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BARSIC [14]
2 years ago
7

You are the operations manager for an airline and you are considering a higher fare level for passengers in aisle seats. How man

y 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.
Mathematics
1 answer:
Georgia [21]2 years ago
7 0

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

<h3>What is a confidence interval of proportions?</h3>

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

#SPJ1

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