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There was a question asked here around why the t-test is appropriate for hypothesis testing linear regression coefficients: Why is a T distribution used for hypothesis testing a linear regression coefficient?. The answers on the page focus on demonstrating that if you take the deviation of the estimated coefficient, $\hat{\beta}$ from the true coefficient, $\beta$ and then divide by the residual sum of squares (RSS), then the distribution of that number is a t-distribution.

But why should I take that for granted? Why use that particular test statistic and not another? Is that particular test statistic special? Is it uniformly most powerful (UMP) among its peers?

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    $\begingroup$ What don't you want to take for granted? It's unclear what you're asking about - after all, the math is correct, so... $\endgroup$
    – jbowman
    Jan 21, 2023 at 20:42
  • $\begingroup$ If you assume the specific test statistic, then the t distribution follows from it, sure. But why that particular test statistic? What if I want to use a different test statistic that I think will be better? For example, don't divide by residual sum of squares. Or take the cube of the deviation in the numerator. $\endgroup$
    – ryu576
    Jan 21, 2023 at 21:50
  • $\begingroup$ @ryu576, To make your query (+1) more general and interesting, you can wonder why the ANOVA test is used and how it is superior to any other test. $\endgroup$ Jan 31, 2023 at 6:19

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Why use that particular test statistic and not another?

The t-test coincidences with the likelihood-ratio test and therefore has as good properties in being a powerful test.

Is it uniformly most powerful (UMP) among its peers?

Yes if you consider an alternative point hypothesis (think of Neyman's and Pearson's theorems), but not necessarily for composite hypotheses.

See for example the difference in power of one-sided versus two-sided tests. None of these tests are dominant everywhere

power

Image from the question: https://stats.stackexchange.com/a/548241/164061

However, for a one-sided composite hypothesis, e.g. $H_0: \beta = 0$ versus $H_a: \beta > 0$ the t-test is UMP among unbiased tests. You can argue this by considering that of all hypothesis tests with a simple alternative hypotheses inside the region of the one-sided composite hypothesis, the t-test is the same and UMP.

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  • $\begingroup$ Sorry, it took me a while to get to this. I'm awarding the bounty, but have a follow up question - the most interesting part to me is the last line "You can argue this by considering that of all the hypothesis tests with a simple alternative hypotheses inside the region of the one-sided composite hypothesis, the t-test is the same and UMP". Could you please elaborate on this? $\endgroup$
    – ryu576
    Jan 28, 2023 at 9:40
  • $\begingroup$ @ryu576 You can regard the most powerful test when the hypothesis are point hypothesis $H_0: \theta = 0$ and $H_a: \theta = \theta_a$. For any value $\theta_a > 0$ on one side, the most powerful test is the same one-sided t-test, therefore it is UMP in the entire region and for the composite hypotheses $H_0: \theta = 0$ and $H_a: \theta > 0$ $\endgroup$ Jan 28, 2023 at 12:39
  • $\begingroup$ Ok, I'm happy to restrict attention to point hypotheses. For those, why is the test statistic corresponding to the t-test UMP? Do you have any pointers for what I should read so I can convince myself of this? $\endgroup$
    – ryu576
    Jan 30, 2023 at 6:39
  • $\begingroup$ @ryu576 The t-test is equivalent to a likelihood ratio test and the latter is most powerful (smallest type II error) among all tests with a same given type I error. Neyman and Pearson discussed this case and several others in the 30's. For references see this question about the Neyman Pearson lemma, which is used to prove the theorem that the likelihood ratio test is most powerful. $\endgroup$ Jan 30, 2023 at 7:16

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