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 you want to be 95% confident that the sample percentage is within 4.2 percentage points of the true population percentage.

Answer 1

Answer:

545 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:

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

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 you want to be 95% confident that the sample percentage is within 4.2 percentage points of the true population percentage.

We dont know the true proportion, so we use $\pi = 0.5$, which is when the largest sample size will be needed. We have to find n for which $M = 0.042$. So

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

$0.042 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.042\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.042}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.042})^2$

$n = 544.4$

Rouding up

545 randomly selected air passengers must be surveyed.