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I'm trying to study for a test in my AP Statistics course. My lecturer spent the majority of the unit going over the various proportion tests. On my review, I was presented with the following question:

Suppose that you wanted to estimate p, the true proportion of students at your school who have a tattoo with 98% confidence and a margin of error no more than 0.10. How many students should you survey?

What I'm not understanding is what should be substituted for p. In the given problem, no value for p is given, but yet I need to find n using the following formula:

$$ ME = (z\ast )(\sqrt{\frac{p(1-p)}{n}}) $$

How can I determine a value for n?

Zulfe
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3 Answers3

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The first thing you need to do is to solve your formula for $n$. Squaring both sides and then solving for $n$ ultimately gives $n=\frac{z^2p(1-p)}{ME^2}$ Now here is the thing. Since you do not know the value of $p$, you need to ensure that any possible value of $p$ is "accounted" for. Looking at $p(1-p)$, this is a parabola with a maximum for $p=0.5$ (the vertex is located at $0.5,0.25)$ In other words, your sample size should be at its greatest IF the $p$ in your population turns out to be $0.5$. Therefore, with an unknown proportion, your formula becomes $n=\frac{0.25z^2}{ME^2}$ For a 98% confidence you can read the $z$ from a Standard Normal Table, the ME is said to 0.10 and thus you can fill in the data in your formula. In case of rounding, you should round up to the next whole. Hope this helps a bit...

imranfat
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    Nice teaching at what I judge to be an appropriate level--without giving away the final answer.(+1) BTW: Among pollsters, the inverse of this general problem is that the 95% margin of error (assuming random sampling from the appropriate population) is $1/\sqrt{n}$. This uses your idea to protect against the most variable case, when $p = 1/2,$ and then uses 2 instead of 1.96. – BruceET Mar 09 '16 at 21:05
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Just replace z* with 2.33 (since P(X)>|2.33|=2% when X is std normal -- a standard distributional assumption for proportions -- implying 98% confidence), and solve for n. Your answer will be in terms of p, the sample proportion of students with a tattoo which is only known ex-post.

PS Some rep would be appreciated (need 10 to psot an image!)

Markus
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You can't solve for $n$ with the information given. Rearranging the expression you can get: $$n=\frac{(z^*)^2p(1-p)}{ME^2}$$ but $z^*$ is a function of $n$, and $p$ is unknown. If you knew $p$ then you could solve for $n$ by trial and error using a binary search.

MaxW
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  • In what way does $(z^*)^2 = 5.412$ vary with $n$? Are you thinking of the t distribution instead of the normal? In R, qnorm(.99)^2 returns 5.411894, or use printed normal tables. – BruceET Mar 09 '16 at 20:59
  • Yes, Student's T rather than the normal distribution for $z^{*}$. – MaxW Mar 09 '16 at 21:06
  • And what role does Student's t distribution play here? – BruceET Mar 10 '16 at 04:20
  • @BruceET - Looking at the problem more closely you don't need to use Student T. For p=0.5, which has highest variance in p, n > 30, so standard normal approximation is fine. – MaxW Mar 10 '16 at 22:49