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 0.38. 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,

= 0.38

1 - = 1 - 0.38 = 0.62

margin of error = E = 0.04

At 90% confidence level z

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table ( see the 0.05 value in standard normal (z) table corresponding z value is 1.645 )   

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.645 / 0.04)2 * 0.38 * 0.62

=398.46

Sample size = 399