Complete 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 99% confident that the sample percentage is within 3.5 percentage points of the true population percentage.
Answer:
1,355 air passengers must be surveyed
Step-by-step explanation:
Note that the population proportion is not stated in the question, it will be assumed to be 50%. i.e. p = o.5
Error Margin, E = 3.5% = 0.035
Confidence Level, CL = 90% = 0.99
Therefore, Significance Level, [tex]\alpha = 0.01[/tex]
At [tex]\alpha = 0.01[/tex], [tex]Z_{crit} = Z_{\alpha/2} = Z_{0.005} = 2.576[/tex]
The number of randomly selected air passengers that must be surveyed:
This is the sample size and is given by the equation:
[tex]n = \frac{(Z_{\alpha/2})^2 p(1-p)}{E^2} \\n = \frac{2.576^2 * 0.5 *(1-0.5)}{0.035^2}\\n = 1354.24\\n = 1355[/tex]
