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 99% confident that the sample percentage is within 5.5 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 = 549
(round up to the nearest integer.)
b. assume that a prior survey suggests that about 34% of air passengers prefer an aisle seat.
n =
(round up to the nearest integer.)

Explanation:

Step1: Recall the formula for sample size

The formula for sample size \(n\) when estimating a proportion is \(n=\frac{z^{2}\cdot p(1 - p)}{E^{2}}\). For a \(99\%\) confidence level, the \(z -\)score \(z = 2.576\) (from standard normal distribution tables). The margin of error \(E=0.055\).

Part a

Step2: Assume \(p = 0.5\) (when nothing is known)

Substitute \(z = 2.576\), \(p=0.5\), \(1 - p = 0.5\), and \(E = 0.055\) into the formula.

$$n=\frac{(2.576)^{2}\times0.5\times0.5}{(0.055)^{2}}$$

$$n=\frac{6.635776\times0.25}{0.003025}$$

$$n=\frac{1.658944}{0.003025}\approx548.4$$

Since we round up to the nearest integer, \(n = 549\) (already given as correct for part a).

Part b

Step3: Use \(p=0.34\) and \(1 - p = 0.66\)

Substitute \(z = 2.576\), \(p = 0.34\), \(1 - p=0.66\), and \(E = 0.055\) into the formula.

$$n=\frac{(2.576)^{2}\times0.34\times0.66}{(0.055)^{2}}$$

First, calculate \((2.576)^{2}=6.635776\), then \(0.34\times0.66 = 0.2244\)

$$n=\frac{6.635776\times0.2244}{0.003025}$$

$$n=\frac{1.490068}{0.003025}\approx492.6$$

Answer:

\(n = 493\) (rounded up to the nearest integer for part b)