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.

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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.