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 9595​% confident that the sample percentage is within 2.52.5 percentage points of the true population percentage. complete parts​ (a) and​ (b) below.

Answer 1

Recall that margin of error is given by:

$M=z_{\alpha/2}\sqrt{ \frac{p(1-p)}{n}} \\ \\ \Rightarrow\frac{M^2}{\left(z_{\alpha/2}\right)^2}= \frac{p(1-p)}{n} \\ \\ \Rightarrow n= \frac{\left(z_{\alpha/2}\right)^2p(1-p)}{M^2}$

Given that you want to be 95​% confident that the sample percentage is within 2.5 percentage points of the true population percentage.

This means that $z_{\alpha/2}=1.96$ and M = 2.5% = 0.025.

Part A:

When nothing is known about the percentage of persengers who prefer aisle seat, we make use of p = 50% = 0.5.

Thus,

$n=\frac{1.96^2\times0.5\times0.5}{(0.025)^2}= \frac{0.9604}{0.000625} =1,537$

Therefore, 1,537 randomly selected air passengers must be​ surveyed to be 95​% confident that the sample percentage is within 2.5 percentage points of the true population percentage.



Part B:

Given that recent surveys surgest that about 38% of all air passengers prefer an aisle seat, thus p = 38% = 0.38

Thus,

$n=\frac{1.96^2\times0.38\times0.62}{(0.025)^2}= \frac{0.9051}{0.000625} =1,449$

Therefore, 1,449 randomly selected air passengers must be​ surveyed to be 95​% confident that the sample percentage is within 2.5 percentage points of the true population percentage.