Question

According to a Harris Poll in 2009, 72% of those who drive and own cell phones say they use them to talk while they are driving. If you wish to conduct a survey in your city to determine what percent of the drivers with cell phones use them to talk while driving, how large a sample should be if you want your estimate to be within 0.02 with 95% confidence.

Answer 1

Answer:

We need a sample of at least 1937.

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is:

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

For this problem, we have that:

$\pi = 0.72$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

How large a sample should be if you want your estimate to be within 0.02 with 95% confidence.

We need a sample of at least n.

n is found when M = 0.02. So

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

$0.02 = 1.96\sqrt{\frac{0.72*0.28}{n}}$

$0.02\sqrt{n} = 1.96\sqrt{0.72*0.28}$

$\sqrt{n} = \frac{1.96\sqrt{0.72*0.28}}{0.02}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.72*0.28}}{0.02})^{2}$

$n = 1936.16$

Rounding up to the nearest number.

We need a sample of at least 1937.