How I should compute the standard error of PR?

#1
Hi,
When there are two independent groups and proportion ratio (PR) is:
PR = p1 / p2

How I should compute the standard error of PR? Should I use the logarithms of the proportion ratios? If yes, I found two different formula for it, which one is correct?

SE = (1/p1 + 1/p2 - 1/n1 - 1/n2) ^1/2

SE = ((1-p1)/n1p1) + (1-p2/n2p2)) ^1/2


Why do they say SE(ln(RR)) instead of SE((RR))?
( I attached a tutorial )
 

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