If "use the square root of the pooled variances" is true what is wrong with the example below:

> xx<-c(2,3,4)

> var(xx)

[1] 1

> sd(xx)

[1] 1

> mean(xx)

[1] 3

> yy<-c(9,10,11)

> mean(yy)

[1] 10

> var(yy)

[1] 1

> zz<-c(2,3,4,9,10,11)

> mean(zz)

[1] 6.5

> var(zz)

[1] 15.5

> sd(zz)

[1] 3.937004

What you are providing is the variance for the

**combined data set **(S^2) ---- which is not an average of the two separate variances (1, 1).

In fact, you really don't even need the data to obtain your result (S^2=15.5). All that is needed are the two sample sizes (3, 3), the two means (3, 10), and the two variances (1, 1) of the indivdual data sets and then the variance for the

**combined data ** (Variance = 15.5) can be obtained as:

\( s^{2}=\frac{n_{x}^{2}s_{x}^{2}+n_{y}^{2}s_{y}^{2}-n_{y}s_{x}^{2}-n_{y}s_{y}^{2}-n_{x}s_{x}^{2}-n_{x}s_{y}^{2}+n_{y}n_{x}s_{x}^{2}+n_{y}n_{x}s_{y}^{2}+n_{x}n_{y}\left ( \bar{X}-\bar{Y} \right )^{2}}{\left (n_{x}+n_{y}-1 \right )\left ( n_{x}+n_{y} \right )} \)

where you can see in the far right-hand side of the numerator how the square of the difference between the two means will play a role in the computation of the variance for the combined data.

Now, if you want the variance for three (or more) combined data sets, then all you need to do is just keep applying the equation I provided above separately as you combine the data sets one at a time...e.g. combine 1 & 2 and then (1, 2) & 3 ....and so on.... for any number of data sets. Obviously, as you progress you would also need the means of combined data -- which are, of course, easy to obtain.

I would also note that the original poster is asking two

**different** questions. The first question asks for an average of the three variances (or standard deviations) and the second question asks for the variance (or standard deviation) for the combined data....these are different calculations. My first post addresses the first question and this (second) post addresses the second question.