Solved – Comparing two proportions: Which standard error formula to use and when

proportion;standard error

When comparing 2 proportions, I've seen 2 ways the standard error can be calculated.

The first I've seen is:
$$
SE = \sqrt{SE_1^2 + SE_2^2}
$$

The second one is:
$$
SE = \sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}
$$

Under what circumstances would I use the first one vs the second one?

What I'm trying to accomplish is calculate the z-score and confidence that $p_2$ is greater than $p_1$.

Edit, here are examples of sites that seem to talk about the same thing but use different SE formulas:

http://www.kean.edu/~fosborne/bstat/06d2pop.html
https://onlinecourses.science.psu.edu/stat100/node/57

Best Answer

I found my answer. It comes down to pooling or unpooling.

You would use $$SE = \sqrt{SE_1^2 + SE_2^2}$$ if you believe the variances are not similar and $$SE = \sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}$$ if you believe variances are similar.

Some additional literature can be found here:

http://mathforum.org/kb/message.jspa?messageID=725796

Thanks @whuber for the inspiration.

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