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

9999​%

confident that the sample percentage is within

5.55.5

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.

nequals=nothing

​(Round up to the nearest​ integer.)

Answer 1

Solution :

Given that,

= 0.5 ( assume 0.5)

1 - = 1 - 0.5= 0.5

margin of error = E =5.5 % = 0.055

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 / 0.055)2 * 0.5* 0.5

=548.41

Sample size = 549