Solution. (a) Chicken: \(35/70 = 0.50\) against \(5/50 = 0.10\), \(\RR = 5.0\). Potato salad: \(22/60 = 0.37\) against \(18/60 = 0.30\), \(\RR = 1.22\). The chicken is the likely vehicle; the weak association with the salad may come from guests who ate both.

(b) \(\text {SE}(\log \widehat {\RR }) = \sqrt {1/35 - 1/70 + 1/5 - 1/50} = 0.441\), so the interval is \(5.0\,e^{\pm 0.864} = (2.1,\ 11.9)\). By Proposition 4.4.1, \(X^2 = 120(35 \times 45 - 5 \times 35)^2/(70 \times 50 \times 40 \times 80) = 21.0\), \(P < 0.0001\).

(c) The overall attack rate is \(40/120 = 0.333\), so \(\text {PAF} = (0.333 - 0.10)/0.333 = 0.70\). By Miettinen, \(P_c = 35/40 = 0.875\) and \(\text {AF}_E = 4/5 = 0.8\), giving \(0.875 \times 0.8 = 0.70\). Seven in ten cases would not have occurred without the chicken; the other cases arose from the background or from another source.

Solution. (a) Observed: \((90 \times 140)/(110 \times 60) = 1.91\). Corrected exposed counts: \((90 - 0.05 \times 200)/0.80 = 100\) among cases and \((60 - 10)/0.80 = 62.5\) among controls, so the corrected odds ratio is \((100 \times 137.5)/(100 \times 62.5) = 2.20\).

(b) Nondifferential misclassification of a binary exposure biases the odds ratio towards \(1\) (Theorem 2.4.3), so the corrected value lies further from \(1\).

(c) If exposure is recalled at interview after diagnosis, cases may report it more completely than controls (recall bias), giving a higher sensitivity among cases. The bias could then be away from the null, and the correction must use separate sensitivities and specificities for cases and controls, which needs a validation study within each group.

Solution. (a) Rates per \(1000\) person-years: under \(50\), \(1.5\) and \(0.625\) (ratio \(2.4\)); \(50\) and over, \(10\) and \(5\) (ratio \(2.0\)). Crude: \((52/12{,}000)/(34/20{,}800) = 2.65\). The exposed are older (\(33\%\) of their person-time over \(50\), against \(23\%\) of the unexposed), and death rates rise with age, so age inflates the crude ratio.

(b) The numerator is \(12 \times 16{,}000/24{,}000 + 40 \times 4800/8800 = 8.0 + 21.82 = 29.82\), and the denominator \(10 \times 8000/24{,}000 + 24 \times 4000/8800 = 3.33 + 10.91 = 14.24\), so \(\widehat {\text {IRR}}_{\text {MH}} = 2.09\). The variance of its log is \(\sum _k D_k T_{1k}T_{0k}/T_k^2\) divided by the product of numerator and denominator: \((22 \times 8000 \times 16{,}000/24{,}000^2 + 64 \times 4000 \times 4800/8800^2)/(29.82 \times 14.24) = 0.0489\), \(\text {SE} = 0.221\), interval \((1.36,\ 3.23)\). The test: observed \(52\) exposed deaths, expected \(36.42\), variance \(20.76\), so \(X^2 = (52 - 36.42)^2/20.76 = 11.7\).

(c) Expected deaths among the exposed at the unexposed rates: \(8000 \times 0.000625 + 4000 \times 0.005 = 5 + 20 = 25\), so \(\text {SMR} = 52/25 = 2.08\).

(d) The stratum ratios, \(2.4\) and \(2.0\), are close; Woolf’s homogeneity statistic is \(0.13\) on \(1\) degree of freedom. A summary of about \(2.1\) is appropriate, and all the methods give it.

Solution. (a) \(\hat \psi = 36/18 = 2.0\). McNemar: \((36 - 18)^2/54 = 6.0\), \(P = 0.014\). The exact Clopper–Pearson interval for \(\pi = \psi /(1 + \psi )\) from \(36\) of \(54\) is \((0.525,\ 0.789)\), which transforms to \((1.11,\ 3.74)\).

(b) By Theorem 6.3.1 the discordant pairs are binomial with \(\pi = 2/3\) under the alternative and \(\tfrac 12\) under the null. The one-sample version of Theorem 6.4.1 needs \[ n_d = \frac {\bigl (1.960 \times 0.5 + 1.282\sqrt {(2/3)(1/3)}\bigr )^2}{(2/3 - 1/2)^2} = 90.3 , \] so \(91\) discordant pairs, and about \(91/0.35 = 260\) pairs in all.

Solution. (a) By Theorem 7.4.2 with \(\theta = 1\), \(1.8 = (1 + 2\pi _1)/(1 + 2 \times 0.2)\), so \(1 + 2\pi _1 = 2.52\) and \(\pi _1 = 0.76\): the confounder would have to be nearly four times as common among the exposed. Whether that is plausible depends on what the confounder is.

(b) No. By Theorem 7.4.4, a confounder that explains a risk ratio of \(1.8\) must have \(R_C \geq 1.8\). Directly, \(1.8 = (1 + 0.5\pi _1)/(1 + 0.5\pi _0)\) would need \(\pi _1 \geq 1.6\) even with \(\pi _0 = 0\), which is impossible.

Solution. The apparent prevalence is \(0.032\). By Theorem 2.2.4, \(\hat p = (0.032 + 0.99 - 1)/0.89 = 0.0247\). By Theorem 3.4.1, \(\hat I = \hat p/\{(1 - \hat p)\bar D\} = 0.0247/(0.9753 \times 4) = 0.0063\) per person-year, about \(6.3\) per \(1000\). Ignoring the misclassification would give \(0.032/(0.968 \times 4) = 8.3\) per \(1000\), a third too high.

Solution. (a) By (6.1), \(q_1 = 1.5 \times 0.3/(1 + 0.3 \times 0.5) = 0.391\). Theorem 6.4.1 with \(k = 2\) gives \(316\) cases and \(632\) controls.

(b) \(316\) cases at \(2\) per \(100{,}000\) person-years need about \(16\) million person-years: a million people for sixteen years. The case-control study reaches the same cases by finding them where they already are, which is why it is the design of choice for rare diseases.

Solution. Each matched set is drawn from a risk set at one moment and one age. Within such a set, the chance that the case is exposed rather than one of the controls depends on the exposed and unexposed hazards at that moment, in the ratio \(\lambda _1/\lambda _0\), exactly as in the proof of Theorem 4.6.1, applied to an instant instead of a period. If the rate ratio is constant across ages and times, every set is a stratum with the same odds ratio \(\lambda _1/\lambda _0\), and by Theorem 10.1.3 the Mantel–Haenszel estimator over many small strata is consistent for it (Problem 10.8.3 gives its form for \(1{:}2\) sets). The controls are sampled from people at risk, not from people who remained disease-free to the end, so the rarity of the disease plays no part. Matching on age adjusts for it, provided the analysis keeps the sets.