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. ± 2.492 years.
B. ± 1.711 years.
C. ± 2.326 years.
D. ± 2.797 years.

C. ± 2.326 years.

Step-by-step explanation:

We have the standard deviation for the sample. So we use the t-distribution to solve this question.

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

In which is the standard deviation of the population and n is the size of the sample.

In this question:

So

The correct answer is:

C. ± 2.326 years.

question:

could we have options?