# What is the justification for using the sample mean in confidence intervals?

In the set-up for the classical CLT we have that

$$\frac{\sqrt{n}}{\sigma}(\bar{X}_n-\mu)\to^d N(0,1)$$

as $$n\to \infty$$, which gives rise to the $$1-\alpha$$ asymptotic confidence interval formula for $$\bar{X}_n$$:

$$[\mu-\frac{\sigma}{\sqrt{n}}z_{\alpha/2},\mu+\frac{\sigma}{\sqrt{n}}z_{\alpha/2}]$$

In practice $$\mu$$ and $$\sigma$$ are unknown, and so we replace those quantities with the sample mean $$\hat{\mu}$$ and the sample standard deviation $$\hat{\sigma}$$ respectively.

My question is what is the rigorous justification for this?

Since the sample standard deviation $$\hat{\sigma}$$ is a consistent estimator for $$\sigma$$, an application of Slutsky theorem gives

$$\frac{\sqrt{n}}{\hat{\sigma}}(\bar{X}_n-\mu)\to^d N(0,1)$$

so I understand that replacing $$\sigma$$ with $$\hat{\sigma}$$ is justified. However the result

$$\frac{\sqrt{n}}{\hat{\sigma}}(\bar{X}_n-\hat{\mu})\to^d N(0,1)$$ is clearly false, and in fact

$$P(\bar{X}_n \in [\hat{\mu}-\frac{\hat{\sigma}}{\sqrt{n}}z_{\alpha/2},\hat{\mu}+\frac{\hat{\sigma}}{\sqrt{n}}z_{\alpha/2}])=1$$ for all $$n$$.

Am I missing someting?

• The part that is "clearly false" and its sequel are irrelevant, because the defining condition for a confidence interval is that it cover the true mean, not the estimated one.
– whuber
Sep 27, 2020 at 12:07
• @whuber Yes you are right. I messed up the definition of a confidence interval... Sep 27, 2020 at 12:49
• Your first "confidence" interval actually is a probability interval, see here. I can't get what application of Slutsky theorem you are speaking of, and $\sqrt{n}(\overline{X}_n-\mu)/\hat{\sigma}\sim t_{n-1}$. The part that is "clearly false" does not make sense to me, because $\hat\mu=\overline{X}_n$. Sep 27, 2020 at 13:11

As whuber said, the definition of a confidence interval is that it must cover the true mean $$\mu$$, not the sample mean $$\bar{X}_n$$.