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 95​% 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=  

Assume that a prior survey suggests that about 32% of air passengers prefer an aisle seat.

n=

Answer 1

Solution :

Given that,

margin of error = E = 0.025

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

(a)

= 0.5

1 - = 1 - 0.5 = 0.5

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.96 / 0.025)² * 0.5 * 0.5

= 1537

sample size = n = 1537

(b)

= 0.32

1 - = 1 - 0.32 = 0.68

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.96 / 0.025)² * 0.32 * 0.68

= 1338

sample size = n = 1338