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 = □
(round up to the nearest integer.)

Explanation:

Step1: Determine 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, the \(z\) - value \(z_{\alpha/2}=z_{0.05}\approx1.645\).

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

When nothing is known about the proportion \(p\), we use \(p = 0.5\) (this gives the maximum value of \(p(1 - p)\)). The formula for the sample size \(n\) in estimating a proportion is \(n=\frac{z_{\alpha/2}^{2}\times p(1 - p)}{E^{2}}\), where \(E = 0.015\) (since \(1.5\) percentage points \(=0.015\)).
Substitute \(z_{\alpha/2}=1.645\), \(p = 0.5\), and \(E=0.015\) into the formula:

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Answer:

\(2963\)