Test for differences in amount of events per invididual betwen two groups

Question

I have two groups of people, those who finished a treatment with a drug and those who didn't. Also, I have the amount of adverse reactions each individual had to the drug. An individual could have from none to 3 or 4 adverse reactions. I need to test if there is a difference in the amount of adverse reactions individuals had between those who finished and those who didn't finish the treatment. For instance:

	Finished treatment	Didn't finish treatment
Number of Individuals	142	10
Number of Adverse events	231	12

So, in this case, the expected amount of adverse events per individual for individuals who finished treatment would be 231/142 = 1.62 while the expected amount of adverse events per individual for individuals who didn't finished treatment would be 12/10 = 1.2 How can I test whether this difference is significant?

I thought about a chi-square or a fisher exact test, but I don't think that's right, since only the column margin makes sense (summing in a row makes sense but summing in a column doesn't).

Welcome to CV! Does it make sense to model these numbers as Poisson? You can check this if you still have the original numbers of adverse events per individual in the larger group. Look at the ratio variance:mean if this is between 0.8 and 1.25, you can safely assume that. — Ute, Jul 24 '23 at 21:39
Hi Ute, thanks for your answer and your help. All the information I have is in that table unfortunately. Anyways, if I could model these numbers as Poisson what sort of test could I use? — Andrés Rabinovich, Jul 25 '23 at 20:38

Ute · Accepted Answer · 2023-07-25T23:48:52.250

You are completely right, although your table may look like a contingency table, it is not, and chi-square or Fisher's exact test are not suitable to compare the two groups.

If it is justified to assume that the number of adverse events per individual has a Poisson distribution, you can use a test for comparison of Poisson rates instead.

The most common situation where you would test rates is when you observe rare events in two time intervals of length $t_1$ and $t_2$. In your case, you would replace time interval length by group size. Several suggestions for such a test can be found as answers in this thread. For R users, it is easy: poisson.test, first argument x: vector of number of adverse events, second argument T: vector of population sizes.

If you have the original counts, you should check if the Poisson assumption is satisfied. This is commonly done by looking at the index of dispersion: variance divided by mean. If this index is larger than 1, you have overdispersed data. For your purpose, you should be ok if it does not exceed 1.3 in the larger population. ^{(I did a quick simulation check)}
The recent paper by Mijburg and Visagie: "An overview of goodness-of-fit tests for the Poisson distribution(2020, South African Statistical Journal, 54 (2) 207-230) is perhaps useful here.

In your case, the difference between the two groups is not significant - you would need about five times as many persons in the smaller group to detect the difference between rates 1.62 and 1.2 (at the usual $\alpha$ = 0.05). If the data were overdispersed, the observed difference would be even less significant, because of larger variance.

Thank you very much Ute for your time and your complete answer! Makes sense to use a test for comparison of Poisson rates then. — Andrés Rabinovich, Jul 27 '23 at 14:44

Test for differences in amount of events per invididual betwen two groups

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