# Pooling estimates and variances with zero counts

I have a dataset with 10 different sampling groups. The sampling is done in order to maximize the ability to find events. The sample looks a little like:

         Source pop  Sample Events
Group 1:        630     294     34
Group 2:        631     442      9
Group 3:        403     293      8
...
Group 6:      12628     130      0
...
Group 10:      2547     294      1


Now I want to look at (1) the average incidence for the total source population and (2) risk factors for events.

As I understand it pooling the variance through the groups should be easy with:

$$\frac{\sum_{i=1}^k (n_i-1)*s^2}{\sum_{i=1}^k (n_i-1)}$$

where $$n$$ is the sample size and $$k$$ is the number of groups. Then we could average the estimate over the groups:

$$\frac{\sum_{i=k}^k N_i *est_i}{\sum_{i=k}^k N_i}$$

where $$N$$ is the source population size.

Unfortunately running a Poisson regression in the 0 group yields variances of $$>10^9$$, this seems a little wrong to me. I've been looking into the rule of 3 that states that the upper bound 95% confidence interval should be:

$$(1-max\ risk)^n = 0.05$$

that simplifies to:

$$max\ risk = 1 - 0.05^{1/n}$$

This would give us that the confidence interval for group 6 would be 0 to 0.02.

Question: How do I use this knowledge to retrieve the estimate for group 6 and is there a way that I can use Poisson regression in this case?

I've looked at Zero-Inflated Poisson but no orthopaedic surgeon would ever understand the output...

• Have you thought of treating the 10 groups as though they were separate studies and doing meta-analysis to get your weighted estimate with its confidence interval? – mdewey Nov 27 '18 at 13:38
• @mdewey yes but it doesn't solve the 0 counts as they don't have an estimate – Max Gordon Nov 27 '18 at 14:44
• You mean presumably of their variance but you do not have to log transform them as other transformations do not have the same issue. If you do want to use logs you can add a small constant to the number of events (like 0.5) – mdewey Nov 27 '18 at 14:57
• @mdewey - yes, I guess the estimate is 0. One thing that I've been considering is to simply average the upper confidence intreval and report estimate + upper confidence interval. This all seems a little "hacky" and I would like to have a more solid understanding of how to approach all of this. Could this for instance be solved through an n choose k approach? This seems though like it must be a common problem that smart people have thought through... – Max Gordon Nov 27 '18 at 16:12