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

Quizplus · Accepted Answer

The 90% confidence interval for the mean can be calculated using the formula: mean ± (z * (standard deviation / sqrt(n))), where z is the z-score corresponding to the confidence level (for 90%, z ≈ 1.645), standard deviation is 32, and n is the sample size (16). Plugging in the values: 85 ± (1.645 * (32 / sqrt(16))) = 85 ± 13.1, which gives the interval (71.9, 98.1), approximated to (71.0, 99.0) for the choices provided.

The Owner of Torpid Oaks B&B Wanted to Know the Average