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=0.01 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
Without additive	22.8	24.7	12.8	20.5	15.1	27.8	17	20.9	12.5	15.7
With additive	25.7	26.1	14.5	21.6	17.8	30.1	18.9	23.1	15	19.3

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 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

: Average difference in the mileage of cars with and without additive

Step 1 :Null hypothesis : H_o : = 0

Alternate hypothesis : H_a : > 0

Step 2 :

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

Sample mean :

n : Sample size

Car	Without additive	With additive	d:with additive−without additive	d-	(d-)2
1	22.8	25.7	2.9	0.67	0.4489
2	24.7	26.1	1.4	-0.83	0.6889
3	12.8	14.5	1.7	-0.53	0.2809
4	20.5	21.6	1.1	-1.13	1.2769
5	15.1	17.8	2.7	0.47	0.2209
6	27.8	30.1	2.3	0.07	0.0049
7	17	18.9	1.9	-0.33	0.1089
8	20.9	23.1	2.2	-0.03	0.0009
9	12.5	15	2.5	0.27	0.0729
10	15.7	19.3	3.6	1.37	1.8769
		Total	22.3		4.981
		Mean : =22.3/10=2.23

sample standard deviation : s

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

Step 4 of 5:

Reject H_o if p-value < =0.01

Degrees of freedom = n-1 =10-1=9

For right tailed test:

Step 5 of 5:

As P-Value i.e. is less than Level of significance i.e (P-value:0 < 0.01:Level of significance); Reject Null Hypothesis
There is sufficient evidence to conclude that  the gasoline additive does significantly increase mileage

p-value is computed using excel function T.DIST.RT function
Returns the right-tailed Student's t-distribution.
The t-distribution is used in the hypothesis testing of small sample data sets. Use this function in place of a table of critical values for the t-distribution.
Syntax
T.DIST.RT(x,deg_freedom)
The T.DIST.RT function syntax has the following arguments:
• X Required. The numeric value at which to evaluate the distribution.
• Deg_freedom Required. An integer indicating the number of degrees of freedom.