From my understanding, when we construct a confidence interval for a sample mean with a sample size of n, we try to estimate the standard deviation of the sampling distribution
$$σ_{\overline{x}} = \frac{σ}{\sqrt{n}}$$
with the standard error
$$SE_{\overline{x}} = \frac{s}{\sqrt{n}}$$
where s is the sample standard deviation used to approximate the population parameter σ.
Am I correct in assuming that a sample proportion uses the same equation as a sample mean because it is just the mean of variables which are either 1 or 0? I've come across this equation on the internet and on AP statistics at Khan Academy for the standard error of a sample proportion.
$$SE_{\hat{p}} = \sqrt{\frac{\hat{p}(1-\hat{p})}n}$$
However, if we want to use the equation above for $SE_{\overline{x}}$, we first find the sample standard deviation
$$s=\sqrt{\frac{n\hat{p}(1-\hat{p})}{n-1}} $$
Then
$$SE_{\hat{p}} = \frac{s}{\sqrt{n}} = \frac{\sqrt{\frac{n\hat{p}(1-\hat{p})}{n-1}}}{\sqrt{n}}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n-1}}$$
which is not the same as the equation I found on Khan Academy/online. Could someone help explain where I went wrong?
