Let's say I've got 2 different populations (e.g. 35 and 40) with 2 different means and SDs (22.3, 28.0) and n (e.g. 35,40). I don't actually know the individual values. Obviously you can calculate a pooled SD (NOT what I want) it in R with something like this:

n <- c(35,40)
mean <- c(22.3,28.0)
sd <- c(3.2, 4.9)
df <- data.frame(n,mean,sd)

sqrt( sum(df$sd^2 * (df$n - 1)) / (sum(df$n - 1)) )

This gives an SD of around 4.194827.

However, would it be valid to assume these 2 populations are normal, synthesise them using the means and SDs given, and then take an SD of these 2 hypothetical populations? I don't see why not as long as I'm not explicit we've estimated the SD in a paper's methods?

I could do this e.g. with:

> result <- NULL
> for (i in 1:10000) {
+   control <- rnorm(n = 35,mean = 22.3, sd=3.2)
+   active <- rnorm(n = 40,mean = 28, sd=4.9)
+   result <- c(result,sd(c(active,control)))
+ }
> mean(result)
[1] 5.06259

I iterated 10000 times to get a good average.

So

1. Is this a valid calculation?

2. If yes/no, is there a better way of doing it?

A colleague appears to have an excel spreadsheet which gives a very similar number using this formula:

=SQRT((A2*(D2^2)+A2*((C2-G2)^2)+B2*(F2^2)+B2*((E2-G2)^2))/(A2+B2))

Where A, C and E are n, mean, sd of the 1st arm, and B, D, F are the same of the 2nd arm. G is the weighted mean of the 2 groups.

It seems to give the same result as my synthetic data so I assume it is valid?

Thanks