Solution. \(P_c = P_e p_1/p = P_e\RR \,p_0/[p_0\{1 + P_e(\RR - 1)\}] = P_e\RR /\{1 + P_e(\RR - 1)\}\). Then \[ P_c\,\frac {\RR - 1}{\RR } = \frac {P_e\RR }{1 + P_e(\RR - 1)} \cdot \frac {\RR - 1}{\RR } = \frac {P_e(\RR - 1)}{1 + P_e(\RR - 1)} , \] which is Levin’s formula.

Solution. \(\widehat {\OR } = (40 \times 170)/(60 \times 30) = 3.78\) and \(\widehat {\text {AF}}_E = 2.78/3.78 = 0.735\). From Theorem 5.3.1, \(1 - \widehat {\text {PAF}} = (60/100)/(170/200) = 0.706\), so \(\widehat {\text {PAF}} = 0.294\) \((= 0.4 \times 0.735)\). The standard error of \(\log (1 - \widehat {\text {PAF}})\) is \(\sqrt {40/(100 \times 60) + 30/(200 \times 170)} = 0.0869\), so \(1 - \text {PAF}\) lies in \(0.706\,e^{\pm 0.170} = (0.595,\ 0.837)\) and the PAF in \((0.163,\ 0.405)\).

Solution. (a) Relative to the risk \(p_0\) at level none, the population risk is \(p_0(0.5 + 0.3 \times 2 + 0.2 \times 4) = 1.9p_0\), so \(\text {PAF} = 0.9/1.9 = 0.474\). The proportions of cases at the three levels are \(0.5/1.9\), \(0.6/1.9\), \(0.8/1.9\), that is \(0.263\), \(0.316\), \(0.421\). Then \(0.316 \times \tfrac 12 + 0.421 \times \tfrac 34 = 0.158 + 0.316 = 0.474\).

(b) The exposed have risk \(p_0(0.3 \times 2 + 0.2 \times 4)/0.5 = 2.8p_0\), so \(\RR = 2.8\), and with \(P_e = 0.5\) Levin gives \(0.5 \times 1.8/(1 + 0.5 \times 1.8) = 0.9/1.9 = 0.474\). Collapsing does no harm here, because the reference group is unchanged and the collapsed risk ratio is the correct average. It goes wrong if the collapsed ratio is taken from elsewhere (another population with a different mix of light and heavy exposure), or if the question is the effect of reducing heavy exposure to light, which only the separate ratios can answer.

Solution. People with both exposures have risk \(0.05\) against \(0.001\) with neither, so \((0.05 - 0.001)/0.05 = 98\%\) of their cases are attributable to the combination. They make up \(5\%\) of the population and contribute \(0.05 \times 0.05 = 0.0025\) of the population risk of \(0.0077\), a third of all cases. Each of these cases is prevented by removing smoking or by removing the occupational exposure, so it is counted in both PAFs. The sum exceeds \(1\) by the share of cases that either removal would prevent.

Solution. \(\RD = 0.04\), \(\text {NNT} = 25\). \(\text {SE}(\RD ) = \sqrt {0.08 \times 0.92/1000 + 0.12 \times 0.88/1000} = 0.0134\), so \(\RD \in (0.0138,\ 0.0662)\). Inverting, the NNT lies between \(1/0.0662 = 15\) and \(1/0.0138 = 73\). The inversion is valid only because the interval for \(\RD \) excludes \(0\); if it did not, the interval for the NNT would run from a number needed to treat, through infinity, to a number needed to harm.

Solution. By Miettinen’s formula \(\text {PAF} = P_c\,\text {AF}_E\) with \(0 \leq P_c \leq 1\) and \(0 \leq \text {AF}_E \leq 1\) (for \(\RR \geq 1\)), so \(\text {PAF}\) is at most each factor. Equality \(\text {PAF} = \text {AF}_E\) (with \(\text {AF}_E > 0\)) requires \(P_c = 1\). The PAF is close to \(\text {AF}_E\) when almost all cases are exposed, as for lung cancer and smoking, where \(\text {AF}_E\) is about \(0.9\) and the PAF is about \(0.8\) to \(0.9\).