The owner of Torpid Oaks B&B wanted to know the average distance its guests had traveled. A random sample of 16 guests showed a mean distance of 85 miles with a standard deviation of 32 miles. The 90 percent confidence interval (in miles) for the mean is approximately:

A) (71.0,99.0)

B) (71.8,98.2)

C) (74.3,95.7)

D) (68.7,103.2)

Question

Business

Ronald

4 August, 07:05

The owner of Torpid Oaks B&B wanted to know the average distance its guests had traveled. A random sample of 16 guests showed a mean distance of 85 miles with a standard deviation of 32 miles. The 90 percent confidence interval (in miles) for the mean is approximately:

A) (71.0,99.0)

B) (71.8,98.2)

C) (74.3,95.7)

D) (68.7,103.2)

+1

Answers (1)

Know the Answer?

Elvis Hood · Answer 1

A) (71.0,99.0)

Explanation:

To calculate the confidence interval we can use the following formula:

mean distance + / - (t-score x standard deviation) / √sample size

85 + / - (1.753 x 32) / √16

85 + / - 56.06 / 4

85 + / - 14.024

minimum = 85 - 14 = 71

maximum = 85 + 14 = 99