Hedges Correction Factor

Problem

Hedges' small-sample correction multiplies the standardized mean difference by J = 1 - 3 / (4 (n1 + n2) - 9).

Given the two sample sizes, return J.

Hint

The correction approaches one for large samples.

case 1

in: [10,10]

expected: 0.9577464788732395

case 2

in: [5,5]

expected: 0.9032258064516129

case 3

in: [100,100]

expected: 0.9962073324905183

Starter codePython

def hedges_correction_factor(n1, n2):
    # Your code here
    pass

3 test cases · run your code against them in the editor