Answer:
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