Question

An auto company claims that the fuel efficiency of its sedan has been substantially improved. A consumer advocate organization wishes to compare the fuel efficiencies of its sedan between the 2017 and 2019 models. Each of 12 drivers drove both the 2017 and 2019 models on a highway and the fuel efficiency was measured (in miles per gallon). The same driver drove the same path for both the models.

Driver
Model 1 2 3 4 5 6 7 8 9 10 11 12
2019 31.1 32.4 31.3 33.5 31.7 32.0 31.8 29.9 31.0 32.8 32.7 33.8
2017 28.7 32.1 29.6 30.5 31.9 30.9 32.3 33.1 29.6 30.8 31.1 31.6

Use the 0.1 level of significance to test the claim that the average fuel efficiency of the sedan has improved. Assume normal population.

Answer 1

Answer:

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.983 -0}{\frac{1.685}{\sqrt{12}}}=2.021$

$df=n-1=12-1=11$

$p_v =P(t_{(11)}>2.021) =0.0342$

If we compare the the p value with the significance level provided $\alpha=0.1$, we see that $p_v < \alpha$, so then we can reject the null hypothesis. and there is a significant increase in the miles per gallon from 2017 to 2019 at 10% of significance.

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

Let put some notation :

x=test value 2017 , y = test value 2019

x: 28.7 32.1 29.6 30.5 31.9 30.9 32.3 33.1 29.6 30.8 31.1 31.6

y: 31.1 32.4 31.3 33.5 31.7 32.0 31.8 29.9 31.0 32.8 32.7 33.8

Solution to the problem

The system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x \leq 0$

Alternative hypothesis: $\mu_y -\mu_x >0$

Because if we have an improvement we expect that the values for 2019 would be higher compared with the values for 2017

The first step is calculate the difference $d_i=y_i-x_i$ and we obtain this:

d: 2.4, 0.3, 1.7,3,-0.2, 1.1, -0.5, -3.2, 1.4, 2, 1.6, 2.2

The second step is calculate the mean difference  

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{11.8}{12}=0.983$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =1.685$

We assume that the true difference follows a normal distribution. The 4th step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.983 -0}{\frac{1.685}{\sqrt{12}}}=2.021$

The next step is calculate the degrees of freedom given by:

$df=n-1=12-1=11$

Now we can calculate the p value, since we have a right tailed test the p value is given by:

$p_v =P(t_{(11)}>2.021) =0.0342$

If we compare the the p value with the significance level provided $\alpha=0.1$, we see that $p_v < \alpha$, so then we can reject the null hypothesis. and there is a significant increase in the miles per gallon from 2017 to 2019 at 10% of significance.