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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?

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  • $\begingroup$ 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
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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

$$\frac{\overline{X}-\mu_0}{S/\sqrt{n}}$$

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.

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  • $\begingroup$ @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

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