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Here in the definition 5.3.11 on the page 249 they write "The $p$-value associated with a test is the smallest significance level $α$ for which the null hypothesis is rejected."

My question is what is the order here: $\bar{X}>\bar{x}$ or $\bar{X} <\bar{x}$ and why? I.e. when it is the case that $\alpha$ is the smallest significabnce level, if

$H_0:\mu=\mu_0$ versus $H_1:\mu>\mu_0$

OR on the other hand

$H_0:\mu=\mu_0$ versus $H_1:\mu<\mu_0$ ?

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    $\begingroup$ Is the PDF you link to a legitimate fair use? It would be better in any case to write the complete definition in the question here. $\endgroup$ – David K Oct 30 '18 at 13:06
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A p-value less than or equal to $\alpha$ means you reject the null hypothesis and have evidence that the mean reflects the alternate hypothesis. The alternate hypothesis can be either $H_a:\mu>\mu_0, \text{ or } H_a : \mu < \mu_0\text{ or simply } H_a : \mu\neq \mu_0$ depending on the context of the test.

For example, you have not tested the mean of a population for several years so you want to assess whether it has changed. Because you don't know whether it is larger or smaller, it is appropriate to use $H_a : \mu\neq \mu_0$. In another situation, you may suspect the mean has decreased (test scores seem to be lower than normal) in which case you would test using $H_a : \mu < \mu_0$.

Edit: One can never prove an $H_a$ with a single test but only provide evidence to support it. One can never discount the possibility of a type I error. This is the random chance of obtaining a p-value less than $\alpha$. There also exists the possibility of another variable affecting the result. But with more testing and control of other variables, one can be more confident in the result.

For $H_a:\mu>\mu_0$, this is a result in the right tail to the right of a Z score corresponding to $\alpha$ of a normal distribution $\text{ and } H_a : \mu < \mu_0$, in the left tail $\text{ and } H_a : \mu\neq \mu_0$ which can be in either tail.

enter image description here

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  • $\begingroup$ Why we cannot prove $H_a$ and can only reject $H_0$? A geometric intuition using the normal distribution curve and at the same time the $p$-value would help. $\endgroup$ – user122424 Oct 30 '18 at 18:33
  • $\begingroup$ See edit for additional information. $\endgroup$ – Phil H Oct 30 '18 at 23:57

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