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For t-tests, according to most texts there's an assumption that the population data is normally distributed. I don't see why that is. Doesn't a t-test only require that the sampling distribution of sample means is normally distributed, and not the population?

If it is the case that t-test only ultimately requires normality in the sampling distribution, the population can look like any distribution, right? So long as there is a reasonable sample size. Is that not what the central limit theorem states?

(I'm referring here to one-sample or independent samples t-tests)

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    $\begingroup$ Well, the sample mean as a random variable can only be normal if the single parts are also normal. But you are right: the t-test is asymptotically nonparametric (no normal distribution), but still the within group variances (in the two-sample situation) should be similar and existing. $\endgroup$
    – Michael M
    Mar 11, 2015 at 14:55
  • $\begingroup$ By within-group variances being similar, are you referring to the assumption of the homogeneity of variance? If so, the welch's t-test correct for this, correct? $\endgroup$
    – Peter Nash
    Mar 11, 2015 at 15:08
  • $\begingroup$ Yes, exactly. If the Welch's corrected degrees of freedom go to infinity, then also his procedure would be distribution free (citation needed however...). $\endgroup$
    – Michael M
    Mar 11, 2015 at 15:12

2 Answers 2

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For t-tests, according to most texts there's an assumption that the population data is normally distributed. I don't see why that is. Doesn't a t-test only require that the sampling distribution of sample means is normally distributed, and not the population?

The t-statistic consists of a ratio of two quantities, both random variables. It doesn't just consist of a numerator.

For the t-statistic to have the t-distribution, you need not just that the sample mean have a normal distribution. You also need:

  • that the $s$ in the denominator be such that $s^2/\sigma^2 \sim \chi^2_d$*

  • that the numerator and denominator be independent.

*(the value of $d$ depends on which test -- in the one-sample $t$ we have $d=n-1$)

For those three things to be actually true, you need that the original data are normally distributed.

If it is the case that t-test only ultimately requires normality in the sampling distribution, the population can look like any distribution, right?

Let's take iid as given for a moment. For the CLT to hold the population has to fit the conditions... -- the population has to have a distribution to which the CLT applies. So no, since there are population distributions for which the CLT doesn't apply.

So long as there is a reasonable sample size. Is that not what the central limit theorem states?

No, the CLT actually says not one word about "reasonable sample size".

It actually says nothing at all about what happens at any finite sample size.

I'm thinking of a specific distribution right now. It's one to which the CLT certainly does apply. But at $n=10^{15}$, the distribution of the sample mean is plainly non-normal. Yet I doubt that any sample in the history of humanity has ever had that many values in it. So - outside of tautology - what does 'reasonable $n$' mean?


So you have twin problems:

A. The effect that people usually attribute to the CLT -- the increasingly close approach to normality of the distributions of sample means at small/moderate sample sizes -- isn't actually stated in the CLT**.

B. "Something not so far from normal in the numerator" isn't enough to get the statistic having a t-distribution

**(Something like the Berry-Esseen theorem gets you more like what people are seeing when they look at the effect of increasing sample size on distribution of sample means.)


The CLT and Slutsky's theorem together give you (as long as all their assumptions hold) that as $n\to\infty$, the distribution of the t-statistic approaches standard normal. It doesn't say whether any given finite $n$ might be enough for some purpose.

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    $\begingroup$ For those three things [normality of sample mean, chi-squareness of sample variance, and independence of the two] to be actually true, you need that the original data are normally distributed. Are you saying that only the Normal has those three properties? I'm not contending the statement is false, just curious if that's what you're saying. $\endgroup$
    – Andrew M
    May 14, 2015 at 7:05
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    $\begingroup$ @AndrewM Certainly only the normal has all three together. In addition the first or the third alone are enough to imply the normal -- the third characterizes the normal (Lukacs, 1942), and for finite numbers of independent random variables, only the normal has the first (Cramér’s decomposition theorem). It's conceivable that there's some other way to get the second, but I'm not aware of one. $\endgroup$
    – Glen_b
    May 14, 2015 at 9:09
  • $\begingroup$ @AndrewM in respect of the second, the work of Ahsanullah (1987,1989) may be relevant. $\endgroup$
    – Glen_b
    May 14, 2015 at 9:33
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    $\begingroup$ Thanks for those references @Glen_b ! I was not aware of the Lukacs result, and Cramer's decomposition theorem as stated is rather stronger than the version I had on the top of my head ($X \sim $ Normal iff $AX \sim $ Normal, for all matrices $A$). $\endgroup$
    – Andrew M
    May 14, 2015 at 13:44
  • $\begingroup$ @AndrewM The difference is the result you quote there doesn't rely on independence, while Cramer's result does. They're both useful in their place. $\endgroup$
    – Glen_b
    May 15, 2015 at 2:15
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Just got an interview question related to this and had the same question in my head (while providing my answers). Did a little bit of reading and found the following explanation (excerpt from here) provides the best intuition for me:

According to the central limit theorem, the distribution of sample mean values tends to follow the normal distribution regardless of the population distribution if the sample size is large enough [2]. For this reason, there are some books which suggest that if the sample size per group is large enough, the t-test can be applied without the normality test. Strictly speaking, this is not true. Although the central limit theorem guarantees the normal distribution of the sample mean values, it does not guarantee the normal distribution of samples in the population. The purpose of the t-test is to compare certain characteristics representing groups, and the mean values become representative when the population has a normal distribution. This is the reason why satisfaction of the normality assumption is essential in the t-test. Therefore, even if the sample size is sufficient, it is recommended that the results of the normality test be checked first. Wellknown methods of normality testing include the Shapiro–Wilks test and the Kolmogorov–Smirnov test.

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  • $\begingroup$ As discussed in many threads on this site, this is not good advice. Moreover, the basis of this opinion is wrong on several counts: the mean is representative for any symmetric distribution, not just Normal ones; the Normality assumption is not "essential;" a bigger issue is the extent to which skewness of the underlying distribution translates into a non-Student t distribution of the sampling statistic; and formal Normality testing is invalid and almost useless. $\endgroup$
    – whuber
    May 20, 2021 at 12:36

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