Question

Data on household vehicle miles of travel (VMT) are compiled annually by the Federal Highway Administration and are published in the National Household Travel Survey, Summary of Travel Trends. Independent random samples of $15$midwestern households and $14$southern households provided the following data on last year's VMT, in thousands of miles.

At the $5\%$significance level, does there appear to be a difference in last year's mean VMT for midwestern and southern households? (Note: ${\overline{x}}_1=16.23,s_1=4.06,{\overline{x}}_2=17.69$, and $s_2=4.42$.)

Short answer

The presented data do not provide adequate evidence to infer that there is a difference in last year's mean VMT between Midwestern and Southern households at the significance level of $5\%$.

Step-by-step solution

Given Information

Vehicle miles traveled (VMT) data for two populations of Midwestern and Southern families is presented as a sample.

${\overline{x}}_1=16.23,s_1=4.06$

${\overline{x}}_2=17.69,s_2=4.42$

The significance level is$5\%$.

Explanation

Population $1$: Midwestern households, ${\overline{x}}_1=16.23,s_1=4.06$, and $n_1=15$.

Population $2$: Southern households, ${\overline{x}}_2=17.69,s_2=4.42$, and $n_2=14$.

The major goal is to determine whether there is a difference in mean VMT between Midwestern and Southern homes from the previous year.

Define null and alternate hypotheses.

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

Alternate hypotheses:$H_a:{\mu }_1\ne {\mu }_2$

Hypotheses is two-tailed.

We decided significance level as$5\%$

Calculation

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

$\Rightarrow s_p=\sqrt{\frac{(15-1)(4.06{)}^2+(14-1\left)\right(4.42{)}^2}{15+14-2}}$

$\Rightarrow s_p=\sqrt{\frac{14(16.4836)+13(19.5364)}{27}}$

$\Rightarrow s_p=4.237$

Test statistic,$t_0=\frac{{\overline{x}}_1-{\overline{x}}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

$\Rightarrow t_0=\frac{16.23-17.69}{4.237\sqrt{\frac{1}{15}+\frac{1}{14}}}$

$\Rightarrow t_0=-0.927$

We determine the critical values

Here,localid="1651300882093" role="math" $$\begin{gathered} \text{df}=n_1+n_2-2 \\ =15+14-2 \\ =27 \end{gathered}$$

$\Rightarrow \mathrm{df}=27$

Using table IV

localid="1651300891246" role="math" $$\begin{gathered} \text{Critical value,}\pm t_{\alpha /2}=\pm t_{0.05/2} \\ =\pm t_{0.025} \\ =\pm 2.052 \end{gathered}$$

Since $t_0=-0.927$, the test statistic does not fall into the two-tailed hypotheses test rejection zone. As a result, null hypotheses are not ruled out.