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**heathrowjohnny** ** 1. ** Show that $\displaystyle T = \frac{\bar{X} - \mu}{S / \sqrt{n}} $ has the $\displaystyle t $ distribution with $\displaystyle n-1 $ degrees of freedom. The population of interest in normal, so that $\displaystyle X_{1}, \ldots, X_{n} $ constitutes a random sample from a normal distribution with both $\displaystyle \mu $ and $\displaystyle \sigma $ unknown.

Now a $\displaystyle t $ distribution with n - 1 degrees of freedom (Mr F adds) is defined as $\displaystyle \frac{Z}{\sqrt{\chi^{2}_{n-1}}} $ where $\displaystyle Z $ is a standard normal variable.

Mr F says: No. $\displaystyle {\color{red}T = \frac{Z}{\sqrt{\chi^{2}_{n-1}{\color{blue}/(n-1)}}}}$ has a t-distribution with n - 1 degrees of freedom.

So $\displaystyle \bar{X} - \mu $ is a standard normal variable,

Mr F says: No. $\displaystyle {\color{red}\bar{X}}$ is normally distributed with mean $\displaystyle {\color{red}\mu}$ and variance $\displaystyle {\color{red}\sigma^2/n}$ so $\displaystyle {\color{red} \sqrt{n}\left( \frac{\bar{X} - \mu}{\sigma}\right)}$ is a standard normal variable.

and $\displaystyle S/\sqrt{n} $ is the square root of a chi-squared

Mr F says: No. How did you get this from the (correct) theorem quoted below?

(we used the fact that $\displaystyle (n-1)S^{2}/ \sigma^{2} \sim \chi^{2}_{n-1} $). Is this correct?