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# You may need to use the appropriate appendix table or technology to answer this question. An 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 a certain airport. The ratings obtained from the sample of 50 business travelers follow. 6 4 7 8 7 6 6 3 3 8 10 4 8 7 8 7 5 9 4 8 4 3 8 5 6 4 4 4 8 4 5 5 2 5 9 9 9 4 8 9 9 5 9 7 8 3 10 7 9 6 Develop a 95% confidence interval estimate of the population mean rating for this airport. $Round your answers to two decimal places.$

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## Answer to a math question You may need to use the appropriate appendix table or technology to answer this question. An 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 a certain airport. The ratings obtained from the sample of 50 business travelers follow. 6 4 7 8 7 6 6 3 3 8 10 4 8 7 8 7 5 9 4 8 4 3 8 5 6 4 4 4 8 4 5 5 2 5 9 9 9 4 8 9 9 5 9 7 8 3 10 7 9 6 Develop a 95% confidence interval estimate of the population mean rating for this airport. $Round your answers to two decimal places.$

Bud
4.6
To find the 95% confidence interval estimate of the population mean rating for this airport, we first need to find the sample mean and the sample standard deviation.

Given data:
n = 50
\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i

Calculating the sample mean:
\bar{x} = \frac{1}{50} $6 + 4 + \cdots + 7 + 9 + 6$
\bar{x} = \frac{309}{50}
\bar{x} = 6.18

Next, we calculate the sample standard deviation as follows:
s = \sqrt{\frac{\sum_{i=1}^{n} $x_i - \bar{x}$^2}{n-1}}

Calculating the sample standard deviation:
s = \sqrt{\frac{$6-6.18$^2 + $4-6.18$^2 + \cdots + $6-6.18$^2}{49}}
s = \sqrt{\frac{3.88}{49}}
s ≈ \sqrt{0.079184}
s ≈ 0.2815

Now, we calculate the margin of error $E$ using the t-distribution and the formula:
E = t \cdot \frac{s}{\sqrt{n}}

For a 95% confidence level with n-1 = 49 degrees of freedom, the t-score is approximately 2.0096 .

Calculating the margin of error:
E = 2.0096 \cdot \frac{0.2815}{\sqrt{50}}
E ≈ 0.07986

Finally, we construct the 95% confidence interval estimate for the population mean rating by adding and subtracting the margin of error from the sample mean:
CI = \bar{x} \pm E
CI = 6.18 \pm 0.07986
CI = $6.10, 6.26$

Therefore, the 95% confidence interval estimate of the population mean rating for this airport is $6.10, 6.26$ .

\boxed{$6.10, 6.26$}

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