I have been wondering about this for a while now and I just could not find an answer to it, so I'd be glad if someone here could help me out!

If I have a simple regression model with a standardized criterion and one predictor, which is a dummy coded variable (two levels), it's the same as calculating a t-test. The t-value from the t-test is the same as the one from the regression model, and also the p-value is exactly the same. But why is the beta-coefficient in my regression model not the same as Cohen's d?

The effect size Cohen's d is the difference between two groups in Standard Deviations. The beta-coefficient in a regression model with dummy coded variables is the deviation of the respective group from a reference group (which for two groups equals the difference between those both groups). Hence Cohen's d and beta should be the same number, or where is my error in reasoning? But the values are completely different (-0.495 vs. -1.137), so I don't think it has to do with pooled vs. unpooled SDs or something like this...


1 Answer 1


Cohen's $d$ can be thought of as semi-standardized, that is, standardized on the dependent variable but not on the independent variable. Recall that the relationship between $\beta$ (a standardized regression coefficient) and $b$, an standardized regression coefficient is: $$ \beta = \frac{b(s_x)}{s_y} .$$ However, Cohen's $d$ is approximately: $$ d = \frac{b}{s_y} $$ with the difference being that the denominator is the pooled within-groups $s$ and not the overall $s$. Thus, Cohen's $d$ will be bigger than the above, proportional to the size of the effect. In your example, the effect is big enough to make a rather noticeable difference.

If the two groups have an equal sample size, then $s_x$ is 0.5. Thus, we would expect the two estimates to be off by about 1/2 plus the effect of the difference between $s$ and $s_{pooled}$.


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