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

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

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 $i = 2,3,4$

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

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)

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 for instance by means of an F-test comparing the sum of squared residuals, which is also called analysis of variance (ANOVA).

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

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

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 $i = 2,3,4$