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I encountered this problem:

Suppose $Y$ has the following PDF $$f(y)=\frac c{(1+y)^2}\cdot I(0\le y\le \theta),\quad\theta>0.$$ A single value $y$ is observed from the distribution.
Derive a formula for a suitable $p$-value for testing the null hypothesis $H_0:\theta = k$ where $k$ is a specified positive constant, against the alternative hypothesis $H_1: \theta>k$.

The answer turned out to be $p=P(Y>y)$.

The reason went like:

If $y$ is large then this is evidence in favour of the alternative hypothesis.

However, according to a post I just read, smaller $p$-values are in favour of $H_0$. In this design, $p$ being smaller means $y$ being larger, which is evidence in favour of $H_1$ i.e. against $H_0$.

What goes wrong?

Vim
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    smaller $p$-values are AGAINST $H_0$. Generally, we say that p-value is too small( < 0.05 or 0.01), so we reject null hypothesis. – user158565 Jun 03 '17 at 02:38
  • @a_statistician ok.. it seems that if $p – Vim Jun 03 '17 at 02:43
  • I think you need to read that link carefully. They said the smaller p values are more convincing for rejecting H_0. – user158565 Jun 03 '17 at 02:48
  • @a_statistician yeah you're right. They're talking about smaller p value implying more convincing evidence *against* the null. Thanks! – Vim Jun 03 '17 at 02:51
  • That's how hypothesis tests work... – Glen_b Jun 03 '17 at 04:24
  • @Glen_b I think I got it now. By the way is there some established consensus on how to choose the right $p$-value? In this example, we choose it to be the right tail to $y$ (of course it's in order to make smaller $p$ in favour of $H_1$. But why can't I choose, say, $p=P(Y>y+1|\theta=k)$? Although I understand that would lead to $p(y=k+1/2)\ne 0$ which seems ridiculous. – Vim Jun 03 '17 at 04:35
  • You ***don't*** choose a p-value. That's something you calculate. It's determined by the null and alternative hypotheses, the test statistic and the data. You should have your hypotheses and test statistic before you have any data, or you're likely to be subject to claims of p-hacking (i.e. cheating, whereupon your p-values mean nothing). What you're talking about there sounds exactly like p-hacking. – Glen_b Jun 03 '17 at 05:45
  • @Glen_b because the question asked me to find a *suitable* $p$-value making me wonder there are many different ways to define a $p$-value (by which I don't mean we choose a particular value but the way p value depends on TS since p value is just a function of TS). It is not like p hacking, since i defined it prior to any observed data. – Vim Jun 03 '17 at 05:48
  • Do you have a test statistic and a rejection rule? – Glen_b Jun 03 '17 at 06:30
  • @Glen_b in this specific problem, the TS should be $y$, and the rejection rule I think is up to us to design. – Vim Jun 03 '17 at 08:58
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    The purpose of my post at https://stats.stackexchange.com/a/130772/919 was precisely to answer this question: it outlines the principles and concepts needed to decide what the critical region ought to be for computing a p-value. – whuber Jun 03 '17 at 15:08
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    @whuber thank you for this wonderful answer, insightful as well as amazingly interesting. – Vim Jun 03 '17 at 16:14

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