Question

Suppose a car rental firm wants to estimate the average number of kilometers traveled per day by each of its cars rented in a certain city. A random sample of 20 cars rented in that city reveals that the sample mean travel distance per day is 85.5 kilometers, with a population standard deviation of 19.3 kilometers. Compute a 99% confidence interval to estimate Q. (2 points) Interpret your answer. (1 point)​

Answer 1

Using the z-distribution, it is found that the 99% confidence interval to estimate Q is (74.4, 96.6). The interpretation is that we are 99% sure that the true mean for all cars rented in the city is between these two values.

What is a z-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• z is the critical value.

• n is the sample size.

• $\sigma$ is the standard deviation for the population.

In this problem, we have a 99% confidence level, hence$\alpha = 0.99$, z is the value of Z that has a p-value of $\frac{1+0.99}{2} = 0.995$, so the critical value is z = 2.575.

The other parameters of the interval are given as follows:

$\overline{x} = 85.5, \sigma = 19.3, n = 20$.

Hence the bounds of the interval are given as follows:

$\overline{x} - z\frac{\sigma}{\sqrt{n}} = 85.5 - 2.575\frac{19.3}{\sqrt{20}} = 74.4$

$\overline{x} + z\frac{\sigma}{\sqrt{n}} = 85.5 + 2.575\frac{19.3}{\sqrt{20}} = 96.6$

The interpretation is that we are 99% sure that the true mean for all cars rented in the city is between these two values.

More can be learned about the z-distribution at https://brainly.com/question/25890103

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