The T distribution arises as the ratio between a Gaussian r.v. and the square root of a Chi-square r.v.: X ~ Gaussian(m, v) S ~ ChiSquare(n) T = X / sqrt

The T distribution arises as the ratio between a Gaussian r.v. and the square root of a Chi-square r.v.: X ~ Gaussian(m, v) S ~ ChiSquare(n) T = X / sqrt(S / n) The sampling distribution of the sample mean is Gaussian whenever the data is Gaussian, or whenever the CLT applies. The sampling distribution of the sample variance is Chi-square whenever the data is Gaussian. But the CLT does not have any effect here. In general there isn't much else that we can say about the sampling distribution of the sample variance. Thus if we want to compute a "sample Z statistic" using those estimated quantities, and we know the data is Gaussian, then we know the sampling distribution of that "sample Z statistic". It's the T distribution. But the assumption of underlying Gaussian data is important here. The CLT doesn't help us derive a T distribution. But it is true that our "sample Z statistic" is asymptotically Gaussian, in which case the T distribution itself is approximately Gaussian. [0] So a "T test" (meaning "test of a difference in means"), using the actual T distribution as the null distribution of the test statistic, is basically never valid on non-Gaussian data. But it's valid (asymptotically) using the Gaussian distribution as the null distribution of the test statistic. That's a lot of reasoning to say: `avg(y) / sqrt(var(y) / n)` could be either T or Gaussian, depending on the assumptions and context. But I would push back on conflating that with `avg(y) / sqrt(σ^2 / n)`. Even if they have the same distribution, they are not the same thing. [0]: If you want a good writeup, see https://stats.stackexchange.com/a/253318/36229 https://stats.stackexchange.com/a/253318/36229. Or look into a good statistics textbook.