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 3.5 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 34​% of air passengers prefer an aisle seat. nequals nothing ​(Round up to the nearest​ integer.)

Answer 1

Solution,

Given that,

a) =  1 - = 0.5

margin of error = E = 0.035

At 90% confidence level

= 1 - 90%

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05  = 1.645

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

= (1.645 / 0.035)2 * 0.5 * 0.5

= 552.25

sample size = n = 553

b) = 0.34

1 - = 1 - 0.34 = 0.66

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

= (1.645 / 0.035)2 * 0.34 * 0.66

= 495.69

sample size = n = 496