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Let's say we want to test the following hypotheses:

$H_0: \mu = 0$ $H_1: \mu > 0$

for a random sample $\{X_1, \dots , X_n\}$ that is normally distributed $X_i \sim N(\mu, 25)$ (so variance is known).

Why is it that we fail to reject $H_0$ if $\bar{X} = -300$ for example? What does 'fail to reject' even mean in a one-sided test? What would be the meaning of such a p-value even?

Darby Bond
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    It means the same as usual, that you do not have the necessary evidence to reject the null in favor of the alternative. Does something about this bother you? – Dave Jul 03 '20 at 21:45
  • Yes, it seems preposterous that we do not have enough to reject the null, even if the data does not support the alternative. – Darby Bond Jul 03 '20 at 21:46
  • +1 For Dave. Also: your null hypothesis is poorly specified for a one-sided test (it should be what the alternative hypothesis is the complement of; so $H_0: \mu \leq 0$ for $H_1 : \mu > 0$). See for example, my answer (and the brief exchange in the comments) here. – Alexis Jul 03 '20 at 21:48
  • @Alexis I see why working with a complement helps with interpretation but if you specify your one sided test with a null that is a range and not a point value (like $\mu = 0$), how do you arrive at a test statistic? – Darby Bond Jul 03 '20 at 21:52
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    Null dist'n of test statistic is based on the $=$ part of $\le$ in $H_0.$ – BruceET Jul 03 '20 at 22:47
  • @DarbyBond Just as BruceET says. – Alexis Jul 03 '20 at 22:53
  • But @BruceET, isn't that a bit arbitrary? If I use $\mu = -300$ for example, in the null distribution, I can argue that I am still testing for $H_0$ ( since $-300 \leq 0$) but I get a different test statistic than using $\mu = 0$. Isn't this a problem? – Darby Bond Jul 03 '20 at 22:57
  • Not sure I follow 'arbitrary': $H_{0} : \mu \leq -300$ implies $t = \frac{\overline{X} - -300}{s_{\overline{X}}}$. Does that make sense? Or are you asking why any value in the range under $H_0$ is not used to create the test statistic? In general if $H_{0}: \mu \leq C$, then $t=\frac{\overline{X}-C}{s_{\overline{X}}}$, and $p = P(T_{df} > t)$ (Flip the inequality sign for $p$ if you also do so for $H_{0}$). – Alexis Jul 03 '20 at 23:00
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    The basic problem here, and perhaps the reason for mutual misunderstanding, is that by limiting the possible hypotheses to $\mu\ge 0$ you are making an assumption under which the outcome $\bar X=-300$ is astronomically unlikely. That calls into question your entire approach, which in light of this outcome is invalidated, and so the issue of what it might mean to reject or not reject $H_0$ becomes irrelevant. – whuber Jul 04 '20 at 15:15
  • Hi @whuber. I absolutely agree with that. So if you have a test where you get an outcome like $\bar{X} = 300$ when you weer testing for $\mu = 0$ you must question whether you were conducting the right test in the first place. – Darby Bond Jul 04 '20 at 15:23
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    An alternative approach--and likely the one assumed by many readers--is that the problem isn't with the test but instead lies in making an overly narrow assumption. One-sided tests are therefore usually interpreted as $H_0:\mu\le 0$ vs $H_1:\mu\gt 0,$ thereby removing the artificial (and in this case counterfactual) limitation to $\mu\ge 0$ without changing the test or its outcome at all. – whuber Jul 04 '20 at 15:34

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