# Reliability of mean of standard deviations

Hi all,

I've a question which probably is going to show my ignorance about statistics :). I have a large set of machines that produce iron bars of certain lengths. For each machine, I have ran experiments and have a list of lengths. From those I can calculate a mean and sample standard deviation. I don't really care about their means and I am mainly focused on the variation. Therefore, I basically only record a sample standard deviation per machine. I think the results of each machine follow a normal distribution. So far so good :)

I now want to combine these variations into a single number. Therefore, I calculate the quadratic average of each machine variation, let's call it X. In the next step, I also would like to give an estimate for the spread around X. What is this number called and what's the best way to calculate it?

Edit: I'll try to clarify with an example. Suppose I measure 3 machines and find that they produce M1: 100 +/- 7 M2: 120 +/- 8 M3: 130 +/- 9

where the numbers behind the +/-'s are the sample standard deviations of observed values on that single machine. As said before, I don't care about the means but only in the spread, so I define {X_1, X_2, X_3} = {7,8,9}. Their quadratic average is X = RMS(X_i) = $\sqrt{194}$ and I think of X as an indication of the average spread of a machine in my park.

Suppose that I would have found {X_1, X_2, X_3} = {3,8,11}. Their quadratic average is the same $\sqrt{194}$, but the spread around it is obviously bigger. My confidence in the correctness of $\sqrt{194}$ as the average spread of a machine should therefore be lower (I'd like to test some more machines, for instance) and I would like to express this in a number.

-
I don't know whether there is an official name for it, but what I do know is that it is extremely difficult to calculate reliably. Of course, my expertise is in the area of finance, where the process is "heteroscedatic" (moves around with time). If your machines behave the same way day in and day out, you have an easier problem... – Igor Rivin Jan 10 '11 at 15:53
It's not clear to me what you mean. In your situation, it is standard to partition the total variance into within-machine and between-machine components, which might be what you are looking for. Under reasonable assumptions and balanced data, there are even nice formulas. Could you write out the expression you used to calculate your X? – B R Jan 10 '11 at 17:47
I've added an example to clarify – Frank Meulenaar Jan 11 '11 at 18:59
This doesn't look like a research-level mathematics question. stats.stackexchange.com is $\rightarrow$ that way. Voting to close. – Nate Eldredge Jan 12 '11 at 0:20
I didn't know that site existed, thanks; close this one please. – Frank Meulenaar Jan 20 '11 at 15:53