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 you want to be 90​% confident that the sample percentage is within 2.5 percentage points of the true population percentage.

Complete parts​ (a) and​ (b) below.

Assume that nothing is known about the percentage of passengers who prefer aisle seats.

N =

Answer 1

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 2.5% = 0.025

At 90% confidence level

= 1 - 0.90 = 0.10

/2 =0.05

Z/2 = 1.645 ( Using z table )

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

= (1.645 / 0.025)2 * 0.5 * 0.5

=1082.41

Sample size =1083