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Math Help - T-distribution

  1. #1
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    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?
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    Quote Originally Posted by heathrowjohnny View Post
    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?
    ..
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    Quote Originally Posted by heathrowjohnny and edited by Mr F View Post
    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.
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