9.6 Problems

Problem 9.6.1. In a published comparison of treatments for kidney stones (Charig et al., 1986), results were reported by stone size. Treatment A succeeded in \(81\) of \(87\) small stones and \(192\) of \(263\) large stones; treatment B in \(234\) of \(270\) small and \(55\) of \(80\) large. Find the success rates by stone size and overall, and explain.

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Solution. Small stones: A \(0.931\), B \(0.867\). Large stones: A \(0.730\), B \(0.688\). Overall: A \(273/350 = 0.780\), B \(289/350 = 0.826\). A is better for both sizes but worse overall. Treatment A was used mostly for large stones (\(263\) of its \(350\)), which are harder to treat, and B mostly for small ones. Stone size is a confounder: it affects success and is associated with the treatment given. The stratified comparison is the valid one.

Problem 9.6.2. A case-control study stratified by sex gives, for men, \(20\) exposed and \(80\) unexposed cases and \(40\) exposed and \(200\) unexposed controls; for women, \(40\), \(20\), \(30\) and \(150\). Compute the stratum odds ratios with intervals, the Woolf test of homogeneity and the crude odds ratio. How should the results be reported?

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Solution. Men: \(\widehat {\OR } = (20 \times 200)/(80 \times 40) = 1.25\), \(v = 0.0925\), interval \((0.69,\ 2.27)\). Women: \(\widehat {\OR } = (40 \times 150)/(20 \times 30) = 10.0\), \(v = 0.115\), interval \((5.1,\ 19.4)\). The pooled log odds ratio has weights \(10.81\) and \(8.70\); \(Q = 20.8\) on \(1\) degree of freedom, \(P < 0.0001\). The crude odds ratio is \((60 \times 350)/(100 \times 70) = 3.0\). The odds ratio is strongly heterogeneous: the exposure is associated with the disease in women but hardly in men. Neither the crude \(3.0\) nor a pooled estimate (about \(3.2\)) describes either sex, and the results should be reported separately.

Problem 9.6.3. Show that if the odds ratio is homogeneous across two strata with different baseline risks and \(\OR \neq 1\), the risk ratio cannot also be homogeneous.

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Solution. Suppose \(\OR = \psi \) and \(\RR = \theta \) in both strata. With baseline risk \(p_0\) and \(p_1 = \theta p_0\), \(\psi = \theta (1 - p_0)/(1 - \theta p_0)\). Solving for \(p_0\): \(\psi - \psi \theta p_0 = \theta - \theta p_0\), so \(p_0(\theta - \psi \theta ) = \theta - \psi \) and \(p_0 = (\psi - \theta )/\{\theta (\psi - 1)\}\), a single value. Two strata with different baseline risks cannot share both \(\psi \) and \(\theta \) unless \(\psi = \theta = 1\).

Problem 9.6.4. Using the weights of Example 9.4.2, find the pooled estimate if the youngest stratum is omitted, and comment on its precision.

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Solution. \(\hat \beta _W = (12.86 \times 0.693 + 10.63 \times 0.901)/23.49 = 0.787\), an odds ratio of \(2.20\), with \(\text {SE} = 1/\sqrt {23.49} = 0.206\). The estimate hardly moves, but the standard error rises from \(0.184\) to \(0.206\), as Theorem 9.3.1 implies: every stratum adds to \(\sum w_k\), however small its weight.

Problem 9.6.5. In a cohort, the risk of a disease is \(0.02\) in people with neither of two exposures, \(0.04\) with exposure A only, \(0.05\) with B only and \(0.12\) with both. Compute the risk ratios, the RERI, and say whether there is interaction on the multiplicative scale and on the additive scale.

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Solution. \(\RR _{10} = 2\), \(\RR _{01} = 2.5\), \(\RR _{11} = 6\). On the multiplicative scale the joint ratio expected without interaction is \(5\), against \(6\) observed, a modest departure. On the additive scale \(\text {RERI} = 6 - 2 - 2.5 + 1 = 2.5\): the joint excess risk (\(0.10\)) is much larger than the sum of the separate excesses (\(0.02 + 0.03 = 0.05\)). There is strong additive interaction; preventing either exposure in people who have both prevents more cases than the separate effects suggest.

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