Question

A researcher wishes to estimate the average number of minutes per day a person spends on the Internet. How large a sample must she select if she wishes to be 99% confident that the population mean is within 10 minutes of the sample mean? Assume the population standard deviation is 42 minutes. Round your final answer up to the next whole number

Answer 1

Solution :

Given that,

standard deviation =s =   =42

Margin of error = E = 10

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58 ( Using z table ( see the 0.005 value in standard normal (z) table corresponding z value is 2.58 )  

sample size = n = [Z/2* / E] 2

n = ( 2.58*42 / 10 )2

n =117.42

Sample size = n =118