Question

NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. In an earlier study, the population proportion was estimated to be 0.32

How large a sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.03? Round your answer up to the next integer.

Answer 1

Solution :

Given that,

= 0.32

1 - = 1-0.32 = 0.68

margin of error = E = 0.03

At 85% confidence level

= 1-0.85% =1-0.85 =0.15

/2 =0.15/ 2= 0.075

Z/2 = Z_{0.075 = 1.44}

Z_/2 = 1.44

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

= (1.44/0.03)² *0.32*0.68

= 501

sample size = n = 501