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?
