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?
Best Answer
The aim of a confidence interval is to estimate an interval for a population parameter using sample information. In this case we are interested in estimating a confidence interval for the population proportion $p$ of adults aged $18$ to $29$ that use the internet. To that end, we draw a sample of size $n=500$ adults with ages in $[18,29]$ and check how many of them are using the internet. This is the sample proportion $\hat{p}$ and it appears that $\hat{p} = 0.93$. A $95\%$ confidence interval for $p$ is then given by (recheck the calculation) $$\left[\hat{p} - z_{1-\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},\hat{p} + z_{1-\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right] = [0.9297, 0.93026],$$ where $\alpha = 0.05$ and hence $z_{1-\alpha/2} = 1.96$. The conclusion is that you are $95\%$ confident that the population proportion $p$ lies in the above interval. Note that is not the same as saying 'the probability that $p$ lies in $[0.9297,0.93026]$ is $0.95$'. The latter statement is invalid since $p$ is a number, so it either lies in the interval or it doesn't.
Exercise B. is completely similar with a different $\alpha = 0.01$.