In: Statistics and Probability
Two fuel additives are being tested to determine their effect on gasoline mileage. Seven cars were tested with additive 1 and nine cars were tested with additive 2. The following data show the miles per gallon obtained with the two additives.
Additive 1 | Additive 2 |
---|---|
17.3 | 17.7 |
17.4 | 18.8 |
20.1 | 21.3 |
15.7 | 20.0 |
18.2 | 22.1 |
17.6 | 19.7 |
17.5 | 18.8 |
19.7 | |
21.2 |
Use α = 0.05 and the MWW test to see whether there is a significant difference between gasoline mileage for the two additives.
Find the value of the test statistic:
W = ?
Find the p-value. (Round your answer to four decimal places.):
p-value = ?
Additive 1 ( X ) | Additive 2 ( Y ) | |||
17.3 | 0.1488 | 17.7 | 4.9382 | |
17.4 | 0.0816 | 18.8 | 1.2593 | |
20.1 | 5.8288 | 21.3 | 1.8983 | |
15.7 | 3.943 | 20 | 0.0061 | |
18.2 | 0.2645 | 22.1 | 4.7428 | |
17.6 | 0.0073 | 19.7 | 0.0494 | |
17.5 | 0.0345 | 18.8 | 1.2593 | |
19.7 | 0.0494 | |||
21.2 | 1.6328 | |||
Total | 123.8 | 10.3085 | 179.3 | 15.8356 |
Mean
Standard deviation
Mean
Standard deviation
To Test :-
H0 :-
H1 :-
Test Statistic :-
t = -3.2784
Test Criteria :-
Reject null hypothesis if
DF = 13
Result :- Reject Null Hypothesis
Decision based on P value
P - value = P ( t > 3.2784 ) = 0.006
Reject null hypothesis if P value <
level of significance
P - value = 0.006 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis
There is sufficient evidence to support the claim that there is a significant difference between gasoline mileage for the two additives.