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NASA is conducting an experiment to find out the fraction of people who black out at...

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.

Homework Answers

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

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