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.
Click on the datafile logo to reference the 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
Develop a 95% confidence interval estimate of the population mean rating for Miami. If required, round your answers to two decimal places. Do not round intermediate calculations.

Answer 1

Answer:

The 95% confidence interval estimate of the population mean rating for Miami is (5.7, 7.0).

Step-by-step explanation:

The (1 - α)% confidence interval for the population mean, when the population standard deviation is not provided is:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}$

The sample selected is of size, n = 50.

The critical value of t for 95% confidence level and (n - 1) = 49 degrees of freedom is:

$t_{\alpha/2, (n-1)}=t_{0.05/2, 49}=2.000$

*Use a t-table.

Compute the sample mean and sample standard deviation as follows:

$\bar x=\frac{1}{n}\sum {x}=\frac{1}{50}\times [6+4+6+...+9+6]=6.34\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{50-1}\times 229.22}=2.163$

Compute the 95% confidence interval estimate of the population mean rating for Miami as follows:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}$

     $=6.34\pm 2.00\times\frac{2.163}{\sqrt{50}}\\\\=6.34\pm 0.612\\\\=(5.728, 6.952)\\\\\approx(5.7, 7.0)$

Thus, the 95% confidence interval estimate of the population mean rating for Miami is (5.7, 7.0).