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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 sample average was 6.34, with a sample standard deviation of 2.163. The ratings obtained from the sample of 50 business travelers follow.2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 10, 10Develop a 95% confidence interval estimate of the population mean rating for Miami.Enter your answer using parentheses and a comma, in the form (n1, n2).

Answer :

ChiKesselman ChiKesselman

Today at 4:06 AM

Answer:

95% Confidence interval: (5.7252 ,6.9548)

Step-by-step explanation:

We are given the following information in the question:

Sample mean = 6.34

Sample standard deviation, s = 2.163

Sample size, n = 50

95% Confidence interval:

[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]

Putting the values, we get,

[tex]t_{critical}\text{ at degree of freedom 49 and}~\alpha_{0.05} = \pm 2.01[/tex]

[tex]6.34 \pm 2.01(\displaystyle\frac{2.163}{\sqrt{50}} ) = 6.34 \pm 0.6148 = (5.7252 ,6.9548)[/tex]

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