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. 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 (to 2 decimals).

The 95% confidence interval estimate of the population mean  is (7.38, 8.46).

Step-by-step explanation:

Before building the confidence interval, we need to find the mean and standard deviation of the sample.

In this sample, there are 50 values.

Mean of the sample:

Sum of all values divided by the number of values.

With the help of a calculator, the mean is of 7.92.

Standard deviation:

Square root of the division of the sum of the difference squared between each value and the mean by the number of the values.

With the help of a calculator, the standard deviation is of 1.8958.

Confidence interval:

We have the confidence interval for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 50 - 1 = 49

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 49 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.01

The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 7.92 - 0.54 = 7.38

The upper end of the interval is the sample mean added to M. So it is 7.92 + 0.54 = 8.46

The 95% confidence interval estimate of the population mean  is (7.38, 8.46).

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