Question

If the Internet provider wanted an estimate of the proportion p hat that would use the Internet rather than the library, with a margin of error of at most 0.01 in a 99% confidence interval, how large a sample size would be required?

Answer 1

Solution,

Given that,

= 0.5

1 - = 0.5

margin of error = E = 0.01

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 / 0.01 )2 * 0.5 * 0.5

= 16589.44

= 16589

sample size = n = 16589