Given a large population (size $n$) separated into some groups (for simplicity of equal size $k$), each population member is assigned a $1$ (true) or a $0$ (false). The population mean is $p$.

Null hypothesis: the distribution is binomial, hence the group means follow a distribution with mean $p$ and variance $kp(1-p)$.

Alternative hypothesis: the difference between the group means is not consistent with a binomial distribution but due to some other reason.

How do I compute a p-value for the null hypothesis?

In my actual data the variance between the group means is much bigger than what one would expect under a binomial distribution. I want to use this p-value as a justification that the difference between groups is not just due to chance.


1 Answer 1


I used a Cramer-von Mises-Test as explained here: https://en.wikipedia.org/wiki/Cram%C3%A9r%E2%80%93von_Mises_criterion

It looks at the difference between the actual cumulative distribution and the expected one. The actual one I have from the observed data and the expected one is the binomial distribution as described in the question. The square difference is integrated over the entire domain. This value is then compared to some tabulated values. Wikipedia is unfortunately not very specific about what to compare to but the stats Software R is. The Cramer-von Mises-Test is integrated in the library goftest and can be computed with cvm.test . This gives a p-value exactly as I wanted.

  • 2
    $\begingroup$ This is a generic goodness of fit test. It will (in principle) reject the null even if the actual distribution is more narrow than the expected (binomial) one. Given that you want to show that it's wider, I am not sure it is an appropriate test. Or at least you have to supplement it with direct demonstration that your actual distribution is wider than binomial. $\endgroup$
    – amoeba
    Feb 12, 2018 at 14:09

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