12.5 Practice Problems
Problem 12.5.1. How many events are needed to detect a hazard ratio of \(0.75\) with \(90\%\) power at the two-sided \(5\%\) level and equal allocation? How does the answer change with allocation \(2:1\) to the new treatment?
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Solution. \(z_{0.025} + z_{0.1} = 1.960 + 1.282 = 3.242\), \((\log 0.75)^2 = 0.0828\), so \(d = 10.51/(0.25 \times 0.0828) = 508\). With \(p = \frac 23\), \(p(1 - p) = \frac 29 = 0.222\) instead of \(0.25\), and \(d = 10.51/(0.222 \times 0.0828) = 572\), about \(12\%\) more.
Problem 12.5.2. A trial needs \(300\) events. In the planned population, \(40\%\) of patients will die during the study and \(10\%\) will be lost to follow-up. How many patients should be recruited?
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Solution. \(300/0.40 = 750\) patients would give the events if no-one were lost; allowing for \(10\%\) losses, \(750/0.9 = 834\). (This simple inflation is approximate; losses also remove patients who would have died.)
Problem 12.5.3. Using the derivation of Theorem 12.1.1, find the power of a trial with \(150\) events and equal allocation to detect a hazard ratio of \(0.7\) at the two-sided \(5\%\) level.
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Solution. \(\sqrt {dp(1 - p)}\,|\log \Delta | = \sqrt {150 \times 0.25} \times 0.357 = 6.12 \times 0.357 = 2.18\). Power \(\approx P(Z > 1.960 - 2.18) = P(Z > -0.22) = \Phi (0.22) = 0.59\).
Problem 12.5.4. Why does a survival trial’s sample size depend on the number of events rather than on the number of patients, and why are many trials stopped at a fixed number of events?
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Solution. The variance of the log-rank statistic, and of the estimated log hazard ratio, is about \(1/\{dp(1 - p)\}\): censored patients contribute to the risk sets but not to this information directly. A trial stopped at a fixed number of events has known power whatever the event rate turns out to be, while one stopped at a fixed date may see fewer events than planned if patients do better than expected.
Problem 12.5.5. In the Mayo model, patient A has bilirubin \(1\) mg/dl and patient B \(8\) mg/dl, with all else equal. What is the hazard ratio of B to A?
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Solution. \(8\) is three doublings of \(1\), so the ratio is \(1.84^3 = 6.2\). Equivalently \(8^{0.880} = e^{0.880\log 8} = 6.2\), where \(0.880 = \log 2.411\) is the coefficient of \(\log (\text {bili})\).
Problem 12.5.6. The PBC trial found no treatment effect, yet the data produced a widely used prognostic model. Explain why a negative trial can still be a good source for such a model, and what limits its use.
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Solution. Randomization does not matter for prognosis when the treatment has no effect, so the trial is a well-documented cohort with careful follow-up, standardized measurements at entry and few missing values. Its limits: the patients were those eligible and willing to enter a trial at one centre in 1974–1984, and later changes in diagnosis and treatment (earlier detection, ursodeoxycholic acid, transplantation) change the baseline risk, so the model should be validated in current patients before use.
Problem 12.5.7. For each of the following, name the method from Table 2 and say why: (a) the probability that a dialysis patient receives a transplant within \(3\) years, when many die first; (b) whether a new drug reduces the rate of hospital admissions, which recur; (c) the median time to relapse in a single group, with \(30\%\) censored.
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Solution. (a) The cumulative incidence of transplantation, with death as a competing risk, since one minus Kaplan–Meier would overstate it. (b) The Andersen–Gill model (or PWP) with robust standard errors, since admissions recur within patients and are correlated. (c) The Kaplan–Meier estimate and its median, with a confidence interval from Greenwood’s formula, since it uses the censored times correctly without assuming a distribution.
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