Confidence interval for the Proportion Question on my Homework:
The Pew Research Center has conducted extensive research on the young adult population (Pew Research website, November 6, 2012). One finding was that 93% of adults aged 18 to 29 use the Internet. Another finding was that 21% of those aged 18 to 28 are married. Assume the sample size associated with both findings is 500.
Round your answers to four decimal places.
A.  Develop a 95% confidence interval for the proportion of adults aged 18 to 29 that use the Internet.
B.  Develop a 99% confidence interval for the proportion of adults aged 18 to 28 that are married.
Hate to say it, missed this day of class and i am having a hard time working this one out. Anyone got any clue?

The number of positive answers to either question is Bernoulli distributed which means that its $\sigma = \sqrt{np(1-p)$.  You don't know what $p$ is but you can estimate it from the fraction observed in the sample.
So for the first problem, $\sigma = \sqrt{500\cdot 0.93 \cdot 0.07}.  Then you can look up in tables for the $z$-test and find that the 95% confidence interval is $\pm 1.96\sigma}$
The second problem is done in the same way, except the 99% CI is at about $2.57 \sigma$.
