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 0.49.

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 90% confidence level with an error of at most 0.04? Round your answer up to the next integer.

Answer 1

Solution :

Given that,

= 0.49

1 - = 1 - 0.49 = 0.51

margin of error = E = 0.04

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= (1.645 / 0.04)² * 0.49 * 0.51

= 422.64 = 423

sample size = 423