How is t-test on non-normal distribution justified? I have heard that when the sample size n is large enough, we can apply the t-test to non-normal distribution due to the CLT. But as far as I can see, the CLT only justify the normality of the sample mean. To apply t-test, we need $S^2(n-1)/\sigma^2$ to follow a $\chi^2$ distribution, where $S^2$ is the corrected sample variance. How does the CLT justify this then? The only thing I can think of is to resample the data as the mean of many batches. That is we create a new data set $\{y_i\}$ from the data set $\{x_i\}$ by defining $y_i = \frac{1}{k}\sum_{j=ki+1}^{ki+k}x_j$. The $\{y_i\}$ will be approximately normal due to CLT, so we can apply the t-test to $\{y_i\}$ instead of $\{x_i\}$. Is this how the t-test is justified for non-normal observation when n is large? The computation would be quite different though.

The CLT can also be applied to the chi-squared distribution. Note that the chi-square asymptotically approaches the normal distribution as the number of degrees of freedom increases. If the normality condition is violated, then this asymptotic may be much slower, but eventually you will still arrive at a normal distribution.
