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I have conducted Pearson's correlation in R and need help interpreting the results.

Pearson's product-moment correlation

data:  A.C$Average.tortuosity and A.C$Area
t = -0.6168, df = 14, p-value = 0.5473
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
 -0.6092369  0.3622606
sample estimates:
       cor 
-0.1626531

I am interested in reporting the r and p values. Would it be fair to say based on the the above output that:

There was a non-significant correlation between the two variables, r (14) = -.16, p > .05.

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  • $\begingroup$ Yes, you're right. The p value that's greater than 0.05 failed to reject the alternative hypothesis and the conclusion is that there is no significant correlation, namely accept the null hypothesis where correlation is equal to zero. $\endgroup$ – David Z Jul 14 '14 at 14:46
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I think it would be better to say that the correlation of the two variables was statistically non-significant. Thus you can't conclude that the correlation is not equal to 0. The p-value is in fact 0.5473.

You can also see that the 95 percent confidence interval has the number 0 which also shows the non-significance of the result.

I never like the wording "non-significant correlation" because it says that there both is a correlation and that it's *non-significant". That's a contradiction -- either there is a statistically significant correlation or (by the properties of a hypothesis test) you fail to show a significant result. If you fail to show the result, you can't conclude that your correlation is different from 0.

Hope that helps!

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  • $\begingroup$ (+1) although the interpretation in the OP could be read as 'the sample correlation is 0.16. Based on the p value of 0.55, we cannot say anything about the true correlation'. This would not be a contradiction. $\endgroup$ – Michael M Jul 14 '14 at 16:08
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You better say that there is no Linear association (or correlation) between the two variables. By the way I cannot see you r value ?

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  • $\begingroup$ Your answer is very short and is rather like a comment. Can you extend your answer please? $\endgroup$ – Ferdi Feb 14 '17 at 14:00

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