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.
 A: This focuses on one-sample t-tests.

as far as I can see, the CLT only justify the normality of the sample mean.

More specifically, under some conditions, in the limit as $n\to\infty$, $Z_n=(\bar{Y}_n-\mu)/(\sigma/\sqrt{n})$ converges in distribution to a standard normal.

To apply t-test, we need $S^2(n−1)/σ^2$
to follow a $χ^2$ distribution

Sure, and you also need that numerator and denominator are independent.

How does the CLT justify this then?

Of itself, it doesn't; it's one piece of the story but several more pieces would be needed, and they don't get you to the t-test statistic actually having a t-distribution (not that you need it, exactly).
Note that $T_n=Z_n / (s/\sigma)$. We could, for example apply Slutsky's theorem to this to show that $T_n$ is also asymptotically normal.
There are conditions for these theorems, so they must hold to be sure to get the result that $T_n$ is asymptotically standard normal.
Then we'd need to show that the error in using $t$ in place of the standard normal would also 'work', asymptotically (which it does; indeed, in the regions people usually worry about it tends to come in more quickly than the other parts, unless you go to the extreme tail - say as a result of a Bonferroni-type adjustment and a lot of comparisons).
So eventually we should expect the distribution of the t-statistic to be well approximated by the t-distribution (under $H_0$). This would suggest that at some sufficiently large sample size, if all the conditions hold, the null distribution t-statistic should be reasonably approximated by a t-statistic, and so a t-test conducted this way should have about the right significance level, $\alpha$.
It doesn't give you a guarantee about how large $n$ might need to be but in recent times there's been progress on getting Berry-Esseen like bounds for t-statistics.
There's also a potential worry about how good the relative power might be (if your sample was large because you're looking to pick up a small effect size, you may need all the power you can get).
Nevertheless, in many "not too nasty" cases the t-test does pretty well. (At the same time, it's not hard to find cases where sample sizes need to be pretty big.)

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?

No, it's not; a potential justification is as above, and then there's practical evidence from simulations to see when it becomes useful under a variety of skewed or heavy-tailed etc possibilities.
However, if you're comfortable with resampling (as this suggests), you can get that significance-level guarantee* at any sample size, across any distribution via a permutation test. You can even use the t-statistic itself as the statistic in that test if you wish; no concerns there with "is n=80 enough?" or whatever.
That doesn't fix the potential power issue, of course.
* to not exceed it; in very small samples you may end up below it, even with a simple null.
A: 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.
