The owner of Limp Pines Resort wanted to know the average age of its clients. A random sample of 25 tourists is taken. It shows a mean age of 46 years with a standard deviation of 5 years. The width of a 98 percent CI for the true mean client age is approximately:
A)± 1.711 years.
B)± 2.326 years.
C)± 2.492 years.
D)± 2.797 years.

Question

Quizplus · Accepted Answer

We are given the sample mean and standard deviation as follows:

Sample mean (x̄) = 46 years 
Standard deviation (s) = 5 years 
Sample size (n) = 25

We are asked to find the width of a 98% confidence interval (CI) for the true mean client age.

We can use the formula for the confidence interval:

CI = x̄ ± z*(s/√n)

where z is the z-score corresponding to the desired level of confidence.

For a 98% confidence level, we find the z-score from the z-table or calculator to be approximately 2.326.

Thus, the width of the confidence interval is:

CI = 46 ± 2.326*(5/√25) = 46 ± 2.492

Therefore, the answer is C, ± 2.492 years.

The Owner of Limp Pines Resort Wanted to Know the Average