I need to find the proper sample size, and I have found formula in the Stanford lectures. But I can’t understand what the “variability” of proportions means in this case.

Consider two proportions : P_hat = 1,2% P(1)_hat - P_hat = 0,2%enter image description here

  • 3
    $\begingroup$ For proportions, this is my guess without having access to the context: Under $H_0$ that the two proportions are equal one can find the pooled success probability $\bar p = (p_1+p_2)/2.$ Then the variance $p(1-p)$ of the null Bernoulli distribution is $\bar p(1-\bar p).$ If this makes no sense, then please provide more context. $\endgroup$
    – BruceET
    May 20, 2019 at 7:10
  • $\begingroup$ Well, there is no other data provided. Thank you for your help $\endgroup$
    – Daria
    May 20, 2019 at 10:18
  • 1
    $\begingroup$ You can determine any moment of any binary distribution using the techniques and formulas at stats.stackexchange.com/questions/294737. They indicate there is no difference in meaning of "variability" in the two settings: both refer to the variance $\sigma^2.$ $\endgroup$
    – whuber
    May 20, 2019 at 12:46
  • $\begingroup$ Does it work for proportions too? And in my case what should I plug into formula? $\endgroup$
    – Daria
    May 20, 2019 at 16:49


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