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Usually with simple hypotheses I will have something like

$$H_0: \beta_1 = 0 | H_A: \beta_1 \ne 0$$

But suppose I have a null hypotheis

$$H_0: \beta_1 = \beta_2 = \beta_3 = 0$$

Question

What is the alternative hypothesis? Is there an assumed one or could there be multiple plausible one's and it is up to the tester to specify?

My econometrics professor is super hand wavy and just said $\text{Not} H_0$ was the alternate. But that seems ridiculous to me.

Possibility 1:

$$H_A: \beta_1 = \beta_2 = \beta_3 \ne 0$$ Possibility 2:

$$H_A: \beta_1 \ne 0, \beta_2 \ne 0, \beta_3 \ne 0$$

There are obviously more possibilities, but these are enough to illustrate my point.

But these are like completely different statements. Does this mean I have to specify the alternate hypothesis and there isn't a given/assumed one unlike simple hypotheses? Does this just mean my professor did a bad job?

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    $\begingroup$ I think you missed a few possibilities there: eg $\beta_1 = \beta_2 = 0, \beta_3 \ne 0$ $\endgroup$ – Kitter Catter May 26 '16 at 1:59
  • $\begingroup$ Yes, you have to specify the alternate. Indeed, the complement of the null is determined by the set of all distributions you are considering, so your professor merely pushed the answer one step back without saying anything informative: you could have replied, "complement of the null in which space of distributions?" The dialog at stats.stackexchange.com/a/130772/919 illustrates some of the thought processes that go into specifying a null and alternate hypothesis. It purposely uses a setting in which the alternate is unconventional. $\endgroup$ – whuber May 26 '16 at 2:07
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The alternative hypothesis is β1≠0 OR β2≠0 OR β3≠0, via De Morgan's Laws. Proving any of those three conditions would disprove the null hypothesis.

The breakdown β1≠β2 OR β2≠β3 OR β3≠0 is mathematically equivalent but might be easier to show significance or design an experiment for.

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  • $\begingroup$ Yes, $\neg\left( (\beta_1 = 0) \wedge (\beta_2 = 0) \wedge (\beta_3 = 0)\right) =( \beta_1 \neq 0) \vee (\beta_2 \neq 0) \vee (\beta_3 \neq 0)$. The two are entirely equivalent, and in my opinion, there's no reason to prefer one way of writing it over the other. $\endgroup$ – Matthew Gunn May 26 '16 at 5:11
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The alternative hypothesis is:
$\beta_{1} \ne 0$ OR
$\beta_{2} \ne 0$ OR
$\beta_{3} \ne 0$ OR
($\beta_{1} \ne 0$ AND $\beta_{2} \ne 0$) OR
($\beta_{0} \ne 0$ AND $\beta_{3} \ne 0$) OR
($\beta_{2} \ne 0$ AND $\beta_{3} \ne 0$) OR
( $\beta_{2} \ne 0$ AND $\beta_{2} \ne 0$ AND $\beta_{3} \ne 0$).

More simply: at least one of the $\beta$s is not equal to $0$.

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    $\begingroup$ Perhaps you're including conditions 4 through 7 to try to be explicit/complete, but from a propositional logic perspective, they're entirely superfluous. $\endgroup$ – Matthew Gunn May 26 '16 at 5:00
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    $\begingroup$ @MatthewGunn From a substantive research perspective, they motivate subsequent post hoc tests. ;) $\endgroup$ – Alexis May 26 '16 at 20:00
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The general alternative to complete equality (a point null) is "at least one beta is not equal to zero", which can be expressed in all manner of alternative ways.

If you have a more specific alternative than that you should give it specifically, and if at all possible choose a test statistic that relates to that particular alternative instead of a much more general one (this will improve power).

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  • $\begingroup$ +1 The reference to power is key: it is a hint as to why the "obvious" logical answer (the alternative is the logical negation of the null) which--even were it well-defined (which it is not, at least until the full set of distributions is specified)--can be exceptionally poor because it can lead to ineffective tests. $\endgroup$ – whuber May 26 '16 at 2:55
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    $\begingroup$ @whuber I don't understand your comment. What is exceptionally poor about the negation of the null? $\endgroup$ – Stan Shunpike May 26 '16 at 3:39
  • $\begingroup$ Stan: Until the full set of distributions--that is, the complete underlying parametric model--has been described, negating the null is at best ambiguous. This is a crucial error. Consider a classical t-test with $H_0:\mu = 0$. When the set of distributions is parameterized by $\mu\in[0,\infty)$, the alternative is one-sided, whereas when $\mu\in(-\infty,\infty)$ the alternative is two-sided. Thus, those who provide merely logical arguments, and suggest thereby that $H_A$ is uniquely determined by $H_0$, are guilty of ignoring such essential distinctions. $\endgroup$ – whuber May 26 '16 at 15:13

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