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 α=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   Without additive   With additive
1   27.2   29.7
2   13.7   15.7
3   12.5   14.1
4   13.4   16.4
5   14.7   17.8
6   25.5   27.8
7   10.3   12.1
8   13.9   15
9   14.5   17.3
10   25   28

Step 1 of 5: State the null and alternative hypotheses for the test.

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 H0. Round the numerical portion of your answer to three decimal places.

Step 5 of 5: Make the decision for the hypothesis test. ( reject or fail to reject null hypothesis)

Answer 1

H₀: < 0

H₁: > 0

= (2.5 + 2 + 1.6 + 3 + 3.1 + 2.3 + 1.8 + 1.1 + 2.8 + 3)/10 = 2.32

s_d = sqrt(((2.5 - 2.32)^2 + (2 - 2.32)^2 + (1.6 - 2.32)^2 + (3 - 2.32)^2 + (3.1 - 2.32)^2 + (2.3 - 2.32)^2 + (1.8 - 2.32)^2 + (1.1 - 2.32)^2 + (2.8 - 2.32)^2 + (3 - 2.32)^2)/9) = 0.68

The test statistic t = ( - D)/(s_d/)

                             = (2.32 - 0)/(0.68/)

                             = 10.789

At alpha = 0.05, the critical value is t_0.95,9 = 1.833

Reject H₀, if t > 1.833

Since the test statistic value is greater than the critical value(10.789 > 1.833), so we should reject the null hypothesis.