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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?

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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 $\alpha = 0.01$.

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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 $2.57 \sigma$.

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  • $\begingroup$ hmm not sure i follow $\endgroup$ – Jake Mar 31 '16 at 23:48

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