Calculation of 95% Confidence Interval for an Incidence Rate Difference Suppose that I have two pre-computed Incidence Rates with 95% CI, calculated from the same sample. Let's say
IR1: 10.0% [8.0-12.0%]
IR2: 21.0% [15.0-27.0%].
If I want to calculate an IR difference, without having the number underlying this IR, how can I compute the corresponding 95% Confidence Interval for the difference?
My guess is that I need to extract the Standard Error from each of the confidence interval, trying to estimate the SE for the difference. But I do not know how.
Thank you in advance.

So a variance estimator for the the incidence rate is
$$ \widehat{Var}(\widehat{IR_1}) = \frac{A_1}{T_1^2}$$
where $A$ is the number of events and $T$ is time. A variance estimator for the incidence rate difference is
$$ \widehat{Var}(\widehat{IR_1} - \widehat{IR_2}) = \frac{A_1}{T_1^2} + \frac{A_2}{T_2^2}$$
Under the assumption of $IR_1$ and $IR_2$ being independent, we can estimate the variance of the difference as the sum of the variances.
So to calculate the variance for the incidence rate difference, you need to first calculate the variance for each (which you can do via algebra remembering $LCL = IR - \alpha * SE$ where $\alpha=1.96$ for two-sided 95% confidence intervals). After you have both variances, add them together then take the square root.
Note: this process is more involved for the incidence rate ratio. Since we estimate the variance of the log incidence rate, you would need to use the delta method.
