Calculate 95% CI for ratio of odds ratio How can I calculate the 95% CI for the ratio of two odds ratios, given:

*

*OR in test group = OR 21.85 (95% CI 3.08, 155.12)

*OR in control group = OR 27.15 (95% CI 11.64, 63.31)

 A: The standard approach to calculating confidence intervals for odds ratios is to treat them as log-normally distributed. Your data are consistent with this, specifically,

*

*In the test group, log parameters $\hat{\mu}_T=3.08$ and $\hat{\sigma}_T=1$ are consistent with an estimated odds ratio of $\exp(\hat{\mu}_T)\approx 21.76$ and a confidence interval from $\exp(\hat{\mu}_T-1.96\times \hat{\sigma}_T)\approx 3.06$ to $\exp(\hat{\mu}_T+1.96\times \hat{\sigma}_T)\approx 154.47$.

*In the control group, log parameters $\hat{\mu}_C=3.30$ and $\hat{\sigma}_C=0.43$ are consistent with an estimated odds ratio of $\exp(\hat{\mu}_C)\approx 27.11$ and a confidence interval from $\exp(\hat{\mu}_C-1.96\times \hat{\sigma}_C)\approx 11.67$ to $\exp(\hat{\mu}_C+1.96\times \hat{\sigma}_C)\approx 62.98$.

Of course, some rounding errors are present, but it looks like a safe bet to proceed on this assumption.
Now, the ratio of two independent log-normals is again log-normal, where log-means are subtracted from each other and log-variances add up. So we can calculate the expectation of the ratios $\frac{\text{OR}_T}{\text{OR}_C}$ as well as the confidence interval straightforwardly:
$$ \begin{align*}
\exp(\hat{\mu}_T-\hat{\mu_C}) &\approx 0.80 \\
\exp\big((\hat{\mu}_T-\hat{\mu_C})-1.96\times \sqrt{\hat{\sigma}_T^2+\hat{\sigma}_C^2}\big)
&\approx 0.10 \\
\exp\big((\hat{\mu}_T-\hat{\mu_C})+1.96\times \sqrt{\hat{\sigma}_T^2+\hat{\sigma}_C^2}\big)
&\approx 6.78.
\end{align*} $$
