6.5 Practice Problems

Problem 6.5.1. Two groups have the following data (months): group A: \(4, 7^+, 9\); group B: \(2, 5, 8^+\). Compute the log-rank statistic.

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Solution. Death times \(2\), \(4\), \(5\), \(9\). At \(2\): \(n = 6\), \(n_A = 3\), \(d = 1\) (B): \(E_A = 0.5\), \(V = 0.25\). At \(4\): \(n = 5\), \(n_A = 3\), \(d = 1\) (A): \(E_A = 0.6\), \(V = 0.24\). At \(5\): \(n = 4\), \(n_A = 2\), \(d = 1\) (B): \(E_A = 0.5\), \(V = 0.25\). At \(9\): \(n = 1\) (only A’s \(9\) remains), \(E_A = 1\), \(V = 0\). \(O_A = 2\), \(E_A = 2.6\), \(V = 0.74\), \(\chi ^2 = 0.36/0.74 = 0.49\), \(p = 0.48\).

Problem 6.5.2. In a trial, the treated group had \(30\) deaths against \(40.2\) expected, and the control group \(50\) deaths against \(39.8\) expected. Estimate the hazard ratio of treated to control and interpret it.

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Solution. \(\widehat {\text {HR}} = (30/40.2)/(50/39.8) = 0.746/1.256 = 0.59\). The treated patients’ hazard of death was about \(59\%\) of the controls’, a \(41\%\) reduction at any time, if the hazards are proportional.

Problem 6.5.3. Why is the log-rank test called a rank test, and what does it ignore?

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Solution. It depends on the data only through the order of the event and censoring times and the group labels: which subjects are at risk at each event time and which group each event belongs to. Changing the times while keeping their order (for example by measuring in weeks instead of days, or by any increasing transformation) does not change the statistic. It ignores how far apart the times are.

Problem 6.5.4. When should a stratified log-rank test be used instead of the ordinary one? Give an example.

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Solution. When a factor strongly affects survival and is unevenly distributed between the groups, or when the trial was stratified by it, as by centre or disease stage. In the veteran trial, cell type has a large effect on survival; stratifying by it compares the regimens within cell types and prevents an imbalance of cell types between arms from creating or hiding a difference.

Problem 6.5.5. Explain why the log-rank test can miss a real difference when survival curves cross, and what to report instead.

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Solution. The statistic sums \(d_{1j} - E_{1j}\) over all times. If group \(1\) does worse early and better late, the early terms are positive and the late ones negative, and they can cancel. Report the curves themselves, survival at clinically relevant times with confidence intervals, a restricted mean survival time, or a pre-specified weighted test if the timing of the effect was predicted.

Problem 6.5.6. Show that at an event time with \(d_j = 1\) the variance \(V_j\) equals \(p_j(1 - p_j)\) with \(p_j = n_{1j}/n_j\), and interpret.

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Solution. With \(d_j = 1\), \(\frac {n_j - d_j}{n_j - 1} = 1\) and \(V_j = p_j(1 - p_j)\), the variance of a Bernoulli variable: with one death, the question is only whether it falls in group \(1\), which under \(H_0\) happens with probability \(p_j\), the group’s share of those at risk.

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