UkusakazaL

2021-08-04

The average annual mileage of a car is 23,500 kilometers, with a standard deviation of 3900 kilometers, according to a sample of 100 Virginians who own cars.
Assume that the measurement distribution is roughly normal.
a) Create a 99 percent confidence interval for the typical Virginian vehicle's annual mileage in kilometers.
b) If we assume that Virginians drive an average of 23,500 kilometers per year, what can we say with 99 percent certainty about the potential extent of our error?

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.

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