Solution. (a) \(315.4 - 37.5 = 277.9\) deaths per \(10{,}000\) houses in seven weeks. If the Southwark and Vauxhall houses had instead received Lambeth water, and the ratio reflects the effect of the water, about \(278\) deaths per \(10{,}000\) houses would have been avoided. The difference measures the burden of the exposure on an absolute scale; the ratio measures the strength of the association, which is what bears on causation. A ratio of \(8.4\) on a tiny baseline could be of little public-health importance; here both are large.

(b) Let \(h_1, h_0\) be the average number of residents per house in the two groups. Deaths per person are deaths per house divided by \(h_1\) or \(h_0\), so the ratio per person is \(8.4 \times h_0/h_1\). The two ratios agree when \(h_1 = h_0\). They would differ if, say, the cheaper company supplied more crowded houses. Snow’s argument that the supplies were intermixed street by street makes a large difference in occupancy unlikely.

Solution. (a) A case-control study (sampling on disease). Risk is not estimable from the data alone; the odds ratio is. If the number of births in the source population is known, the controls are a known fraction of the non-cases and Proposition 1.4.5 recovers risks. (b) A historical cohort study: exposure (employment) is defined at the start and deaths are followed forward from records. Risks and rates are estimable. (c) A cross-sectional study. It estimates the prevalence of bronchitis among smokers and non-smokers, not the risk of developing it. (d) A community (cluster-randomised) trial. Risks and rates are estimable, with the village as the unit of randomisation, which the analysis must take into account.

Solution. \(\OR = \RR \cdot (1 - p_0)/(1 - p_1)\). If \(p_1 \geq p_0\) then \(\RR = p_1/p_0 \geq 1\) and \(1 - p_0 \geq 1 - p_1\), so the second factor is at least \(1\) and \(\OR \geq \RR \). If \(p_1 \leq p_0\) both inequalities reverse: \(\RR \leq 1\) and the factor \((1 - p_0)/(1 - p_1) \leq 1\), so \(\OR \leq \RR \).

Solution. (a) \(P(Y^1 = 1) = 4/8\) and \(P(Y^0 = 1) = 3/8\), so the causal risk ratio is \(4/3\). Observed: the exposed (persons 1–4) show \(Y^1 = 1, 1, 1, 0\), a risk of \(3/4\); the unexposed (5–8) show \(Y^0 = 1, 0, 0, 0\), a risk of \(1/4\). The observed risk ratio is \(3\).

(b) Only person 3 has \(Y^1 \neq Y^0\); the exposure causes that person’s disease. Persons 1, 2 and 5 would be ill either way. Three of the four doomed people (1, 2, 5) are in the exposed group, so the exposed would have had risk \(2/4\) even without exposure, against \(1/4\) for the unexposed. Exchangeability fails, and the observed ratio mixes the effect of the exposure with the higher baseline risk of the exposed: the association is confounded.

Solution. \(r_j = 0.02(1 - x_j) + 0.04x_j = 0.02 + 0.02x_j\), giving \(0.022\), \(0.028\) and \(0.034\). These lie exactly on \(r = 0.02 + 0.02x\), so the ecological risk ratio is \((0.02 + 0.02)/0.02 = 2\), the individual risk ratio. The baseline risk is the same in every region and so is the effect, which is the condition of the last proposition. In Example 1.5.4 the baseline risk rose with the proportion exposed, and the regression attributed that rise to the exposure.

Solution. The controls are a sample of \(\bar D\), the cases of \(D\), so \(a/m_1\) and \(c/m_0\) estimate \(q_1\) and \(q_0\), and by Theorem 1.4.2 the sample odds ratio estimates the disease odds ratio \(p_1(1 - p_0)/\{p_0(1 - p_1)\} = \RR \,(1 - p_0)/(1 - p_1)\), which tends to \(\RR \) as \(p_1, p_0 \to 0\). With \(p_1 = 0.02\), \(p_0 = 0.01\): \(\RR = 2\) and \(\OR = 2 \times 0.99/0.98 = 2.02\). With \(p_1 = 0.4\), \(p_0 = 0.2\): \(\RR = 2\) and \(\OR = 2 \times 0.8/0.6 = 2.67\). The approximation is good only for a rare disease.