Question

What sample size is required to ensure the 99% confidence interval has a width no greater than 20 when sampling from a population with standard deviation of 30?

Answer 1

Solution :

Given that,

standard deviation =s =   =30

Margin of error = E = width/2=20/2=10

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z_/2 = 2.58

sample size = n = [Z_/2* / E] ²

n = ( 2.58* 30/ 10)²

n =59.9

Sample size = n =60