I have three groups of data, each with a binomial distribution (i.e. each group has elements that are either success or failure). I do not have a predicted probability of success, but instead can only rely on the success rate of each as an approximation for the true success rate. I have only found this [question](https://stats.stackexchange.com/questions/80123/test-if-two-samples-of-binomial-distributions-comply-with-the-same-p), which is close but does not seem to exactly deal with the this scenario.

To simplify down the test, let's just say that I have 2 groups (3 can be extended from this base case).

| Group | Trials $n\_i$ | Successes $k\_i$ | Percentage $p\_i$ |
| --- | --- | --- | --- |
| Group 1 | 2455 | 1556 | 63.4% |
| Group 2 | 2730 | 1671 | 61.2% |

I don't have an expected success probability, only what I know from the samples.

The success rate of each of the sample is fairly close. However my sample sizes are also quite large. If I check the CDF of the binomial distribution to see how different it is from the first (where I'm assuming the first is the null test) I get a very small probability that the second could be achieved.

In Excel:

> 
> 1-BINOM.DIST(1556,2455,61.2%,TRUE) = 0.012
> 
> 
> 

However, this does not take into account any variance of the first result, it just assumes the first result is the test probability.

Is there a better way to test if these two samples of data are actually statistically different from one another?