Solution. \(\pi _1 = 0.10\), \(\pi _0 = 0.05\), \(\bar \pi = 0.075\), \(z_\beta = 1.282\). Then \begin {gather*} \sigma _0 = \sqrt {2 \times 0.075 \times 0.925} = 0.3725, \\ \sigma _1 = \sqrt {0.10 \times 0.90 + 0.05 \times 0.95} = 0.3708, \\ n = \frac {(1.960 \times 0.3725 + 1.282 \times 0.3708)^2}{0.05^2} = 581.3 \end {gather*} so \(582\) in each group after rounding up, \(1164\) in all.

Solution. \(\delta = 0.05\), \(\sigma _0 = \sqrt {2 \times 0.125 \times 0.875} = 0.468\), \(\sigma _1 = \sqrt {0.1275 + 0.09} = 0.466\). The power is \(\Phi \{(0.05\sqrt {500} - 1.96 \times 0.468)/0.466\} = \Phi (0.431) = 0.67\). The study has only a two-in-three chance of detecting a risk ratio of \(1.5\).

Solution. With \(q_1 = 0.333\) and \(q_0 = 0.2\), the power formula of Theorem 6.4.1 with \(n = 100\) gives \(0.57\) for \(k = 1\), \(0.76\) for \(k = 3\) and \(0.84\) for \(k = 10\). Tripling the controls raises the power substantially; going from \(3\) to \(10\) adds much less, as Proposition 6.2.1 predicts. Even with unlimited controls the power would stay below \(0.9\), because the \(100\) cases fix the first term of the variance.

Solution. Exposed cases: \(n_{11} + n_{10} = 18 + 30 = 48\); exposed controls: \(n_{11} + n_{01} = 18 + 12 = 30\). The crude odds ratio is \((48 \times 70)/(52 \times 30) = 3360/1560 = 2.15\). It is closer to \(1\) than the matched estimate. Pairs are alike in the matching variable, which here is associated with exposure, and ignoring the matching treats the controls as if they were a random sample of the base when they have been drawn towards the cases’ exposure distribution.

Solution. \(\pi = 1.5/2.5 = 0.6\), so \(D = (1.960 + 1.282)^2/\{0.24\,(\log 1.5)^2\} = 10.51/(0.24 \times 0.1644) = 266\) cases. Each pair of exposed and unexposed person-years yields \(0.004 + 0.006 = 0.010\) cases, so about \(26{,}600\) person-years are needed in each group. A modest effect on a rare outcome needs a large cohort.

Solution. With \(q_1 = q_0 = q\) the variance is \((1 + 1/k)/\{m_1 q(1 - q)\}\), and with \(k = 1\) and \(m_1'\) cases it is \(2/\{m_1' q(1 - q)\}\). These are equal when \(m_1' = 2m_1/(1 + 1/k) = 2m_1 k/(k + 1)\). With four controls per case, \(100\) cases are worth \(160\) cases with one control each: extra controls substitute for cases, but at most \(100\) cases with unlimited controls are worth \(200\) matched one-to-one.