# One weird trick for eye-balling a means comparison

Often, I’m in a seminar or reading a paper and I want to quickly see if the the difference in two means is likely to be due to chance or not. This comparison requires computing the standard error of the difference in means, which is $SE(\Delta) = \sqrt{SE_1^2 + SE_2^2}$, where $SE_1$ is the standard error of the first mean and $SE_2$ is the standard error of the second mean. (Let’s call the difference in means $\Delta$.)

Squaring and taking square roots in your head (or on paper for that matter) is a hassle, but if the two standard errors are about the same, we can approximate this as $SE(\Delta) \approx \frac{3}{2} \times SE_1$, which is a particularly useful approximation. The reason is that the 95% CI for $\Delta$ is $4 \times SE(\Delta) = 6 SE_1$ (i.e., 6 of our “original” standard errors).  As such, we can construct the 95% CI for the difference Greek-geometer style, by taking the origin CI, diving it into 4ths and then adding one more SE to each end.

The figure below illustrates the idea – we’re comparing A & B and so we construct a confidence interval for the difference between them, that is 6 SE’s in height. And we can easily see if that CI includes the top of B. ## What if the SE’s are different?

Often the means we compare don’t have the same standard error, and so the above approximation would be poor. However, so long as the standard errors are not so different, we can compute a better approximation without any squaring or taking square roots.  One approximation for the true standard error that’s fairly easy to remember is: $\sqrt{SE_1^2 + SE_2^2} \approx \frac{3}{2}SE_1 + \frac{2}{3}(SE_1 - SE_2)$.

This is just the Taylor series approximation of the correct formula about $SE_1 - SE_2 \approx 0$ (and using $\sqrt{2} \approx 3/2$ and $1/\sqrt{2} \approx 2/3$).