Solution. Men: \(n = 340\), \(R = 4000/340 = 11.76\), \(S = 3200/340 = 9.41\). Women: \(n = 240\), \(R = 6000/240 = 25.0\), \(S = 600/240 = 2.5\). So \(\widehat {\OR }_{\text {MH}} = 36.76/11.91 = 3.09\), with weights \(9.41/11.91 = 0.79\) on men’s \(1.25\) and \(0.21\) on women’s \(10.0\). The stratum odds ratios are strongly heterogeneous (\(Q = 20.8\)), and the average \(3.09\) describes neither sex. The Mantel–Haenszel estimator assumes a common odds ratio; when that fails, it estimates a weighted average whose weights depend on the design, not a population parameter of interest.

Solution. Men: \(n = 600\), \(a n_0/n = 20 \times 400/600 = 13.33\), \(b n_1/n = 10 \times 200/600 = 3.33\). Women: \(n = 700\), \(40 \times 300/700 = 17.14\), \(15 \times 400/700 = 8.57\). So \(\widehat {\RR }_{\text {MH}} = 30.48/11.90 = 2.56\). The variance numerator is \(\sum (m_1 n_1 n_0/n^2 - ab/n)\): men \(30 \times 200 \times 400/600^2 - 200/600 = 6.667 - 0.333 = 6.333\), women \(55 \times 400 \times 300/700^2 - 600/700 = 13.469 - 0.857 = 12.612\); total \(18.945\). Then \(\Var (\log \widehat {\RR }_{\text {MH}}) = 18.945/(30.48 \times 11.90) = 0.0522\), \(\text {SE} = 0.229\), interval \((1.64,\ 4.01)\). For the risk difference, \(\sum (an_0 - bn_1)/n = (13.33 - 3.33) + (17.14 - 8.57) = 18.57\) and \(\sum n_1 n_0/n = 133.3 + 171.4 = 304.8\), so \(\widehat {\RD }_{\text {MH}} = 0.061\). The crude risk ratio is \((60/600)/(25/700) = 2.80\); sex confounds the crude association slightly, since women are more often exposed and at higher baseline risk.

Solution. Each set is a stratum with \(n_k = 3\). If the case is exposed, \(a = 1\), \(b = 0\), \(c = j\), \(d = 2 - j\), so \(R = (2 - j)/3\) and \(S = 0\). If the case is unexposed, \(a = 0\), \(b = 1\), \(c = j\), \(d = 2 - j\), so \(R = 0\) and \(S = j/3\). Summing, the factors \(\tfrac 13\) cancel. With the data, the numerator is \(15 \times 2 + 10 \times 1 + 5 \times 0 = 40\) and the denominator \(8 \times 1 + 2 \times 2 = 12\), so \(\widehat {\OR }_{\text {MH}} = 3.33\). Sets in which the case and both controls share an exposure status contribute nothing.

Solution. The rates per \(100{,}000\) (smokers, non-smokers) are \(61\) and \(11\); \(240\) and \(112\); \(720\) and \(490\); \(1469\) and \(1083\); \(1918\) and \(2120\). The differences are \(50\), \(128\), \(230\), \(386\) and \(-202\). The excess rises with age up to \(74\) and reverses in the oldest group, which is small and contains survivors (smokers susceptible to heart disease have died earlier). The additive and multiplicative pictures differ: the relative effect falls steadily with age, while the number of extra deaths per smoker grows through middle and early old age. Which matters depends on the question.

Solution. \(d_k = n_{0k} - b_k\) and \(c_k = n_{1k} - a_k\), with \(a_k\), \(b_k\) independent. So \(E(a_k d_k) = n_{1k}p_{1k}\,n_{0k}(1 - p_{0k})\) and \(E(b_k c_k) = n_{0k}p_{0k}\,n_{1k}(1 - p_{1k})\). Their ratio is \(p_{1k}(1 - p_{0k})/\{p_{0k}(1 - p_{1k})\} = \psi \), so \(E(a_k d_k) - \psi E(b_k c_k) = 0\). Dividing by the fixed \(n_k\) gives \(E(R_k - \psi S_k) = 0\), and the consistency argument of Theorem 10.1.3 goes through.