Hi-
The Fisher Z transform can be calculated in terms of a Pearson r as either:
Z = 0.5*log((1+r)/(1-r))
or
Z = atanh(r),
where the latter is the inverse of the hyperbolic tangent function. They *should* both give similar results up to a large number of decimal places even for a high correlation.
Agreed, I don't see where the \sqrt() factor would come into play.
In terms of your specific question, your Z = 1.62 corresponds to a Pearson correlation of:
r = np.tanh(1.6) = 0.92167,
which *is* pretty high. If you are taking an average time series from a region of interest, your expected correlation with any time series within even that region would depend a lot on the size of the region and its homogeneity/noise. It would be good to doublecheck the calculation but it might not be beyond the realm of possibility to get the value you are seeing.
--pt
