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 confidence interval 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

To calculate the width of the confidence interval, we can use the formula:

width = 2 * z * (standard deviation / √n)

where z is the critical value for the 98% confidence interval (which can be found using a standard normal distribution table or calculator), standard deviation is 5 years (given), and n is the sample size of 25 (given).

For a 98% confidence interval, the critical value is approximately 2.326.

So, plugging in the values, we get:

width = 2 * 2.326 * (5 / √25) ≈ 2.492

Therefore, the width of the 98% confidence interval for the true mean client age is approximately ± 2.492 years.

The correct answer is (C).

The Owner of Limp Pines Resort Wanted to Know the Average