Question

7.1.35-T Question Help 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 95% confident that the sample percentage is within 3.5 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. (Round up to the nearest integer.)​

Answer 1

Answer:

You must survey 784 air passengers.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

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

This means that $\pi = 0.5$, which is when the largest sample size will be needed.

Within 3.5 percentage points of the true population percentage.

We have to find n for which M = 0.035. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.035 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.035\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.035}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.035})^2$

$n = 784$

You must survey 784 air passengers.