Answer to Question #347827 in Statistics and Probability for teya

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

The critical value for $\alpha = 0.01, df=n-1=19$ degrees of freedom is $t_c​=z_{1−α/2;n−1}=2.860935.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$ $$=(85.5- 2.860935\times\dfrac{19.3}{\sqrt{20}},$$ $$85.5+ 2.860935\times\dfrac{19.3}{\sqrt{20}})$$

$$=(73.153, 97.847)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $73.153 < \mu < 97.847,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(73.153, 97.847).$