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

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.

i can't see the picture : /