Confidence and proportion You wish to estimate,with $99\%$ confidence, the proportion of Canadian drivers who want the speed limit raised to $130$ kph. Your estimate must be accurate to within $5\%$. How many drivers must you survey,if your initial estimate of the proportion is $0.60$?
I know that $99\%$ is $2.575$ but i dont know how to set up the problem. I don't think that $130$ kph even has anything to do with the problem. I think i am over thinking this question.
 A: The estimated interval for $p$ is 
$$ \left[\hat p-z_{\left( 1-\frac{\alpha}{2}\right) }\cdot \sqrt{\frac{\hat p \cdot (1- \hat p)}{n}} ; \hat p+z_{\left( 1-\frac{\alpha}{2}\right) }\cdot \sqrt{\frac{\hat p \cdot (1- \hat p)}{n}}\right] $$
Thus $z_{\left( 1-\frac{\alpha}{2}\right) }\cdot \sqrt{\frac{ \hat p \cdot (1- \hat p)}{n}}$ has to  be $\leq 0.025 (=\frac{5\%}{2})$
The given values are: $\hat p=0.6; \alpha=1-0.99=0.01$
Therefore the (in-)equation is 
$ z_{ 0.995 }\cdot \sqrt{\frac{0.6 \cdot 0.4}{n}} \leq 0.025$
$ 2.575\cdot \sqrt{\frac{0.24}{n}} \leq 0.025$
The only remaining work is to solve the inequality for n.
Solving for n
$ \sqrt{\frac{0.24}{n}} \leq \frac{0.025}{2.575}$
$ \frac{0.24}{n} \leq \left( \frac{0.025}{2.575} \right) ^2 $
$ \frac{n}{0.24} \geq \left( \frac{2.575}{0.025} \right) ^2 $
The inequality sign turns around, if you take the reziprocals on both sides.
$n \geq \left( \frac{2.575}{0.025} \right) ^2 \cdot 0.24 $
$n \geq 2546.16$
Thus you have to survey 2,547 drivers.
