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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

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    $\begingroup$ 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 '19 at 7:10
  • $\begingroup$ Well, there is no other data provided. Thank you for your help $\endgroup$ – Maria Lavrovskaya May 20 '19 at 10:18
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    $\begingroup$ 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 '19 at 12:46
  • $\begingroup$ Does it work for proportions too? And in my case what should I plug into formula? $\endgroup$ – Maria Lavrovskaya May 20 '19 at 16:49

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