Answer:
We need to contact 542 employees.
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:
98% confidence level
So , z is the value of Z that has a pvalue of , so .
a) How many randomly selected employers must we contact in order to create an estimate in which we are 98% confident with a margin of error of 5%
We dont know the proportion, so we use , which is when we are going to need the largest sample size.
We have to find n when M = 0.05. So
Rounding up
We need to contact 542 employees.