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

9090​%

confident that the sample percentage is within

1.51.5

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.)

b. Assume that a prior survey suggests that about

3232​%

of air passengers prefer an aisle seat.

nequals=nothing

​(Round up to the nearest​ integer.)

Answer 1

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E =1.5 % = 0.015

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= (1.645 / 0.015)2 * 0.5 * 0.5

=3006.69

Sample size = 3007

(B)

Solution :

Given that,

= 0.32

1 - = 1 - 0.32 = 0.68

margin of error = E =1.5 % = 0.015

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= (1.645 / 0.015)2 * 0.32 * 0.68

=2617.02

Sample size =2617