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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

User Rcpfuchs
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1 Answer

5 votes

Answer:

We need a sample of at least 752 students.

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{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

The margin of error of the interval is:


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

90% confidence level

So
\alpha = 0.1, z is the value of Z that has a pvalue of
1 - (0.1)/(2) = 0.95, so
Z = 1.645.

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
\pi = 0.5.


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


0.03 = 1.645\sqrt{(0.5*0.5)/(n)}


0.03√(n) = 1.645*0.5


√(n) = (1.645*0.5)/(0.03)


(√(n))^(2) = ((1.645*0.5)/(0.03))^(2)


n = 751.67

Rounding up

We need a sample of at least 752 students.

User ANAS AJI MUHAMMED
by
8.2k points

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