It seems to me that, after having a predetermined level of significance, the Neyman-Pearson approach only requires to see if the observed x lies in the rejection region or not. In this case, how to relate it to p-value? Or is it an alternative way to judge if x is rejected or not by using p-values? After reading some references about NP and Fisher's treatment, I am confused.

Thank you for any suggestion.

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    $\begingroup$ It may help you to read my answer here: When to use Fisher and Neyman-Pearson framework? As for how the rejection region relates to p-values, the RR is defined as that region where $p<\alpha$. $\endgroup$ – gung Oct 22 '13 at 1:51
  • $\begingroup$ It is an excellent explaniation. So under Neyman's setting, does it requre any property of T(x), so that p<α is equivalent to x belongs to the rejection region? $\endgroup$ – user28363 Oct 22 '13 at 1:58
  • $\begingroup$ I'm not sure I totally follow your question, but there is no property of T(x) being in the rejection region except that p<a. $\endgroup$ – gung Oct 22 '13 at 2:04
  • $\begingroup$ My question is that, how does p<alpha guarantee that x is rejected. Because Neyman-Pearson does not necessrily require p-value in the procedure, so I think this p-value has to be defined first. $\endgroup$ – user28363 Oct 22 '13 at 3:55
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    $\begingroup$ Yes, p<a guarantees that x is rejected. The N-P approach is based on maintaining a maximum type I error rate, so alpha comes 1st, you figure out what T(x) is when p=a, & that defines the rejection region. $\endgroup$ – gung Oct 22 '13 at 4:15

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