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

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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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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