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

Question

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.

Answer 1

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

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

Answer 2

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 $$y = \beta_0+ \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3 + \beta_4 x_4 + \epsilon$$

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

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

2. 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)