Interpretation of $N$ and $p$ in Pearson’s correlation test? 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?
 A: 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 $|r| >  0.481.$]
