Answer to Question #258494 in Statistics and Probability for Jac

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 3,900 kilometers. Assume the distribution of measurements to be approximately normal.

a) Construct a 99% confidence interval for the average number of kilometers an automobile is driven annually in Virginia.

b) 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?

Answer 1

$n=100$

$\bar{x}=23500$

$s=3900$

a.

To compute a 99% confidence interval for the sample mean$(\bar{x})$ we shall apply the students t distribution since the population variance is unknown. A 99% confidence interval is given as,

$C.I=\bar{x}\pm t_{1-\alpha/2,n-1}*(s/\sqrt{n})$ where $t_{1-\alpha/2,n-1}$ is the table value at $\alpha=1\%$ with $(n-1 )$ degrees of freedom.

The table value is given as,

$t_{1-0.001/2,100-1}=t_{0.995,99}=2.626405$

Now,

$C.I=23500\pm 2.626405*(3900/\sqrt{100})$

$C.I=23500\pm2.626405*390=23500\pm 1024.298$

Therefore, a 99% confidence interval for the sample mean is given as,

$C.I=(222475.7, \space 24524.3 )$

b.

The standard error is given by,

$SE=t_{0.995,99}*(s/\sqrt{n})$

$SE=2.626405*(3900/10)=1024.298$

We can assert with 99% confidence that the possible size of our error will not exceed 1024.298 kilometers and that  the true population average of the kilometers driven by the car owners fall inside the confidence interval.