# n-1, Part 2

Previously, I examined how when calculating the standard deviation of a sample, if we divide by \(n\), we obtain a biased estimate. The problem is exacerbated when the sample size is small, as in the typical psychology experiment. To correct for this, one should instead divide by \(n - 1\). However, the more interesting question is what to do when you’re computing the standard deviation of two separate samples.

When performing a meta-analysis, you might take information from two samples, each of which may be characterized by its mean, standard deviation, and sample size. Combining means is relatively straightforward - just compute a weighted average. For standard deviation, my colleague found this suggestion to do a similar thing for standard deviation:

\[\sigma = \sqrt{\frac{n_1 \sigma_1^2 + n_2 \sigma_2^2 + n_1(\mu_1 - \mu)^2 + n_2(\mu_2 - \mu)^2}{n_1 + n_2}}\]However, you’ll note that there’s no correction being applied here. Perhaps it’s taken care of, since each of these standard deviations is, presumably, itself the result of a correction. First, let’s draw a bunch of random samples from the same underlying population (*M* = 100, *S* = 10). Each sample will consist of 15-50 observations (uniformly distributed).

Now, what happens if we take the simple weighted standard deviation, as described in the equation above?

That looks pretty close. It’s slightly high, I suppose, at an average combined sd of 10.14. But there’s an inconsistency here. The wikipedia page which the above stack exchange answer links to gives a slightly modified formula (which I’ve rewritten slightly for consistency with the above equation:

\[\sigma = \sqrt{\frac{[n_1 - 1] \sigma_1^2 + [n_2 - 1] \sigma_2^2 + n_1(\mu_1 - \mu)^2 + n_2(\mu_2 - \mu)^2}{n_1 + n_2 - 1}}\]Do you spot the difference? In the numerator, the sample standard deviations are weighted by \(n - 1\), while in the denominator the whole thing is divided by \(n_1 + n_2 - 1\). What’s this going to do to the combined standard deviation? It should make it slightly smaller. We’re effectively reducing the size of what’s in the numerator, which reduces the overall quantity, *and* reducing the size of what’s in the denominator, which has the same effect. Not coincidentally, the first method of calculation lead to a slight overestimation. What happens with the new method?

We get a slightly (*very* slightly) smaller combined standard deviation. But, happily enough, the average of all these standard deviations (10.06) is almost exactly lined up with what we know the population standard deviation to be (*S* = 10).