I am calculating an incidence risk (r): number of cases of a disease in a population over one year (c) divided by the total mid-year population (N).

$$ r = \frac{c}{N} $$

Let's assume that c is a precise measure, with no uncertainty around it.

N, however, is a point estimate. I also have a 95% CI around it:

$$ N=125 $$

$$ (u(N),v(N))=(100,150) $$

Can I incorporate this uncertainty into the risk, in order to calculate a 95% CI around r?

This is not something I've ever done. After some online searching, I've come across error propagation, that I think would apply here - error propagation for a quotient.

$$ \sigma_r = |r| \sqrt{(\frac{\sigma_c}{c})^2 + (\frac{\sigma_N}{N})^2 + 2\frac{\sigma_{cN}}{cN}} $$

However, I'm not sure:

  1. If this is appropriate to use in this circumstance
  2. How to go from a confidence interval to sigma_{N} and sigma_{cN}.

1 Answer 1

  1. In your case, $\sigma_c$ and $\sigma_{cN}$ equal zero, as $c$ is measured without error. But actually, there's a simpler approach...

  2. Since the function $1/N$ is monotonic, a valid 95% confidence interval of $1/N$ is just equal to a 95% confidence interval of $N$ run through the function $1/N$. So your 95% CI of $N$ would be $(1/150, 1/100)$, and the associated 95% CI of $c/N$ is $(c/150, c/100)$.


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