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 90​% confident that the sample percentage is within 4.5 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=? ​

(Round up to the nearest​ integer.)

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

n=?

(Round up to the nearest​ integer.)

Answer 1

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 4.5% = 0.045

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 / 0.045)2 * 0.5 * 0.5

=334.07

Sample size =334

(B)

Solution :

Given that,

= 0.36

1 - = 1 - 0.36= 0.64

margin of error = E = 4.5% = 0.045

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 / 0.045)2 * 0.36 * 0.64

=307.88

Sample size =308