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I have conducted a Pearson's r and have $$r = 0.328$$ and $$p = 0.110$$

I understand that this presents a moderate positive correlation but the $p$ value suggests that I am unable to reject the null of there being no correlation.

Is this merely reported as such or am I required to conduct a further test, please?

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  • $\begingroup$ Do you mean Spearman rank correlation (as in the tag) or Pearson correlation (as in the question)? $\endgroup$ Commented Jan 21, 2017 at 18:03
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    $\begingroup$ There is no further test. The P-value is the result of the pertinent testing. What the correlation means is the next question and should be considered using a scatter plot to see if it is just weak scatter or something more obvious is messing up the relationship. $\endgroup$
    – Nick Cox
    Commented Jan 21, 2017 at 18:10
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    $\begingroup$ If this has nothing to do with Spearman's rho you should remove the tag. $\endgroup$ Commented Jan 21, 2017 at 18:12
  • $\begingroup$ @Michael Chernick Isn't the question and the answer the same, no matter which correlation? $\endgroup$
    – Bernhard
    Commented Jan 21, 2017 at 18:41
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    $\begingroup$ No Pearson's correlation coefficient measures degree of linearity while Spearman's measure closeness to monotonicity. So for a given set of bivariate data one could be significant and the other not. $\endgroup$ Commented Jan 21, 2017 at 18:45

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If by "r" you mean what I would call the "regular" correlation, that is normalized covariance, then for linear regression:

$\hat {\beta} = {\rm cor}(Y_i, X_i) \cdot \frac{ {\rm SD}(Y_i) }{ {\rm SD}(X_i) }$

So you could say that it's already accounted for in the parameter estimate.

The p-value is the only thing you should pay attention to when it comes to hypothesis testing, at least if you accept that the conditions under which it is valid are met. If there is a very strong correlation and the hypothesis is rejected you should consider getting more data, or getting a less noisy sampling if possible.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – whuber
    Commented Jan 22, 2017 at 16:46

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