6.5 Sample size for comparing rates
Proposition 6.5.1. To detect an incidence rate ratio \(\theta \) with two-sided level \(\alpha \) and power \(1 - \beta \), when the exposed share of person-time is \(T_1/(T_1 + T_0)\), the total number of cases needed is approximately \[ D = \frac {(z_{\alpha /2} + z_\beta )^2}{\pi (1 - \pi )(\log \theta )^2}, \qquad \pi = \frac {\theta T_1}{\theta T_1 + T_0} . \]
Proof. By Theorem 4.5.1, given \(D\) cases, \(D_1 \sim \text {Bin}(D, \pi )\), and \(\log \widehat {\text {IRR}} = \operatorname {logit}\hat \pi - \log (T_1/T_0)\). By the delta method \(\Var (\operatorname {logit}\hat \pi ) \approx 1/\{D\pi (1 - \pi )\}\). A level-\(\alpha \) test with power \(1 - \beta \) against \(\log \theta \) needs the standard error to equal \(\log \theta /(z_{\alpha /2} + z_\beta )\); solving for \(D\) gives the formula. □
The design question is then turned into person-time, since the number of cases expected per person-year of follow-up determines how long, or how large, the cohort must be.
Example 6.5.2. To detect a rate ratio of \(2\) with equal person-time in the two groups (\(\pi = 2/3\)), at \(\alpha = 0.05\) and power \(0.80\), \[ D = \frac {(1.960 + 0.842)^2}{(2/9)(\log 2)^2} = 73.5 , \] so about \(74\) cases are needed. If the unexposed rate is \(5\) per \(1000\) person-years, every \(1000\) person-years in each group yield \(10 + 5 = 15\) cases, so \(74\) cases need about \(4900\) person-years in each group. The formula uses the variance at the alternative; a simulation of the conditional test gave power \(0.83\), so the formula errs slightly on the safe side here.
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