# 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.

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.
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The estimated interval for p is
$\left[\stackrel{^}{p}-{z}_{\left(1-\frac{\alpha }{2}\right)}\cdot \sqrt{\frac{\stackrel{^}{p}\cdot \left(1-\stackrel{^}{p}\right)}{n}};\stackrel{^}{p}+{z}_{\left(1-\frac{\alpha }{2}\right)}\cdot \sqrt{\frac{\stackrel{^}{p}\cdot \left(1-\stackrel{^}{p}\right)}{n}}\right]$
Thus ${z}_{\left(1-\frac{\alpha }{2}\right)}\cdot \sqrt{\frac{\stackrel{^}{p}\cdot \left(1-\stackrel{^}{p}\right)}{n}}$ has to be $\le 0.025\left(=\frac{5\mathrm{%}}{2}\right)$
The given values are: $\stackrel{^}{p}=0.6;\alpha =1-0.99=0.01$
Therefore the (in-)equation is
${z}_{0.995}\cdot \sqrt{\frac{0.6\cdot 0.4}{n}}\le 0.025$
$2.575\cdot \sqrt{\frac{0.24}{n}}\le 0.025$
The only remaining work is to solve the inequality for n.
Solving for n
$\sqrt{\frac{0.24}{n}}\le \frac{0.025}{2.575}$
$\frac{0.24}{n}\le {\left(\frac{0.025}{2.575}\right)}^{2}$
$\frac{n}{0.24}\ge {\left(\frac{2.575}{0.025}\right)}^{2}$
The inequality sign turns around, if you take the reziprocals on both sides.
$n\ge {\left(\frac{2.575}{0.025}\right)}^{2}\cdot 0.24$
$n\ge 2546.16$
Thus you have to survey 2,547 drivers.