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

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.

## 1 Answer

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.

• If I understood well, so the answer would depend on whether the confidence interval was derived by z-based method or t-based method. Thank you for your answer. – Senna May 14 '20 at 14:41
• That gets into why we use z-based intervals and t-based confidence intervals. If we know the population variance, we don't bother with t-based confidence intervals, and the z-based interval has its width determined by $\sigma^2/2$. When we don't know the population variance (pretty much always), we estimate the population variance by $s^2$ and use t-based confidence intervals to account for the uncertainty surrounding the estimate (i.e. accounting for the fact that our estimate might be a bad estimate). – Dave May 14 '20 at 14:51