Tell me more ×
Cross Validated is a question and answer site for statisticians, data analysts, data miners and data visualization experts. It's 100% free, no registration required.

I'm sorry if my terminology is wrong, I'm making it up.

I have hundreds of objects counted simultaneously in two treatments, each measured three times (not in pairs). e.g.:

            A1  A2  A3  B1  B2  B3
 Object 1   2   4   3   5   8   1
 Object 2   0   0   0   0   0   0
 Object 3   20  15  12  1   0   3
 Object 4   0   6   2   0   200 250
 Object 5   60  64  70  10  0   5

I want to look for differences between A and B. The problem is that some objects in some samples give zero or very low values even when other samples (for the same object and treatment) give very high counts (e.g. 0,200,250). There is a good reason to believe the high counts but that the low counts are due to obstruction by unobserved phenomena.

I know I could use some kind of non-parametric t-test and that it would very conservative, due to the high variance. What I would like to do is perhaps model the unobserved phenomena as an obstruction parameter that differs between treatments . I have a vague inkling that a G-Test or maximum-likelihood might be the way to go but I don't know where to start.

This probably won't help but is what I am thinking:-

p(counts(A,B)) = sum(p(counts(A) - obstructA * p(obstructA) )  + sum(p(counts(B) - obstructB * p(obstructB) ) 
share|improve this question
Is there a big distinction between 0 and 1? I mean could 0 be a real value, and could the obstruction or whatever give small values like 1 or only 0? – Karl Sep 27 '11 at 19:09
We think there is no big distinction and that 0 and 1 are similar. – Dave Gerrard Sep 28 '11 at 9:24
Darn; that makes it harder. You could think of the outcomes as mixture of some small values (due to the obstruction) and then a distribution of "real" values whose mean is the parameter you're looking for. It would be easier if the "obstruction" component were strictly a point mass at 0. I'll think some more. – Karl Sep 28 '11 at 16:24

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.