2
$\begingroup$

You are given a "blackbox":

$RSS = (Y - X\beta)'(Y-X\beta)$

for any linear model $Y=X\beta+\epsilon$

You have $n=60$ observations, and two predictor variables. You want to test

$H_0: \beta_1 = \beta_2$ in the model:

$Y = \beta_0+\beta_1X_1+\beta_2X_2+\epsilon$

Describe how you would accomplish this using the "black box"?

My professor has outlined changing

$H_0: \beta_1 = \beta_2$ to be $\theta = \beta_1 - \beta_2 = 0$, so $H_0: \theta = 0$.

Then rewrite the linear regression (scalar) form as:

$Y = \beta_0 + (\beta_2+\theta) X_1 + \beta_2 X_2 + \epsilon$

I don't understand where we take the algebra from here...

$\endgroup$
  • $\begingroup$ Collect your $\beta_2$ terms together, and get $\theta$ by itself... and you have something that looks like a regression model, but with a new "$X$" variable, a function of two variables you currently have. If you construct that new variable, you merely have to fit the model and test a coefficient. $\endgroup$ – Glen_b Jun 30 '13 at 22:31
  • $\begingroup$ Incidentally, this is 'standard bookwork' and should almost certainly have the self-study tag. You should also read its tag wiki info. I'd suggest replacing either of your last two tags which aren't especially relevant. $\endgroup$ – Glen_b Jun 30 '13 at 22:35
  • $\begingroup$ I'm still having a problem developing that linear model. Do we ultimately perform a t-test or an F-Test on the "Reduced" model ? I'm trying to reduce my model but am currently only at: Y = B0 + B2*[[Theta]*X1 + X2] + ei How do I get a new X out of this? Have I done the algebra wrong? $\endgroup$ – simplemts Jul 2 '13 at 18:41
  • $\begingroup$ Your algebra is wrong. You should not have any products of coefficients! Split the $X_1$ term into two terms. Then collect the two terms in $\beta_2$ together. Let $X_3 = X_1 + X_2$. You will have an ordinary regression equation. You do a t-test on the term which has $\theta$ by itself. $\endgroup$ – Glen_b Jul 2 '13 at 23:32
  • $\begingroup$ (ctd) ... However, the same test can also be done as an F-test, by comparing nested models, one with $\theta$ in it, the other without. $\endgroup$ – Glen_b Jul 2 '13 at 23:42
1
$\begingroup$

In the interests of making sure this question has an answer (and given the comments I made satisfied the OP):

$Y = \beta_0 + (\beta_2+\theta) X_1 + \beta_2 X_2 + \epsilon$

Split the $X_1$ term into two terms.

$Y = \beta_0 +\theta X_1+\beta_2 X_1 + \beta_2 X_2 + \epsilon$

Then collect the two terms in $β_2$ together.

$Y = \beta_0 +\theta X_1+\beta_2 (X_1 + X_2) + \epsilon$

Let $X_3=X_1+X_2$.

$Y = \beta_0 +\theta X_1+\beta_2 X_3 + \epsilon$

You have an ordinary regression equation.

You can do a t-test on the term which has $θ$ by itself. However, the same test can also be done as an F-test, by comparing nested models, one with $θ$ in it, the other without.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.