In this situation, there are different formula for the mean and standard error, but the same logic and procedure for solving problems remains the same.
Demonstration of these ideas is shown in an example.
and SE() =
The shape will be approximately normal for sufficiently large samples. This is reasonable to assume if each sample individually meets the rule of thumb for the approximation.
The sampling distribution for the difference in sample proportions will be approximately normal with a mean of 0 and a standard deviation of = .0663. The z-score is
z = (.16 - 0) / .0663 = 2.41
Since the area to the right of 2.41 under the standard normal curve is .0080, the probability is less than 1% that the difference would be this large.
Bret Larget, email@example.com