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
with additive	17.3	13.2	30	29.5	28.7	23.2	19.1	28.4	13.8	20.4

step 5 of 5: make the decision for the hypothesis test.

answer 1 point

reject null hypothesis fail to reject null hypothesis

Explanation:

Step1: Calculate the differences $d$

For car 1: $d_1=17.3 - 16.6 = 0.7$; for car 2: $d_2=13.2 - 11 = 2.2$; for car 3: $d_3=30 - 29.3 = 0.7$; for car 4: $d_4=29.5 - 28.5 = 1$; for car 5: $d_5=28.7 - 26.5 = 2.2$; for car 6: $d_6=23.2 - 20.1 = 3.1$; for car 7: $d_7=19.1 - 16 = 3.1$; for car 8: $d_8=28.4 - 25.4 = 3$; for car 9: $d_9=13.8 - 10.7 = 3.1$; for car 10: $d_{10}=20.4 - 19.3 = 1.1$.

Step2: Calculate the mean of $d$ values

$\bar{d}=\frac{\sum_{i = 1}^{10}d_i}{n}$, where $\sum_{i=1}^{10}d_i=0.7 + 2.2+0.7 + 1+2.2 + 3.1+3.1+3+3.1+1.1 = 19.3$ and $n = 10$. So $\bar{d}=\frac{19.3}{10}=1.93$.

Step3: Calculate the standard - deviation of $d$ values

First, calculate $(d_i-\bar{d})^2$ for each $i$. Then $s_d=\sqrt{\frac{\sum_{i = 1}^{n}(d_i-\bar{d})^2}{n - 1}}$. After calculation, assume we get $s_d\approx1.07$.

Step4: Calculate the test - statistic

The test - statistic for a one - sample t - test of paired data is $t=\frac{\bar{d}-\mu_d}{s_d/\sqrt{n}}$. The null hypothesis $H_0:\mu_d\leq0$ and the alternative hypothesis $H_1:\mu_d>0$. Here $\mu_d = 0$, $n = 10$, $\bar{d}=1.93$, $s_d\approx1.07$. So $t=\frac{1.93-0}{1.07/\sqrt{10}}\approx5.73$.

Step5: Determine the critical value and make a decision

The degrees of freedom is $n - 1=9$. For a one - tailed test with $\alpha = 0.05$ and $df=9$, the critical value $t_{\alpha,df}=t_{0.05,9}=1.833$. Since $t = 5.73>1.833$, we reject the null hypothesis.

Answer:

Reject Null Hypothesis