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For the sample proportion $\hat{p}$, standard error is known as $\sqrt\frac{\hat{p}(1-\hat{p})}{n}$.

I wonder whether this is the unbiased estimator for the standard deviation of the sample proportion, which is $\sqrt\frac{p(1-p)}{n}$ where $p$ is the true proportion.

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  • $\begingroup$ An estimate of the (large sample) standard error of $\hat p$ is $\sqrt\frac{\hat{p}(1-\hat{p})}{n}$, because standard error of $\hat p$ is just the s.d. of $\hat p$. $\endgroup$ – StubbornAtom Mar 16 at 15:57
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Suppose that $X\sim \text{Bin}(n,p)$, so that $\hat{p}=X/n$. It follows that $$E\hat{p}=np/n=p; \quad {E\hat{p}^2}=n^{-2} EX^2=\frac{np(1-p)+n^2p^2}{n^2}=\frac{p(1-p)}{n}+p^2.$$ So $$ E\frac{\hat{p}(1-\hat{p})}{n}=\frac{1}{n}E\hat{p}-\frac{1}{n}E\hat{p}^2=\frac{p(1-p)}{n}-\frac{p(1-p)}{n^2} $$ By Jensen's inequality (square root function is concave), it follows that $$ E\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\leq\sqrt{E\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{p(1-p)}{n}-\frac{p(1-p)}{n^2}}<\sqrt{\frac{p(1-p)}{n}} $$ as long as $p\neq 0,1$.

It follows that the estimator is not unbiased.

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  • $\begingroup$ Thanks! Then, what would be the unbiased estimator? Moreover, is there any reason to use the standard error after replacing $p$ in the s.d. with $\hat{p}$? $\endgroup$ – user3509199 Mar 16 at 15:59

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