Question

Editors preparing a report on the economy are trying to estimate the percentage of business that plan to hire additional employees in the next 60 days. They are willing to accept a margin of error of 7% but want 99% confidence.

How many randomly selected employers will they need to contact?

Answer 1

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 7% = 0.07

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (2.576 / 0.07)² * 0.5 * 0.5

= 338.56

sample size = 339

339 randomly selected employers will they need to contact .