standard deviation in calculating confidence intervals

Question

I asked a question earlier and realized the question I asked wasn't a good representation of what I meant to ask. (Why use the standard deviation of sample means for a specific sample?)

Now that I've come across a specific example in the lesson itself, I'll try again.

"Construct 80%, 90%, 95%, and 99% confidence intervals for the population mean if the standard eviation of the population is 900. Use the following sample data: $n = 100, \bar{x} = 425$"

80%: $425 \pm 1.28 (\frac{900}{\sqrt{100}}) = 425 \pm 115$ or $310$ to $540$

Why is it in this particular example we are using $\sigma_\bar{x} = \frac{\sigma}{\sqrt{n}}$ when $\sigma_\bar{x}$ is the standard deviation of the distribution of sample means? Since we were given a specific sample mean it makes it sound like we need the standard deviation of that specific sample rather than the standard deviation of the sample mean distribution.

I might just be missing some information about confidence intervals which is why I don't understand. The online system we are using in my introduction to statistics course isn't the best.

Answer 1

Since we were given a specific sample mean it makes it sound like we need the standard deviation of that specific sample rather than the standard deviation of the sample mean distribution.

You are confusing a random variable and its realization. The general form of an $100(1-\alpha)\%$ confidence interval for the population mean $\mu$ is $$ I=\bar{x}\pm z_{\alpha/2}\sigma_{\bar{x}}. $$ Here $\bar{x}$ is random, and $\sigma_{\bar{x}}$ is not. Namely, $\sigma_{\bar{x}}$ is the standard deviation (also called standard error) of $\bar{x}$. Notice that $I$ is a random interval. Given the data, you compute a realization of the confidence interval by plugging-in the observed value of $\bar{x}$ and the deterministic value of $\sigma_{\bar{x}}$.

If $x_1,\ldots,x_n$ is a random sample obtained from the population by sampling independently with replacement, then the standard deviation of the sample mean $\bar{x}$ is $$ \sigma_{\bar{x}}^2=Var\left(\frac{1}{n}\sum_{i=1}^nx_i\right)=\frac{1}{n^2}\sum_{i=1}^nVar(x_i)=\frac{1}{n^2}\sum_{i=1}^n\sigma^2=\frac{\sigma^2}{n}, $$ and thus $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$.

If $x_1,\ldots,x_n$ is obtained by simple random sampling, then $x_i$ and $x_j$ are not independent, and it can be shown that $$ \sigma_{\bar{x}}^2=\frac{\sigma^2}{n}\left(1-\frac{n-1}{N-1}\right), $$ where $N$ is the population size. If $n$ is small compared to $N$, then $\frac{n-1}{N-1}\approx0$ and therefore $\sigma_{\bar{x}}\approx\frac{\sigma}{\sqrt{n}}$.