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 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 business travelers follow. 2 6 7 8 9 9 9 10 10 10 10 9 4 5 6 6 8 7 9 10 9
6 5 7 6 8 4 2 10 9 9 10 10 9 8 7 5 9 9 3 6 2 9
7 10 7 9 9 9 9

Required:
Develop a confidence interval estimate of the population mean rating for Miami.

6.8622 ; 8.1778

Step-by-step explanation:

Given the data:

2 6 7 8 9 9 9 10 10 10 10 9 4 5 6 6 8 7 9 10 9

6 5 7 6 8 4 2 10 9 9 10 10 9 8 7 5 9 9 3 6 2 9

7 10 7 9 9 9 9

Assume a confidence interval of 95%

Using calculator :

The mean of the distribution (m) = 7.52

Standard deviation (s) = 2.314

Sample size),

Obtaining the t score ;

Degree of freedom (df) = (n-1) = 50 - 1 = 49

t(1-α/2), 49 = t(0.025, 49) = 2.01 ( t distribution table)

Margin of Error :

2.01 * s/√n = 2.01 * 2.314/7.0710678

Margin of Error (E) = 0.6578

Confidence interval:

(Mean - E) ≤μ≤ (mean + E)

(7.52 - 0.6578) ≤μ≤ (7.52 + 0.6578)

6.8622 ≤μ≤ 8.1778

step-by-step explanation:

the answer is c. 23/3