# Question on the chi-square test and Poisson distribution for significance testing

A quick question to check some reasoning: consider the chi-square test for a 2x2 table. For sake of argument, there's 500 patients in each group / category (male and female) and 10% of the total have some condition - if that consists of 60 men and 40 women with the illness, a chi-squared test gives a statistic of 4.444, significant at $$\alpha < 0.05$$ indicating a difference between men and women. Let's consider what happens if this disease were Poisson distributed, with expected value $$\lambda = 50$$ cases per group. Then we can simulate the probability of finding $$k$$ cases in a subset of people, and it looks like this: The reason this is throwing me a little is that the areas under the tails are more than I'd expect at $$\alpha = 0.05$$; for example, a chi-square test suggests you'd find significance if men had $$k\leq 40$$ and $$k\geq 59$$ cases, but the sum of the area under the curve between $$0-40$$ and $$59-100$$ seems quite substantial (quick calculation suggests one has about ~18% chance of lying in this region). I'm wondering if the reason for this is that I can't just consider one group in isolation, and have to multiply probability - for example, if I see 60 cases in men, I must see 40 cases in women? Or is there some reason Poisson assumptions (or similar) don't hold here? Any insight welcome, just to make sure I properly understand what's going on....

## 1 Answer

Correct, you cannot just look at one group in isolation.

If the number of illnesses in men (and women) is Poisson with mean 50, then there is about 9% chance of observing 60 or more illnesses among men. There is also about a 9% chance that there will be 40 or fewer illnesses among women. It is very rare that both of these will happen (about 0.8%). There is also 0.8% chance of 60 or more women and 40 or fewer men. So, those add up to 1.6% so far. But, there are many other combinations that would result in a significant chi-square test; e.g. 70 vs. 48, etc. When you add up all those, you will get close to 5%. If you know the formula for the chi-square statistic, then for each x (illness in men) between 0 and 500 find all the corresponding values of y (illness in women) that would make the statistic significant. Then add the probabilities of all those events P[X=x and Y=y] from x=0,...,500.

• Perfect, thanks! – DRG Feb 19 at 17:59