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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.

3SAT
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mariah
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1 Answers1

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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.

callculus42
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  • I think i solved for n. I got 96. would that be the right answer? – mariah Jun 17 '15 at 19:37
  • @mariah Not really. I have made an edit, which shows the calculations. – callculus42 Jun 18 '15 at 04:08
  • But all of this is assuming a normal distribution, is it not? – mathreadler Jun 18 '15 at 08:57
  • @mathreadler Not necessarily. If a sample size is greater than 30 you can use the central limit theorem. – callculus42 Jun 18 '15 at 09:02
  • Got to be easy to find counter-examples to that. Say a very split population with opinions as a sum of two normal distributions very far apart and each having low variance. – mathreadler Jun 18 '15 at 09:11
  • @mathreadler Im not sure, what do you mean. Im not talking about two normal distributions. The CLT is about the sum of Independent and identically distributed variables. – callculus42 Jun 18 '15 at 09:15
  • Yes. If those identically distributed variables are sums of low variance gaussians far apart from each other I think it can take more than 30 self-convolutions to get something close to a gaussian. I'm not on my computational computer right now, but will get back to you when I've tried some. – mathreadler Jun 18 '15 at 10:46
  • @mathreadler The value of 30 is only a rule of thumb, right. But here the value of n is much bigger. Therefore I don´t see a real problem. – callculus42 Jun 18 '15 at 10:54
  • Yep. That should do it. – mathreadler Jun 18 '15 at 11:06