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 $4\times1$ 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.
 A: 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
$$T =
    \begin{pmatrix}
    0 & 1 & -1 & 0 & 0 \\
    0 & 1 & 0 & -1 & 0\\
    0 & 1 & 0 & 0 & -1 \\
      \end{pmatrix}
$$
$$ \mathbf{\beta} =
    \begin{pmatrix}
    \beta_0\\
    \beta_1\\
    \beta_2\\
    \beta_3\\
    \beta_4
    \end{pmatrix}
$$
A: Compare 


*

*model 1 which assumes all the coefficients for $x_1$, $x_2$, $x_3$, and $x_4$ are equal $$y = \beta_0 + \beta(x_1+x_2+x_3+x_4) + \epsilon$$
versus


*

*model 2  which assumes all the coefficients for $x_1$, $x_2$, $x_3$, and $x_4$ are free (not equal) $$y = \beta_0+\beta(x_1+x_2+x_3+x_4) + \beta_2^\prime x_2 + \beta_3^\prime x_3 + \beta_4^\prime x_4 + \epsilon$$ or differently parameterized 
$$y = \beta_0+ \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3 + \beta_4 x_4 + \epsilon$$

You can do this comparison, for instance, by means of 


*

*an  F-test comparing the sum of squared residuals, which is also called analysis of variance (ANOVA). (as jbowman noted in the comments)

*In your case you seem to want to do three t-tests $$\begin{array}{rcl} \beta_i^\prime &=& \beta_i-\beta \\ &=&  \beta_i-\beta_1 \\ &=& 0 \end{array}$$ for $i = 2,3,4$
I think that you need to make sure that the $\beta_i^\prime$ are independent. (or, I do not really know what this T-matrix and the method applied to it is)
