Question

An advertising company wishes to estimate the mean household income for all single working professionals who own a foreign automobile. If the advertising company wants a 95​% confidence interval estimate with a margin of error of plus or minus 2600​, what sample size is needed if the population standard deviation is known to be ​$27500​?

Answer 1

Solution :

Given that,

standard deviation = =27500

margin of error = E = 2600

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

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

= (( 1.960 * 27500) / 2600)2

=430

Sample size =430