Solution. The rates per \(1000\) are \(2\) and \(10\) (exposed), \(1\) and \(5\) (unexposed): the ratio is \(2\) in both groups. Crude: \((40/8000)/(35/23{,}000) = 3.29\). Combined standard \(25{,}000\) and \(6000\), weights \(0.806\) and \(0.194\): \(\text {DSR}_1 = 3.548\), \(\text {DSR}_0 = 1.774\), \(\text {CMF} = 2.00\). SMR: \(E = 5000 \times 0.001 + 3000 \times 0.005 = 20\) and \(O = 40\), so \(\text {SMR} = 2.00\). Mantel–Haenszel: numerator \(10 \times 20{,}000/25{,}000 + 30 \times 3000/6000 = 8 + 15 = 23\), denominator \(20 \times 5000/25{,}000 + 15 \times 3000/6000 = 4 + 7.5 = 11.5\), ratio \(2.00\). With an exactly homogeneous ratio every method recovers it.

Solution. Concordant pairs can be fitted perfectly by their intercepts and carry no information about \(\beta \). For a discordant pair with the case exposed, the likelihood is \(\psi u/\{(1 + \psi u)(1 + u)\}\) with \(u = e^{\alpha _i}\); its derivative in \(u\) vanishes when \(1 - \psi u^2 = 0\), so \(u = \psi ^{-1/2}\), and the maximum is \(\psi /(1 + \sqrt \psi )^2\). For a pair with the control exposed, the likelihood is \(u/\{(1 + u)(1 + \psi u)\}\), maximised at the same \(u\), with value \(1/(1 + \sqrt \psi )^2\). The profile log-likelihood is therefore \(n_{10}\log \psi - 2(n_{10} + n_{01})\log (1 + \sqrt \psi )\). With \(\phi = \sqrt \psi \), its derivative is \(2n_{10}/\phi - 2(n_{10} + n_{01})/(1 + \phi )\), zero when \(n_{10}(1 + \phi ) = (n_{10} + n_{01})\phi \), that is \(\phi = n_{10}/n_{01}\). So \(\hat \psi = (n_{10}/n_{01})^2\), the square of the consistent conditional estimate.

Solution. The expected deaths among the exposed at unexposed rates are \(T_{1k}r_{0k}\): \(12{,}000 \times 0.000267 = 3.2\), \(15{,}000 \times 0.0009 = 13.5\), \(10{,}000 \times 0.00325 = 32.5\) and \(3000 \times 0.0075 = 22.5\), total \(71.7\). The weights are \(0.045\), \(0.188\), \(0.453\) and \(0.314\). The \(50\)–\(64\) stratum carries the most: it has a large share of the exposed person-time and a fairly high death rate, so it produces the most expected deaths. The youngest stratum, with the most person-time in all but a very low rate, counts for little.

Solution. The SMR interval treats the non-smokers’ rates as known exactly. Its width reflects only the Poisson variation in the \(630\) deaths among smokers, while the non-smokers contribute just \(101\) deaths and their rates are much less precise. The Mantel–Haenszel variance includes both sources. A corrected SMR variance adds the variance of the expected count: \(\Var (\log \text {SMR}) \approx 1/O + \Var (E)/E^2\), with \(\Var (E) = \sum _k T_{1k}^2 D_{0k}/T_{0k}^2\) from the Poisson variances of the reference rates.

Solution. Agreement between methods shows that the result does not depend on the choice of weights, which is reassuring only if the stratum ratios are similar. In Example 13.3.1 they range from \(5.7\) to \(0.9\), and the methods agree because they all weight the older strata heavily. A different population with more young people would give different summaries from every method. The appropriate check of robustness is a test of homogeneity, which fails here, and the appropriate report is the age-specific ratios.