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Find the minimum sample size needed to estimate, within two percent, the percentage of US voters who intend to vote Republican in the next election. Use 90% confidence and assume a previous poll indicates 42% intend to vote Republican.

90 percent confidence = 1.645 thats all i get then im stumped. which formula can i plug any info into?

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3 Answers 3

Let $\hat{p}$ be the sample proportion of Republican voters, and let $p$ be the true proportion of Republican voters. You would like $P(|\hat{p}-p| \le 0.02)$ to be $90\%$.

The mean of the random variable $\hat{p}$ is $p$, and the standard deviation of $\hat{p}$ is $\sqrt{\frac{p(1-p)}{n}}$, where $n$ is the sample size.

A not unreasonable estimate for $p$ is that it is about $0.42$. So a not unreasonable estimate for the standard deviation of $\hat{p}$ is $\sqrt{\frac{(0.42)(0.58)}{n}}$.

By the normal approximation to the binomial, we want $1.645$ "standard deviation units" to be about $0.02$. In symbols, $$1.645 \sqrt{\frac{(0.42)(0.58)}{n}}\approx 0.02.$$ Now you can use some algebra to solve for $n$. You may have been supplied a ready-rolled formula for $n$. But you can also use the information here to roll your own.

Comment: If one expects a probability to be not very far from $0.5$, the unknown term $\sqrt{p(1-p)}$ can be assumed to be about $0.5$. Because of our prior knowledge about the rough value of $p$, we used $\sqrt{(0.42)(0.58)}$ instead. This really makes very little difference, since $\sqrt{(0.42)(0.58)}$ is about $0.4936$, which is awfully close to $0.5$.

It would be sensible not to use the $0.42$ information at all, but I imagine you are expected to use it. Please note that if $p$ is very far from $0.5$, then $\sqrt{p(1-p)}$ will not be close to $0.5$.

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The standard deviation of the sample proportion is is $\sqrt{pq/n\ {}}$ where $n$ is the sample size, $p$ is the proportion who will vote Republican, and $q=1-p$ is the proportion who will not. The largest that $pq=p(1-p)$ can be is when $p=1/2$.

Does any of that look familiar to you?

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yes. but thats not one of the formulas in this text. i cant even figure out the anything with this.. –  jen Oct 22 '11 at 1:33
    
oh yeah actually.. thanks –  jen Oct 22 '11 at 1:42

You probably want to use an equation like

this equation

http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval#Normal_approximation_interval

Edit: added a formula that might appear in the textbook

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now we're talking. thank you –  jen Oct 22 '11 at 1:42
    
i got it! within 2 percent adds a .02 on the bottom half of a different equation solving for n n = 1.645 squared times .42 times .58 over .02 = 1648 people.... yay thanks anyway!! –  jen Oct 22 '11 at 1:53
    
@jen that's the same equation... –  opt Oct 22 '11 at 2:12
    
yeah.. i just didnt recognize it... –  jen Oct 22 '11 at 2:26

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