How to calculate the relative risk based on two independent confidence intervals Medicine A cures 30% of patients (95% CI: 17 to 45).
Medicine B cures 15% of patients (95% CI: 10 to 20).
So I can divide 30% by 15% and say that medicine A is twice as likely to cure the patients compared to medicine B, right?
My question is: how would I make this same calculation for the confidence intervals? 
I want to say that medicine A is twice as likely (95% CI: X to Y) to cure the patient as medicine B. Conceptually, do I just divide the CIs or is there something else to do?
 A: You can use the Delta method to obtain an approximate distribution of your relative risk, as shown by that link. Then you can define a pivot and use this to obtain a CI.
I understand that there might be some confusion regarding the use of the Delta method, so here are a few simple steps that show how to construct an approximate CI for the relative risk. 


*

*Estimate the RR from the data

*Find the natural log of RR: $\log(RR)$

*The confidence coefficient is from the standard
normal distribution: 1.96 for a 95% confidence
interval


Now you need the standard error. Using the Delta method for sample sizes $n$ and $m$ with probabilities $p$ and $q$ respectively, this is found to be 
$$SE=\sqrt{\frac{1-p}{pn}+\frac{1-q}{qm}}$$
Of course you need to replace the unknown quantities with your estimates, let's denote them by $\widehat{p}$ and $\widehat{q}$. You might notice that this is the second approximation we are using.
Now that you have the formula, compute the standard error: $SE$


*Calculate the lower and upper limits on the log
scale: $\log(RR) scale: \log(RR)
±
1.96  \times SE \log(RR)$

*Exponentiate! 
You can find plenty such information throughout the internet and the above steps are taken from here. We all have Fisher to thank for these approximations!
