I have a question concerning a claim I read in statistics books concerning the applicability of the t-distribution to compute confidence intervals for large $n$ if the data is not normally distributed (but has finite variance).
The statement is that for small $n$ one should check the Gaussian assumption on the data, but if $n$ is large enough then the t-distribution can be used to compute the confidence intervals. The idea seems to be that the distribution of the mean will converge to a Normal distribution by the central limit theorem.
Here is what I don't understand: A t-distributed RV can be characterized as a Normal RV divided by a $\chi$-distributed RV with $n-1$ degrees of freedom. I see that the distribution of the mean will converge to a Normal distribution, but what about the distribution of the standard error? If the original data is not Normal, why should the standard error follow a $\chi$-distribution? Therefore, why should the standardized RV have a t-distribution?
The only way I could understand this is that for large $n$
the distribution of the mean converges to a Normal distribution
the t-distribution converges to a Normal distribution
so basically for large $n$ one could use a Gaussian distribution anyway for the confidence interval. However, this somehow does not make the "detour" over the t-distribution.
Or is there another mathematical reason why the mean divided by the standard error should have a t-distribution? Am I missing something? Thanks for your help.