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. 3 7 8 9 10 10 10 9 9 9 9 8 5 6 7 7 9 8 10 9 8 7 6 8 7 9 5 3 9 10 10 9 9 8 9 8 6 10 8 4 7 3 10 8 9 8 10 10 8 8 Develop a 95% confidence interval estimate of the population mean rating for Miami. Round your answers to two decimal places.

Answer 1

Sample size = n = 50

Sample mean = = 7.92

Standard deviation = s = 1.9149

We have to construct 95% confidence interval.

Formula is

Here E is a margin of error.

Degrees of freedom = n - 1 = 50 - 1 = 49

Level of significance = 0.05

tc = 2.010 ( Using t table)

So confidence interval is ( 7.92 - 0.544216 , 7.92 + 0.544216) = > ( 7.38 , 8.46)