) 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

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 , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

The margin of error is:

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

This means that

99% confidence level

So , z is the value of Z that has a pvalue of , so .

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

Rounding up

n has to be at least 16577.

4444444444444 i think