Question 3 4 pts 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 $95 \%$ confident that the sample percentage is within 5.4 percentage points of the true population percentage.
Solution
Step 1 ：We are given that we want to be 95% confident that the sample percentage is within 5.4 percentage points of the true population percentage. This means that our margin of error (E) is 0.054.
Step 2 ：We are also given that the z-score for a 95% confidence level is 1.96.
Step 3 ：Since we don't have an estimated proportion of the population, we'll use 0.5 as it maximizes the sample size. So, p = 0.5.
Step 4 ：We can use the formula for sample size in a proportion to find the required sample size. The formula is: \(n = \frac{{Z^2 \cdot p \cdot (1-p)}}{{E^2}}\)
Step 5 ：Substituting the given values into the formula, we get: \(n = \frac{{(1.96)^2 \cdot 0.5 \cdot (1-0.5)}}{{(0.054)^2}}\)
Step 6 ：Solving the above expression, we find that n = 330.
Step 7 ：So, the operations manager for the airline must survey at least \(\boxed{330}\) randomly selected air passengers.
