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 1.5 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.
n = 3007
(round up to the nearest integer.)
b. assume that a prior survey suggests that about 32% of air passengers prefer an aisle seat.
n =
(round up to the nearest integer.)

Explanation:

Step1: Find the z - value

For a 90% confidence level, the significance level \(\alpha=1 - 0.90 = 0.10\), and \(\alpha/2=0.05\). Using the standard normal distribution table or a calculator, \(z_{\alpha/2}=z_{0.05}\approx1.645\). The margin of error \(E = 0.015\) (since 1.5 percentage points \(=0.015\)).

Step2: Use the formula for sample size when \(p\) is known

The formula for sample size \(n\) when estimating a proportion is \(n=\dfrac{z_{\alpha/2}^{2}\cdot p(1 - p)}{E^{2}}\). Given \(p = 0.32\) (from the prior survey), \(1-p=1 - 0.32=0.68\), \(z_{\alpha/2}=1.645\), and \(E = 0.015\).
Substitute the values into the formula:

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Answer:

\(n = 2626\)