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.

4	8	9	10	1	1	1	8	8	8	8	7	6
7	8	8	10	9	1	8	7	8	7	9	8	10
6	4	8	1	1	8	8	7	10	9	7	1	7
5	8	4	1	9	8	9	1	1	7	7

Develop a 95% confidence interval estimate of the population mean rating for Miami (to 2 decimals).

( , )

Answer 1

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   3.0132
Sample Size ,   n =    50
Sample Mean,    x̅ = ΣX/n =    6.3200

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   49          
't value='   tα/2=   2.0096   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   3.0132   / √   50   =   0.4261
margin of error , E=t*SE =   2.0096   *   0.4261   =   0.8563
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    6.32   -   0.856333   =   5.4637
Interval Upper Limit = x̅ + E =    6.32   -   0.856333   =   7.1763
95%   confidence interval is (   5.46   < µ <   7.18   )