Solution. With two groups the model is saturated: the score equations make the fitted number of cases among the exposed equal \(a\) and among all subjects equal \(a + b\), so the fitted risks are \(a/n_1\) and \(b/n_0\). Then \(\hat \alpha = \operatorname {logit}(b/n_0) = \log (b/d)\) and \(\hat \alpha + \hat \beta = \log (a/c)\), so \(\hat \beta = \log (ad/bc)\). The information matrix for \((\alpha , \beta )\) is \(\begin {pmatrix} w_0 + w_1 & w_1 \\ w_1 & w_1 \end {pmatrix}\) with \(w_1 = n_1\hat p_1(1 - \hat p_1) = ac/n_1\) and \(w_0 = bd/n_0\). Its inverse has \((2, 2)\) entry \((w_0 + w_1)/(w_0 w_1) = 1/w_0 + 1/w_1\), which is \(n_0/(bd) + n_1/(ac) = 1/b + 1/d + 1/a + 1/c\).

Solution. By Proposition 12.1.2 the odds ratio is \(e^{10 \times 0.047} = e^{0.47} = 1.60\), with interval \(e^{10(0.047 \pm 1.96 \times 0.012)} = (1.26,\ 2.02)\). By Theorem 12.2.1 this is a valid estimate of the population odds ratio, although the study sampled on disease.

Solution. By Theorem 12.2.1, \(\alpha ^* = -5 + \log (1/0.01) = -5 + 4.605 = -0.395\). From a fitted \(-0.40\) the population intercept is \(-0.40 - 4.605 = -5.005\), and with \(\hat \beta = 0.8\) the risks are \(1/(1 + e^{5.005}) = 0.0067\) and \(1/(1 + e^{4.205}) = 0.0147\). This is Proposition 1.4.5 in model form.

Solution. \(e^{1.136 - 0.309k}\) for \(k = 0, \dots , 4\): \(3.11\), \(2.29\), \(1.68\), \(1.23\), \(0.91\). A single score uses one degree of freedom to test and describe a steady decline of the log rate ratio with age, which is what the stratum estimates suggest; four indicators would reproduce the five stratum ratios exactly (the saturated model) and test heterogeneity of any kind with four degrees of freedom, with less power against a trend. The residual deviance of \(1.5\) on \(3\) degrees of freedom shows that the linear trend describes the data adequately.

Solution. With \(\psi = e^\beta \), Proposition 12.3.1 gives (a) \(\psi /(\psi + \psi + 1) = \psi /(2\psi + 1)\); (b) \(1/(1 + 2\psi )\); (c) \(1/3\), free of \(\psi \), so the set carries no information. In general a set with an exposed case and \(j\) exposed controls contributes \(\psi /\{(j + 1)\psi + (2 - j)\}\), and one with an unexposed case \(1/\{j\psi + (3 - j)\}\). Maximising the product gives the conditional estimate; the Mantel–Haenszel estimate of Problem 10.8.3 is a non-iterative approximation to it.