= Solution
Let $Y_I,Y_C$ be attendance totals and $T_I,T_C$ the observed <person-time at risk>. Under independent <Poisson distributions> for the totals, the estimated <incidence rate ratio> is $\widehat R=(Y_I/T_I)/(Y_C/T_C)$. Treating the exposures as known, the supplied large-count approximation, or the <delta method>, gives
$$
\log\widehat R=\log Y_I-\log Y_C+\log T_C-\log T_I,\qquad \widehat{\operatorname{Var}}(\log\widehat R)=Y_I^{-1}+Y_C^{-1}.
$$
Independence adds the log-count variances, while fixed exposure terms add no variance. Form the normal <confidence interval> on the log scale and exponentiate:
$$
\boxed{\widehat R\exp\{\pm1.96\sqrt{Y_I^{-1}+Y_C^{-1}}\}.}
$$
This specifies the calculation without evaluating it. Individual <randomization> makes a child the assignment unit, but an independent Poisson count model remains an additional assumption about outcomes, not a consequence of randomization alone.
Back to article page
