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=0.1 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 21.9 13.9 22.7 15.2 11 19.6 17.5 19.2 16.9 27.4 With additive 22.2 16.7 25.5 18.6 12.2 21.2 20.3 21.2 18 29.1

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.

Answer 1

Solution:

We have to test if the gasoline additive does significantly increase mileage.

d=(gas mileage with additive)−(gas mileage without additive).

Significance level = α = 0.1

Step 1) State the null and alternative hypotheses for the test.

$H_{0}: \mu_{d}=0$

Vs

$H_{1}: \mu_{d}>0$

Step 2) Find the value of the standard deviation of the paired differences.

$\bar{x}_{d}=\frac{\sum d_{i}}{n}$

$S_{d}=\sqrt{\frac{\sum d_{i}^{2}-(\sum d_{i})^{2}/n}{n-1}}$

Cars	Without additive	With additive	d=with-without	d^2
1	21.9	22.2	0.3	0.09
2	13.9	16.7	2.8	7.84
3	22.7	25.5	2.8	7.84
4	15.2	18.6	3.4	11.56
5	11	12.2	1.2	1.44
6	19.6	21.2	1.6	2.56
7	17.5	20.3	2.8	7.84
8	19.2	21.2	2	4
9	16.9	18	1.1	1.21
10	27.4	29.1	1.7	2.89
			$\sum d_{i} =19.7$	$\sum d_{i}^{2} =47.27$

Thus

$\bar{x}_{d}=\frac{\sum d_{i}}{n} = \frac{19.7}{10} = 1.97$

$S_{d}=\sqrt{\frac{\sum d_{i}^{2}-(\sum d_{i})^{2}/n}{n-1}}$

$S_{d}=\sqrt{\frac{47.27 -(19.7)^{2}/10}{10-1}}$

$S_{d}=\sqrt{\frac{47.27 -38.809 }{9}}$

$S_{d}=\sqrt{\frac{8.461 }{9}}$

$S_{d}=\sqrt{ 0.940111 }$

$S_{d}=0.969593$

Step 3) Compute the value of the test statistic.

$t = \frac{\bar{x}_{d}-\mu _{d}}{S_{d}/\sqrt{n}}$

$t = \frac{1.97-0}{0.969593 /\sqrt{10}}$

$t = \frac{1.97}{0.969593 / 3.162278 }$

$t = \frac{1.97}{0.306612 }$

Step 4) Determine the decision rule for rejecting the null hypothesis H0.

Df = n - 1 = 10 - 1 = 9

Right tail area = one tail area = 0.10

From t table, we get t critical value = 1.383

Thus decision rule is:

Reject null hypothesis H0, if t test statistic value > t critical value = 1.383, otherwise we fail to reject H0.

Step 5) Make the decision for the hypothesis test.

Since t test statistic value = 6.425 > > t critical value = 1.383, we reject null hypothesis H0.

Thus it can be concluded, from the data, that the gasoline additive does significantly increase mileage.