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I've been reading around about hypothesis testing. I don't understand why the following one sided tests are equivalent:

$H_0:\mu \leq \mu_0$; $H_a:\mu > \mu_0$

and

$H_0:\mu = \mu_0$; $H_a:\mu > \mu_0$

Any thoughts?

Edit: I think I understand why they are equivalent. Anything that's rejected by the second hypothesis will be also rejected by the first (at least for Z tests and T tests you learn about in a first course in stats). Maybe a better question to ask is -- are there scenarios where these two inferences are not equivalent?

yoshi
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    It might help to recall the definition of a p-value for a composite null hypothesis as the *maximum* p-value over all constituent simple nulls: see [Is the p-value still uniformly distributed when the null hypothesis is composite?](http://stats.stackexchange.com/q/58929/17230). So you have to give the null its best shot, & in some cases - including the usual z tests & t-tests for the mean of normally distributed observations - how to do so is quite clear. – Scortchi - Reinstate Monica Apr 08 '16 at 15:35
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    Sorry, I don't have enough reputation, thus a useful link as answer: http://www.quantdec.com/envstats/notes/class_13/tests.htm See also [here](https://stats.stackexchange.com/q/45825/209028) – Christoph May 24 '19 at 11:29

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I do not think those are equivalent and in fact I believe one of them

H0:μ=μ0; Ha:μ>μ0 is incorrect.

Philosophically, the 'rules' for forming the Ho and the Ha are that they be a-mutually exclusive and b-exhaustive, and so I think technically that form of the null is incorrect because it's not exhaustive (e.g. it omits the result in which (using your single sample example) the obtained mean is actually significantly lower).

Pragmatically, you are correct there aren't any cases where anything that's rejected by the second version of the hypothesis wont be also rejected by the first version - because the critical value for rejection region would go in the tail corresponding to the alternative hypothesis, leaving the entire other part of the distribution in the zone of the null. But the fact that the practical implication is invariant doesn't make the expression of the hypothesis correct (for the reason stated above, that it fails one of the rules of hypothesis formation).

Marina_ANOVA
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