Question

What sample size would be required so that the width of the 99% confidence interval would be at most 0.04 units wide?

Answer 1

Solution,

Given that,

=  1 - = 0.5  

margin of error = E = width / 2 = 0.04 / 2 = 0.02

At 99% confidence level

= 1 - 99%

= 1 - 0.99 =0.01

/2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= (2.576 / 0.02)² * 0.5 * 0.5

= 4147.36

sample size = n = 4148