As the title says, my question is can the null hypothesis ($H_0$) be more true than the alternative hypothesis ($H_1$) or vice versa?

For example, if the statistical power is really high (extremely let us say $0.99$) then by the definition of statistical power, the type 2 error is $0.01$, which means that we will reject $H_1$ (false null hypothesis) almost always, i.e., $H_0$ is more likely to be true.

But does it mean that $H_0$ is more true than $H_1$? It sounds a bit odd to me, so I ask this question here (sorry if this question is not a proper question in here).

If we cannot say "$H_0\;/\;H_1$ is more true than $H_1\;/\;H_0$", can you explain why we can't say this?

  • $\begingroup$ Within a Bayesian framework, saying that $H_0$ is more probable is perfectly fine if you are willing to use a prior assigning a point mass to the point null hypothesis, see stats.stackexchange.com/a/442587/77222 $\endgroup$ Nov 25, 2020 at 21:23
  • $\begingroup$ Hi, I marked it as a duplicate of two questions that ask very similar, but differently stated questions. They seem to answer your question. If this is not the case, please edit your question to clarify what was not answered by the linked threads. $\endgroup$
    – Tim
    Nov 25, 2020 at 21:31
  • $\begingroup$ Statistical power does not mean what you claim it says, so reviewing that concept would be one good place to start getting a good answer to this question. $\endgroup$
    – whuber
    Nov 25, 2020 at 21:58
  • $\begingroup$ George Box wrote, famously and correctly: "All models are wrong but some are useful." Neither the $H_0$ nor the $H_1$ will ever be really true, and as such none of them will be more true than the other. The best that we can ever say is that the data is more compatible with the $H_0$ way of thinking about reality than with the $H_1$ way. Note by the way that normally the $H_1$ isn't a single distribution but a continuous set, and a test will normally have high power against distributions in $H_1$ far from $H_0$ but low power against those close to $H_0$. $\endgroup$ Nov 25, 2020 at 22:00


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