Question

The Environmental Protection Agency (EPA) publishes fuel economy values that are known to be good estimates of the fuel economy a typical driver will achieve under average driving conditions. One of the fuel economy values the EPA publishes is a combined estimate, which represents a combination of city driving (55%) and highway driving (45%). A large car rental company has 16 Ford Explorers in their fleet. They collected fuel economy data from these cars and calculated the sample mean to be 23.29 mpg and sample standard deviation as 0.78 mpg.
Required:
The car rental company is willing to assume that the conditions are met for constructing a confidence interval. Use the collected data to construct a 90% confidence interval for the mean combined fuel economy for Ford Explorers.

Answer 1

Answer:

The 90% confidence interval for the mean combined fuel economy for Ford Explorers is between 22.95 and 23.63 mpg.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$. So we have T = 1.7531

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.7531\frac{0.78}{\sqrt{16}} = 0.34$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 23.29 - 0.34 = 22.95 mpg

The upper end of the interval is the sample mean added to M. So it is 23.29 + 0.34 = 23.63 mpg

The 90% confidence interval for the mean combined fuel economy for Ford Explorers is between 22.95 and 23.63 mpg.