Gasoline Additive. This exercise shows what can happen when a hypothesis-testing procedure designed for use with independent samples is applied to perform a hypothesis test on a paired sample. The gas mileages, in miles per gallon (mpg), of 10 randomly selected cars, both with and without a new gasoline additive, are shown in the following table.

1. Apply the paired $t$-test to decide, at the $5\%$significance level, whether the gasoline additive is effective in increasing gas mileage.
2. Apply the pooled $t$-test to the sample data to perform the hypothesis test.
3. Why is performing the hypothesis test the way you did in part (b) inappropriate?
4. Compare your result in parts (a) and (b).

Given in the question that, the gas mileages, in miles per gallon (mpg), of 10 randomly selected cars, both with and without a new gasoline additive.

We must use the paired $t$-test to determine whether the gasoline addition is effective in enhancing gas mileage at the $5\%$ significance level.

Purpose of paired $t$-test, To compare two population means, ${\mu }_1$and ${\mu }_2$using a hypothesis test.

Null hypothesis,

$H_0:{\mu }_1={\mu }_2$

Alternative hypothesis:

$H_a:{\mu }_1<{\mu }_2$

Where ${\mu }_1$and ${\mu }_2$represent the average of all cars with and without the additional gasoline additive, respectively.

The test will be run at a significance level of $5\%$. So $\alpha =0.05.$

Statistical test will be:

$t=\frac{\overline{d}}{s_{d}l\sqrt{n}}$

We create the table below to determine the above test statistic.

$\begin{array}{|cccc|}\hline \text{Population 1}& \text{Population 2}& \text{Difference}{d}_i& d_i^2\\ 25.7& 24.9& 0.8& 0.64\\ 20.0& 18.8& 1.2& 1.44\\ 28.4& 27.7& 0.7& 0.49\\ 13.7& 13.0& 0.7& 0.49\\ 18.8& 17.8& 1.0& 1\\ 12.5& 11.3& 1.2& 1.44\\ 28.4& 27.8& 0.6& 0.36\\ 8.1& 8.2& -0.1& 0.01\\ 23.1& 23.1& 0.0& 0.00\\ 10.4& 9.9& 0.5& 0.25\\ \text{Sum}& & 6.6& 6.12\\ \hline\end{array}$

From the above table, we know that:

$$\begin{gathered} n=10, \\ \sum d_i=6.6 \\ \sum d_i^2=6.12 \end{gathered}$$

Therefore,

$$\begin{gathered} \overline{d}=\frac{\sum d_i}{n} \\ =\frac{{\displaystyle 6.6}}{{\displaystyle 10}} \\ =0.66 \end{gathered}$$

Let's find $s_d$

$$\begin{gathered} s_d=\sqrt{\frac{\sum d_i^2-{\left(\sum d_i\right)}^2/n}{n-1}} \\ =\sqrt{\frac{6.12-\frac{(6.6{)}^2}{10}}{10-1}} \\ =\sqrt{0.196} \\ =0.4427 \end{gathered}$$

The test statistic's value will be:

$$\begin{gathered} t=\frac{\overline{d}}{s_d/\sqrt{n}} \\ =\frac{0.66}{0.4427/\sqrt{10}} \\ =\frac{0.66}{0.14} \\ =4.71 \end{gathered}$$

The degree of freedom will be:

$$\begin{gathered} df=n-1 \\ =10-1 \\ =9 \end{gathered}$$

The $t_a$table reveals that,

$df=9$

$t_a=t_{0.05}=1.833$

The rejection region is depicted in the diagram below.

The test statistic value is $t=4.71$, which is in the rejection range. As a result, we reject$H_0$

As a result, the test results are statistically significant at the $5\%$level.

Given in the question that,

Let's consider the Null and Alternative hypothesis:

$$\begin{gathered} H_0={\mu }_1={\mu }_2 \\ {H}_a={\mu }_1>{\mu }_2 \end{gathered}$$

Where ${\mu }_1$and ${\mu }_2$represent the average of all cars with and without the additional gasoline additive, respectively. The hypothesis is right-tailed in this case. The test will be run at a significance threshold of $5\%$, so $\alpha =0.05$. Calculate the test statistic's value:

$t=\frac{\overline{x_1}-\overline{x_2}}{s_p\sqrt{\frac{1}{m_1}+\frac{1}{m_2}}}$

Where,

$s_p=\sqrt{\frac{\left(n_1-1\right)s_1^2+\left(n_2-2\right)s_2^2}{n_1+n_2-2}}$

According to the information,

$$\begin{gathered} n_1=10 \\ \overline{x_1}=18.91 \\ {s}_1=7.47 \\ {n}_2=10 \\ \overline{x_2}=18.25 \\ {s}_2=7.42 \end{gathered}$$

The value for the pooled standard information will be:

$$\begin{gathered} s_p=\sqrt{\frac{\left(n_1-1\right)s_1^2+\left(n_2-2\right)s_2^2}{n_1+n_2-2}} \\ =\sqrt{\frac{{\displaystyle (10-1)(7.47{)}^2+(10-2\left)\right(7.42{)}^2}}{{\displaystyle 10+10-2}}} \\ =7.445 \end{gathered}$$

The test statistic will be:

$$\begin{gathered} t=\frac{\overline{x_1}-\overline{x_2}}{{}^{s_P\sqrt{\frac{1}{P_1}}+\frac{1}{{}^{M_2}}}} \\ =\frac{18.91-18.25}{7.445\sqrt{\frac{1}{10}+\frac{1}{10}}} \\ =0.20 \end{gathered}$$

Then, compute the critical value as follow:

First, we have to find the degree of freedom:

$$\begin{gathered} df=n_1+n_2-2 \\ =10+10-2 \\ =18 \end{gathered}$$

The $t_a$table reveals that,

$$\begin{gathered} df=18 \\ {t}_a=t_{0.05}=1.734 \end{gathered}$$

The test statistic's value is $t=0.20$, which is within the acceptable range. As a result, we are unable to reject $H_0$. As a result, at the $5\%$level, the testing results are not statistically meaningful.

Given in the question that,

$\alpha =5\%$

We need to figure out why you executed the hypothesis test the way you did in component (b).

We know that, we can reject the null hypothesis if $P$value$\le a$.

Here, the given samples are not independent, the hypothesis test should be performed using the paired t test rather than the pooled -test.

Given in the question that, to refer answers from part (a) and (b) . Then, we have to compare the result in parts (a) and (b).

Parts (a) and (b) provide distinct outcomes.

There is enough evidence to infer that the gasoline addition increases gas mileage using the paired $t$-test.

There is insufficient data to infer that the gasoline addition increases gas mileage using a pooled $t$-test.