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. 7 7 3 8 4 4 4 5 5 5 5 4 9 10 9 9 8 10 4 5 4 10 10 10 11 4 9 7 5 4 4 5 5 4 3 10 10 4 4 8 7 7 4 9 5 9 4 4 4 4 Develop a 95% confidence interval estimate of the population mean rating for Miami. Round your answers to two decimal places. ( , )

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	7	0.5476
	7	0.5476
	3	10.6276
	8	3.0276
	4	5.1076
	4	5.1076
	4	5.1076
	5	1.5876
	5	1.5876
	5	1.5876
	5	1.5876
	4	5.1076
	9	7.5076
	10	13.9876
	9	7.5076
	9	7.5076
	8	3.0276
	10	13.9876
	4	5.1076
	5	1.5876
	4	5.1076
	10	13.9876
	10	13.9876
	10	13.9876
	11	22.4676
	4	5.1076
	9	7.5076
	7	0.5476
	5	1.5876
	4	5.1076
	4	5.1076
	5	1.5876
	5	1.5876
	4	5.1076
	3	10.6276
	10	13.9876
	10	13.9876
	4	5.1076
	4	5.1076
	8	3.0276
	7	0.5476
	7	0.5476
	4	5.1076
	9	7.5076
	5	1.5876
	9	7.5076
	4	5.1076
	4	5.1076
	4	5.1076
	4	5.1076
Total	313	299.62

Mean X̅ = Σ Xi / n
X̅ = 313 / 50 = 6.26
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 299.62 / 50 -1 ) = 2.4728

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 50- 1 ) = 2.01 ( Critical value from t table )
6.26 ± t(0.05/2, 50 -1) * 2.4728/√(50)
Lower Limit = 6.26 - t(0.05/2, 50 -1) 2.4728/√(50)
Lower Limit = 5.56
Upper Limit = 6.26 + t(0.05/2, 50 -1) 2.4728/√(50)
Upper Limit = 6.96
95% Confidence interval is ( 5.56 , 6.96 )