Recently we've been learning hypothesis testing in class, and I have an idea about incorporating probability integral transformation into it, which I think have some decent advantages over the current methodology. Let me explain it with an example:

Consider a trivial one-sample $z$-test, which has $H_0: \mu \leq \mu_0$ versus $H_1: \mu > \mu_0$. If we take the conventional path, a good test statistic would be

$$z = \frac{\bar x - \mu_0}{\sigma / \sqrt{n}}$$

with the region of rejection being

$$z \geq z_{1-\alpha}$$

Now, my idea is to use the following test statistic

$$\mathbb{z} = \Phi \left( \frac{\bar x - \mu_0}{\sigma / \sqrt{n}} \right)$$

where $\Phi$ is the CDF of the standard normal distribution (which is the distribution of $z$). Then the region of rejection will become

$$\mathbb{z} \geq 1-\alpha$$

This is because $\mathbb{z}$ has a uniform distribution on the interval $[0, 1]$.

Basically, I'm feeding whatever test statistic into the CDF of its distribution, and the resulting test statistic would have a standard uniform distribution (here is why). In my humble opinion, this method has several advantages:

  • The region of rejection in most hypothesis testing problems (including $z$-test, $t$-test, $\chi^2$ test and $F$-test, among others) would be similar, i.e. $[0, \alpha]$, $[1-\alpha, 1]$, or $[0, \frac{\alpha}{2}] \cup [1-\frac{\alpha}{2}, 1]$.
  • The calculations of $p$-values are also greatly simplified. For example, it's simply $1 - \mathbb{z}$ in this case.
  • Since CDFs are non-decreasing, the "shape" of the region of rejection won't change.
  • With modern technologies like MATLAB or R, calculating $\mathbb{z}$ is as convenient as, if not even more convenient than, finding $z_{1-\alpha}$.

I know this is by no means a huge improvement over the current implementation, but I think it can make hypothesis testing easier. Someone must have thought about the same method before I do, so why isn't it widely used?

  • $\begingroup$ Well, as you point out, that is tantamount to using p-values, so it is quite widely used. $\endgroup$ Commented May 14, 2018 at 9:40

1 Answer 1


I could be mistaken, but it looks to me like you're literally describing the procedure for calculating the p-value corresponding to a z-statistic (your new $z$ being equal to $1-p$). I need hardly point out that p-values are indeed widely used.

Transforming test statistics (e.g. $F$, $t$, $\chi^2$, etc.) to p-values is useful because it renders them into a common space with an intuitive interpretation, from which you can directly draw inferences about your hypotheses. But it's also useful to retain and report the original statistic, because it has a bit more information about how you arrived at your conclusion, and thus is more transparent to your audience.

  • $\begingroup$ so I reinvented the $p$-value? sounds cool :) $\endgroup$
    – nalzok
    Commented May 14, 2018 at 11:38

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