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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. 6, 4, 6, 8, 7, 7, 6, 3, 3, 8, 10, 4, 8 7, 8, 7, 5, 9, 5, 8, 4, 3, 8, 5, 5, 4 4, 4, 8, 4, 5, 6, 2, 5, 9, 9, 8, 4, 8 9, 9, 5, 9, 7, 8, 3, 10, 8, 9, 6 Develop, a, 95%, confidence, interval, estimate, of, the, population, mean, rating, for, Miami.

Respuesta :

Using the t-distribution, it is found that the 95% confidence interval for the population mean rating for Miami is (5.73, 6.95).

What is a t-distribution confidence interval?

The confidence interval is:

[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]

In which:

  • [tex]\overline{x}[/tex] is the sample mean.
  • t is the critical value.
  • n is the sample size.
  • s is the standard deviation for the sample.

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

Considering the given sample, the other parameters are given as follows:

[tex]\overline{x} = 6.34, s = 2.16, n = 50[/tex]

Hence, the bounds of the interval are:

[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 6.34 - 2.0096\frac{2.16}{\sqrt{50}} = 5.73[/tex]

[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 6.34 + 2.0096\frac{2.16}{\sqrt{50}} = 6.95[/tex]

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

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