Final answer:
To survey air passengers with a given confidence level and margin of error, use the formula for sample size with the Z-value for the desired confidence level. For part (a), without knowing preferences, use 0.5 for p. For part (b), where 31% of passengers are assumed to prefer aisle seats, use 0.31 for p.
Step-by-step explanation:
To determine how many randomly selected air passengers you must survey, you need to consider the desired level of confidence and the margin of error. For part a, where nothing is known about the percentage of passengers who prefer aisle seats, we use the formula for sample size in estimating a proportion:
n = (Z^2 * p * (1 - p)) / E^2
Where:
- Z is the Z-value from the standard normal distribution for the desired confidence level
- p is the estimated proportion of success
- E is the desired margin of error
Since we don't have a prior estimate for p, we use 0.5 as it maximizes the required sample size. For a 90% confidence level, the Z-value is approximately 1.645. Plugging in the values to the formula, we get:
n = (1.645^2 * 0.5 * 0.5) / 0.055^2
Solving this, you'll need to round up to the nearest integer.
For part b, where it is assumed that 31% of air passengers prefer an aisle seat, we use the same formula with p = 0.31:
n = (1.645^2 * 0.31 * 0.69) / 0.055^2
Again, calculate the value and round up to the nearest integer for the sample size.