Final answer:
To be 95% confident with a 1.5 percentage point margin of error, 1068 randomly selected passengers must be surveyed if nothing is known about aisle seat preference. If a prior survey indicated a 31% preference, then 753 passengers must be surveyed.
Step-by-step explanation:
You are the operations manager for an airline considering a higher fare level for passengers in aisle seats, and you need to know how many randomly selected air passengers must you survey to be 95% confident that the sample percentage is within 1.5 percentage points of the true population percentage.
Part A
To calculate the sample size needed when nothing is known about the preference for aisle seats, we use the formula for the sample size of a proportion:
n = (Z^2 * p * (1-p)) / E^2
Where:
- n is the sample size
- Z is the z-score corresponding to the confidence level (1.96 for 95% confidence)
- p is the estimated proportion (0.5 if unknown)
- E is the desired margin of error (0.015 for 1.5 percentage points)
Plugging the numbers in:
n = (1.96^2 * 0.5 * (1-0.5)) / 0.015^2
n = 1067.1111, which we round up to 1068.
So, we need to survey at least 1068 air passengers to meet the criteria.
Part B
When assuming a prior survey suggests 31% of air passengers prefer an aisle seat, p changes from 0.5 to 0.31. The new calculation would be:
n = (1.96^2 * 0.31 * (1-0.31)) / 0.015^2
n = 752.3718, which we round up to 753.
So, under this assumption, we would need to survey at least 753 passengers.