Question

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.

Answer 1

C. ± 2.326 years.

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
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.98)/(2) = 0.01`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.01 = 0.99`, so
`z = 2.326/tex]</p><p><strong>Now</strong>, find the width of the interval</p><p>[tex]W = z*(\sigma)/(√(n))`

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

In this question:

`\sigma = 5, n = 25`

So

`W = z*(\sigma)/(√(n))`

`W = 2.326*(5)/(√(25))`

The correct answer is:

C. ± 2.326 years.