Determining Sample Size for a Desired Margin of Error 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?
 A: 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... 
A: 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!)
A: 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.
