# The owner of a large equipment rental company wants to make a rather quick estimate of the average number of days a piece of ditch-digging equipment i

The owner of a large equipment rental company wants to make a rather quick estimate of the average number of days a piece of ditch-digging equipment is rented out per person per time. The company has records of all rentals, but the amount of time required to conduct an audit of all accounts would be prohibitive. The owner decides to take a random sample of rental invoices. Fourteen different rentals of ditch-diggers are selected randomly from the files, yielding the following data. She wants to use these data to construct a $99\mathrm{%}$ confidence interval to estimate the average number of days that a ditch-digger is rented and assumes that the number of days per rental is normally distributed in the population. Use only the appropriate formula and/or statistical table in your textbook to answer this question. Report your answer to 2 decimal places, using conventional rounding rules.
DATA: 3 1 3 2 5 1 2 1 4 2 1 3 1 1
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Derrick

Step 1
The formula for $99\mathrm{%}$ confidence interval to estimate the average number of days that a ditch-digger rented is given below:
$CI=\stackrel{―}{X}±{t}_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)$
Step 2
The sample mean and standard deviation can be obtained using Excel:
Enter the data into Excel file.
Enter $“=AVERAGE\left(A1:A14\right)\text{"}$ to get the sample mean as 14.
Enter $“=STDEV.S\left(A1:A14\right)\text{"}$ to get the sample variance as 30.
That is,$\stackrel{―}{X}=2.14,s=1.30\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}n=14$
Step 3
The degrees of freedom corresponding to t distribution is $n-1.$
From the table of student "t" distribution, the critical value corresponding to 13 degrees of freedom is 3.01.
Step 4
The $99\mathrm{%}$ confidence interval to estimate the average number of days that a ditch-digger is (1.09, 3.19).
$CI=\stackrel{―}{X}±{t}_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)$
$=2.14±3.01\left(\frac{1.30}{\sqrt{14}}\right)$
$=2.14±3.01\left(0.35\right)$
$=2.14±1.05$
$=\left(1.09,3.19\right)$