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=0.05α=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	1	2	3	4	5	6	7	8	9	10
Without additive	29.1	17.5	22.9	18.9	11.7	14.5	19.5	12.5	27.8	9.4
With additive	30.2	19	26.2	20	12.9	17.8	20.3	13.8	29.2	11

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 Null Hypothesis Fail to Reject Null Hypothesis

Answer 1

S. No	with	without	diff:(d)=x1-x2	d2
1	30.2	29.1	1.1	1.21
2	19	17.5	1.5	2.25
3	26.2	22.9	3.3	10.89
4	20	18.9	1.1	1.21
5	12.9	11.7	1.2	1.44
6	17.8	14.5	3.3	10.89
7	20.3	19.5	0.8	0.64
8	13.8	12.5	1.3	1.69
9	29.2	27.8	1.4	1.96
10	11	9.4	1.6	2.56
	total	=	Σd=16.6	Σd2=34.74
	mean dbar=	d̅     =	1.6600
	degree of freedom =n-1                            =		9
	Std deviaiton SD=√(Σd2-(Σd)2/n)/(n-1) =		0.893433
	std error=Se=SD/√n=		0.2825
	test statistic            =	(d̅-μd)/Se         =	5.876

Step 4 of 5 :  

0.05 level with right tail test and n-1= 9 df, critical t=			1.833	from excel: 1*t.inv(0.05,9)
Decision rule: reject Ho if test statistic t>1.833

Step 5 of 5:

since test statistic falls in rejection region we reject null hypothesis