It is possible to find point estimate of population mean and population variance when confidence interval of population mean is given?

Question

Let's say that somehow $100(1-\alpha)\%$ confidence interval of population mean $\mu$ is known as $(a,b)$ and the number of samples is $n$. Is it possible to infer point estimates of population mean and population variance from this information? In this case, the assumption is that the population follows normal distribution.

One idea is that because confidence interval of population mean can be calculated if we know sample mean $\overline{x}$ and population variance $\sigma^{2}$: $$\overline{x}-z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\leq\mu\leq\overline{x}+z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$$ , we can set $a=\overline{x}-z_{\alpha/2}\frac{\sigma}{n}$, $b=\overline{x}+z_{\alpha/2}\frac{\sigma}{n}$ and solve for $\overline{x}$ and $\sigma$. Certainly, in this case, $\overline{x}$ can be treated as point estimate of population mean. However, what about $\sigma^{2}$? Is this "true" population variance or is this just "point estimate" of population variance? I am really confused about how $\sigma^{2}$ should be interpreted in this case.

Answer 1

You can derive the $\bar{x}$ and $\sigma^2$ that generated that confidence interval, yes. Knowing the sample size and $\alpha$-level is critical, however, and you cannot solve the problem without that information.

The z-based confidence interval implies a known variance that is used in calculating the confidence interval, so when you use the width to solve for variance, you are solving for the true variance $\sigma^2$, not an estimate $s^2$. If the confidence interval is t-based, then you would be solving for $s^2$.

The width of a z-based confidence interval does not depend on the data, since you know the population variance. When you know a parameter, you don't bother to estimate it.