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	14.8	24.2	18.8	11.7	23.2	19.5	14.1	23.4	9.8	13.9
With additive	17.2	25.8	19.1	12.9	26	20.9	16.1	24.8	11.3	14.5

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

The mean of the difference = 1.52, and the standard deviation of the difference S_d = 0.754

= 0.10 and the degrees of freedom = number of pairs - 1 = 10 - 1 = 9

Let be the difference in scores of the 2 populations (With additive - Without Additive)

____________________________

The Hypothesis:

H0: = 0 : The gasoline addictive is not effective in increasing mileage.

Ha: > 0: The gasoline addictive was effective in increasing gas mileage

_______________________________

The Test Statistic: Since sample size is small, and population std. deviation is unknown, we use the students t test.

t = 6.375 (6.38 if required to 2 decimal places)

The p value: (Right tailed) at t = 6.375, degrees of freedom = 9 is 0.0001

(4) The Rejection Region:

The Critical value: (Right tailed) for = 0.10, df = 9 is +1.383

If t test is > 1.383, Reject H0

If p value is < (0.10), Reject H0.

(5) The Decision:   Since t observed (6.375) is > 1.383, We Reject H0.

Also since P value (0.0001) is < (0.10) , We Reject H0.

(v) The Conclusion: There is sufficient evidence at the 90% significance level to conclude that the gasoline addictive was effective in significantly increasing gas mileage

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Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)².

#No	Without	With	Difference
1	14.8	17.2	2.4
2	24.2	25.8	1.6
3	18.8	19.1	0.3
4	11.7	12.9	1.2
5	23.2	26	2.8
6	19.5	20.9	1.4
7	14.1	16.1	2
8	23.4	24.8	1.4
9	9.8	11.3	1.5
10	13.9	14.5	0.6

#	Difference	Mean	(X-Mean)2
1	2.4	1.52	0.7744
2	1.6	1.52	0.0064
3	0.3	1.52	1.4884
4	1.2	1.52	0.1024
5	2.8	1.52	1.6384
6	1.4	1.52	0.0144
7	2	1.52	0.2304
8	1.4	1.52	0.0144
9	1.5	1.52	0.0004
10	0.6	1.52	0.8464

n	10
Sum	15.2
Mean	1.52
SS	5.116
Variance	0.5684
Std Dev	0.7540