# 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)
• Is that all the information you are given? Or do you have access to the data as well or anything else? Jun 12, 2020 at 15:05
• I have only this information Jun 12, 2020 at 21:19
• In that case, Stephan's approach is the right one (and would be fine otherwise too, but I think if you had the data, you could run a correctly specified logistic regression to identify that parameter). Jun 13, 2020 at 16:37

• 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 $$\exp(\hat{\mu}_C+1.96\times \hat{\sigma}_C)\approx 62.98$$.
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*}