I am trying to calculate the sample size for a trial in which I wish to compare two rare proportions. I know that in the control group, the proportion is 0.07, while in the treatment group the proportion is expected to be 0.04. Using the Fisher's Exact Test for two proportions, I get that in order to achieve a power of 80%, for the 5% significance level, I need 971 subjects from each group.

I tried calculating the sample size for using the Odds Ratios. The OR, according to the proportions mentioned above, is 1.8. I calculated the sample size, and got 545 per group, for the same power and significance level.

I don't understand why I need so many less subjects to test the hypothesis that the OR=1, vs testing the hypothesis that P1-P2=0 ?

Why would anyone use proportion difference then ? And people do !

Can I simply use the logistic regression to test for the OR instead of using Fisher's exact test ? Will regulatory bodies such as the FDA accept it ? It feels like cheating in a way.

Can you please help me to put things in order here ? Thank you !

P.S Sample sizes were calculated using a sample size software, so I take it they are correct

  • $\begingroup$ You don't state your methods of sample size estimation. Sample size estimates usually involve some approximation, however, and whatever you are doing might not be as accurate for low rates and small differences. Bear in mind that Fisher's exact test is best and for large samples all these tests are quite similar, should give very close results, and should require the same sample sizes to give those results. $\endgroup$ Jun 11, 2017 at 14:36

1 Answer 1


I believe that the software you are using is incorrect. As David Smith states, for large sample sizes, all appropriate tests for this situation give very similar results. Using the "pwr" package in R estimates the N needed for p1=.07, p2=.04, alpha=.05, and power=.80 to be approximately 890. This is essentially what I get when I simulate the odds ratio (this could also be done analytically but simulating it was fast and easy). Here is the R code and results:

> ## N needed per group for .80 power with p1 = .07 and p2 = .04
> library(pwr)
> h <- 2*asin(sqrt(.07)) - 2*asin(sqrt(.04))
> pwr.2p.test(h=h,sig.level=.05,power=.8)

Difference of proportion power calculation for binomial distribution (arcsine transformation) 

          h = 0.1328108
          n = 889.9585
  sig.level = 0.05
      power = 0.8
alternative = two.sided

NOTE: same sample sizes

> ## Simulate power for odds ratio method with N = 890
> a <- rbinom(1000000,890,.07)
> c <- rbinom(1000000,890,.04)
> b <- 890-a
> d <- 890-c
> lgor <- log((a*d)/(b*c))
> se <- sqrt(1/a+1/b+1/c+1/d)
> z <- abs(lgor/se)
> prop.table(table(z>1.96))

   FALSE     TRUE 
0.206293 0.793707 

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