Confused about the normality assumption of a t-test

I am highly confused about the normality assumption for using a $$t$$-test. So, as far as I've understood it when you have got $$n$$ iid normally distributed random variables $$X_i$$ you can calculate the $$t$$-statistic given by $$\frac{\bar X - \mu}{S \sqrt{n}},$$ where $$\bar X$$ is the sample mean, $$\mu$$ is the distribution mean and $$S$$ is the squareroot of the sample variance.

Under the above assumptions this $$t$$-statistic will be distributed according to a Student's $$t$$-distribution.

However, in a lot of practical references I've looked at, the normality assumption on the variables $$X_i$$ seems to be dropped often for the alternative assumption that your sample size $$n$$ is large. I can only assume that this is motivated by the central limit theorem which guarantees that for large $$n$$ the sample mean $$\bar X$$ is distributed according to a normal distribution. But as I see it normality of $$\bar X$$ is something entirely different than normality of $$X_i$$. I get the feeling that it should just simply be wrong to assume a $$t$$-distribution for a $$t$$-statistic which is based on not normally distributed variables, even if the sample size is large.

If you want to rely on the central limit theorem wouldn't it make more sense to compute a statistic like $$\frac{\bar X - \mu}{S'},$$ where $$S'$$ is an unbiased estimator of the standard deviation and argue that for large samples this is distributed according to a normal distribution with mean $$0$$ and standard deviation $$1$$?

• The entire point--both historically and now--to the t-test is its application to small $n,$ not large $n.$ The CLT tells us little about small $n$ (and gives a relatively poor test.)
– whuber
Feb 7 '20 at 16:50
• I would be cautious about ignoring the normality assumption based on the sample size being "large". The sample size needed to result in a normal distribution of the t statistic depends on the underlying distribution of the data, so there's no magic cutoff value. (E.g. it's bad advice to say "If the sample size is greater than 30, you can ignore the normality assumption".) In addition, if the data distribution is very far from normal --- or at least far from symmetric --- a t test may not be what is desired anyway. Feb 8 '20 at 13:27

Why the student distribution to test equality of mean (Gaussian setting):

The student distribution $$T(k)$$ arises from the situation where:

$$T \sim U/V$$

where $$U \sim N(0,)$$ and $$V \sim \chi^2(k)$$ and are independent.

This is typically the situation you encountered in our initial setting:

if $$X$$ iid from normal distribution then $$\bar X - \mu \sim N(0,\sigma^2/n)$$ and $$(n-1) S^2 / \sigma^2 \sim \chi^2(n-1)$$, you can show they are independent and thus the ratio $$T$$ follows a student distribution.

What happens when relaxing gaussianity of the $$X$$ but appealing for CLT:

When relaxing gaussianity of the $$X$$ but appealing for CLT, you consider the approximation $$\bar{X} \sim N(\mu,\sigma^2/n)$$. So basically, the numerator relationship $$\bar X - \mu \sim N(0,\sigma^2/n)$$ still applies.

• Thank you for your answer. Could you maybe expand a little on the "convolved" reasoning for the denominator, i.e. why the sample variance is roughly $\chi^2$-distributed for large samples? Feb 7 '20 at 11:27