Question

How large a sample should be taken if the population mean is to be estimated with 99% confidence to within $75? The population has a standard deviation of $893. (Round your answer up to the next whole number.)

You may need to use the appropriate table in Appendix B to answer this question.

Answer 1

Answer :

Given data :-

The population has a standard deviation = 893

S = 893

margin of error = 75

E = 75

Confidence interval = 99%

= 99/100

C = 0.99

Now we need to find out the  How large a sample should be taken if the population n = ?

we know that

n = z(alpha/2)^2 S^2 / E^2

Where,

S = 893

E = 75

z(alpha/2) => alpha = 1-C

= 1-0.99

  alpha = 0.01

z(alpha/2) = 0.01/2

z(alpha/2) = 0.005

by using the z normal table technology

z(alpha/2) = 2.5758

let,

n = z(alpha/2)^2 S^2 / E^2

=> n = ( 2.5758² * 893²)/75²

=> n = ( 6.6347*797449)/5625

=> n = (5290834.88)/5625

=> n = 940.59

rounding up the answer

n = 941

the large random sample n = 941