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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...

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  • $\begingroup$ 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
    Commented Jun 30, 2013 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
    Commented Jun 30, 2013 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
    Commented Jul 2, 2013 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
    Commented Jul 2, 2013 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
    Commented Jul 2, 2013 at 23:42

1 Answer 1

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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.

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