# Statistics with possibly dependent events

I expect this is a completely standard statistics question.

My biologist cohort is doing an experiment in which data comes from cells and each "slide" consists of 20 or so cells. Slides are certainly independent from each other, but we don't know whether cells on the same slide are independent from one another. Conservatively one may take an average of all the cells on each slide and treat these means as trials, but since slides are expensive and time-consuming to run, one does not get enough data to draw conclusions in this way, and one is tempted to (and people in her department do) take each cell as a separate trial.

Question: What is the correct statistical test to run in this situation, i.e. in which trials naturally come in groups and trials from the same group may or may not be independent? Is it correct to test for independence first and then to act accordingly?

EDIT: Perhaps a more reasonable request is the following. What is the appropriate way to test whether the cells on each slide act independently (abstractly, trials from the same groups)? Do I seek an ANOVA? As I understand it, an ANOVA would be more useful if I wanted to prove that the cells on individual slides do not act independently, which I hope is not the case.

-

For example one model could be that if Xt and Xt-1 are neighboring cells then Xt=r Xt-1 +et where et is an independent error term with variance independent of t and say 0 < r < 1 and t goes from 1 to n (i.e. there are n cells on a slide). Then r is the correlation between Xt and Xt-1. Let s$^2$ be the variance of Xt. The collection of cells on the slide form a stationary series. Then if r is close to 0, Var(sample mean) is close to s$^2$/n. If r is close to 1 Var(sample mean) is close to s$^2$. In general it is a particular function of r. In such a case applying the independence formula s$^2$/m =var(mean) and solving for m gives you the number of independent sample you would need to have the same variance as for your correlated sample.