# Why does normal distribution turn into t distribution when variance is unknown?

• I
Suppose ##X## ~ N(##\mu##,##\sigma^2##). Then ##\bar{X}## ~ N(##\mu##,##\frac{\sigma^2}{n}##), where ##\bar{X}## is the random variable for sample mean for samples of size ##n##.

But when the population variance ##\sigma^2## is unknown and the sample size ##n## is small, ##\bar{X}## no longer follows a normal distribution but instead follows a t distribution, such that ##T=\frac{\bar{X}-\mu}{S/\sqrt{n}}## ~ t##_{n-1}##, where ##s^2=\frac{n}{n-1}\times##sample variance##=##the unbiased estimator of the population variance and ##n-1## is the degree of freedom of the t distribution.

My question is why does the distribution of ##\bar{X}## changes just because we do not know the population variance? Shouldn't the population variance still be some fixed value ##\sigma^2## (it's just that it's unknown to us at the moment), and thus making ##\bar{X}## follow a normal distribution still: ##\bar{X}## ~ N(##\mu##,##\frac{\sigma^2}{n}##)? It seems that objective reality (the specific distribution of ##\bar{X}##) changes according to subjective knowledge (whether we know ##\sigma^2## or not). And this I find puzzling.

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Stephen Tashi
We should first observe that you are making some statements that are only true when ##X## has normal distribution.

But when the population variance ##\sigma^2## is unknown and the sample size ##n## is small, ##\bar{X}## no longer follows a normal distribution but instead

That is not correct. ##T## and ##\bar{X}## are different random variables. ( A "statistic" is defined as a random variable that is a function of the values in a sample. ##T## and ##\bar{X}## are statistics. )

It isn't the distribution of ##\bar{X}## that changes when the sample size is small. Instead it is the choice of which statistic people prefer to use when doing statistical tests.

When the sample size is large, people approximate ##\sigma^2## (the population variance) by the sample variance of the particular sample they have, in the belief that the sample variance computed from a large sample is probably close to the population variance. They assume ##\bar{X}## has distribution ##N(\mu,s^2/n)##.

When sample size is small there is less reason to believe that that the sample variance of a particular sample is close to the population variance. So people use ##T-##tests to make decisions. The distribution of ##T## is not the same as a normal distribution. However, this does not change the fact that ##\bar{X}## still has distribution ##N(\mu,\sigma^2/n)## (provided ##X## has a normal distribution).

StatGuy2000, Happiness and FactChecker
The mean has normal distribution as it was said.

As for the t-statistic T, you first need to see that the sample standard deviation is a random variable (it contains the sum of squares of X, hence it could never be a "fixed value" as you suggested), and that it has a chi-distribution (or equivalently, sample variance has a chi-squared distribution). Check the 2nd post here for a proof: https://stats.stackexchange.com/que...bution-of-variance-a-chi-squared-distribution

Afterward, you can see that the t-statistic T is a ratio of a normal random variable with a chi random variable. You can check why this means that it has a t-student distribution here: https://stats.stackexchange.com/que...sqrt-chi2s-s-gives-you-a-t-distribution-proof, where his W equals your s^2 and his s equals your n.

If the population's variance was known, you would use that instead of S^2, and the only random variable in T would be the mean, with a normal distribution. In that case, T would have normal distribution as well.

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Happiness
Stephen Tashi