Determining Sample Size. In Exercises 31–38, use the given data to find the minimum sample size required to estimate a population proportion or percentage.

Airline Seating

You are the operations manager for American Airlines, and you are considering a higher fare level for passengers in aisle seats. You want to estimate the percentage of passengers who now prefer aisle seats. How many randomly selected air passengers must you survey? Assume that you want to be 95% confident that the sample percentage is within 2.5 percentage points of the true population percentage.

a. Assume that nothing is known about the percentage of passengers who prefer aisle seats.

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

(based on a 3M Privacy Filters survey).

The percentage of passengers who prefer aisle seats is to be estimated.

The sample size needs to be determined. The following values are given:

The margin of error is equal to 0.025.

The confidence level is equal to 95%.

Let $\hat{p}$denote the sample proportion of passengers who prefer aisle seats.

Let $\hat{q}$denote the sample proportion of passengers who prefer aisle seats.

Here, nothing is known about the percentage to be estimated.

The formula for finding the sample size is as follows:

$n=\frac{{\left[z_{\frac{\alpha }{2}}\right]}^{2}0.25}{E^2}$

The confidence level is equal to 95%. Thus, the level of significance is equal to 0.05.

The value of $z_{\frac{\alpha }{2}}$for $\alpha =0.05$from the standard normal table is equal to 1.96

Substituting the required values, the following value of the sample size is obtained:

$$\begin{gathered} n=\frac{{\left(1.96\right)}^2\times 0.25}{{\left(0.025\right)}^2} \\ =1536.64 \\ \approx 1537 \end{gathered}$$

Hence, the required sample size is 1537.

b.

The value of $\hat{p}$is given to be equal to:

$$\begin{gathered} \hat{p}=38\% \\ =\frac{38}{100} \\ =0.38 \end{gathered}$$

The value of $\hat{q}$is given to be equal to:

$$\begin{gathered} \hat{q}=1-\hat{p} \\ =1-0.38 \\ =0.62 \end{gathered}$$

The formula for finding the sample size is as follows:

$n=\frac{{\left[z_{\frac{\alpha }{2}}\right]}^2\hat{p}\hat{q}}{E^2}$

Substituting the required values, the following value of the sample size is obtained:

$$\begin{gathered} n=\frac{{\left(1.96\right)}^2\times 0.38\times 0.62}{{\left(0.025\right)}^2} \\ =1448.13 \\ \approx 1448 \end{gathered}$$

Hence, the required sample size is 1448.