I'm working on a basic lecture about inferential statistics and I've tried to find a reliable and convergent answer in previous posts on Cross-validated. However, after reading several threads (like this one here) and/or here), I'm still a bit unsure of the correct answer. Due to different and divergent answers, I would like to check if everyone can agree with this statement: "*As t-tests is a special case of linear models, particularly regression, one of its assumptions is the normality of the residuals. Therefore, we should not check the normality of the raw scores of both groups grouped, but the residuals of the model or the normality of each group individually / separately *".

My question is due to divergent answers. On CrossValidated, one highly upvoted answer (from @thomas-levine) says that "T-test does not assume that the population is normally distributed"

first opinion

However, the second opinion was written by @toneloy, which I particularly agree with, says that T-test assumes the population is normally distributed.

second opinion

Then we have a third answer that says that normality assumption refers to sampling distribution.

third opinion

Answers based on high-quality books or papers are valuable. Thank you.


1 Answer 1


The way to reconcile the opposing views is the following:

1) The t-test does assume normality.

2) The t-test is robust to deviations from said normality.

For #1, we do need normality in order for the test statistic to have the t distribution that we assume. However, #2 kicks in by slowing the data themselves to be non-normal but allowing for the test statistic to converge to the assumed distribution. This is where we get the “$30=\infty$“ joke, that a sample size of 30 tends to be enough to get sufficiently close to the assumed distribution (though you may not need nearly that many, and 30 may not be nearly enough).

  • $\begingroup$ "However, #2 kicks in by slowing the data themselves to be non-normal but allowing for the test statistic to converge to the assumed distribution" ... you can invoke CLT + Slutsky to show it converges to a normal (rather than a t). What do we have above that? Outside of specific cases, perhaps, does anything show that it's going to eventually be closer to a t than it is to a normal? (in which case 'converges to a t' might work) $\endgroup$
    – Glen_b
    Commented Mar 29, 2020 at 23:11

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