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I am trying to calculate the required sample size for a trial, where my outcome is proportions (for each subject I am measuring if the subject had an event or not).

I tried two ways and got different results, I wish to understand the reasons. I'll explain.

If my null hypothesis is: $H_{0}: p=0.1$ and the alternative is: $H_{1}: p\neq 0.1$ , and the value of p under the alternative hypothesis is 0.05, then to achieve 80% power, I need 231 subjects, using the two sided binomial test.

However, if my null hypothesis is: $H_{0}: p_{1}-p_{2}=0$ and the alternative is:$H_{1}: p_{1}-p_{2}\neq0$ , and if I assume that $p_{1}=0.05$ and $p_{2}=0.1$ (same difference), then using two.sided Fisher's exact test, I will need 474 subjects in each group to achieve 80% power.

My intuition say that approximately a single group hypothesis will require half of the sample size required for a two group scenario (different tests, so approximately). 231 subjects differ from 474.

How can I explain the fact that hypothesis on a single proportion requires a much smaller sample size, given the same effect (difference), same significance level (5%) and same power (80%) ?

Thank you.

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  • $\begingroup$ Aren't these tests getting at different things though? The one proportion will tell you for a time the effect is not .1, but not really compare to anything. $\endgroup$ – Kitter Catter May 15 at 14:09
  • $\begingroup$ Yeah, they don't do the same thing, but I would expect the sample size to be roughly similar. Does it has something to do with the fact that in two groups you have two variances ? $\endgroup$ – user3275222 May 15 at 18:20

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