Cohen's d (Effect Size)

d=xˉ1−xˉ2spd = \frac{\bar{x}_1 - \bar{x}_2}{s_p}

Worked example: means 105 vs 100 with sp 10 → d = 0.5 — press Try an example to run it live, then adjust anything.

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Cohen's d (Effect Size) explained

dx̄1x̄2sp

A p-value tells you whether a difference is detectable; Cohen's d tells you whether it is big. Divide the gap between two means by the pooled standard deviation and you get the effect in standard-deviation units, comparable across studies and instruments. A treatment group averaging 105 against a control at 100 with sp = 10 gives d = 0.5. Jacob Cohen proposed the rough benchmarks of 0.2, 0.5 and 0.8 for small, medium and large in his 1969 power-analysis handbook — and spent much of the rest of his career warning people not to apply them mechanically, since a d of 0.1 on a mortality outcome can matter far more than a d of 1.0 on a lab task.

The trap this fixes is the significance illusion. With 10,000 subjects per arm, a d of 0.04 is highly significant and completely uninteresting; with 12 subjects, a d of 0.9 may fail to reach significance and still be the most important result in the paper. Report both. Reversed, the formula recovers scale: an intervention reported at d = 0.8 that moved a mean from 48 to 52 implies a pooled standard deviation of 4/0.8 = 5 in the original units.

Cohen's d (Effect Size) formula

d=xˉ1−xˉ2spd = \frac{\bar{x}_1 - \bar{x}_2}{s_p}
Where
  • dd= Cohen's d
  • xˉ1\bar{x}_1= Mean of group 1
  • xˉ2\bar{x}_2= Mean of group 2
  • sps_p= Pooled standard deviation

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