Unoccupied seats on flights mean lost revenue for airlines. Suppose a large airl

Question

Unoccupied seats on flights mean lost revenue for airlines. Suppose a large airline wants to estimate its meannumber of unoccupied seats per flight over the last year. They take a random sample for 225 flights andcalculate the average number of unoccupied seats to be 11.6. Assume the standard deviation for the numberof unoccupied seats per flight for this airline is known and equal to 4.103.

(a) Find and interpret a 90% confidence interval for, the mean number of unoccupied seats per flight forthis particular airline. Note write down both of the short and formal interpretation of the confidenceinterval.

(b) Suppose the airline wants their 90% confidence interval to have a margin of error of no more than 0.25.How large must the sample size be in order to satisfy this requirement?

Explanation / Answer

(a)

Above 90% CI indicates, there are 90% chance that for a sample of 225 flights of the airlines, the mean will fall within the range of (11.1501 , 12.0499)

b)

Hence required sample size = 729

CI for 90% n 225 mean 11.6 z-value of 90% CI 1.6449 std. dev. 4.103 SE = std.dev./sqrt(n) 0.27353 ME = z*SE 0.44992 Lower Limit = Mean - ME 11.15008 Upper Limit = Mean + ME 12.04992 90% CI (11.1501 , 12.0499 )

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