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

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


$$SE_{\hat{p}} = \frac{s}{\sqrt{n}} = \frac{\sqrt{\frac{n\hat{p}(1-\hat{p})}{n-1}}}{\sqrt{n}}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n-1}}$$

which is not the same as the equation I found on Khan Academy/online. Could someone help explain where I went wrong?

  • $\begingroup$ While perhaps not quite a duplicate, there's some related discussion in this question $\endgroup$ – Glen_b Aug 16 '18 at 18:16
  • $\begingroup$ These formulas apply only when $n$ is so large that the difference between $n$ and $n-1$ doesn't matter. For better alternatives with smaller $n,$ look at these posts: stats.stackexchange.com/search?q=clopper . $\endgroup$ – whuber Aug 16 '18 at 21:19

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