Question # 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?

Confidence intervals 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 $$\displaystyle{99}\%$$ confidence interval for the average number of kilometers an automobile is driven annually in Virginia.
b) What can we assert with $$\displaystyle{99}\%$$ 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? 2021-08-17

a) The $$\displaystyle{99}\%$$ 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 $$\displaystyle{\left({n}\right)}={100}$$
Critical value:
From the standard normal distribution table, for the $$\displaystyle{99}\%$$ confidence level the critical values is $$\displaystyle{\left({z}^{{\cdot}}\right)}{2.58}$$.
The confidence interval formula for the population mean is,
$$\displaystyle{C}.{I}.=\overline{{{x}}}\pm{z}^{{\cdot}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}}$$
Substitute mean $$\displaystyle={23},{500}$$, standard deviation $$\displaystyle{\left(\sigma\right)}$$ is 3,900 and sample size $$\displaystyle{\left({n}\right)}={100}$$.
$$\displaystyle{C}.{I}.=\overline{{{x}}}\pm{z}^{{\cdot}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}}$$
$$\displaystyle={23},{500}\pm{2.58}{\frac{{{3.900}}}{{\sqrt{{{100}}}}}}$$
$$\displaystyle={23},{500}\pm{2.58}{\left({390}\right)}$$
$$\displaystyle={23},{500}\pm{1006.2}$$
$$\displaystyle={\left({23},{500}-{1006.2},\ {23},{500}+{1006.2}\right)}$$
$$\displaystyle={\left({22},{493.8},\ {24},{506.2}\right)}$$
Thus, the $$\displaystyle{99}\%$$ confidence interval for the average number of kilometres an automobile is between $$\displaystyle{22},{493.8}$$ and $$\displaystyle{24},{506.2}$$.
b) The required formula is obtained below:
$$\displaystyle{M}{E}={z}_{{{\frac{{\alpha}}{{{2}}}}}}{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}$$
Substitute 3,900 for $$\displaystyle\sigma$$, 100 for n and 2.58 for $$\displaystyle{Z}_{{{\frac{{\alpha}}{{{2}}}}}}$$.
$$\displaystyle{M}{E}={2.58}{\left({\frac{{{3},{900}}}{{\sqrt{{{100}}}}}}\right)}$$
$$\displaystyle={2.58}{\left({\frac{{{3},{900}}}{{{10}}}}\right)}$$
$$\displaystyle={2.58}{\left({390}\right)}$$
$$\displaystyle={1006.2}$$
With $$\displaystyle{99}\%$$ confidence that the possible size of our error will not exceed the 1006.2 kilometres.