Question

The Arizona Department of Transportation wishes to survey state residents to determine what proportion of the population would like to increase statewide highway speed to 75 from 65 mph. At least how many residents do they need to survey if they want to be at least 99% confident that the sample proportion is within 0.02 of the true proportion?

Answer 1

Answer:

They need to survey 4145 residents.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

The margin of error is given by:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

At least how many residents do they need to survey if they want to be at least 99% confident that the sample proportion is within 0.02 of the true proportion?

This is n for which $M = 0.02$. As we have no estimate for the proportion, we use $\pi = 0.5$. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.02\sqrt{n} = 2.575*0.5$

$\sqrt{n} = \frac{2.575*0.5}{0.02}$

$(\sqrt{n})^2 = (\frac{2.575*0.5}{0.02})^2$

$n = 4144.1$

Rounding up:

They need to survey 4145 residents.