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

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 , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

The margin of error is:

Sampling error of 0.03.

This means that

99.74% confidence level

So , z is the value of Z that has a pvalue of , so .

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

This means that

What is the sample size

The sample size is n. So

Simplifying by 0.03 both sides

The sample size is 1875.

4 + 5 7/9= 9 7/9

step-by-step explanation: