# Estimate individual variance from pool variance

#### Frameshift

##### New Member
Let's say I have a group of 100 tiny beads, and I want to know the variance in their weight. However, my scale can't weigh one bead at a time, so I weight them in non-overlapping pools of 10 beads each. Then I can calculate the variance among pools, but what I really want to know is the variance among individual beads.

It seems like there should be a simple way to estimate this, but I'm not thinking of it. I'd be willing to assume a normal distribution of individual bead weights and an absence of measurement error at the pool level.

Thanks for any help you can offer!

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#### hlsmith

##### Not a robit
So you have variance of 10 groups of 10 beads and what the variance for the full 100 - which would be the variance of the beads, correct?

#### Frameshift

##### New Member
Yes, I want to know the variance among individual beads for the full 100. I know it should be higher than the variance among pools.

#### hlsmith

##### Not a robit
Is this an actual problem you can conduct. If so, can you perform regroupings or is the scale not sensitive enough?

So, 1-10, 2-11, 3-12, etc.

#### Frameshift

##### New Member
The example I gave is an analogy for a case I'm trying to help someone with. Unfortunately it is not possible to do any regrouping. They only have the data they have, and the samples are not retrievable for further measurement.

#### rogojel

##### TS Contributor
hi,
I think you could use the Central Limit Theorem here which says that indpendently of the distribution of the individual values the mean of a sample of n is normally distributed, with the mean of the normal distribution being the "real" mean of the population and the stddev of the mean beind the stddev of the individual values divided by sqrt(n). So, I would build groups of 5 and calculate the means, build the distribution of the means and estimate the stddev of the means. Times sqrt(5) this will give an estimate for the individual stddev that you are looking for. Then, just to be sure I would repeat the procedure with groups of 10 - if the second estimate is close to the first one you are done.

regards

#### Dason

##### Ambassador to the humans
$$Var(\sum_{i=1}^{10} X_i) = \sum_{i=1}^{10} Var(X_i)$$ and since all of your bead variances are assumed equal we just get that the variance of the sum of 10 is 10 times the individual variance. So estimate the variance of the pooled data and divide by 10.

#### Frameshift

##### New Member
I think these two proposals are correct. I don't have my colleague's data in front of me, but I've been playing with 1000 randomly generated (gaussian) values, that were simulated to have a SD and Var of 1. If I examine pools in groups of 10, I get PoolVar = 0.09 and PoolSD = 0.3. If I use pools of 5, I get PoolVar = 0.2 and PoolSD = 0.45. This all seems pretty close to the expectations given. So I will just multiply the PoolVar by the pool size. Thanks!

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