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.
1 5 6 7 8 8 8 9 9 9 9 10 3 4 5 5 7 6 8 9 10 5 4 6 5 7 3 1 9 8 8 9 9 10 7 6 4 8 10 2 5 1 8 6 9 6 8 8 10 10

Requried:
Develop a 95% confidence interval estimate of the population mean rating for Miami.

Answer 1

Answer:

The 95% confidence interval estimate of the population mean rating for Miami is (6.0, 7.5).

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:

$t_{\alpha/2, (n-1)}=t_{0.05/2, 49}\approx t_{0.025, 60}=2.000$

*Use a t-table.

Compute the sample mean and sample standard deviation as follows:

$\bar x=\frac{1}{n}\sum X=\frac{1}{50}\times [1+5+6+...+10]=6.76\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{49}\times 31.12}=2.552$

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

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

     $=6.76\pm 2.000\times \frac{2.552}{\sqrt{50}}\\\\=6.76\pm 0.722\\\\=(6.038, 7.482)\\\\\approx (6.0, 7.5)$

Thus, the 95% confidence interval estimate of the population mean rating for Miami is (6.0, 7.5).