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

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

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 (s)/(√(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=(1)/(n)\sum X=(1)/(50)* [1+5+6+...+10]=6.76\\\\s=\sqrt{(1)/(n-1)\sum (x-\bar x)^(2)}=\sqrt{(1)/(49)* 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 (s)/(√(n))`

`=6.76\pm 2.000* (2.552)/(√(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).