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For a binomial random variable $Y$ we consider $\sigma$ to be $\sqrt{np(1-p)}$ but for the $\sigma$ (standard error) of a sampling distribution of proportions we get $\sqrt{p(1-p)/n}.$

Why do we take $p/n$ instead of $np$ in the formula for the standard error of the sampling distribution?

whuber
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Nathgun
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    The estimate of the mean for the proportion is Y/n. – Michael R. Chernick Nov 17 '17 at 00:19
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    [Basic properties of variance](https://en.wikipedia.org/wiki/Variance#Basic_properties) -- standard error of $\hat{p}$ is $\text{Var}(\frac{Y}{n}) = \frac{1}{n^2} \text{Var}(Y)=p(1-p)/n$ – Glen_b Nov 17 '17 at 05:53
  • Also see [here](https://stats.stackexchange.com/questions/176757/why-does-the-standard-error-of-hat-p-sqrtp1-p-n?) and [here](https://stats.stackexchange.com/questions/85818/variance-of-sample-proportion-decreases-with-n-but-of-a-count-increases-with-n?) – Glen_b Nov 17 '17 at 05:56

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