# degrees of freedom for linear regression

If you have a reduced model with $H_0:\beta_1 = 1$, $H_a: \beta_1 \neq 1$, then the reduced model is: $$Y_i = 1X_i + \beta_0 + \epsilon_i$$

Are the degrees of freedom for the error term SSE $n-1$?

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You would have $n-1$ degrees of freedom for the error term in the null model.

I think a caveat may be needed if you want to do an F-test. With the usual assumptions about normality, independence, and homoskedasticity, if you were testing the null hypothesis that $\beta_1=0$, then the F-statistic would have an F-distribution if the null hypothesis is true. However, that conclusion relies not only on the fact that the numerator and denominator would each have a chi-square distribution, but also that they are independent. If I can believe some tentative stuff I just did, if the null hypothesis is $\beta_1=1$, then the numerator and denominator in the F-statistic would not be independent. I'm going to need to look at this more closely to see what should be done about it. It should be possible to derive a likelihood-ratio test. With the more usual null hypothesis, that would be the usual F-test. With this null hypothesis, it would be different.

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Huh? The numerator and denominator of the $F$-statistic are still independent in this case (as well as in much more general ones). This follows from the fact that the numerator is a (continuous) function of the unrestricted estimate $\hat{\beta} = (\hat{\beta}_0, \hat{\beta}_1)$ and the denominator is a function of $\hat{\sigma}^2$, which is a function of $Y - X \hat{\beta}$. Since $\mathrm{Cov}(\hat{\beta}, Y - X \hat{\beta}) = 0$, then (under the standard assumptions you've stated) the result follows. – cardinal Sep 10 '11 at 4:08
OK, I'm going to look at this again....... – Michael Hardy Sep 11 '11 at 0:13