Question

You are the operations manager for an airline and you are considering a higher fare level for passengers in aisle seats. How many randomly selected air passengers must you​ survey ? Assume that a prior survey suggests that about 53​% of air passengers prefer an aisle seat. Assume that you want to be 90% confident that the sample percentage is within 4.5 percentage points of the true population percentage. Round the answer to the nearest large whole number!

Answer 1

Solution :

Given that,

= 0.53

1 - = 1 - 0.53 = 0.47

margin of error = E = 4.5% = 0.045

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= (1.645 / 0.045)2 * 0.53 * 0.47

=332.87

n=333

if z value two decimal so answer is

(1.65 / 0.045)2 * 0.53 * 0.47

=335