# A test to find out the variance within groups

I'm working on biological samples and I'm doing classification on these based on weight (linear discriminant analysis). I need to find out whether the variance within the sample is negligible or somehow accounted for in my model. Any ideas on how I can find this out?

To be clear, say I'm using 100 cell cultures to classify but I need to account for the variance within these cell cultures, perhaps 10 tries for each cell culture i.e 100*10 samples. How do I account for this? Can I use a test which will find out that the variance is negligble and that the use of the values in the model will therefore be fine.

I was thinking of ANOVA at first but this only compares whether populations are different, which doesn't really answer my question.

• Just curious, what does it mean a test for a variance within a group? AFAIK, a group is just a sample. Its variance $\sum_i(avg - x_i)/n$ is called the group variance or variance within the group for sure, be definition. It means that you do not need any test to prove or falsify the computed variance. Commented Sep 1, 2016 at 20:37
• With variance within the group I mean the same test on the same cell culture 10 times over - will the test differ even if it's the same cell culture? If I check the variance for these tests, my question is, at what point or variance is it not a good idea to use the values for LDA. Commented Sep 1, 2016 at 20:43
• Simply put, when is the variance "too large". Commented Sep 1, 2016 at 20:44
• But what is a sample in your culture then? If it is a single experiment then call 10 experiments a group or a single sample and compute the variance as per normal. You cannot compute a variance of a single trial because there is $\sum \over n-1$ actually in the denominator. You will get variances of $0 / \infty = 0$ in single trials. Having a group of trials should be fine for ANOVA. Asking if variance is large or not is the same as asking if plant is tall or not. It only depends in relation to other plants. You have to ANOVA to figure this out Commented Sep 1, 2016 at 20:54