Question

Question 14 of 14 Step 2 of 5 01:14:31 A major oil company has developed a new gasoline aditive that is supposed to increase mileage. To test this hypothesis, ten cars are 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 c that the gasoline additive does significantly increase mileage? 0.05 for the test. Assume that the gas mileages (gas mileage with additive-gas mileage without additive). Use a significance level of α are normally distributed for the population of all cars both with and without the additive. Car 10 Without additive 16.6 159 17 20 264 12. 3 20. 28.9 10. | With additive-l 17 | 17.7 | 19.5 | 20.2 | 29.3 | 13.4 17.5 | 23.1 | 29.3 | 11 Copy Data Step 2 of S: Find the value of the standard deviation of the paired differences. Round your answer to two decimal places. m Tables Keypad AnswerHow to Enter) 2 Points

Answer 1

From the given data, values of d = Without additive - With additive are got as follows:

d values = - 0.4, - 1.8, - 2.5, - 0.2, - 2.9, - 0.7, - 2.2, - 3.0, - 0.4, - 0.9

From d values, the following statistics are calculated:

n = 10

$\bar{d}$ = $\sum_{i=1}^{10}$ d_i/10

= - 15/10

= - 1.5

d	d - $\bar{d}$	(d - $\bar{d}$)2
- 0.4	1.1	1.21
-1.8	- 0.3	0.09
- 2.5	- 1.0	1.00
- 0.2	1.3	1.69
- 2.9	- 1.4	1.96
- 0.7	0.8	0.64
- 2.2	- 0.7	0.49
- 3.0	- 1.5	2.25
- 0.4	1.1	1.21
- 0.9	0.6	0.36
Total =		10.9

Sample Variance (s²) is given by:

s² = $\sum_{i=1}^{10}$ (d_i - $\bar{d}$)²/9

= 10.9/9

= 1.2111

So,

Standard Deviation (s) is given by:

s = $\sqrt{}$1.2111

= 1.1005

So,

Answer is:

1.10

Answer 2

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)$d=\text{(gas mileage with additive)}-\text{(gas mileage without additive)}$.  Use a significance level of α=0.05$\alpha =0.05$ 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	16.6$16.6$	15.9$15.9$	17$17$	20$20$	26.4$26.4$	12.7$12.7$	15.3$15.3$	20.1$20.1$	28.9$28.9$	10.1$10.1$
With additive	17$17$	17.7$17.7$	19.5$19.5$	20.2$20.2$	29.3$29.3$	13.4$13.4$	17.5$17.5$	23.1$23.1$	29.3$29.3$	11$11$

Copy Data

Step 3 of 5 :  

Compute the value of the test statistic. Round your answer to three decimal places