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 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. 8 8 4 0 5 5 5 4 4 4 4 3 10 6 10 10 0 8 5 4 3 2 4 7 8 9 10 8 4 5 5 4 4 3 8 9 9 5 3 9 8 8 5 10 4 10 5 5 3 3 develop a confidence interval estimate of the population mean rating for miami. round your answers to two decimal places. ( , )

Answer 1

Using the t-distribution, the 95% confidence interval of the population mean rating for Miami is (4.97, 6.51).

What is a t-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm t\frac{s}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• t is the critical value.

• n is the sample size.

• s is the standard deviation for the sample.

The parameters for this problem are given as follows:

$\overline{x} = 5.74, s = 2.7, n = 50$

The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 50 - 1 = 49 df, is t = 2.0096.

Hence the bounds of the interval are given as follows:

$\overline{x} - t\frac{s}{\sqrt{n}} = 5.74 - 2.0096\frac{2.7}{\sqrt{50}} = 4.97$

$\overline{x} + t\frac{s}{\sqrt{n}} = 5.74 + 2.0096\frac{2.7}{\sqrt{50}} = 6.51$

More can be learned about the t-distribution at https://brainly.com/question/16162795

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