Answer:
They will need to contact 462 people
Explanation:
In this question, we are asked to calculate the number of employers, randomly selected will the editors need to contact.
We proceed as follows;
we note that the confidence level is 99%
For (1−α)=0.99
α=0.01
α /2 = 0.005
From the Standard Normal Table, the required z=0.005 for 99% confidence level is 2.58
We now proceed to calculate the sample size
Mathematically, we can obtain the sample size as follows;
p = 0.5 and q = 1-p = 1-0.5 = 0.5
n = (z^2 * p * q)/(mE)^2 where mE is the margin of error
n = [(2.58)^2 * 0,5 * 0.5]/(0.06)^2
n = 462
