Effect Size Of Improvement

Problem

Compute a standardized effect size for an improvement: (mean_new - mean_old) / pooled_std. A zero pooled standard deviation returns 0.0.

Hint

Standardizing the gain by the pooled spread makes improvements comparable across benchmarks.

case 1

in: [10,8,2]

expected: 1

case 2

in: [5,5,3]

expected: 0

case 3

in: [1,0,0]

expected: 0

case 4

in: [0,2,4]

expected: -0.5

Starter codePython

def effect_size_improvement(mean_new, mean_old, pooled_std):
    # Your code here
    pass

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