In: Statistics and Probability
show your work please
You are the operations manager for American Airlines and you are considering a higher fare level for passengers in aisle seats. You want to estimate the percentage of passengers who now prefer aisle seats. a) How many randomly selected air passengers must you survey? Assume that you want to be 95% confident that the sample percentage is within 2.5 percentage points of the true population percentage. b) Assume that a prior survey suggests that about 38% of air passengers prefer an aisle seat.
final answers:
a) Sample Size | |
b) Sample Size | |
Hypothesis |
Solution,
Given that,
= 1 - = 0.5
margin of error = E = 0.025
At 95% confidence level
= 1 - 95%
= 1 - 0.95 =0.05
/2
= 0.025
Z/2
= Z0.025 = 1.96
sample size = n = (Z / 2 / E )2 * * (1 - )
= (1.96 /0.025 )2 * 0.5 * 0.5
= 1536.64
sample size = n = 1537
b) = 0.38
1 - = 1- 0.38 = 0.62
sample size = n = (Z / 2 / E )2 * * (1 - )
= (1.96 /0.025 )2 * 0.38 * 0.62
= 1448.12
sample size = n = 1449