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Hello this is my first question on this site, but I frequent in Stack Overflow and Ask Different.

I self-teaching statistics using the book Statistical Methods: The Geometric Approach and I'm having a few difficulties applying geometry to statistical problems, specifically with calculating a test statistic as seen in the book's introduction. For the figure below, the author states that you can calculate the test statistic for a given sample - in this case the point $P(13,15)$ - by finding two lengths, A and B, and then finally taking the ratio $\frac{A}{B}$ to find your test statistic.

With the given example point, the author finds a ratio of: $\frac{14\sqrt{2}}{\sqrt{2}} = 14$. I am stuck on figuring out how the author found the lengths for A and B. I tried obvious tricks like Pythagorean Theorem and Law of Sines but I was not able to get anywhere. A point in the right direction with an explanation would be wonderful.

Context to question: context

Figure in question: Figure in question

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  • $\begingroup$ Always best to put to put all pictures and figures in the question, if you can, rather than use links. $\endgroup$
    – The Count
    Jul 15, 2019 at 22:09
  • $\begingroup$ I apologize, I don’t have enough rep on this site to embed photos. I was forced to post links. Perhaps someone can edit it for me so the pictures are directly embedded in the question. $\endgroup$
    – Tom Hood
    Jul 15, 2019 at 22:10
  • $\begingroup$ Oh, I didn't realize that was a thing. I'll do it for you. One sec. $\endgroup$
    – The Count
    Jul 15, 2019 at 22:36
  • $\begingroup$ I can't find the relevant info in the book (I don't own a copy) nor can I find a pdf after a minute or so of trying, sadly, but perhaps someone has a copy. This seems like a context issue. If you put a link to a picture of the whole page and any other relevant book stuff, I'll put that in for you, too. Or you could just quote the book directly. $\endgroup$
    – The Count
    Jul 15, 2019 at 22:43
  • $\begingroup$ I actually own the PDF, but I'm not sure if I'm "allowed" to upload the entire text. I have 4 pages that provide context to the problem, how would you suggest I upload it? $\endgroup$
    – Tom Hood
    Jul 15, 2019 at 23:45

1 Answer 1

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I believe I found an answer to my question.

Considering the vector, $v=[13\;15]^T$, which is represented in the 2 space figure in the question as $P(13,15)$, the value of A and B can be solved like so:

  • $A=14\sqrt{2}$ is found by solving for the length of the projection of $v$ onto the unit vector $U_1=\frac{1}{\sqrt{2}}[1\;1]^T$: $$A=v.U_1=\frac{13+15}{\sqrt{2}}=\frac{28}{\sqrt{2}}=14\sqrt{2} $$
  • $B=\sqrt{2}$ is found by solving for the length of the projection of $v$ onto the unit vector $U_2=\frac{1}{\sqrt{2}}[-1\;1]^T$: $$B=v.U_2=\frac{-13+15}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} $$

In both of these values, $\frac{1}{\sqrt{2}}[1\;1]^T$ is the unit vector at $45^\circ$ to the x-axis and y-axis - the equiangular line between the two axes. The vector $\frac{1}{\sqrt{2}}[-1\;1]^T$ is the perpendicular vector to the unit vector of the equiangular line.

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