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

545 randomly selected air passengers must be surveyed.

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (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{(\pi(1-\pi))/(n)}`

`0.042 = 1.96\sqrt{(0.5*0.5)/(n)}`

`0.042√(n) = 1.96*0.5`

`√(n) = (1.96*0.5)/(0.042)`

`(√(n))^2 = ((1.96*0.5)/(0.042))^2`

`n = 544.4`

Rouding up

545 randomly selected air passengers must be surveyed.