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 90% confident that the sample percentage is within 2.5 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.
N =
Solution :
Given that,
= 0.5
1 -
= 1 - 0.5 = 0.5
margin of error = E = 2.5% = 0.025
At 90% confidence level
= 1 - 0.90 = 0.10
/2
=0.05
Z/2
= 1.645 ( Using z table )
Sample size = n = (Z/2
/ E)2 *
* (1 -
)
= (1.645 / 0.025)2 * 0.5 * 0.5
=1082.41
Sample size =1083
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