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Suppose that the epidemiologist wants to re-estimate the population proportion and wishes for her 95% confidence interval to have a margin of error no larger than 0.04. How large a sample should she take to achieve this? Please carry answers to at least six decimal places in intermediate steps.

User Gdir
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1 Answer

4 votes

Answer:

She needs a sample of at least 385.

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)}

95% confidence level

So
\alpha = 0.05, z is the value of Z that has a pvalue of
1 - (0.05)/(2) = 0.975, so
Z = 1.96.

How large a sample should she take to achieve this?

She needs a sample of size at least n.

n is found when M = 0.04.

We do not know the true population proportion, so we use
\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)}


0.05 = 1.96\sqrt{(0.5*0.5)/(n)}


0.05√(n) = 1.96*0.5

Dividing both sides by 0.05


√(n) = 19.6


(√(n))^(2) = 19.6^(2)


n = 384.2

Rounding up

She needs a sample of at least 385.

User Yohanna
by
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