Solution. \(\hat p = 0.012\). The Wald interval is \[ 0.012 \pm 1.96\sqrt {0.012 \times 0.988/250} = 0.012 \pm 0.0135 , \] that is \((-0.0015,\ 0.0255)\), which includes impossible negative values. The Wilson interval, from (3.1), is \((0.0041,\ 0.0347)\), and the Clopper–Pearson interval is \((0.0025,\ 0.0347)\). The two good intervals are asymmetric about \(\hat p\), as they should be near a boundary.

With \(x = 0\) the exact upper limit solves \((1 - p)^{250} = 0.025\), giving \(p_U = 1 - 0.025^{1/250} = 0.0146\), close to \(-\ln (0.025)/250 = 3.69/250\). The same argument for a one-sided \(95\%\) limit gives \(-\ln (0.05)/n \approx 3/n\), the “rule of three”.

Solution. With \(D = 0\), (3.2) gives \(\lambda _U = \chi ^2_{2,\,0.975}/(2T) = 7.378/(2 \times 12400) = 2.97\) per \(10{,}000\) person-years; equivalently \(P(D = 0) = e^{-\mu } = 0.025\) gives \(\mu = 3.689\) and \(\lambda _U = 3.689/T\). For \(3.689/T < 10^{-4}\) needs \(T > 36{,}890\) person-years.

Solution. A person is alive and disease-free at time \(u\) with probability \(e^{-(\lambda + \mu )u}\), since the two hazards add. The probability of developing the disease in \((u, u + du]\) is that times \(\lambda \,du\). Integrating, \[ \int _0^t \lambda e^{-(\lambda + \mu )u}\,du = \frac {\lambda }{\lambda + \mu }\bigl (1 - e^{-(\lambda + \mu )t}\bigr ) . \] With the given values: \(0.2(1 - e^{-1}) = 0.2 \times 0.632 = 0.126\), against \(1 - e^{-0.2} = 0.181\) without competing deaths. The second figure is the risk in a hypothetical population where nobody dies of anything else; the first is the proportion who actually develop the disease. Both are legitimate, but they answer different questions, and a risk reported in an elderly population must say which is meant.

Solution. The denominator \(8\) treats all eight workers as observed for five years, but they contributed only \(25\) of the possible \(40\) person-years. Workers 3, 5 and 8 entered late, worker 4 moved away and worker 7 died, so their disease status at five years is unknown; treating them as disease-free underestimates the risk. Under a constant hazard the five-year risk is \(1 - e^{-0.08 \times 5} = 1 - e^{-0.4} = 0.33\).

Solution. Prevalence odds are \(0.004 \times 4 = 0.016\) in smokers and \(0.002 \times 8 = 0.016\) in non-smokers, so both prevalences are \(0.016/1.016 = 1.57\%\) and the prevalence ratio is \(1\). Smoking doubles the incidence but halves the duration, and a cross-sectional study, which sees prevalence, would find no association. By Theorem 3.4.1 the prevalence odds ratio equals the incidence rate ratio times the ratio of mean durations, here \(2 \times \tfrac 12\).

Solution. Deaths in B are \(40 \times 1.5 + 60 \times 18 = 60 + 1080 = 1140\), a crude rate of \(11.4\) per \(1000\), still twice town A’s \(5.6\). The crude rate weights the specific rates by the town’s own age distribution: \(0.4 \times 1.5 + 0.6 \times 18\) for B against \(0.8 \times 2 + 0.2 \times 20\) for A. Town B puts \(60\%\) of its weight on the high old-age rate, town A only \(20\%\), and that difference outweighs B’s lower rates in both groups. The crude comparison reverses the direction of the age-specific comparisons.