Question

The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maximum possible rating is 10. Suppose a simple random sample of 50 business travelers is selected and each traveler is asked to provide a rating for the Miami International Airport. The ratings obtained from the sample of 50 business travelers follow.

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	6	0.1156
	4	5.4756
	6	0.1156
	8	2.7556
	7	0.4356
	7	0.4356
	6	0.1156
	3	11.1556
	3	11.1556
	8	2.7556
	10	13.3956
	4	5.4756
	8	2.7556
	7	0.4356
	8	2.7556
	7	0.4356
	5	1.7956
	9	7.0756
	5	1.7956
	8	2.7556
	4	5.4756
	3	11.1556
	8	2.7556
	5	1.7956
	5	1.7956
	4	5.4756
	4	5.4756
	4	5.4756
	8	2.7556
	4	5.4756
	5	1.7956
	6	0.1156
	2	18.8356
	5	1.7956
	9	7.0756
	9	7.0756
	8	2.7556
	4	5.4756
	8	2.7556
	9	7.0756
	9	7.0756
	5	1.7956
	9	7.0756
	7	0.4356
	8	2.7556
	3	11.1556
	10	13.3956
	8	2.7556
	9	7.0756
	6	0.1156
Total	317	229.22

Mean X̅ = Σ Xi / n
X̅ = 317 / 50 = 6.34
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 229.22 / 50 -1 ) = 2.1629

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 50- 1 ) = 2.01
6.34 ± t(0.05/2, 50 -1) * 2.1629/√(50)
Lower Limit = 6.34 - t(0.05/2, 50 -1) 2.1629/√(50)
Lower Limit = 5.73
Upper Limit = 6.34 + t(0.05/2, 50 -1) 2.1629/√(50)
Upper Limit = 6.95
95% Confidence interval is ( 5.723 , 6.95 )