Question

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.

Answer 1

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

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