Question

A supplier of a certain car parts chain of stores wants to estimate the average length of time car owners plan to keep their cars. From the past it is known that the standard deviation of the length of times car owners plan to keep their cars is 6.5 years. A random sample of 25 car owners results the average length of time of 7.2 years. Calculate and interpret a 99% confidence interval for the average length of time all car owners plan to keep their cars.

Answer 1

Answer:

The 99% confidence interval for the average length of time all car owners plan to keep their cars is between 3.85 years and 10.55 years.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.005 = 0.995$, so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.575*\frac{6.5}{\sqrt{25}} = 3.35$

The lower end of the interval is the sample mean subtracted by M. So it is 7.2 - 3.35 = 3.85 years

The upper end of the interval is the sample mean added to M. So it is 7.2 + 3.35 = 10.55 years

The 99% confidence interval for the average length of time all car owners plan to keep their cars is between 3.85 years and 10.55 years.