Suppose that we are facing a hypothesis: $$H_0: \mu\geq 0 \quad vs \quad H_1: \mu\geq c$$ where $\mu$ is the parameter of interest and $c$ is a unknown paraneter.

Here, we only have the distribution of an estimator for $\mu$.

Is there any testing method that can be applied to this situation?

  • 1
    $\begingroup$ The big problem here is that if $\mu\geq c$, both these hypotheses are true. Rather than two mutually exclusive hypotheses, $H_1$ is simply nested within $H_0$. You can hardly reject $H_0$ in favour of $H_1$ when $H_1$ being true guarantees $H_0$ being true. $\endgroup$
    – Glen_b
    Commented Sep 22, 2022 at 0:51
  • $\begingroup$ I guess you are considering the case where $c$ is positive. I made some misleading points. I am thinking about the case where $c$ is negative so that $H_0$ does not nest $H_1$. How about this case? $\endgroup$
    – M.C. Park
    Commented Sep 22, 2022 at 1:35
  • 2
    $\begingroup$ I was indeed assuming $c$ to be positive. But if it is negative, $H_0$ is nested within $H_1$; this is less of a problem (in that now there are at least some values of $\mu$ for which you would wish to reject $H_0$ in favour of $H_1$, those in the set-difference), but there's still the basic logical/philosophical issue of what you want to have happen when both hypotheses are true. Is there a practical situation you're considering? If so there may be clues in it to a better approach. $\endgroup$
    – Glen_b
    Commented Sep 22, 2022 at 1:39
  • $\begingroup$ Thank you for the comment. I did not realize that both $H_0$ and $H_1$ could be true if $c$ is negative. Thank you for your contribution. $\endgroup$
    – M.C. Park
    Commented Sep 22, 2022 at 1:47
  • 1
    $\begingroup$ If $c$ is $-2$, say, then $\mu\geq c$ includes all the values on the number line to the right of (i.e. greater than) $-2$ (along with $-2$ itself). Clearly that includes $0$ and all values greater than $0$. Did you intend $H_1: \mu\leq c$ for some $c<0$? If so, perhaps my original attempt at an answer might be resurrected. $\endgroup$
    – Glen_b
    Commented Sep 22, 2022 at 1:48


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