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

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.