Question

) Beth wants to determine a 99 percent confidence interval for the true proportion p of high school students in the area who attend their home basketball games. Out of n randomly selected students she finds that that exactly half attend their home basketball games. About how large would n have to be to get a margin of error less than 0.01 for p

Answer 1

Answer:

n has to be at least 16577.

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:

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

Out of n randomly selected students she finds that that exactly half attend their home basketball games.

This means that $\pi = 0.5$

99% confidence level

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

About how large would n have to be to get a margin of error less than 0.01 for p

We have to find n for which M = 0.01. So

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

$0.01 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.01\sqrt{n} = 2.575*0.5$

$\sqrt{n} = \frac{2.575*0.5}{0.01}$

$(\sqrt{n})^2 = (\frac{2.575*0.5}{0.01})^2$

$n = 16576.6$

Rounding up

n has to be at least 16577.