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)
  • 3
    $\begingroup$ Is that all the information you are given? Or do you have access to the data as well or anything else? $\endgroup$
    – doubled
    Jun 12, 2020 at 15:05
  • $\begingroup$ I have only this information $\endgroup$ Jun 12, 2020 at 21:19
  • 1
    $\begingroup$ 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). $\endgroup$
    – doubled
    Jun 13, 2020 at 16:37

1 Answer 1


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:

$$ \begin{align*} \exp(\hat{\mu}_T-\hat{\mu_C}) &\approx 0.80 \\ \exp\big((\hat{\mu}_T-\hat{\mu_C})-1.96\times \sqrt{\hat{\sigma}_T^2+\hat{\sigma}_C^2}\big) &\approx 0.10 \\ \exp\big((\hat{\mu}_T-\hat{\mu_C})+1.96\times \sqrt{\hat{\sigma}_T^2+\hat{\sigma}_C^2}\big) &\approx 6.78. \end{align*} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.