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Given y = Bo +B1X1 +B2X2 +B3X3+ B4X4 + e

Question asks: Use general linear hypothesis to show how to test: a) Ho: B1=B2=B3=B4 b) Ho: B1=B2, B3=B4

Given the solution in the photo, I'm seeking an explanation to the solution to questions part a) and b) above.

Specifically, what is the reason for the matrix lay-out in this solution in both solutions a and b?

enter image description here

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  • $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Commented Sep 23 at 14:27
  • $\begingroup$ Is that better? $\endgroup$
    – Jaymie
    Commented Sep 23 at 14:40
  • $\begingroup$ Off the top of my head, I don't see why $\mathbf c = \{0, 0, 0\}$ isn't what we want in part (a) ? $\endgroup$
    – Ben Bolker
    Commented Sep 23 at 15:00
  • $\begingroup$ Math typesetting will make this easier to read. More information: math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – Sycorax ♦
    Commented Sep 25 at 0:44

2 Answers 2

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Most likely, there's a typo in the solutions you have (for one thing, the matrices shape are not coherent, and $\beta$ in the $c$ matrix is undefined).

For part a), $c$ should be $\{0,0,0\}$, so that you are testing $\beta_1 - \beta_2 = 0$ and $\beta_2 - \beta_3 = 0$ and $\beta_3 - \beta_4 = 0$, which is another way of writing $\beta_1 = \beta_2 = \beta_3 = \beta_4$

For part b), you are testing $\beta_1 - \beta_2 = 0$ and $\beta_3 - \beta_4 = 0$

EDIT : for further explanation, if you do the matrix product Tβ, for instance in case a), you have $\beta_0 \times 0+\beta_1 \times 1+\beta_2 \times (−1)+\beta_3 \times0+\beta_4 \times0$ for the first coefficient, then
$\beta_0 \times0+\beta_1 \times 0+\beta_2 \times 1+\beta_3 \times(−1)+\beta_4 \times0$ for the second coefficient, then
$\beta_0 \times0+\beta_1 \times 0+\beta_2 \times 0+\beta_3 \times1+\beta_4 \times(−1)$ for the last coefficient

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  • $\begingroup$ That's very interesting. It isn't the first "typo" I've come across in this book and solutions set which, frankly, I'm not impressed with. Can you explain why the matrices are laid out in the solution the way they are and the values 0,0,1,00 etc. It's just not clicking with me. $\endgroup$
    – Jaymie
    Commented Sep 23 at 15:24
  • $\begingroup$ If you do the matrix product $T\beta$, for instance in case a), you have $\beta_0 \times 0 + \beta_1 \times 1 + \beta_2 \times (-1) + \beta_3 \times 0 + \beta_4 \times 0$ for the first coefficient, then $\beta_0 \times 0 + \beta_1 \times 0 + \beta_2 \times 1 + \beta_3 \times (-1) + \beta_4 \times 0$ for the second coefficient, then $\beta_0 \times 0 + \beta_1 \times 0 + \beta_2 \times 0 + \beta_3 \times 1 + \beta_4 \times (-1)$ for the last coefficient $\endgroup$
    – afloy
    Commented Sep 23 at 15:31
  • $\begingroup$ Thank you. I think that gets me on my way. $\endgroup$
    – Jaymie
    Commented Sep 23 at 16:54
  • $\begingroup$ @afloy, can you add that comment to your answer? $\endgroup$
    – Ben Bolker
    Commented Sep 23 at 17:04
  • $\begingroup$ Thank you for all that contributed. I did some reading and checking out YouTube videos and was able to use those and the information provided here to piece together this puzzle. It was interesting to see that different people had different approaches to this problem. @Ben Bolker, in all my research, one thing was consistent and that is that the Matrix C = {0,0,0} -- the solution book is incorrect. What a disappointment, especially when one is doing self-study. $\endgroup$
    – Jaymie
    Commented Sep 24 at 15:52
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In case anyone is interested in the gory details, this is where I landed with the question.

enter image description here

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