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Bumek [7]
2 years ago
8

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
Mathematics
1 answer:
Lubov Fominskaja [6]2 years ago
4 0

Answer:

The required sample size for the new study is 801.

Step-by-step 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{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

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

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

0.03\sqrt{n} = 1.96\sqrt{0.25*0.75}

\sqrt{n} = \frac{1.96\sqrt{0.25*0.75}}{0.03}

(\sqrt{n})^2 = (\frac{1.96\sqrt{0.25*0.75}}{0.03})^2

n = 800.3

Rounding up:

The required sample size for the new study is 801.

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