Solution. \(\hat p_1 = 0.08\) and \(\hat p_0 = 0.03\). \(\widehat {\RR } = 2.67\), with \[ \text {SE}(\log \widehat {\RR }) = \sqrt {1/32 - 1/400 + 1/18 - 1/600} = 0.287 \] and interval \((1.52,\ 4.68)\). \(\widehat {\RD } = 0.050\), with \[ \text {SE}(\widehat {\RD }) = \sqrt {0.08 \times 0.92/400 + 0.03 \times 0.97/600} = 0.0152 \] and interval \((0.020,\ 0.080)\). \(\widehat {\OR } = (32 \times 582)/(18 \times 368) = 2.81\), with \[ \text {SE}(\log \widehat {\OR }) = \sqrt {1/32 + 1/18 + 1/368 + 1/582} = 0.302 \] and interval \((1.56,\ 5.08)\).

Solution. The odds of remaining free are \((1 - p_i)/p_i\), the reciprocal of the disease odds, so their ratio is the reciprocal of \(\OR \). With \(p_1 = 0.2\) and \(p_0 = 0.1\), \(\RR = 2\) but the risk ratio for remaining free is \(0.8/0.9 = 0.889\), not \(0.5\). The same data are a doubling of the risk of disease and an \(11\%\) reduction in the chance of staying well. The risk ratio depends on which outcome is called the event; the odds ratio does not (\(2.25\) one way, \(1/2.25\) the other).

Solution. With the correction, \(\widehat {\OR } = (0.5 \times 40.5)/(6.5 \times 20.5) = 0.152\) and \(\text {SE}(\log \widehat {\OR }) = \sqrt {2 + 1/6.5 + 1/20.5 + 1/40.5} = 1.49\), so the interval is \(0.152\,e^{\pm 2.92} = (0.008,\ 2.8)\). The point estimate suggests strong protection, but the interval includes values well above \(1\): six cases cannot distinguish a protective exposure from a moderately harmful one. The point estimate depends strongly on the arbitrary \(\tfrac 12\) and should not be reported alone.

Solution. Cases: \(120\) exposed and \(160\) unexposed, \(280\) in all. Controls: \(1120\), of whom a share \(20{,}000/100{,}000 = 0.2\) are exposed, so \(224\) exposed and \(896\) unexposed. The odds ratio is \((120 \times 896)/(160 \times 224) = 3.0 = 6/2\), the rate ratio, as Theorem 4.6.1 says.

Solution. \(P(a = 6) = \binom {10}{6}\binom {12}{2}/\binom {22}{8} = 210 \times 66/319{,}770 = 13{,}860/319{,}770 = 0.0433\). The mean is \(n_1 m_1/n = 10 \times 8/22 = 3.64\) and the variance \(n_1 n_0 m_1 m_0/\{n^2(n - 1)\} = 10 \times 12 \times 8 \times 14/(484 \times 21) = 1.32\). The observed \(6\) is \((6 - 3.64)/\sqrt {1.32} = 2.05\) standard deviations above the mean.

Solution. The odds ratio of cases against the subcohort is \((150/200)/(250/750) = 2.25\). The subcohort’s exposure odds estimate \(N_1/N_0\), so the odds ratio estimates \[ \frac {A_1/A_0}{N_1/N_0} = \frac {A_1/N_1}{A_0/N_0} , \] the ten-year risk ratio. The controls represent the whole cohort at the start, including people who later become cases, so the comparison is between the cases and the people at risk at baseline, which is exactly what a risk ratio compares. No approximation is involved.