It is common to see $H_0$ stated as the inexistence of a difference between two data groups. The rejection of this hypothesis is always in favor of the alternative one that there is a difference. This makes a lot of sense, beginning from the fact that the name "null" suggests exactly this. However, I have the doubt: what would be the complications of stating $H_0$ as there being an effect? Could it be used to gather evidence in favor of the alternative hypothesis that the difference is 0?

I have a sense of why this would the answer to the latter question would be "no" (starting from the fact that we cannot prove a negative), but I'm looking for a more rooted response.

  • 1
    $\begingroup$ You can have a simple null hypothesis that the effect is $4$ or whatever. Or a composite null hypothesis that the effect is $\le 4$, or one with effect $\ge 4$ or one with the effect in the interval $[-4,4]$. What would usually make no sense is a null hypothesis that the effect is $\not=0$ if it would be difficult or impossible to reject such a hypothesis $\endgroup$
    – Henry
    Nov 9, 2021 at 22:27
  • $\begingroup$ Does this answer your question? While I addressed correlation, it could be any other parameter. // Note however that the null does not have to be zero. Your null could be $H_0: \theta = 4$, for example $\endgroup$
    – Dave
    Nov 9, 2021 at 22:32
  • $\begingroup$ A number of posts on site address this question. $\endgroup$
    – Glen_b
    Nov 10, 2021 at 2:08
  • $\begingroup$ Some other similar posts, of the many: stats.stackexchange.com/questions/264614/…, stats.stackexchange.com/questions/179527/…, stats.stackexchange.com/questions/444298/… $\endgroup$ Nov 10, 2021 at 14:53
  • $\begingroup$ The op's question is a great one. If we are testing a screw in a bridge for safety, when a faulty screw would mean that the portrait of the president who inaugurated the bridge would fall to the sea eventually, proceed in the usual way, i.e., we want to test whether the screw is faulty assuming that it is not. But if the failure of the screw implies that the bridge would collapse, then the null has to be that the screw is faulty, not the reverse. And likewise since there is so much clinical verbiage these days, a new procedure/treatment must be tested as if it were 'bad', and its goodness must $\endgroup$ Nov 16, 2021 at 23:26

1 Answer 1


You are, perhaps, thinking about a 'negativist' null hypothesis of the general form $\text{H}_{0}^{-}\text{: }|\theta|\ge \Delta$, with $\text{H}_{\text{A}}^{-}\text{: }|\theta|< \Delta$, as used in tests for equivalence, and where $\Delta$ is the smallest effect size that you care about a priori to the test. (As expressed here, the equivalence region ($-\Delta, \Delta$) is symmetric, although it need not be. For example, one could express an asymmetric equivalence range $\text{H}^{-}_{0}\text{: }\theta \le \Delta_{\text{lower}}\textbf{ OR } \theta \ge \Delta_{\text{upper}}$, where $|\Delta_{\text{lower}}| \ne \Delta_{\text{upper}}$. One place this is sometimes done is when $\theta$ measures a relative difference like an odds ratio or relative risk, and where $\Delta_{\text{lower}} = \frac{1}{\Delta_{\text{upper}}}$.)

In plain language, for a one-sample or two-sample test the null is "There is an effect of magnitude at least as large as $\Delta$, and the alternative is "The magnitude of the effect is less that $\Delta$." The omnibus form of this null hypothesis would be "There is an effect of magnitude at least $\Delta$ between every group."

Note: For a continuous distribution we cannot simply invert the null hypothesis from the two-sided test for difference (i.e. $\text{H}_{0}^{+}\text{: }\theta = 0$) to be $\text{H}_{0}^{+}\text{: }\theta \ne 0$, because we would have to find (probabilistic) evidence in favor of $\text{H}_{\text{A}}\text{: }\theta = 0$, but the probability of a continuously distributed variable (e.g., normal, $t$, etc.) exactly equaling a specific number is 0 (i.e. $P(X = c) = 0$), and so you would never reject such an inverted null (i.e. $\text{H}_{0}\text{: }\theta \ne 0$ is no good).

Selected References

Anderson, S., & Hauck, W. W. (1983). A new procedure for testing equivalence in comparative bioavailability and other clinical trials. Communications in Statistics—Theory and Methods, 12(23), 2663–2692.

Reagle, D. P., & Vinod, H. D. (2003). Inference for negativist theory using numerically computed rejection regions. Computational Statistics & Data Analysis, 42(3), 491–512.

Wellek, S. (2010). Testing Statistical Hypotheses of Equivalence and Noninferiority (Second Edition). Chapman and Hall/CRC Press.

See also: [equivalence], equivalence test, [tost]

  • 1
    $\begingroup$ This type of testing is also used in futility analyses in clinical drug development. Rather than looking for strong evidence to reject a null hypothesis of no treatment effect (or low end-of-study power), we retain a null hypothesis that the treatment does work (or there is sufficient end-of-study power) based on weak evidence to the contrary. $\endgroup$ Nov 9, 2021 at 22:57
  • $\begingroup$ @GeoffreyJohnson Yup! A really cool heritage of clinical/pharmacological biostats. Propagating to other statistical sciences now as well... seeing it in psychology, I published with it in Demography, and PLoS ONE, and teach my biostats students to help avoid confirmation bias by combining inferences from both tests for difference and tests for equivalence (i.e. to form 'relevance test' conclusions). $\endgroup$
    – Alexis
    Nov 9, 2021 at 23:00
  • $\begingroup$ Say, @GeoffreyJohnson I wonder if you might be a good person to have a go at answering my question Is there a generalized concept of noncentrality of a distribution?? There are some answers there which are helpful, but none really land it for me (I have Fisher's original derivation of the non-central $t$ on hold with ILL… need to go pick that up… maybe it will help :). $\endgroup$
    – Alexis
    Nov 9, 2021 at 23:14
  • 1
    $\begingroup$ Thank you, Alexis! Pretty clear. $\endgroup$ Nov 18, 2021 at 20:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.