Chapter 13.3, Problem 21E

a.

To determine

To find:The regression equation for the data.

Answer to Problem 21E

The regression equation for the data is $y=8.872-0.0068x_1+0.7163x_2+2.03x_3$ .

Explanation of Solution

Given information: The data is shown below.

$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$
31.3	50	19	4.0	37.4	60	30	5.0	24.0	60	24	4.0	51.0	80	34	7.5
56.9	90	38	8.0	43.3	60	26	7.0	36.2	50	21	7.0	34.3	60	22	2.5
43.1	70	28	6.5	36.3	70	25	7.5	26.5	50	17	2.0	31.5	60	24	5.0
41.5	70	25	5.5	38.4	70	31	5.5	47.1	80	34	8.5	33.2	60	23	4.0
39.0	60	26	6.5	41.5	60	27	7.5	38.1	70	27	5.5	39.2	60	29	6.5
40.9	70	29	5.0	36.l	60	23	6.0	33.5	60	24	2.5	46.7	70	27	7.5
35.9	60	23	5.5	38.5	60	23	6.0	43.6	70	27	10.0	30.4	70	32	4.0
43.5	70	28	5.5	42.4	60	24	9.0	41.0	80	32	6.5	43.2	60	25	5.5
47.9	80	34	6.5	46.5	70	31	5.5	50.2	50	26	9.0	30.6	60	26	3.5
33.8	70	26	4.5	43.1	80	32	6.0	34.4	50	22	4.0	43.3	70	28	7.5
41.1	70	26	8.0	50.8	60	26	10.0	47.9	80	34	6.5	32.4	60	22	5.5
38.7	70	26	8.0	44.2	70	28	4.5	46.2	50	21	10.0	35.5	60	22	4.5

Calculation:

The MINITAB is shown below,

  

  Figure-1

From Figure-1 it is clear that the regression equation is $y=8.872-0.0068x_1+0.7163x_2+2.03x_3$

b.

To determine

To find: The value of variable $y$ .

Answer to Problem 21E

The value of variable $y$ is $42.201$ .

Explanation of Solution

Given information: The data is shown below.

$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$
31.3	50	19	4.0	37.4	60	30	5.0	24.0	60	24	4.0	51.0	80	34	7.5
56.9	90	38	8.0	43.3	60	26	7.0	36.2	50	21	7.0	34.3	60	22	2.5
43.1	70	28	6.5	36.3	70	25	7.5	26.5	50	17	2.0	31.5	60	24	5.0
41.5	70	25	5.5	38.4	70	31	5.5	47.1	80	34	8.5	33.2	60	23	4.0
39.0	60	26	6.5	41.5	60	27	7.5	38.1	70	27	5.5	39.2	60	29	6.5
40.9	70	29	5.0	36.l	60	23	6.0	33.5	60	24	2.5	46.7	70	27	7.5
35.9	60	23	5.5	38.5	60	23	6.0	43.6	70	27	10.0	30.4	70	32	4.0
43.5	70	28	5.5	42.4	60	24	9.0	41.0	80	32	6.5	43.2	60	25	5.5
47.9	80	34	6.5	46.5	70	31	5.5	50.2	50	26	9.0	30.6	60	26	3.5
33.8	70	26	4.5	43.1	80	32	6.0	34.4	50	22	4.0	43.3	70	28	7.5
41.1	70	26	8.0	50.8	60	26	10.0	47.9	80	34	6.5	32.4	60	22	5.5
38.7	70	26	8.0	44.2	70	28	4.5	46.2	50	21	10.0	35.5	60	22	4.5

Calculation:

From Figure-1 it is clear that the regression equation is $y=8.872-0.0068x_1+0.7163x_2+2.03x_3$

Substitute the values in above equation.

  $\begin{array}{c}y=8.872-0.0068\left(50\right)+0.7163\left(30\right)+2.03\left(6\right)\\ =42.201\end{array}$

Thus, the value of variable $y$ is $42.201$ .

c.

To determine

To find: The confidence interval.

Answer to Problem 21E

The confidence intervalis $\left(36.934,47.47\right)$ .

Explanation of Solution

Given information: The data is shown below.

$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$
31.3	50	19	4.0	37.4	60	30	5.0	24.0	60	24	4.0	51.0	80	34	7.5
56.9	90	38	8.0	43.3	60	26	7.0	36.2	50	21	7.0	34.3	60	22	2.5
43.1	70	28	6.5	36.3	70	25	7.5	26.5	50	17	2.0	31.5	60	24	5.0
41.5	70	25	5.5	38.4	70	31	5.5	47.1	80	34	8.5	33.2	60	23	4.0
39.0	60	26	6.5	41.5	60	27	7.5	38.1	70	27	5.5	39.2	60	29	6.5
40.9	70	29	5.0	36.l	60	23	6.0	33.5	60	24	2.5	46.7	70	27	7.5
35.9	60	23	5.5	38.5	60	23	6.0	43.6	70	27	10.0	30.4	70	32	4.0
43.5	70	28	5.5	42.4	60	24	9.0	41.0	80	32	6.5	43.2	60	25	5.5
47.9	80	34	6.5	46.5	70	31	5.5	50.2	50	26	9.0	30.6	60	26	3.5
33.8	70	26	4.5	43.1	80	32	6.0	34.4	50	22	4.0	43.3	70	28	7.5
41.1	70	26	8.0	50.8	60	26	10.0	47.9	80	34	6.5	32.4	60	22	5.5
38.7	70	26	8.0	44.2	70	28	4.5	46.2	50	21	10.0	35.5	60	22	4.5

Calculation:

The MINITAB output is shown below.

  

  Figure-2

From Figure-2 it is clear that the confidence interval is $\left(36.934,47.47\right)$ .

d.

To determine

To find: The prediction interval.

Answer to Problem 21E

The prediction intervalis $\left(32.888,51.517\right)$ .

Explanation of Solution

Given information: The data is shown below.

$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$
31.3	50	19	4.0	37.4	60	30	5.0	24.0	60	24	4.0	51.0	80	34	7.5
56.9	90	38	8.0	43.3	60	26	7.0	36.2	50	21	7.0	34.3	60	22	2.5
43.1	70	28	6.5	36.3	70	25	7.5	26.5	50	17	2.0	31.5	60	24	5.0
41.5	70	25	5.5	38.4	70	31	5.5	47.1	80	34	8.5	33.2	60	23	4.0
39.0	60	26	6.5	41.5	60	27	7.5	38.1	70	27	5.5	39.2	60	29	6.5
40.9	70	29	5.0	36.l	60	23	6.0	33.5	60	24	2.5	46.7	70	27	7.5
35.9	60	23	5.5	38.5	60	23	6.0	43.6	70	27	10.0	30.4	70	32	4.0
43.5	70	28	5.5	42.4	60	24	9.0	41.0	80	32	6.5	43.2	60	25	5.5
47.9	80	34	6.5	46.5	70	31	5.5	50.2	50	26	9.0	30.6	60	26	3.5
33.8	70	26	4.5	43.1	80	32	6.0	34.4	50	22	4.0	43.3	70	28	7.5
41.1	70	26	8.0	50.8	60	26	10.0	47.9	80	34	6.5	32.4	60	22	5.5
38.7	70	26	8.0	44.2	70	28	4.5	46.2	50	21	10.0	35.5	60	22	4.5

Calculation:

From Figure-12 it is clear that the $\left(32.888,51.517\right)$

e.

To determine

To find: The percentage of variation in variable $y$ .

Answer to Problem 21E

The percentage of variation in variable $y$ is $70.2\%$ .

Explanation of Solution

Given information: The data is shown below.

$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$
31.3	50	19	4.0	37.4	60	30	5.0	24.0	60	24	4.0	51.0	80	34	7.5
56.9	90	38	8.0	43.3	60	26	7.0	36.2	50	21	7.0	34.3	60	22	2.5
43.1	70	28	6.5	36.3	70	25	7.5	26.5	50	17	2.0	31.5	60	24	5.0
41.5	70	25	5.5	38.4	70	31	5.5	47.1	80	34	8.5	33.2	60	23	4.0
39.0	60	26	6.5	41.5	60	27	7.5	38.1	70	27	5.5	39.2	60	29	6.5
40.9	70	29	5.0	36.l	60	23	6.0	33.5	60	24	2.5	46.7	70	27	7.5
35.9	60	23	5.5	38.5	60	23	6.0	43.6	70	27	10.0	30.4	70	32	4.0
43.5	70	28	5.5	42.4	60	24	9.0	41.0	80	32	6.5	43.2	60	25	5.5
47.9	80	34	6.5	46.5	70	31	5.5	50.2	50	26	9.0	30.6	60	26	3.5
33.8	70	26	4.5	43.1	80	32	6.0	34.4	50	22	4.0	43.3	70	28	7.5
41.1	70	26	8.0	50.8	60	26	10.0	47.9	80	34	6.5	32.4	60	22	5.5
38.7	70	26	8.0	44.2	70	28	4.5	46.2	50	21	10.0	35.5	60	22	4.5

Calculation:

From Figure-1 it is clear that the percentage of variation in variable $y$ is $70.2\%$ .

f.

To determine

To find:Whether the given model is useful for prediction.

Answer to Problem 21E

The model is useful in prediction.

Explanation of Solution

Given information: The data is shown below.

$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$
31.3	50	19	4.0	37.4	60	30	5.0	24.0	60	24	4.0	51.0	80	34	7.5
56.9	90	38	8.0	43.3	60	26	7.0	36.2	50	21	7.0	34.3	60	22	2.5
43.1	70	28	6.5	36.3	70	25	7.5	26.5	50	17	2.0	31.5	60	24	5.0
41.5	70	25	5.5	38.4	70	31	5.5	47.1	80	34	8.5	33.2	60	23	4.0
39.0	60	26	6.5	41.5	60	27	7.5	38.1	70	27	5.5	39.2	60	29	6.5
40.9	70	29	5.0	36.l	60	23	6.0	33.5	60	24	2.5	46.7	70	27	7.5
35.9	60	23	5.5	38.5	60	23	6.0	43.6	70	27	10.0	30.4	70	32	4.0
43.5	70	28	5.5	42.4	60	24	9.0	41.0	80	32	6.5	43.2	60	25	5.5
47.9	80	34	6.5	46.5	70	31	5.5	50.2	50	26	9.0	30.6	60	26	3.5
33.8	70	26	4.5	43.1	80	32	6.0	34.4	50	22	4.0	43.3	70	28	7.5
41.1	70	26	8.0	50.8	60	26	10.0	47.9	80	34	6.5	32.4	60	22	5.5
38.7	70	26	8.0	44.2	70	28	4.5	46.2	50	21	10.0	35.5	60	22	4.5

Calculation:

The null hypothesis is, the model is not useful for prediction and the alternative hypothesis is, the model is useful in prediction.

From Figure-1 it is clear that the p value is less than the level of significance of $0.01$ .

Hence, the null hypothesis is rejected.

Thus, the model is useful in prediction.

g.

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.

$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$	$y$	$\left(x_1\right)$	$\left(x_2\right)$	$\left(x_3\right)$
31.3	50	19	4.0	37.4	60	30	5.0	24.0	60	24	4.0	51.0	80	34	7.5
56.9	90	38	8.0	43.3	60	26	7.0	36.2	50	21	7.0	34.3	60	22	2.5
43.1	70	28	6.5	36.3	70	25	7.5	26.5	50	17	2.0	31.5	60	24	5.0
41.5	70	25	5.5	38.4	70	31	5.5	47.1	80	34	8.5	33.2	60	23	4.0
39.0	60	26	6.5	41.5	60	27	7.5	38.1	70	27	5.5	39.2	60	29	6.5
40.9	70	29	5.0	36.l	60	23	6.0	33.5	60	24	2.5	46.7	70	27	7.5
35.9	60	23	5.5	38.5	60	23	6.0	43.6	70	27	10.0	30.4	70	32	4.0
43.5	70	28	5.5	42.4	60	24	9.0	41.0	80	32	6.5	43.2	60	25	5.5
47.9	80	34	6.5	46.5	70	31	5.5	50.2	50	26	9.0	30.6	60	26	3.5
33.8	70	26	4.5	43.1	80	32	6.0	34.4	50	22	4.0	43.3	70	28	7.5
41.1	70	26	8.0	50.8	60	26	10.0	47.9	80	34	6.5	32.4	60	22	5.5
38.7	70	26	8.0	44.2	70	28	4.5	46.2	50	21	10.0	35.5	60	22	4.5

The null hypothesis is, there is no relationship between $y$ and $x_1,x_2,x_3$ and the alternative hypothesis is, there is relationship between $y$ and $x_1,x_2,x_3$ .

For the variable $x_1$ ,

From Figure-1 it is clear that the p value is $0.954$ which is greater than the level of significance of $0.05$ .

Hence, the null hypothesis is not rejected.

Thus, there is no linear relationship between $y$ and $x_1$ .

For the variable $x_2$ ,

From Figure-1 it is clear that the p value is $0.008$ which is less than the level of significance of $0.05$ .

Hence, the null hypothesis is rejected.

Thus, there is alinear relationship between $y$ and $x_2$ .

For the variable $x_3$ .

From Figure-1 it is clear that the p value is $0$ which is less than the level of significance of $0.05$ .

Hence, the null hypothesis is rejected.

Thus, there is a linear relationship between $y$ and $x_3$ .