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

358 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 given by:

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

98% confidence level

So
`\alpha = 0.02`, z is the value of Z that has a pvalue of
`1 - (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{(\pi(1-\pi))/(n)}`

`0.051 = 2.327\sqrt{(0.22*0.78)/(n)}`

`0.051√(n) = 2.327√(0.22*0.78)`

`√(n) = (2.327√(0.22*0.78))/(0.051)`

`(√(n))^2 = ((2.327√(0.22*0.78))/(0.051))^2`

`n = 357.2`

Rounding up

358 randomly selected air passengers must be surveyed.