In: Statistics and Probability
A major oil company has developed a new gasoline additive that is supposed to increase mileage. To test this hypothesis, ten cars are randomly selected. The cars are driven both with and without the additive. The results are displayed in the following table. Can it be concluded, from the data, that the gasoline additive does significantly increase mileage?
Let d=(gas mileage with additive)−(gas mileage without additive)d=(gas mileage with additive)−(gas mileage without additive). Use a significance level of α=0.05α=0.05 for the test. Assume that the gas mileages are normally distributed for the population of all cars both with and without the additive.
Car | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | |||
|
16.6 | 11 | 29.3 | 28.5 | 26.5 | 20.1 | 16 | 25.4 | 10.7 | 19.3 | |||
|
17.3 | 13.2 | 30 | 29.5 | 28.7 | 23.2 | 19.1 | 28.4 | 13.8 | 20.4 |
State the null and alternative hypotheses for the test
Find the values of the two sample proportions, pˆ1 and pˆ2. Round to 3 decimal places
Compute the weighted estimate of p, p‾. Round to 3 decimal places
Compute the value of the test statistic. Round to 2 decimal places
Determine the decision rule for rejecting the null hypothesis H0. Round to 3 decimal places [ (Reject H0 if (t or absolute value of t) is (< or >) (value) ]
Make a decision to reject or fail to reject the null hypothesis.
Following table shows the calculations:
Without additive | With additive | d | (d-mean)^2 |
16.6 | 17.3 | 0.7 | 1.7424 |
11 | 13.2 | 2.2 | 0.0324 |
29.3 | 30 | 0.7 | 1.7424 |
28.5 | 29.5 | 1 | 1.0404 |
26.5 | 28.7 | 2.2 | 0.0324 |
20.1 | 23.2 | 3.1 | 1.1664 |
16 | 19.1 | 3.1 | 1.1664 |
25.4 | 28.4 | 3 | 0.9604 |
10.7 | 13.8 | 3.1 | 1.1664 |
19.3 | 20.4 | 1.1 | 0.8464 |
Total | 20.2 | 9.896 |
Sample size: n =8
Now,
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