of dying aged \(62\) is that of surviving to \(62\) times \(q_{62}\): \(0.990 \times 0.989 \times 0.012 = 0.0117\).