Question

you are the operations manager for an airline and you are considering their higher fare for passengers and I'll seats how many randomly selected are passengers must you serve a assume that you want to be 99% confident that the sample percentage is within 3.5 percentage points of the population percentage assume that nothing is known about the percentage of Passenger who prefer aisle seats

Answer 1

Solution:

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 3.5% = 0.035

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

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

= (2.576 / 0.035)2 * 0.5 * 0.5

= 1354

n = sample size =1354