Question

You are interested in polling U.S. employees to see what proportion​ "play hooky", that​ is, call in sick at least once a year when they simply need time to relax. Obtain a sample size that will ensure a margin of error of at most 0.01 for a​ 99% confidence interval.

Confidence level​ = nothing

target proportion​ = nothing

width​ = nothing

n​ = nothing

Answer 1

Solution :

Given that,

= 1 - = 0.5

margin of error = E = 0.01

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 / 0.01)² * 0.5 * 0.5

= 16589

sample size = 16589