1 answer

A major oil company has developed a new gasoline additive that is supposed to increase mileage....

Question:

A major oil company has developed a new gasoline additive that is supposed to increase mileage. To test this hypothesis, ten

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). Use a significance level of a = 0.01 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 26.5 13.7 14.1 Without additive With additive 27.2 30.2 19.1 21.9 13.8 14.5 21.4 22 16.1 16.4 19 | 28.7 19.3 29.7 26.8 14.8 15.1 Copy Data Step 2 of 5: Find the value of the standard deviation of the paired differences. Round your answer to two decimal places.

Answers

Number With Additive Without Additive Difference (di - \bar{d})^2
30.2 27.2 3 3.5721
21.9 19.1 2.8 2.8561
14.5 13.8 0.7 0.1681
22 21.4 0.6 0.2601
16.4 16.1 0.3 0.6561
19.3 19 0.3 0.6561
29.7 28.7 1 0.0121
26.8 26.5 0.3 0.6561
14.8 13.7 1.1 1E-04
15.1 14.1 1 0.0121
Total 210.7 199.6 11.1 8.849

Mean of paired difference

\bar{d} = \Sigma d_{i}/n = 11.1 / 10 = 1.11

Standard deviation of paired difference

S_{d} = \sqrt{\Sigma (d_{i} - \bar{d})^2 / n-1}
S_{d} = \sqrt{ 8.849 / 10-1} = 0.9916\, \, \approx\, 0.99


.

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