### STANDARD ERROR OF A PROPORTION:

The proportion is the mean of a random variable, $$X,$$ that is $$x_i =1$$ when the individual has a trait, and $$x_i = 0$$ otherwise. The standard deviation involves the expression $$\displaystyle \sum_{i=1}^n (x_i-\bar x)^2$$.

Let’s suppose there are $$m$$ $$x_i = 1$$’s (and $$n-m$$ $$x_i = 0$$’s) among the $$n$$ subjects. Then, the mean corresponds to the proportion $$\bar x = \hat p = \frac{m}{n}$$ and $$x_i-\bar x$$is equal to $$1-\frac{m}{n}$$ for $$m$$ observations and $$0-\frac{m}{n}$$ for $$n-m$$ observations. Hence,

\begin{align}\sum_{i=1}^n (x_i-\bar x)^2 &= m\left(1-\frac{m}{n}\right)^2 + (n-m)\left(0-\frac{m}{n}\right)^2\\[2ex] &=m\left(1-2\frac{m}{n}+\frac{m^2}{n^2}\right) +(n-m)\left(\frac{m^2}{n^2}\right)\\[2ex] &=m-2\left(\frac{m^2}{n}\right)+\frac{m^3}{n^2}+\frac{m^2}{n}-\frac{m^3}{n^2}\\[2ex] &=m-\frac{m^2}{n}\\[2ex] &=m\left(1-\frac{m}{n}\right)\\[2ex] &=n\hat p(1-\hat p) \end{align}.

The sample variance is,

$\mathrm{var}=\frac{\sum_{i=1}^n (x_i-\bar x)^2}{n}=\hat p(1-\hat p)$

The sample proportion is the mean of $$n$$ observations, so the standard error of the proportion is calculated like the standard error of the mean: the SD divided by the square root of the sample size or

$\mathrm{SE}=\sqrt{\frac{\hat p(1-\hat p)}{n}}$

References:

The Standard Error of a Proportion