std.deviation of a binomial random variable vs std error of a distribution of sample proportions

Asked Nov 17 '17 at 00:08

Active Nov 17 '17 at 00:24

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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?

edited Nov 17 '17 at 00:24

whuber

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asked Nov 17 '17 at 00:08

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 -- 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 and here – Glen_b Nov 17 '17 at 05:56

std.deviation of a binomial random variable vs std error of a distribution of sample proportions

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