A binomial distribution has the standard deviation $\sqrt{np(1-p)}$. When you find the standard error of $\bar x$, you divide the standard deviation be $\sqrt n$. Analogously, shouldn't I simply take $\sqrt{np(1-p)}$ and divide by $n$? Where does the $n$ go in the actual formula?
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5And what do you get if you divide $\sqrt{np(1-p)}$ by $n$? – Jake Westfall Oct 13 '15 at 15:04
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If $X$ is a binomial random variable with parameters $n$ and $p$ then its expected value is $E(X)=np$ and its variance $V(X)=np(1-p)$.
Therefore, by some simple rules for linear transformations of random variables $E\left(\frac{X}{n}\right)=p$ and $V\left(\frac{X}{n}\right)=\frac{np(1-p)}{n^2}=\frac{p(1-p)}{n}$.
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A binomial distribution has the standard deviation $\sqrt{np(1−p)}$. When you find the standard error of $\bar{x}$, you divide the standard deviation by $\sqrt{n}$.
You are confusing the Bernoulli with the Binomial. A binomial random variable is a sum of Bernoulli rvs with common mean $p$. $\hat{p}$ is a sample average of iid Bernoulli random variables with common mean $p$ and standard deviation $\sqrt{p(1-p)}$. Therefore, that is the standard deviation you divide by root n to get the standard error of the sample proportion.
AdamO
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