Wikipedia has a Fisher transform of the Spearman rank correlation to an approximate z-score. Perhaps that z-score is the difference from null hypothesis (rank correlation 0)?

This page has the following example:

`4, 10, 3, 1, 9, 2, 6, 7, 8, 5 5, 8, 6, 2, 10, 3, 9, 4, 7, 1 rank correlation 0.684848 "95% CI for rho (Fisher's z transformed)= 0.097085 to 0.918443"`

How do they use the Fisher transform to get the 95% confidence interval?

**Answer**

In a nutshell, a 95% confidence interval is given by

\tanh(\operatorname{arctanh}r\pm1.96/\sqrt{n-3}),

where r is the estimate of the correlation and n is the sample size.

Explanation: The Fisher transformation is arctanh. On the transformed scale, the sampling distribution of the estimate is approximately normal, so a 95% CI is found by taking the transformed estimate and adding and subtracting 1.96 times its standard error. The standard error is (approximately) 1/\sqrt{n-3}.

**EDIT**: The example above in Python:

```
import math
r = 0.684848
num = 10
stderr = 1.0 / math.sqrt(num - 3)
delta = 1.96 * stderr
lower = math.tanh(math.atanh(r) - delta)
upper = math.tanh(math.atanh(r) + delta)
print "lower %.6f upper %.6f" % (lower, upper)
```

gives

```
lower 0.097071 upper 0.918445
```

which agrees with your example to 4 decimal places.

**Attribution***Source : Link , Question Author : dfrankow , Answer Author : dfrankow*