Since my first representation of the rank statistic as paired was incorrect, here is the histogram produced by the simulation
when . It is obviously much closer to zero than previously.
> lm(log(memean)~log(enn)) Call: lm(formula = log(memean) ~ log(enn)) Coefficients: (Intercept) log(enn) -1.162 1.499
meaning that the mean is in rather than in or :
> summary(lm(memean~eth-1)) Coefficients: Estimate Std. Error t value Pr(>|t|) eth 0.3117990 0.0002719 1147 <2e-16 ***
with a very good fit.