Question

the manager of a local recreational center wanted to determine the average amount each consumer spent on traveling to and from the center. on the basis of the findings, the manager was planning on setting the entrance fee. the management noted that customers living at the other side of town had to travel about 15 miles and spent about 20 cents per mile. the manager wanted to be 95% confident of the findings and did not want the margin of error to exceed /- 10 cents. what sample size should the manger use to determine the average travel expenditure

Answer 1

Answer:

The manager should use a sample size of 12 to determine the average travel expenditure

Step-by-step explanation:

Spent about 20 cents per mile. Distance of 15 miles

We have the mean per a distance, so we use the Poisson distribution. In this distribution, the standard deviation is the square root of the mean. So

$\mu = 20*15 = 300, \sigma = \sqrt{300} = 17.32$

Confidence interval:

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

$\alpha = \frac{1 - 0.95}{2} = 0.025$

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

That is z with a pvalue of $1 - 0.025 = 0.975$, so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

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

What sample size should the manger use to determine the average travel expenditure?

This is n for which M = 10. So

$M = z\frac{\sigma}{\sqrt{n}}$

$10 = 1.96\frac{17.32}{\sqrt{n}}$

$10\sqrt{n} = 1.96*17.32$

$\sqrt{n} = \frac{1.96*17.32}{10}$

$(\sqrt{n})^2 = (\frac{1.96*17.32}{10})^2$

$n = 11.52$

Rounding up

The manager should use a sample size of 12 to determine the average travel expenditure