 # A random sample of 100 automobile owners in the state of Virginia shows that an automobile is driven on average 23,500 kilometers per year with a standard deviation of 3900 kilometers. Construct a 99\% confidence interval for the average number of kilometers an automobile is driven annually in Virginia. What can we assert with 99\% confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year? UkusakazaL 2021-08-04 Answered
A random sample of 100 automobile owners in the state of Virginia shows that an automobile is driven on average 23,500 kilometers per year with a standard deviation of 3900 kilometers.
Assume the distribution of measurements to be approximately normal.
a) Construct a $99\mathrm{%}$ confidence interval for the average number of kilometers an automobile is driven annually in Virginia.
b) What can we assert with $99\mathrm{%}$ confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year?
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a) The $99\mathrm{%}$ confidence interval for the average number of kilometres an automobile is driven is obtained below:
The value of mean is 23,500 kilometres, population standard deviation is 3,900 and sample size $\left(n\right)=100$
Critical value:
From the standard normal distribution table, for the $99\mathrm{%}$ confidence level the critical values is $\left({z}^{\cdot }\right)2.58$.
The confidence interval formula for the population mean is,
$C.I.=\stackrel{―}{x}±{z}^{\cdot }\frac{\sigma }{\sqrt{n}}$
Substitute mean $=23,500$, standard deviation $\left(\sigma \right)$ is 3,900 and sample size $\left(n\right)=100$.
$C.I.=\stackrel{―}{x}±{z}^{\cdot }\frac{\sigma }{\sqrt{n}}$
$=23,500±2.58\frac{3.900}{\sqrt{100}}$
$=23,500±2.58\left(390\right)$
$=23,500±1006.2$

Thus, the $99\mathrm{%}$ confidence interval for the average number of kilometres an automobile is between $22,493.8$ and $24,506.2$.
b) The required formula is obtained below:
$ME={z}_{\frac{\alpha }{2}}\left(\frac{\sigma }{\sqrt{n}}\right)$
Substitute 3,900 for $\sigma$, 100 for n and 2.58 for ${Z}_{\frac{\alpha }{2}}$.
$ME=2.58\left(\frac{3,900}{\sqrt{100}}\right)$
$=2.58\left(\frac{3,900}{10}\right)$
$=2.58\left(390\right)$
$=1006.2$
With $99\mathrm{%}$ confidence that the possible size of our error will not exceed the 1006.2 kilometres.