I am trying to reproduce the calculations for sample size in the following article :

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n = sample size required in each group

p1 = proportion of deaths in group Azithromycin

p2 = proportion of deaths in group Placebo

p1-p2 = clinically significant difference = 0.15 (or 15%)

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How to determine p1 and p2 according to the information provided?

  • $\begingroup$ Can you please add the source reference? Also seems to me the expected proportion of deaths in group Placebo is 0.04 (40/1000) based on the text but generally speaking, we need two equations for two unknowns and you have only one equation. $\endgroup$
    – usεr11852
    Jan 25 at 12:45
  • $\begingroup$ p1 is an assumed value. Probably plausibilized on the basis of other studies. $\endgroup$
    – geek45
    Jan 25 at 13:09

1 Answer 1


From the text $p_2=0.04$, $p_1=0.04\times0.85$, assuming the 15% reduction in mortality corresponds to a relative not an absolute difference (which it couldn't be), and the 40 per 1000 is the baseline rate.

If I plug this into the pwr package in R, I get:

# Calculate number of data points needed per group
sizepergroup = pwr::pwr.2p.test(n=NULL, 

# Account for 5% loss to follow-up and multiply by number of groups:
sizepergroup / 0.95 * 2

[1] 32656.54

Which agrees with the text.

If the question was about where $p_1$ comes from (thanks to @geek45 for pointing this out), that is, why did they power for a 15% difference in mortality, then that information is not present in the text. It might be that they felt this was the smallest important difference, or a plausible difference based on previous studies.

In any case the authors have calculated that this is the smallest effect size that the study can detect with reasonable power and have made the decision that the ability to detect a smaller effect would not be worth the required increase in resource.

  • 1
    $\begingroup$ +1 but please explain the need to divide $n$ by 0.95. As. that is not obvious in your answer. (yes, it is to account for the follow-up loss but can be easily lost) $\endgroup$
    – usεr11852
    Jan 25 at 14:13
  • 1
    $\begingroup$ @usεr11852 ive added comments to the code $\endgroup$ Jan 25 at 14:25

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