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You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately 32%. You would like to be 90% confident that your estimate is within 4% of the true population proportion. How large of a sample size is required?

User DominiCane
by
7.6k points

1 Answer

4 votes

Answer:

We need a sample size of at least 369.

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 is:


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

For this problem, we have that:


\pi = 0.32

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 size is required?

We need a sample size of at least n.

n is found when
M = 0.04. So


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


0.04 = 1.645\sqrt{(0.32*0.68)/(n)}


0.04√(n) = 1.645√(0.32*0.68)


√(n) = (1.645√(0.32*0.68))/(0.04)


(√(n))^(2) = ((1.645√(0.32*0.68))/(0.04))^(2)


n = 368.02

Rounding up

We need a sample size of at least 369.

User Reda Maachi
by
7.8k points

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