In simple linear regression, $t = \frac{\hat\beta_1 - \beta_1}{\hat\sigma \sqrt{S_{xx}}} $ is the test statistic for the null hypothesis $H_0 : \beta_1 = 0$. How can I express $t^2$ as an F-distribution, i.e. as the ratio of two independent chi-squared random variables divided by their respective degrees of freedom?

So far I have $t^2 = \frac{\hat\beta_1^2}{\hat\sigma^2 / S_{xx}}$ since $\beta_1 = 0$ and then I've shown that the denominator $\hat\sigma^2 = \frac{\sum (y_i - (\hat\beta_0 + \hat\beta_1x_1)^2 }{n} = \frac{\sum e_i^2}{n}$ is a chi-squared distribution divided by its degrees of freedom, since $e_i$ is the $i$th error and the $e_i$s are independent, normally distributed random variables. But I have no idea how to show that the numerator $\hat\beta_1^2 S_{xx}$ is also a chi-squared distribution... how could I go about doing that?

  • $\begingroup$ Is this a homework question? If so, please add the homework tag... otherwise let us know it's not. $\endgroup$ – jbowman Feb 5 '12 at 3:47
  • $\begingroup$ As a hint - what is the distribution of $\hat{\beta}_1$? What is the relationship between that distribution and the $\chi^2$ distribution (if any)? $\endgroup$ – jbowman Feb 5 '12 at 3:49
  • $\begingroup$ Not homework, it's from the textbook I'm using to try to teach myself linear regression. $\endgroup$ – user24366 Feb 5 '12 at 4:39
  • $\begingroup$ Well, $\hat\beta_1$ ~ $N(0, \sigma^2/S_{xx})$ since $\beta_1 = 0$, and I know that the $\chi^2$ distribution with $k$ degrees of freedom is the sum of the squares of $k$ standard normal random variables... I was thinking that $\hat\beta_1^2$ by itself is a chi-squared random variable with one degree of freedom, but it has a non-standard variance. And how does $S_{xx}$ play into everything? $\endgroup$ – user24366 Feb 5 '12 at 4:50
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    $\begingroup$ Note that $n\hat\sigma^2/\sigma^2$ is a sum of standard normal random variables, while $n\hat\sigma^2$ is a sum of $N(0,\sigma^2)$-random variables. And $\hat\beta_1\sqrt{S_{xx}}/\sigma$ is distributed as $N(0,1)$. $\endgroup$ – B R Feb 5 '12 at 5:59

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