I am trying to figure out why, for finding confidence intervals, we use the following standard error

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

in the formula

$$ \overline{X} \pm z_{(\alpha / 2)} \times SE$$

For the past week I have been cramming into my head that for confidence intervals of $100(1-\alpha)\%$ we use the following formula

$$ \overline{X} \pm z_{(\alpha / 2)}\frac{\sigma}{\sqrt{n}}~~~ \therefore ~~~ SE = \frac{\sigma}{\sqrt{n}}$$

From looking at the answer from this derivation of SE for Binomials my assumption is that $k = n~~ \therefore ~~\sigma_{\overline{X}} = \sqrt{p(1-p)}$

How would one infer $k = n$ from the following question?

For a 90% confidence interval for the population proportion, p, if the sample proportion, p', is 0.4 and the sample size is n = 100 then the error term, E, is ______ (to the nearest 0.001).

  • $\begingroup$ There is the distinction between the standard deviation $\sigma$ of a population and the standard deviation of a sample mean of size $n$ drawn from the (larger) population. I suspect that is the issue which is confusing you, but I'm not sure exactly what question you've formulated. The conclusion $k=n$ is tied to the sample size being $n$. $\endgroup$ – hardmath Mar 12 '18 at 23:29
  • $\begingroup$ Lots of previous Q&A's on this site dealing with this topic (including this). Also other sites. There are several styles of CI for binomial $p.$ I think the one you're being asked to use is $p' \pm 1.645\sqrt{\frac{p'(1-p')}{n}}.$ Look for this formula in your text or notes. This is the most commonly used style of CI, but (according to link earlier) not the best. $\endgroup$ – BruceET Mar 13 '18 at 4:45
  • 1
    $\begingroup$ $SD(p') = SD(X/n) = \sqrt{\frac{p(1-p)}{n}},$ where $X$ is nr Successes in $n$ trials, and $p$ is population proportion. Because $SD(p')$ involves unknown $p,$ it is necessary to estimate $SD(p')$ when making the CI. The estimate is $\sqrt{\frac{p'(1-p')}{n}},$ with $p'$ used instead of unknown $p.$ This CI assumes binomial $X$ has nearly a normal distribution. $\endgroup$ – BruceET Mar 13 '18 at 4:58

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