Question

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

Answer 1

Answer:

The sample size is 1875.

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:

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

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 $1 - \frac{0.0026}{2} = 0.9987$, 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

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

$0.03 = 3\sqrt{\frac{0.25*0.75}{n}}$

$0.03\sqrt{n} = 3\sqrt{0.25*0.75}$

Simplifying by 0.03 both sides

$\sqrt{n} = 100\sqrt{0.25*0.75}$

$(\sqrt{n})^2 = (100\sqrt{0.25*0.75})^2$

$n = 1875$

The sample size is 1875.