Respuesta :
A confidence interval estimate of the population mean rating for Miami is:
Confidence interval:
(Mean - E) ≤ μ ≤ (mean + E)
(7.52 - 0.6578) ≤ μ ≤ (7.52 + 0.6578)
6.8622 ≤μ≤ 8.1778
Now, According to the question:
Given the data:
2 6 7 8 9 9 9 10 10 10 10 9 4 5 6 6 8 7 9 10 9
6 5 7 6 8 4 2 10 9 9 10 10 9 8 7 5 9 9 3 6 2 9
7 10 7 9 9 9 9
Assume a confidence interval of 95%
Using calculator :
The mean of the distribution (m) = 7.52
Standard deviation (s) = 2.314
Sample size),
Obtaining the t score ;
Degree of freedom (df) = (n-1) = 50 - 1 = 49
t(1-α/2), 49 = t(0.025, 49) = 2.01 ( t distribution table)
Margin of Error :
2.01 * s/√n = 2.01 * 2.314/7.0710678
Margin of Error (E) = 0.6578
Confidence interval:
(Mean - E) ≤ μ ≤ (mean + E)
(7.52 - 0.6578) ≤ μ ≤ (7.52 + 0.6578)
6.8622 ≤μ≤ 8.1778
Learn more about Confidence Interval at:
https://brainly.com/question/2396419
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The complete question is this:
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.
2 6 7 8 9 9 9 10 10 10 10 9 4 5 6 6 8 7 9 10 9
6 5 7 6 8 4 2 10 9 9 10 10 9 8 7 5 9 9 3 6 2 9
7 10 7 9 9 9 9
Required:
Develop a confidence interval estimate of the population mean rating for Miami.
