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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)
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    $\begingroup$ 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
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    $\begingroup$ 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

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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:

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

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