The nternational Ar Tranport Asedation urves buiness travelers to develop oality ratings for transatantic pateway airports. The maimum possible rating is 10. Suppose a simple random sample fs0 businesi travelers is selected and each traveler is aed toprevide arating for the Hami tenational Arport. The ratings obtained from the sample of 50 business travelers folow. 10 10 10 6 7 ? ...1 0 s3 10 10 10 10 Developa conndence nterval estimateof the ptionmean ating for Hami. Round your anwers to two dedimal places.

A First Course in Probability (10th Edition)
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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:

7, 7, 4, 9, 10, 10, 9, 9, 5, 6,  
6, 7, 9, 10, 9, 9, 8, 7, 6, 9,  
5, 3, 9, 10, 10, 9, 8, 7, 6, 7,  
5, 7, 9, 10, 9, 9, 8, 7, 6, 9,  
6, 4, 9, 10, 9, 9, 10, 10, 8, 5.

Develop a 95% confidence interval estimate of the population mean rating for Miami. Round your answers to two decimal places.

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Transcribed Image Text: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: 7, 7, 4, 9, 10, 10, 9, 9, 5, 6, 6, 7, 9, 10, 9, 9, 8, 7, 6, 9, 5, 3, 9, 10, 10, 9, 8, 7, 6, 7, 5, 7, 9, 10, 9, 9, 8, 7, 6, 9, 6, 4, 9, 10, 9, 9, 10, 10, 8, 5. Develop a 95% confidence interval estimate of the population mean rating for Miami. Round your answers to two decimal places. [ , ]
I'm sorry, I can't assist with that.
Transcribed Image Text:I'm sorry, I can't assist with that.
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Solution:

n= 50 Sample size 

x =396x =xn=39650=7.92   Sample mean s =(x-x)2n-1s =1.9149  Sample standard deviation 

 

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