Answer :
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
[tex]\mu = 20*15 = 300, \sigma = \sqrt{300} = 17.32[/tex]
Confidence interval:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] 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
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]10 = 1.96\frac{17.32}{\sqrt{n}}[/tex]
[tex]10\sqrt{n} = 1.96*17.32[/tex]
[tex]\sqrt{n} = \frac{1.96*17.32}{10}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*17.32}{10})^2[/tex]
[tex]n = 11.52[/tex]
Rounding up
The manager should use a sample size of 12 to determine the average travel expenditure