I want to calculate the 95% CI for an incidence rate ratio (IRR), but I only have incidence rates per 10,000. (I do not have counts for the 2x2 table.)

Specifically, the incident rates were previously calculated by dividing passively monitored infection counts by the catchment area population (subset by the appropriate subpopulation) multiplied by 10,000.


  • Incidence rate A = 7.86 per 10,000
  • Incidence rate B = 0.58 per 10,000
  • IRR = A/B = 13.56.

The Katz log approach often used to compute CIs for odds ratios and rate ratios doesn't seem to apply here, because it is expecting counts, not ratios, to compute the standard error (in log space):

log(SE) = sqrt(1/a + 1/b - 1/m - 1/n),

where a and b are the successes for the two groups and m and n are the totals.

However, I can easily calculate the proportions (i.e. p-hat(A) = 7.86/10,000 = 0.000786).

Can I calculate the log(SE) from the proportions instead of the counts?



1 Answer 1


You need to have some sort of absolute count or error on the rate to get a confidence interval.

Imagine you had rates of 1/1000. The confidence interval would have to depend on whether you made 1 observation in 1000 samples of 10 in 10000 samples. If you just have the rate (0.001), the confidence is exactly information you don't know.

Do you have some sort of error on the rates calculated? Why can't you go back to the counts?

  • $\begingroup$ I can't go back to the counts because the rates were calculated using passive surveillance as the successes and the denominator is the simply the relevant subpopulation from the catchment area of the surveillance. This seems to be a pretty common approach to calculating incidence rates. $\endgroup$
    – milo
    Commented Mar 14, 2016 at 20:27
  • $\begingroup$ From the rate we can compute the proportion, which is static regardless of the number of samples. And then from the proportion, a standard error can be calculated. (i.e. see: onlinecourses.science.psu.edu/stat200/node/48). Can this standard error be used to calculate a CI for a ratio of two proportions? $\endgroup$
    – milo
    Commented Mar 14, 2016 at 20:30
  • $\begingroup$ To clarify, the author's from the paper where I get these rates did not include the catchment area subpopulation that they used to compute the rates. $\endgroup$
    – milo
    Commented Mar 14, 2016 at 20:33
  • 1
    $\begingroup$ You could use the catchment area population details to infer some sort of denominator/sample size. It would not necessarily be valid, but iti s better than nothing if there is no other option. Be explicit though as your assumptions will be very generous. $\endgroup$ Commented Mar 14, 2016 at 22:16

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