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In this paper, the authors report an estimate of $r = 0.86,\, N = 28,\, p < 0.001,$ using Pearson’s correlation test. The parameter $r$ (or $\rho$) is clear to me, but how are $N$ and $p$ derived and to be interpreted?

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In the notation you quote from this paper $r$ is the sample correlation; it estimates the population correlation $\rho.$ $N$ is the number of $(x,y)$-pairs upon which the test is based. Many texts would use $n$ because this is a sample.

This is a test of $H_0: \rho = 0$ against $H_1: \rho \ne 0$.

Under the assumption that $H_0$ is true and that the data are bivariate normal, this test computes the test statistic $t = r\sqrt{\frac{n-2}{1-r^2}}$, which has Student's t distribution with $n -2$ degrees of freedom, The symbol $p$ is for the P-value of the test. Roughly speaking, this is the probability that the value of $|r|$ is farther from 0 than the observed value ($r = 0.86$, in the example you provide).

Specifically, in this example we have $t = 0.86\sqrt{26/(1 - 0.86^2)} = 8.593$ The exact P-value is the probability that $P(T > 8.593) + P(T < -8.593),$ where $T$ is a random variable having Student's t distribution with $DF = 26.$ Using statistical software the sum of these probabilities is found to be $4.51 \times 10^{-9}.$

The reported P-value $< 0.001,$ indicates that the actual P-value is less than 0.001. (Possibly, this is the best value available in printed tables available to the author.) In any case, one feels safe believing that the $\rho = 0$ is not the true population correlation.

Notes: (1) I have no idea whether the assumption of normality is valid. Given the rather high value $r = 0.86$ and with a look at the scatterplots provided, I guess the assumption is safe. [The Wikipedia article on 'Pearson correlation' mentions alternate approaches in case data are not normal, but you do not have the original data, so you could not apply these methods.] (2) Although it is not obvious from the formulas, this test is exactly the same as the test that the slope of the regression line ($\beta$ in the paper to which you refer) of y on x (or of x on y) is 0.

Addendum (suggested by an off-site reviewer after acceptance of this Answer): Suppose you want to know the exact values of $r$ that lead to rejection of $H_0$ at significance level 0.001 = 0.1%. From printed t tables with $DF = n-2 = 28-2= 26,$ we see that $t_{.0005} = 3.7454$ cuts probability 0.0005 from the upper tail of the density function of the t distribution, so the $P(|T|>3.7454) = 0.001.$ Solving the the formula above for $r$ we see that, $t_{.0005} = 3.7454$ corresponds to $r = t/\sqrt{n-2+t^2} = 0.592,$ so we would reject at level 0.001 for any $r$ with $|r| > 0.592.$ [For the 1% level, $t_{.005} = 2.7969$ and we reject for $|r| > 0.481.$]

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  • $\begingroup$ A minor point: the $p$ value is the probability of observing $|r|$ farther from $0$ than the observed value assuming $H_0$ is true. Depending on the true value of $\rho$, the true probability that $|r|$ is farther from $0$ than the observed value will differ. $\endgroup$
    – user88319
    Aug 19, 2015 at 16:09
  • $\begingroup$ Thanks. I didn't doubt you understood, but enough people have problems understanding null hypothesis tests that I thought it would be good to make the assumptions explicit. $\endgroup$
    – user88319
    Aug 19, 2015 at 16:18
  • $\begingroup$ @Strants: Agreed. Mentioned assumption of $H_0$ in previous sentence. Will re-arrange clauses so there is less chance of misinterpretation. (almost-simultaneous comments) $\endgroup$
    – BruceET
    Aug 19, 2015 at 16:19

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