I was hoping for a bit of help.

I have two very different sized samples of binomial distribution, and I want to figure out if they are significantly different. A previous question posted on here has been helpful, but the answer is different from what I have been advised + I have the added complication of widely different sized samples, so I was hoping to get a bit more input.

An example of the two samples:

Group 1: Successes = 10000, Trials = 40000
Group 2: Successes = 150000, Trials = 3500000

Like the previous posted question, I have no predicted probability of success.

My first approach had been using Chi-squared (producing a nice familiar p-value).

However, I was told that I should use a binomial test of proportions, due to the widely different sample sizes. I should use the larger sample (group 2) as my control, and then create confidence intervals for group 1 to see if they capture the control.

Nevertheless, reading around the topic, it appears statistical hypothesis testing is potentially more appropriate (and again the linked question has that as the proposed answer), thus making use of a z-score. But I'm unsure if my samples meet the requirements for a normal distribution (although large enough, they do not have proportions near 0.5).

So what should I use? I feel like I'm back to chi-squared, but I don't feel confident enough to simply discard all the points above.


$\chi^2$-test on a 2x2 contingency table is completely equivalent to proportion comparison, so go ahead with the approach you understand better.

Check https://www.dummies.com/education/math/statistics/how-to-compare-two-population-proportions/ for a quick review on how to compare proportions.

On the $\chi^2$-test of independence: https://www.statisticssolutions.com/non-parametric-analysis-chi-square/

Anyway, if the data looks like that you presented, you don't even have to do the math! There is statistically significant difference with a p-value of virtually 0!

  • $\begingroup$ Excellent thanks! and I know what you mean, my data really looks like this and it feels silly when it's obviously different, but a significant p-value makes everyone sleep safely at night. $\endgroup$ – SorenK Apr 2 '19 at 16:33

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