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 99% confident that the sample percentage is within 2.5 percentage points of the true population percentage. Complete parts a and b below.

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

n=__

b. Assume that a prior survey suggests that about 36% of air passengers prefer an aisle seat.

n=__

Answer 1

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E =2.5 % = 0.025

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.576 ( Using z table )

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

= (2.576 / 0.025)2 * 0.5* 0.5

=2654.3

Sample size = 2655

(B)

Solution :

Given that,

= 0.36

1 - = 1 - 0.36 = 0.64

margin of error = E =2.5 % = 0.025

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.576 ( Using z table )

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

= (2.576 / 0.025)2 * 0.36* 0.64

=2446.21

Sample size = 2447