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 38​% of air passengers prefer an aisle seat. Assume that you want to be 91% confident that the sample percentage is within 0.3 percentage points of the true population percentage.

Answer 1

Solution :

Given that,

= 0.38

1 - = 1 - 0.38 = 0.62

margin of error = E = 0.3% = 0.003

At 91% confidence level the z is ,

Z/2 = Z0.045 =1.695 ( Using z table )

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

= (1.695 / 0.003)2 * 0.38 * 0.62

= 75209.41

Sample size =75210