In: Statistics and Probability
A car company is considering a new engine filter for its new
hybrid automobile line. But it does not want to switch to the new
brand unless there is evidence that the new filter can improve fuel
economy for the vehicle (miles per gallon). The experimental design
is set up so that each of the 10 cars drive the same course twice -
once with the old filtration system and once with the new version.
The data collected is shown below:
Car # | Current Filter | New Filter |
1 | 7.6 | 7.3 |
2 | 5.1 | 7.2 |
3 | 10.4 | 6.8 |
4 | 6.9 | 10.6 |
5 | 5.6 | 8.8 |
6 | 7.9 | 8.7 |
7 | 5.4 | 5.7 |
8 | 5.7 | 8.7 |
9 | 5.5 | 8.9 |
10 | 5.3 | 7.1 |
Conduct a hypothesis testing to determine whether the new filtration system is superior. Provide the p-value from your analysis.
Current Filter | New Filter | Difference |
7.6 | 7.3 | 0.3 |
5.1 | 7.2 | -2.1 |
10.4 | 6.8 | 3.6 |
6.9 | 10.6 | -3.7 |
5.6 | 8.8 | -3.2 |
7.9 | 8.7 | -0.8 |
5.4 | 5.7 | -0.3 |
5.7 | 8.7 | -3 |
5.5 | 8.9 | -3.4 |
5.3 | 7.1 | -1.8 |
Sample mean of the difference using excel function AVERAGE(), x̅d = -1.4400
Sample standard deviation of the difference using excel function STDEV.S() sd = 2.2406
Sample size, n = 10
Null and Alternative hypothesis:
Ho : µd = 0
H1 : µd < 0
Test statistic:
t = (x̅d)/(sd/√n) = (-1.44)/(2.2406/√10) = -2.0323
df = n-1 = 9
p-value :
Left tailed p-value = T.DIST(-2.0323, 9, 1) = 0.0363
Decision:
p-value < 0.05, Reject the null hypothesis
Conclusion:
There is enough evidence to conclude that the new filtration system is superior at 0.05 significance level.