• 
• $\exp(1.4345+1.96 \times 0.5346) \approx 11.97\exp(1.4345-1.96 \times 0.5346) \approx 1.472$

In R

> exp(summary(m)$coefficients["DSH",1] + 
+     qnorm(c(0.025,0.5,0.975)) * summary(m)$coefficients["DSH",2])
[1]  1.472098  4.197368 11.967884

Here is the derivation using the delta method. Let's look at the familiar -Table below.

Suppose that where is defined as in the table above.

The Odds Ratio is defined as We want to derive the variance of . The multivariable version of the delta method is: Where is the gradient vector. That is: We want to estimate Let the function be The gradient is The variance covariance matrix for a multinomial distribution with categories is $$ \Sigma=\frac{1}{n}\left( \begin{array}{cccc} \left(1-p_{11}\right) p_{11} & -p_{11} p_{12} & -p_{11} p_{21} & -p_{11} p_{22} \\ -p_{11} p_{12} & \left(1-p_{12}\right) p_{12} & -p_{12} p_{21} & -p_{12} p_{22} \\ -p_{11} p_{21} & -p_{12} p_{21} & \left(1-p_{21}\right) p_{21} & -p_{21} p_{22} \\ -p_{11} p_{22} & -p_{12} p_{22} & -p_{21} p_{22} & \left(1-p_{22}\right) p_{22} \\ \end{array} \right) $$ Then equals Now we need which equals: Substituting the MLEs for finally yields So the approximative standard error for the relative risk on the log-scale is So an approximative two-sided confidence interval of level for the relative risk on the original scale is

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:

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