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 a prior survey suggests that about 22​% of air passengers prefer an aisle seat. Assume that you want to be 98% confident that the sample percentage is within 5.1 percentage points of the true population percentage.

Answer 1

Answer:

358 randomly selected air passengers must be surveyed.

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 given by:

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

98% confidence level

So $\alpha = 0.02$, z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$, so $Z = 2.327$.

Assume that a prior survey suggests that about 22​% of air passengers prefer an aisle seat.

This means that $\pi = 0.22$

How many randomly selected air passengers must you​ survey?

We need a sample of n, and n is found for which $M = 0.051$. So

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

$0.051 = 2.327\sqrt{\frac{0.22*0.78}{n}}$

$0.051\sqrt{n} = 2.327\sqrt{0.22*0.78}$

$\sqrt{n} = \frac{2.327\sqrt{0.22*0.78}}{0.051}$

$(\sqrt{n})^2 = (\frac{2.327\sqrt{0.22*0.78}}{0.051})^2$

$n = 357.2$

Rounding up

358 randomly selected air passengers must be surveyed.