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	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 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.

REJECT H0 if: CHOOSE ONE: t or l t I > or < ___________.

Step 5 of 5: Make the decision for the hypothesis test.

Answer 1

: Average difference in mileage between Gas with additive and Gas without additive

Step 1 of 5.

Null hypothesis :H_o :   = 0

Alternate Hypothesis : H_A : > 0

Right Tailed test

Step 2 of 5: Value of the standard deviation of the paired differences. Round your answer to two decimal places

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

Sample mean of paired differences:

standard deviation of the paired differences : s

n : sample size = 10

Car	Without additive	With additive	d	(d-d)	(d-d)2
1	29.1	30.2	1.1	-0.56	0.3136
2	17.5	19	1.5	-0.16	0.0256
3	22.9	26.2	3.3	1.64	2.6896
4	18.9	20	1.1	-0.56	0.3136
5	11.7	12.9	1.2	-0.46	0.2116
6	14.5	17.8	3.3	1.64	2.6896
7	19.5	20.3	0.8	-0.86	0.7396
8	12.5	13.8	1.3	-0.36	0.1296
9	27.8	29.2	1.4	-0.26	0.0676
10	9.4	11	1.6	-0.06	0.0036
		Total	16.6		7.184
		= 16.6/10	1.66

s=0.8934

Step 3 of 5: Compute the value of the test statistic

Step 4 of 5:

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

For Right Tailed test;

Critical value = t_0.05,9 = 1.883

For Right Tailed Test : Reject null hypothesis if value of test statistic : t is greater than Critical Value ; t_0.05,9

Reject H_o : t > 1.883

Step 5 of 5

As t: 5.8757 > critical value : 1.8831 ; Reject null hypothesis :H_o

There is sufficient evidence to conclude that Average difference in mileage between Gas with additive and Gas without additive(d:Gas with additive - Gas without additive) is greater than '0'

i.e There is sufficient evidence to conclude that new gasoline additive increases the mileage.