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 5.5 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 31​% of air passengers prefer an aisle seat. nequals nothing

Answer 1

Answer:

A)n= 703.96

B)n= 602.308

Step by step Explanation:

Given that you want to be 99% confident that the sample percentage is within 3.1 percentage points of the true population percentage.

Then z/2 = 1.645

And M = 3.1% = 0.031

A)Nothing is known therefore,

p = q = 0.50

E=0.031

For 90% confidence, z = 1.645

n = (zα/₂)²(p)(1-p)/M²

n= 1.645²× 0.5 × 0.5/0.031²

n= 703.96

Therefore, 703.96randomly selected air passengers must be​ surveyed to be 99%

B)we know that recent surveys surgest that about 38% of all air passengers prefer an aisle seat, thus p = 35% = 0.35

n = (zα/₂)²(p)(1-p)/M²

n= (1.645²× 0.31 × 0.69)/0.031²

n= 602.308

Hence, 602.308 randomly selected air passengers must be​ surveyed to be 90% confident that the sample percentage is within 3.1 percentage points of the true population percentage.