An advertising firm wants to confirm estimates that 24% of households will buy something on the Internet next year. What sample size will produce a margin of error of 9% with 90% confidence level

A sample size of 61 will produce a margin of error of 9% with 90% confidence level

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 of the confidence interval is:

For this problem, we have that:

90% confidence level

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

What sample size will produce a margin of error of 9% with 90% confidence level

This is n when M = 0.09. So

A sample size of 61 will produce a margin of error of 9% with 90% confidence level

the least common denominatore of 7 and 2 is 9

step-by-step explanation:

7+2=9