119k views
1 vote
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 an allowable sampling error (E) of 0.03 and a confidence level of 99.74%. Of secondary data indicated that 25% of all adults had used the Internet for such a purpose, what is the sample size

User Kay Am See
by
7.8k points

1 Answer

7 votes

Answer:

The sample size is 1875.

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)}

Sampling error of 0.03.

This means that
M = 0.03

99.74% confidence level

So
\alpha = 0.0026, z is the value of Z that has a pvalue of
1 - (0.0026)/(2) = 0.9987, so
Z = 3.

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

This means that
\pi = 0.25

What is the sample size

The sample size is n. So


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


0.03 = 3\sqrt{(0.25*0.75)/(n)}


0.03√(n) = 3√(0.25*0.75)

Simplifying by 0.03 both sides


√(n) = 100√(0.25*0.75)


(√(n))^2 = (100√(0.25*0.75))^2


n = 1875

The sample size is 1875.

User Tameshwar
by
7.9k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories