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It seems that if a one-sided Z-test returns a p-value $>1 - \alpha$, you have enough evidence of absence to reject the alternative hypothesis in favor of the null hypothesis, because if you were to swap the null and alternative hypotheses and repeat the test, that p-value would be $1 - $ the original, and hence $<\alpha$. This conclusion may need to be slightly modified for hypothesis tests with asymmetric distributions, but with some transformation should still hold.

In contrast, it is hard to see how a two-sided Z-test such as $H_0: \mu = 0, H_A: \mu \neq 0$ could ever provide any effective evidence of absence, because it compares an infinitesimal slice of the number line to two infinite intervals. As tests return very high p-values, evidence of absence will shrink the infinite intervals to finite intervals still sandwiching the null hypothesis, but the null hypothesis will always remain infinitesimal in comparison to the alternative hypothesis.

Is this conclusion correct? Can extraordinarily high p-values establish a null hypothesis for a one-sided test but not a two-sided test?

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    $\begingroup$ How do you swap null and alternative hypotheses? $\endgroup$
    – T.E.G.
    Commented Oct 25, 2023 at 5:35
  • $\begingroup$ @T.E.G. For example, if the original test was $H_0: \mu > 0, H_A: \mu \leq 0$, you would perform the test $H_0: \mu \leq 0, H_A: \mu > 0$. $\endgroup$
    – user10478
    Commented Oct 25, 2023 at 16:53
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    $\begingroup$ Swapping these hypotheses and having a null statement without some form of equality ($H_0 : \mu > 0$) is not straightforward. See this question. $\endgroup$
    – T.E.G.
    Commented Oct 25, 2023 at 17:25
  • $\begingroup$ @T.E.G. Interesting, the main answer at the link is a bit beyond my ability to entirely comprehend. Is this concern mainly a formality (like how the partial derivatives $u_{xt}$ and $u_{tx}$ are almost always interchangeable but there are exotic exceptions in advanced contexts), or does it actually lead to meaningfully wrong answers in practice? $\endgroup$
    – user10478
    Commented Oct 25, 2023 at 19:31
  • $\begingroup$ I am not sure. This issue is related to the composite hypotheses and how to test them. As far as I see, this has its own complications. For example, see this answer. $\endgroup$
    – T.E.G.
    Commented Oct 25, 2023 at 21:29

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The comments link to a rather abstruse mathematical argument. The OP said it is "beyond my ability to entirely comprehend" and I agree.

But, if you want to "switch hypotheses" or show evidence of equality rather than difference, their are methods called tests of equivalence. The simplest and most common is TOST - two one sided t-tests (I have linked to its tag, for you to browse). In particular, see this thread

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