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A researcher would like to estimate the population proportion of adults living in a certain town who have at least a high school education. No more information is available about its value. How large of a sample size is needed to estimate it to within 0.15 of the true value with 99% confidence?

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Answer:

We would need a sample of size at least 74.

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)}

99% confidence level

So
\alpha = 0.01, z is the value of Z that has a pvalue of
1 - (0.01)/(2) = 0.995, so
Z = 2.575.

How large of a sample size is needed to estimate it to within 0.15 of the true value with 99% confidence?

We need a sample of size at least n

n is found when
M = 0.15

We don't know the proportion, so we estimate
\pi = 0.5, which is the case for which we are going to need the largest sample size.


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


0.15 = 2.575\sqrt{(0.5*0.5)/(n)}


0.15√(n) = 2.575*0.5


√(n) = (2.575*0.5)/(0.15)


(√(n))^(2) = ((2.575*0.5)/(0.15))^(2)


n = 73.67

Rounding up

We would need a sample of size at least 74.

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