Question

You are an operations manager for American Airlines and you are considering a higher fare level for passengers with aisle seats. You want to estimate the percentage of passengers who now prefer aisle seats. How many randomly selected air passengers must you survey if you want to be 99% confident that the sample percentage is within 4.5 percentage points of the true population percentage? Assume that a recent survey suggests that about 34% of air passengers prefer an aisle seat.

Answer 1

Solution:

Given that:

E= 4.5% = 0.045

= 0.34

1 - = 1 - 0.34= 0.66

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.025 = 2.576

sample size = ( Z/2 / E)2 * * (1 - )     

   = ( 2.576/ 0.045)2 *0.34 *0.66

                   = 735.3

                   = 735

n = 735