Solution. \(d = 5\) events and \(T_{\mathrm {tot}} = 123\) months, so \(\hat \lambda = 5/123 = 0.0407\) per month, s.e. \(0.0407/\sqrt 5 = 0.0182\). The median is \(\log 2/\hat \lambda = 17.1\) months. On the log scale, \(\log \hat \lambda \pm 1.96/\sqrt 5\) gives \(\lambda \in (0.0169, 0.0977)\), and since the median is \(\log 2/\lambda \) the interval for the median is \((7.1, 41.0)\) months: very wide, because only five deaths were observed.

Solution. For a censored subject we know only that \(T_i > t_i\), an event of probability \(S(t_i)\). Writing the joint probability of the observed data as \(\prod f^{\delta }S^{1 - \delta }\) times a factor for the censoring times assumes that censoring is independent of \(T_i\) (non-informative) and that its distribution does not involve \(\theta \), so that this factor can be dropped.

Solution. With \(k\) fixed, \(\ell = \sum \delta _i(\log k + (k - 1)\log t_i) - dk\log \lambda - \lambda ^{-k}\sum t_i^k\). Then \(\partial \ell /\partial \lambda = -dk/\lambda + k\lambda ^{-k - 1}\sum t_i^k = 0\) gives \(\lambda ^k = \sum t_i^k/d\). For \(k = 1\) this is the exponential result \(T_{\mathrm {tot}}/d\).

Solution. With the same number of parameters, AIC prefers the larger log-likelihood: the Weibull, by \(2 \times 7.0 = 14\) in AIC, a large difference. A likelihood ratio test is not appropriate, because neither model is a special case of the other; AIC, or a graphical comparison, is used instead.

Solution. \(k = 1/1.25 = 0.8\) and \(\lambda = e^{5.2} = 181.3\). Since \(k < 1\) the hazard decreases with time. The median is \(\lambda (\log 2)^{1/k} = 181.3 \times 0.693^{1.25} = 181.3 \times 0.632 = 114.6\).

Solution. The interval is \(\hat \lambda e^{\pm 1.96/\sqrt d}\), so the ratio of the limits is \(e^{3.92/\sqrt d} \leq 1.5\), that is \(\sqrt d \geq 3.92/\log 1.5 = 9.67\), \(d \geq 94\) deaths, whatever the number of patients.