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I need some clarification. I am doing a chart review. Plan to test significance with fisher exact as its two interventions, one yes/no outcome. Presumed prevalence in the population is 15%, but includes both interventions, restrictive and liberal. Would like to see that restrictive has about 10% and liberal stay around 15% for clinical significance which is a 30% reduction. Closest approximation to the restrictive in literature is 9%. I did one test for proportions to calculate sample using a known proportion p0 (9%) and 15% p1 for the other. Which gave me a reasonable number of 207. I plugged that in to a confidence interval calculater too and much the same answer. However i then tried using two proportions of 15 vs 9 in a calculator assuming it is not a known proportion and it gave me a sample of 460 for each group. Where am I going wrong?

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If you fix one of the proportions, you remove the effect of its contribution to the variance of the difference. If you're going to ignore part of the variance, it should be little surprise you'd need a smaller sample size to pick up a particular population difference.

Dealing with the proportion in the retrictive group as a random variable - an estimate of the population proportion rather than as an exact population value will naturally require a larger sample size.

Since you don't actually know that it's 0.09 or 0.10 (or whatever else it might be), you should not be treating it as perfectly known. Maybe it's really 0.08 or 0.11.

(I have not checked any of your calculations)

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  • $\begingroup$ Which takes me back to a sample size of 460 per group. What about the calculation using confidence interval and the proportion in my current population of 15%? This also gave me a low sample of 217 $\endgroup$
    – Mna73
    May 13 '19 at 5:11
  • $\begingroup$ Again, you're leaving out the variance of one of the proportions (i.e. the exact same error you made before) -- you need to calculate a CI for the difference in proportions; $\endgroup$
    – Glen_b
    May 13 '19 at 8:23

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