# sample sizes.. which formulas to use?

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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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

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