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 90​% confident that the sample percentage is within 1.5 percentage points of the true population percentage. Complete parts​ (a) and​ (b) below. Assume that nothing is known about the percentage of passengers who prefer aisle seats.

Answer 1

Using the z-distribution, it is found that 3,007 passengers must be surveyed.

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

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

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

The margin of error is given by:

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

In this problem, we have a 90% confidence level, hence$\alpha = 0.9$, z is the value of Z that has a p-value of $\frac{1+0.9}{2} = 0.95$, so the critical value is z = 1.645.

In this problem, we desired a margin of error of M = 0.015, with no prior estimate, hence $\pi = 0.5$, then we solve for n to find the minimum sample size.

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

$0.015 = 1.645\sqrt{\frac{0.5(0.5)}{n}}$

$0.015\sqrt{n} = 1.645(0.5)$

$\sqrt{n} = \frac{1.645(0.5)}{0.015}$

$(\sqrt{n})^2 = \left(\frac{1.645(0.5)}{0.015}\right)^2$

n = 3007.

More can be learned about the z-distribution at brainly.com/question/25890103

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