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I want to show that two variables are NOT correlated. When I run spearman on it, I get a very large p-value and a correlation of about -0.1.

So, I accept the null hypothesis? I'm right?

kjetil b halvorsen
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    Your null is that $\rho \neq 0$ and your alternative that $\rho = 0$. – Firebug Jul 26 '16 at 20:11
  • How does this question differ substantively from [test for no corelation](http://stats.stackexchange.com/questions/225760/test-for-no-corelation)? – Glen_b Jul 27 '16 at 00:02
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    @Firebug how would that work? You can't compute the distribution of the test statistic under the null. – Glen_b Jul 27 '16 at 00:03
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    @Glen_b I know it doesn't simply work like that, but the question is "what is my hypothesis". It was more like a tongue-in-cheek comment. – Firebug Jul 27 '16 at 11:11
  • What OP could actually test is if the correlation is smaller than a reasonably small number. Say, $\text{H}_0 : |\rho| \geq 0.1$. Also, you can't really accept the null like that, only fail to reject. Better yet would be calculating confidence intervals imo. – Firebug Jul 27 '16 at 11:26
  • Okay - so, calculating a CI around the correlation coefficent? –  Jul 27 '16 at 12:10
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    @Firebug ... that last comment almost looks like you're working your way toward reinventing equivalence tests – Glen_b Jul 27 '16 at 23:23

1 Answers1

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You can't. A null hypothesis of $\theta \ne\theta_0$ does not make sense in a hypothesis test. In order to do a hypothesis test, you must compare the test statistic to some distribution under the null hypothesis. When the null hypothesis is $\ne$, then that distribution could be anything, so the p-value could be anything.$^{\dagger}$

What you can do is show that that correlation is within some tolerance that you deem to be practically the same as zero correlation. Perhaps you would find it acceptable is the correlation is $\pm 0.05$. You can test that the correlation is greater in magnitude that $0.05$. This gets at the equivalence testing that Firebug mentioned in the comment. The most straightforward form of this is two one-sided tests (TOST).

$^{\dagger}$Maybe it can't be anything, but it does not have a clear value.

Dave
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