Null hypothesis stated as "there is an effect"

Question

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.

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]