Confidence interval, proportions

[pinto89a]

I a survey, out of 530 women and 509 men, 10.4% of the women and 21.6% of the men answered yes to a question.

Calculate a 99% CI for the difference between two times the population proportion women and the population proportion men who answered "yes".

Answer from book: (-0.091;0.075)

What I did was
p hat 0 = (530 * 0.208 + 509 * 0.216)/ (509 + 530)
and then find
s^2 (p hat x - p hat y) = P hat 0 * (1- p hat 0) * (1/509 + 1/530)
and finally -0008 +- the square root of that

But the answer I get is wrong and I can't see where I made a mistake. Any help please?

 

[kurt.physics]

Your mistake is in calculating the variance of the difference between proportions. You have calculated the variance of a single proportion, but not the difference between two proportions. To calculate the variance of the difference between two proportions, you need to use the formula: Var(p1-p2) = p1(1-p1)/n1 + p2(1-p2)/n2 Where n1 and n2 are the sample sizes of the two groups. In your case, n1 = 530 and n2 = 509.