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 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 29% of air passengers prefer an aisle seat.

n = ?

​(Round up to the nearest​ integer.)

Answer 1

Solution :

Given that,

a) = 0.50

1 - = 1-0.50 = 0.50

margin of error = E = 0.045

At 99% confidence level

= 1-0.99% =1-0.99 =0.01

/2 =0.01/ 2= 0.005

Z/2 = Z_{0.005 = 2.576}

Z/2 = 2.576

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (2.576/0.045)² *0.50*0.50

= 819

sample size = n = 819

b)

= 0.29

1 - = 1-0.29 = 0.71

margin of error = E = 0.045

At 99% confidence level

= 1-0.99% =1-0.99 =0.01

/2 =0.01/ 2= 0.005

Z/2 = Z_{0.005 = 2.576}

Z/2 = 2.576

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (2.576/0.045)² *0.29*0.71

= 675

sample size = n = 675