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A researcher wants to estimate the percentage of all adults that have used the Internet to seek pre-purchase information in the past 30 days, with a tolerable sampling error (E) of 0.03 and a confidence level of 95%. If prior secondary data indicated that 25% of all adults had used the Internet for such a purpose, what is the required sample size for the new study

User Max Bolingbroke
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1 Answer

19 votes
19 votes

Answer:

The required sample size for the new study is 801.

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)}

25% of all adults had used the Internet for such a purpose

This means that
\pi = 0.25

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.

What is the required sample size for the new study?

This is n for which M = 0.03. So


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


0.03 = 1.96\sqrt{(0.25*0.75))/(n)}


0.03√(n) = 1.96√(0.25*0.75)


√(n) = (1.96√(0.25*0.75))/(0.03)


(√(n))^2 = ((1.96√(0.25*0.75))/(0.03))^2


n = 800.3

Rounding up:

The required sample size for the new study is 801.

User James Roeiter
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