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 3.5 percentage points of the true population percentage. Complete parts​ (a) and​ (b) below.

Question content area bottom

Part 1

**a.** Assume that nothing is known about the percentage of passengers who prefer aisle seats.nequals=enter your response here

​(Round up to the nearest​ integer.)

Part 2

**b.** Assume that a prior survey suggests that about 38​% of air passengers prefer an aisle seat.

To determine how many randomly selected air passengers must be surveyed, we'll use the formula for the sample size \(n\) in estimating proportions. The formula is:

\[
n = \frac{Z^2 \cdot p \cdot (1 - p)}{E^2}
\]

Where:
- \(Z\) is the Z-score corresponding to the confidence level.
- \(p\) is the estimated proportion of the population (if known), or \(0.5\) if no prior knowledge is available (this provides the maximum variability).
- \(E\) is the margin of error (expressed as a decimal).

Given that the confidence level is 90%, the corresponding Z-score is 1.645. The margin of error is 3.5 percentage points, so \(E = 0.035\).

### Part (a) - Nothing is known about the percentage of passengers who prefer aisle seats:
Here, we assume maximum variability by setting \(p = 0.5\).

\[
n = \frac{(1.645)^2 \cdot 0.5 \cdot (1 - 0.5)}{(0.035)^2}
\]

\[
n = \frac{2.706 \cdot 0.5 \cdot 0.5}{0.001225} = \frac{0.6765}{0.001225} \approx 552.57
\]

Since sample size must be an integer, **round up to 553**.

Thus, you must survey **553** passengers if nothing is known about the percentage of aisle seat preference.

### Part (b) - Assume prior survey suggests 38% prefer an aisle seat:
Now, we use \(p = 0.38\).

\[
n = \frac{(1.645)^2 \cdot 0.38 \cdot (1 - 0.38)}{(0.035)^2}
\]

\[
n = \frac{2.706 \cdot 0.38 \cdot 0.62}{0.001225} = \frac{0.6367}{0.001225} \approx 519.02
\]

Rounding up, you must survey **520** passengers if prior knowledge indicates that 38% prefer aisle seats.

---

Would you like further details on how the Z-score is derived or anything else?

### Related Questions:
1. What does the margin of error signify in a sample survey?
2. How is the Z-score determined for various confidence levels?
3. Why does assuming maximum variability involve setting \(p = 0.5\)?
4. How would the required sample size change if the confidence level was increased to 95%?
5. What impact does decreasing the margin of error have on the sample size?

### Tip:
Always round up the sample size to ensure the margin of error is met or exceeded, rather than rounding down.

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 3.5 percentage points of the true population percentage. Complete parts (a) and (b) below.

Part 1
**a.** Assume that nothing is known about the percentage of passengers who prefer aisle seats.
(Round up to the nearest integer.)

Part 2
**b.** Assume that a prior survey suggests that about 38% of air passengers prefer an aisle seat.

Learn how to calculate the required sample size for a survey to estimate the percentage of air passengers who prefer aisle seats. This problem involves using confidence intervals and proportion estimation formulas. The solution provides step-by-step calculations for both unknown and known prior proportions, making it ideal for statistics students or business managers.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation