Why do we use the t-distribution when the population standard deviation is not known? And similarly why the normal distribution when standard deviation is known?

  • $\begingroup$ If $X ~ N(\mu, \sigma^2)$ the the sum of $n$ independent samples of $X$ has an $N(n\mu, n\sigma^2)$ distribution and thanks to the Central Limit Theorem the same is approximately the case for non-normally distributed random variables with the same mean and standard deviation for large $n$. So the mean of $n$ independent samples of $X$ has an $N(\mu, {\sigma^2/n})$ distribution, leading to the position when $\sigma$ is known $\endgroup$ – Henry Apr 13 '17 at 13:36

The $t$ distribution arises in the following situation. You have a normally distributed population with unknown mean $\mu$ and standard deviation $\sigma$. You draw $n$ independent samples $X_i$ from this population, and then you want to test a hypothesis about the mean of the population using this sample. The null hypothesis for this test is $\mu=\mu_0$. (The alternative hypothesis varies depending on context.)

The test statistic for this is


where $S$ is the sample standard deviation. Now if $S$ were just a fixed number then this ratio would be normally distributed. Moreover, if it were exactly $\sigma$ then this would be $N(0,1)$ distributed under the null hypothesis (which is why we use the normal distribution for hypothesis tests for the mean when the true standard deviation is known). But in fact $S$ is not fixed, it depends on the sample. The distribution of this ratio, assuming the null hypothesis, is called the Student's $t$ distribution with $n-1$ degrees of freedom.

Intuitively, the difference is that $S$ is occasionally much smaller than $\sigma$, and it is these cases that cause the $t$ distribution to have a longer tail than the normal distribution, especially when the number of degrees of freedom is small.

  • $\begingroup$ @Henry I revised a little bit. The point I was trying to convey is that the $t$ distribution just is the distribution of the ratio $\frac{\overline{X}-\mu}{S/\sqrt{n}}$, if $X_i$ are iid $N(\mu,\sigma)$ variables. It isn't guaranteed to be normal because $S$ is not fixed and in fact it just isn't normal, period. $\endgroup$ – Ian Apr 13 '17 at 13:38
  • $\begingroup$ More generally, the key idea here is that of a pivotal quantity. $\endgroup$ – r.e.s. Apr 13 '17 at 13:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.