Solved – Testing the general linear hypothesis: $H_0: beta_1 = beta_2 = beta_3 = beta_4 = beta$

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Again, we are testing the linear hypothesis;

$H_0: beta_1 = beta_2 = beta_3 = beta_4 = beta$

for the model,

$$y = beta_0 + beta_1x_1+beta_2x_2+beta_3x_3+beta_4x_4+epsilon$$

I know how to solve this for testing if two betas are equal but I don't quite understand the equality of all four. I imagine this is simpler than I'm envisioning but I can't seem to piece it together.

I know I can develop a matrix $T$ of 1's and 0's and multiply it by my vector of prediction coefficients $mathbf{beta}$

The problem is that I'm not entirely sure how to correctly construct the T matrix to test this hypothesis. More specifically, I'm not confident in the output vector.

Here is what I did:

$$T =
begin{pmatrix}
0 & 1 & 0 & 0 & 0 & -1 \
0 & 0 & 1 & 0 & 0 & -1 \
0 & 0 & 0 & 1 & 0 & -1 \
0 & 0 & 0 & 0 & 1 & -1 \
end{pmatrix}
$$

$$ mathbf{beta} =
begin{pmatrix}
beta_0\
beta_1\
beta_2\
beta_3\
beta_4\
beta
end{pmatrix}
$$
NOTE: The $beta$ to the left of the equality should be bold. I'm not implying an equality between the left side of the equality with the sixth element in the beta vector.

When I multiply these I get a $4times1$ vector of 0's. Is this the proper way to set up the test? I don't need to actually test this hypothesis. I just need to properly setup T and beta. Thanks in advance.

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Best Answer

Based on your comment, you do not have idea what four $beta$s should be, and just want to test if they are the same.

It equals to $beta_1 = beta_2 = beta_3 =beta_4$, and can be write in different ways. One of them is:

$beta_1 = beta_2$

$beta_1 = beta_3$

$beta_1 = beta_4$

Based on these 3 equations, the $T$ matrix is $$T = begin{pmatrix} 0 & 1 & -1 & 0 & 0 \ 0 & 1 & 0 & -1 & 0\ 0 & 1 & 0 & 0 & -1 \ end{pmatrix} $$

$$ mathbf{beta} = begin{pmatrix} beta_0\ beta_1\ beta_2\ beta_3\ beta_4 end{pmatrix} $$

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