How to calculate the standard error of a proportion using weighted data? I know the "textbook" estimate of the standard error of a proportion is $SE=\sqrt{\frac{p(1-p)}{n}}$, but does this hold up when the data are weighted?
 A: Yes, this formula generalizes in a natural way.

Standardize the (positive) weights $\omega_i$ so they sum to unity.
In a simple random sample $X_1, \ldots, X_n$ where each $X_i$ independently has a Bernoulli$(p)$ distribution and weight $\omega_i$, the weighted sample proportion is
$$\bar X = \sum_{i=1}^n \omega_i X_i.$$
Since the $X_i$ are independent and each one has variance $\text{Var}(X_i) = p(1-p)$, the sampling variance of the proportion therefore is
$$\text{Var}(\bar X) = \sum_{i=1}^n \text{Var}(\omega_i X_i) =  p(1-p)\sum_{i=1}^n\omega_i^2.$$
The standard error of $\bar X$ is the square root of this quantity.  Because we do not know $p(1-p)$, we have to estimate it.  Although there are many possible estimators, a conventional one is to use $\hat p = \bar X$, the sample mean, and plug this into the formula.  That gives
$$\text{SE}(\bar X) = \sqrt{\bar X(1-\bar X) \sum_{i=1}^n \omega_i^2}.$$

For unweighted data, $\omega_i = 1/n$, giving $\sum_{i=1}^n \omega_i^2 = 1/n$.  The SE becomes $\sqrt{p(1-p)/n}$ and its estimate from the sample is $\sqrt{\bar X(1-\bar X)/n}$.  These are the familiar formulas, showing that the calculation for weighted data is a direct generalization of them.
A: It depends what you mean by `weighted'.
If you mean sampling weights from a survey design there isn't a simple extension of the $p(1-p)/n$ rule except under some quite strong additional assumptions (independent sampling of individuals, with weight independent of the value of $X$).
For example, suppose you had stratified sampling stratified on the value of $X$. The variance of $\bar X$ would be zero.  More realistically, you might have stratified sampling on a variable very close to $X$ -- $X$ measured with error, in some sense -- and the variance would be small. 
In the other direction, if you had cluster sampling, or independent sampling uncorrelated with the value of $X$, you'd get a higher variance than the $p(1-p)/n$ rule would give.
