Chapter 13.1, Problem 16E

a.

To determine

To find:The value of $b_1$ .

Answer to Problem 16E

The value of $b_1$ is $-0.1540$ .

Explanation of Solution

Given information: The data is shown below.

x	12	17	3	17	16	11	14	9
y	13	14	16	13	14	14	13	14

Calculation:

The MINITAB is shown below,

  

  Figure-1

From Figure-1 it is clear that the value of $b_1$ is $-0.1540$ .

b.

To determine

To find: The value of $s_e$ .

Answer to Problem 16E

The value of $s_e$ is $0.716717$ .

Explanation of Solution

Given information: The data is shown below.

x	12	17	3	17	16	11	14	9
y	13	14	16	13	14	14	13	14

Calculation:

From Figure-1 it is clear that the value of $s_e$ is $0.716717$ .

c.

To determine

To find: The value of squares for $x$ .

Answer to Problem 16E

The value of squares for $x$ is $159.9$ .

Explanation of Solution

Given information: The data is shown below.

x	12	17	3	17	16	11	14	9
y	13	14	16	13	14	14	13	14

Calculation:

The table is shown below.

$x$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
12	-0.38	0.1444
17	4.62	21.3444
3	-9.38	87.9844
17	4.62	21.3444
16	3.62	13.1044
11	-1.38	1.9044
14	1.62	2.6244
9	-3.38	11.4244
		${\left(x-\overline{x}\right)}^2=159.9$

Thus, the value of squares for $x$ is $159.9$ .

d.

To determine

To find:The value of standard error of $b_1$ and $S_b$ .

Answer to Problem 16E

The value of standard error of $b_1$ and $S_b$ is $0.0567$ .

Explanation of Solution

Given information: The data is shown below.

x	12	17	3	17	16	11	14	9
y	13	14	16	13	14	14	13	14

Calculation:

The value of

  $\begin{array}{c}{s}_b=\frac{s_e}{\sqrt{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}}\\ =\frac{0.716717}{\sqrt{159.9}}\\ =0.0567\end{array}$

Thus, the value of standard error of $b_1$ and $S_b$ is $0.0567$ .

e.

To determine

To find:The critical value.

Answer to Problem 16E

The critical value is $2.447$ .

Explanation of Solution

Given information: The data is shown below.

x	12	17	3	17	16	11	14	9
y	13	14	16	13	14	14	13	14

Calculation:

The degree of freedom is,

  $\begin{array}{c}dof=n-2\\ =8-2\\ =6\end{array}$

The critical value is $2.447$

f.

To determine

To find:The margin of error.

Answer to Problem 16E

The margin of error is $0.1387$ .

Explanation of Solution

Given information: The data is shown below.

x	12	17	3	17	16	11	14	9
y	13	14	16	13	14	14	13	14

Calculation:

The margin of error is,

  $\begin{array}{c}ME=t_{\frac{\alpha }{2}}\times s_b\\ =2.447\times 0.0567\\ =0.1387\end{array}$

Thus, the margin of error is $0.1387$ .

g.

To determine

To find:The confidence interval for the data.

Answer to Problem 16E

The confidence interval for the data is $\left(-0.2927,-0.0153\right)$ .

Explanation of Solution

Given information: The data is shown below.

x	12	17	3	17	16	11	14	9
y	13	14	16	13	14	14	13	14

Calculation:

The confidence interval is,

  $\begin{array}{c}CI=-0.154\pm 0.1387\\ =\left(-0.2927,-0.0153\right)\end{array}$

Thus, the confidence interval for the data is $\left(-0.2927,-0.0153\right)$ .

h.

To determine

To explain:The test for the hypothesis $H_0:{\beta }_1=0$ versus $H_1:{\beta }_1\ne 0$ .

Explanation of Solution

Given information: The data is shown below.

x	12	17	3	17	16	11	14	9
y	13	14	16	13	14	14	13	14

Calculation:

The test statistics is,

  $\begin{array}{c}t=\frac{b_1}{s_b}\\ =\frac{-0.154}{0.0567}\\ =-2.72\end{array}$

Since, the test statistic is greater than the critical value.

Thus, the null hypothesis is rejected.