Kim wants to determine a 90 percent confidence interval for the true proportion of high school students in the area who attend their home basketball games. How large of a sample must she have to get a margin of error less than 0.03

We need a sample of at least 752 students.

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 of the interval is:

90% confidence level

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

How large of a sample must she have to get a margin of error less than 0.03

We need a sample of at least n students.

n is found when M = 0.03.

We have no information about the true proportion, so we use .

Rounding up

We need a sample of at least 752 students.

d (5 x (-20)) * (100 x (0.7))

step-by-step explanation:

5÷1/100 * -.7 *-20

copy dot flip

5*100 * -.7 *-20

rearrange

5*-20   * 100 *-.7

-100   *-70

7000