In: Statistics and Probability
A tire manufacturer is interested in testing the fuel economy for two different tread patterns. Tires of each tread type were driven for 1000 miles on each of 9
different cars. The mileages, in miles per gallon, were as follows:
| Car | Tread A | Tread B |
|---|---|---|
|
1 |
24.7 |
20.4 |
|
2 |
22.6 |
19.0 |
|
3 |
23.9 |
22.4 |
|
4 |
26.8 |
23.0 |
|
5 |
22.4 |
20.8 |
|
6 |
23.4 |
23.5 |
|
7 |
22.6 |
21.3 |
|
8 |
19.7 |
18.1 |
|
9 |
27.3 |
25.8 |
(a) Construct an 80% confidence interval for the mean difference in fuel economy. Let d represent the mileage of tread pattern A minus the mileage of tread B. Use tables to find the critical value and round the answers to at least two decimal places.
| Car | Tread A | Tread B | d = A - B |
| 1 | 24.7 | 20.4 | 4.3 |
| 2 | 22.6 | 19.0 | 3.6 |
| 3 | 23.9 | 22.4 | 1.5 |
| 4 | 26.8 | 23 | 3.8 |
| 5 | 22.4 | 20.8 | 1.6 |
| 6 | 23.4 | 23.5 | -0.1 |
| 7 | 22.6 | 21.3 | 1.3 |
| 8 | 19.7 | 18.1 | 1.6 |
| 9 | 27.3 | 25.8 | 1.5 |
The sample size is n = 9 . The provided sample data along with the data required to compute the sample mean and sample variance are shown in the table below:
| X | X2 | |
| 4.3 | 18.49 | |
| 3.6 | 12.96 | |
| 1.5 | 2.25 | |
| 3.8 | 14.44 | |
| 1.6 | 2.56 | |
| -0.1 | 0.01 | |
| 1.3 | 1.69 | |
| 1.6 | 2.56 | |
| 1.5 | 2.25 | |
| Sum = | 19.1 | 57.21 |
The sample mean is computed as follows:

Also, the sample variance is

Therefore, the sample standard deviation s is

The number of degrees of freedom are df = 9 - 1 = 8, and the significance level is α=0.2.
Based on the provided information, the critical t-value for α=0.2 and df = 8 degrees of freedom is t_c = 1.397
The 80% confidence for the population mean μ is computed using the following expression

Therefore, based on the information provided, the 80 % confidence for the population mean μ is


CI = (1.45, 2.794)
which completes the calculation.