Editors preparing a report on the economy are trying to estimate the percentage of businesses that plan to hire additional employees in the next 60 days. They are willing to accept a margin of error of 6 % but want 99 % confidence. How many randomly selected employers will they need to contact?

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