Student's T distribution can be derived from the fact that:

$$ \frac{\frac{\bar X - \mu}{\sigma/\sqrt{n}}}{\sqrt{S^2/\sigma^2}} $$

Is an approximately normally distributed variable in the numerator (central limit theorem) with (the square root of) a chi-squared random variable in the denominator. However, does the chi-square result not require that $X$ be normal? If not, what is the reasoning here that allows us to use a chi-square distribution for the denominator?

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    $\begingroup$ To truly get a t-distribution, you need to draw from a normal population to calculate $\bar x$ and $s$. Once you deviate from a perfectly normal population and appeal to the central limit theorem, you also start appealing to Slutsky's theorem. I have heard this combination called the converging together lemma. $\endgroup$
    – Dave
    Commented Jul 27, 2021 at 16:00
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    $\begingroup$ Answer: insert the word "approximately" a second time just before "chi-squared random variable." It also helps to recall these variables must also be "approximately" independent. $\endgroup$
    – whuber
    Commented Jul 27, 2021 at 16:07
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    $\begingroup$ @Him If you want the statistic to be exactly t-distributed when the null is true and the sample size is finite, then you will need the population being sampled to be normally distributed. If you only require the properties of your test to be approximately correc, then you need to think about what degree of approximation is tolerable (it may be necessary to think about more than type I error rates). $\endgroup$
    – Glen_b
    Commented Jul 27, 2021 at 17:20
  • $\begingroup$ @whuber is it common for the sample mean and sample std to be dependent? Is this a population distribution thing or a sampling procedure thing? $\endgroup$
    – Him
    Commented Jul 27, 2021 at 18:07
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    $\begingroup$ They are independent when the data are drawn independently from a common Normal distribution. In other cases, they tend not to be independent. They tend to grow "more" independent as the sample size increases, though. $\endgroup$
    – whuber
    Commented Jul 27, 2021 at 18:42

1 Answer 1


It seems from the comments that the short answer here is for large samples, "no", because the sampling distribution of the mean converges to a normal distribution for large $n$. For small samples, the answer is "maybe" depending on lots of things. It seems from some simulations that one of the biggest factors is the skewness of your population distribution:

All of these are done taking 10000 samples of size n=10. The "theoretical" sampling distributions are normal, chi-square and student's-t respectively.

Uniformly distributed population:

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Beta distributed population with $\alpha = \beta = 0.5$

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Weibull distributed population with $\lambda = 1, k = 0.5$

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As $k$ gets smaller, the t distribution does a worse and worse job for the Weibull distributed population.

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    $\begingroup$ Your discovery of the importance of skewness is right: The $t$-statistic is more sensitive to skewness than kurtosis as detailled here. $\endgroup$ Commented Jul 27, 2021 at 20:43

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