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. 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 confidence interval estimate of the population mean rating for Miami. Round your answers to two decimal places.

So on this case the 95% confidence interval would be given by (5.56;6.96)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

represent the sample mean for the sample  

population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

  (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

(2)  

(3)  

The mean calculated for this case is

The sample deviation calculated

In order to calculate the critical value we need to find first the degrees of freedom, given by:

Assuming a Confidence of 0.95 or 95%, the value of and , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that

Now we have everything in order to replace into formula (1):

So on this case the 95% confidence interval would be given by (5.56;6.96)