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umka2103 [35]
2 years ago
12

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.
Mathematics
1 answer:
erma4kov [3.2K]2 years ago
7 0

Answer:

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

\alpha = \frac{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]Now, find the width of the interval[tex]W = z*\frac{\sigma}{\sqrt{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*\frac{\sigma}{\sqrt{n}}

W = 2.326*\frac{5}{\sqrt{25}}

The correct answer is:

C. ± 2.326 years.

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