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I looked online everywhere on how to use F tests for compound hypothesis testing, but didn't get any instructions on the formula for F testing when $H_0: B_1 = -1$ and $B_2= 0$. enter image description here

I don't need anyone giving me the answers, but I just need to get the formula for doing the F test and likelikhood ratio test (didn't get this from class). Thanks

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You have to compare between two models: (1) Restricted and (2) Full. Where the second model in the picture is the restricted as it omits $\beta_2$ and restricts $\beta_1$ to equal $-1$. $F$ test is defined as follows $$ F = \frac{SSR_r - SSR_F}{SSR_F}\times\frac{n-k}{r}, $$
where $r$ is the number of restricted parameters ($2$ in your case), and $SSR_r$ is the SSR of the restricted model ($0.6788$) and $SSR_F$ of the unrestricted ($0.4277$). $n$ is the number of observations ($22$) and $k$ is the total number of coefficients in the full model ($5$). The calculated value you have to compare to the the $1-\alpha$-th quantile of $F$ distribution with $(r, n-k)$ df.

Where the LR test is defined as $$ LR = \frac{n}{2}\left(\ln(SSR_r) - \ln(SSR_F) \right), $$ such that the computed value you have to compare to the $1-\alpha$-th quantile of $\chi ^2(r)$ distribution.

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  • $\begingroup$ I thought it was $n-(k+1)$ df? Also what is p? Is that supposed to mean k? Also- my notes has the equation for LR as N * the differences, not $n/2$ $\endgroup$ Commented Mar 10, 2017 at 23:25
  • $\begingroup$ 1. If $k$ is only the number of $\beta$ then yes, however in coefficients I meant to include the $\alpha$ as well. 2. I vaguely recall that for the Wald test is $n$ and for the LR is $n/2$, but I'll check it later to be sure. $\endgroup$
    – V. Vancak
    Commented Mar 10, 2017 at 23:48
  • $\begingroup$ Thanks for clarifying, we're pretty much on the same page then. I know how to do these tests, I was just confused because I thought you could only do them for B=0. But you made me realize the new equation put in the restricted model, which allowed me to do the tests- right? $\endgroup$ Commented Mar 12, 2017 at 18:21
  • $\begingroup$ Yes, that's right. $\endgroup$
    – V. Vancak
    Commented Mar 12, 2017 at 19:10
  • $\begingroup$ Hey, hope things are well. I have some more questions about heteroskedasticity/autocorrelation and would really appreciate your help/advice: math.stackexchange.com/questions/2221566/… $\endgroup$ Commented Apr 6, 2017 at 23:03

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