Driving Distances. Refer to Exercise 10.48 and determine a 95% confidence interval for the difference between last year's mean VMTs by midwestern and southern households.

Given in the question that, the vehicle miles travelled (VMT) data for two populations of Midwestern and Southern families are provided as samples in the inquiry.

$$\begin{gathered} {\overline{x}}_1=16.23, \\ {s}_1=4.06 \end{gathered}$$

$$\begin{gathered} {\overline{x}}_2=17.69, \\ {s}_2=4.42 \end{gathered}$$

The significance level is 5%.

Let's consider the Population 1: Midwestern households:

$$\begin{gathered} {\overline{x}}_1=16.23, \\ {s}_1=4.06 \\ {n}_1=15 \end{gathered}$$

Let's consider the Population 2: Southern households,

$$\begin{gathered} {\overline{x}}_2=17.69, \\ {s}_2=4.42 \\ {n}_2=14 \end{gathered}$$

The main goal is to calculate a $95\%$confidence interval for the difference between two population means,${\mu }_1$and${\mu }_2$

The null and alternate hypotheses should be stated.

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

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

Two-tailed hypotheses exist.

Use Table IV to find $t_{\alpha /2}$with $df=n_1+n_2-2$for a confidence level of $1-a$

alpha=0.05 for 95% confidence level,

localid="1651403167353" $$\begin{gathered} \text{df}=n_1+n_2-2 \\ =(15+14-2) \\ =27. \end{gathered}$$

When$\mathrm{df}=27$,using table IV for critical values,

Critical value,

localid="1651403201603" $$\begin{gathered} t_{\alpha /2}=t_{0.05/2} \\ =t_{0.025} \\ =2.052 \end{gathered}$$

Find the confidence interval's endpoints.

Pooled standard deviation:

localid="1651403231074" $s_p=\sqrt{\frac{\left(n_1-1\right)s_1{}^2+\left(n_2-1\right){s_2}^2}{n_1+n_2-2}}$

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

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

$s_p=4.237$$s_p=4.237$

Confidence interval$=\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}·s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

Confidence interval $=(16.23-17.69)\pm 2.052\times 4.237\sqrt{\frac{1}{15}+\frac{1}{14}}$

Confidence interval $=-1.46\pm 3.23$

Confidence interval $=-4.69to1.77$