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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 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.

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

Data: 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

My answer:

Sameple: 50 Mean: 6.34 SD: 2.1629

I got -5.73 and +6.95 so (0.61,13.29)

but it keeps saying that its the wrong answer. Anyone out there have any ideas?

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  • $\begingroup$ This question appears to be missing data $\endgroup$ – Inazuma Mar 31 '16 at 22:48
  • $\begingroup$ OOPS you're right give me one second. Alright i added the data. $\endgroup$ – Jake Mar 31 '16 at 22:49
  • $\begingroup$ The SD of the estimator should be a lot smaller than that. Divide by $\sqrt{50}$ or $\sqrt{49}$ depending on what's done in your course. $\endgroup$ – André Nicolas Mar 31 '16 at 22:54
  • $\begingroup$ @AndréNicolas What do you mean should be? and also its divied by √50 $\endgroup$ – Jake Mar 31 '16 at 22:56
  • $\begingroup$ It looks as if you are using the normal, not the $t$-distribution. Your estimate of the population SD is $2.1629$. So the sample mean has SD this divided by $\sqrt{50}$. $\endgroup$ – André Nicolas Mar 31 '16 at 22:59
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You are doing the arithmetic wrong. For a 95% confidence interval, you need to find the values which are on $\pm 1.96 \times \sigma$ of the mean. I don't know where your -5.73 and +6.95 values are coming from, the confidence interval will be symmetric about the mean on a normal distribution. Try it again and if you are still struggling maybe take a look at this page: http://onlinestatbook.com/2/estimation/mean.html

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  • $\begingroup$ So i did, 6.34 (+,-) 1.96 * (2.1629/√50). And i got (5.74,6.94). which was incorrect. $\endgroup$ – Jake Mar 31 '16 at 23:04
  • $\begingroup$ Do they want you to use $t$-distribution, $49$ degrees of freedom? (If you have not heard of it they don't). $\endgroup$ – André Nicolas Mar 31 '16 at 23:12
  • $\begingroup$ @AndréNicolas Never heard of it. Im certain the answer i got is right, it just might be my online classes website. $\endgroup$ – Jake Mar 31 '16 at 23:21

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