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)d=(gas mileage with additive)−(gas mileage without additive). Use a significance level of α=0.1α=0.1 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
Without additive	25.1	11.8	18.8	13.7	23.6	19.4	16.7	24.7	12.1	20.9
With additive	26	14.31	21.1	15.2	24	20.5	18	25.9	13.8	24.3

Step 2 of 5 : Find the value of the standard deviation of the paired differences. Round your answer to two decimal places.

Step 3 of 5: Compute the value of the test statistic. Round your answer to three decimal places.

Step 4 of 5: Determine the decision rule for rejecting the null hypothesis H0H0. Round the numerical portion of your answer to three decimal places.

Step 5 of 5: Make the decision for the hypothesis test.

Answer 1

Ans:

Car	1	2	3	4	5	6	7	8	9	10
Without additive	25.1	11.8	18.8	13.7	23.6	19.4	16.7	24.7	12.1	20.9
With additive	26	14.31	21.1	15.2	24	20.5	18	25.9	13.8	24.3
d	0.9	2.51	2.3	1.5	0.4	1.1	1.3	1.2	1.7	3.4
mean d=	1.631
sd=	0.8818

1)d-bar=1.631

2)standard deviation of d=0.8818

3)Test statistic:

t=(1.631-0)/(0.8818/sqrt(10))

t=5.849

*(if sd=0.88 is used,then t=5.861)

4)

df=10-1=9

critical t value=1.383

Reject Ho,if t>1.383

5)Reject H₀.

There is sufficient evidence to conclude that the gasoline additive does significantly increase mileage.