1. ## T-distribution

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

Now a $t$ distribution is defined as $\frac{Z}{\sqrt{\chi^{2}_{n-1}}}$ where $Z$ is a standard normal variable. So $\bar{X} - \mu$ is a standard normal variable, and $S/\sqrt{n}$ is the square root of a chi-squared (we used the fact that $(n-1)S^{2}/ \sigma^{2} \sim \chi^{2}_{n-1}$). Is this correct?

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

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

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

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

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

and $S/\sqrt{n}$ 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 $(n-1)S^{2}/ \sigma^{2} \sim \chi^{2}_{n-1}$). Is this correct?
..

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

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

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

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

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

and $S/\sqrt{n}$ 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 $(n-1)S^{2}/ \sigma^{2} \sim \chi^{2}_{n-1}$). Is this correct?
Then:

$T = \frac{Z}{\sqrt{\chi^{2}_{n-1}/(n-1)}}$

$= \frac{\sqrt{n-1}\, Z}{\sqrt{\chi^{2}_{n-1}}}$

$= \frac{\sqrt{n-1} \, \sqrt{n} \, \left( \frac{\bar{X} - \mu}{\sigma}\right)}{\sqrt{(n-1)\, s^2/\sigma^2}}$

$= \frac{\sqrt{n} \, \left( \frac{\bar{X} - \mu}{\sigma}\right)}{\sqrt{s^2/\sigma^2}}$

$= \frac{\sqrt{n} \, \sigma \, \left( \frac{\bar{X} - \mu}{\sigma}\right)}{s}$

$= \frac{\sqrt{n} \, (\bar{X} - \mu)}{s}$

which is the required result.