Find the value of b 1 .

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 11.3, Problem 16E

a.

To determine

Find the value of $b1$ .

a.

Expert Solution

Answer to Problem 16E

The slope $b1$ is –0.154.

Explanation of Solution

Calculation:

The given information is that the sample data consists of 8 values for x and y.

Slope or $b1$ :

$b1=rsysx$

where,

r represents the correlation coefficient between x and y.

$sy$ represents the standard deviation of y.

$sx$ represents the standard deviation of x.

Software procedure:

Step-by-step procedure to find the mean, standard deviation for x and y values using MINITAB is given below:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns of x and y.
• Choose Options Statistics, and select Mean and Standard deviation.
• Click OK.

Output obtained from MINITAB is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 11.3, Problem 16E , additional homework tip 1

Correlation:

$r=1n−1∑(x−x¯sx)(y−y¯sy)$

The table shows the calculation of correlation:

x	y	$x−x¯$	$y−y¯$	$x−x¯sx$	$y−y¯sy$	$(x−x¯sx)(y−y¯sy)$
12	13	–0.38	–0.88	–0.0795	–0.888	0.0706
17	14	4.62	0.12	0.96653	0.12109	0.117
3	16	–9.38	2.12	–1.9623	2.13925	–4.1979
17	13	4.62	–0.88	0.96653	–0.888	–0.8583
16	14	3.62	0.12	0.75732	0.12109	0.0917
11	14	–1.38	0.12	–0.2887	0.12109	–0.035
14	13	1.62	–0.88	0.33891	–0.888	–0.301
9	14	–3.38	0.12	–0.7071	0.12109	–0.0856
Total						–5.1985

Thus, the correlation is

$r=−5.19858−1=−5.19857=−0.743$

$b1=rsysx$

Substitute r as –0.743, $sy$ as 0.991 and $sx$ as 4.78.

$b1=−0.743(0.9914.78)=−0.743(0.207)=−0.154$

Thus, the slope $b1$ is –0.154.

b.

To determine

Find the residual standard deviation $se$ .

b.

Expert Solution

Answer to Problem 16E

The residual standard deviation $se$ is 0.717.

Explanation of Solution

Calculation:

Finding the value of the intercept term before find the residual standard deviation:

Intercept or $b0$ :

$b0=y¯−b1x¯$

$y¯$ represents the mean of y values.

$x¯$ represents the mean of x values.

$b1$ represents the slope coefficient.

Substitute $y¯$ as 13.88, $x¯$ as 12.38 and $b1$ as –0.154.

$b0=13.88−(−0.154)(12.38)=13.88+(0.154)(12.38)=13.88+1.91=15.79$

Thus, the intercept $b0$ is 15.79.

The residual standard deviation $se$ is calculated using the formula,

$se=∑(y−y^)2n−2$

Where,

$∑(y−y^)2$ represents the sum of squares due to error

n represents the sample size.

Thus, the estimated regression equation is $y^=15.79−0.154x$

Use the estimated regression equation to find the predicted value of y for each value of x.

x	y	$y^$	$y−y^$	$(y−y^)2$
12	13	13.942	–0.942	0.88736
17	14	13.172	0.828	0.68558
3	16	15.328	0.672	0.45158
17	13	13.172	–0.172	0.02958
16	14	13.326	0.674	0.45428
11	14	14.096	–0.096	0.00922
14	13	13.634	–0.634	0.40196
9	14	14.404	–0.404	0.16322
Total				3.083

Substitute $∑(y−y^)2$ as 3.083 and n as 8.

$se=3.0838−2=3.0836=0.514=0.717$

Thus, the residual standard deviation $se$ is 0.717.

c.

To determine

Find the sum of squares for x.

c.

Expert Solution

Answer to Problem 16E

The sum of squares for x is 159.88.

Explanation of Solution

Calculation:

The table shows the calculation of sum of squares for x

x	$x−x¯$	$(x−x¯)2$
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
Total		159.88

Thus, the sum of squares for x is 159.88.

d.

To determine

Find the standard error of $b1,sb$ .

d.

Expert Solution

Answer to Problem 16E

The standard error of $b1$ , $sb$ is 0.057.

Explanation of Solution

Calculation:

The standard error of $b1$ is calculated by using the formula:

$sb=se∑(x−x¯)2$

Where,

$se$ represents the residual standard deviation.

$∑(x−x¯)2$ represents the sum of squares due to x.

Substitute $∑(x−x¯)2$ as 159.88 and $se$ as 0.717.

$sb=0.717159.88=0.71712.64=0.057$

Thus, the standard error of $b1$ , $sb$ is 0.057.

e.

To determine

Find the critical value for a 95% confidence interval of $β1$ .

e.

Expert Solution

Answer to Problem 16E

The critical value for a 95% confidence interval of $β1$ is 2.447.

Explanation of Solution

Calculation:

Critical value:

Software procedure:

Step-by-step procedure to find the critical value using MINITAB is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• In Degrees of freedom, enter 6.
• Click the Shaded Area tab.
• Choose Probability and Two tail for the region of the curve to shade.
• Enter the Probability value as 0.05.
• Click OK.

Output obtained from MINITAB is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 11.3, Problem 16E , additional homework tip 2

Thus, the critical value for a 95% confidence interval of $β1$ is 2.447.

f.

To determine

Find the margin of error for a 95% confidence interval of $β1$ .

f.

Expert Solution

Answer to Problem 16E

The margin of error of $b1$ is 0.139.

Explanation of Solution

Calculation:

The margin of error for a 95% confidence interval of $β1$ is calculated using the formula:

$Margin of error=tα2⋅sb1$

Where,

$tα2$ represents the two tailed critical value for a given level of significance.

$sb1$ represents the standard error of $b1$ .

Substitute $tα2$ as 2.447 and $sb1$ as 0.057.

$Margin of error=(2.447)(0.057)=0.139$

Thus, the margin of error of $b1$ is 0.139.

g.

To determine

Construct the 95% confidence interval for $β1$ .

g.

Expert Solution

Answer to Problem 16E

The 95% confidence interval for $β1$ is $(−0.293,−0.015)$

Explanation of Solution

Calculation:

The confidence interval for $β1$ is calculated by using the formula:

$Confidence interval=b1±tα2⋅sb1$

Where,

$b1$ represents the estimated slope coefficient.

$tα2⋅sb1$ represents the margin of error.

$Confidence interval=−0.154±(2.447)⋅(0.057)=−0.154±0.139=−0.293,−0.015$

Thus, the 95% confidence interval for $β1$ is $(−0.293,−0.015)$

h.

To determine

Test the significance of $β1$ using 5% level of significance.

h.

Expert Solution

Answer to Problem 16E

There is a support of evidence to conclude that there is a linear relationship between x and y at 5% level of significance.

Explanation of Solution

Calculation:

The hypotheses used for testing the significance is given below:

Null hypothesis:

$H0:β1=0$

That is, there is no linear relationship between x and y.

Alternate hypothesis:

$H1:β1≠0$

That is, there is a linear relationship between x and y.

Test statistic:

$t=β^1−β1sb1$

Where,

$β^1$ represents the estimated slope coefficient.

$β1$ represents the hypothesized value of slope coefficient.

$sb1$ represents the standard error of slope coefficient.

Substitute $β^1$ as –0.154, $β1$ as 0 and $sb1$ as 0.057.

$t=−0.154−00.057=−0.1540.057=−2.70$

From part e, the critical value is identified as –2.447.

Decision Rule:

If the positive test statistic value is greater than or equal to the positive critical value or less than the negative critical value, then reject the null hypothesis. Otherwise do not reject the null hypothesis.

Conclusion:

The test statistic value is –2.70 and the critical value is –2.447.

The test statistic value is lesser than the critical value.

That is, $−2.70(=test statistic)<−2.447(=critical value)$

Thus, the null hypothesis is rejected.

Hence, there is a support of evidence to conclude that there is a linear relationship between x and y.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Using the accompanying Home Market Value data and associated regression line, Market ValueMarket Valueequals=$28,416+$37.066×Square Feet, compute the errors associated with each observation using the formula e Subscript ieiequals=Upper Y Subscript iYiminus−ModifyingAbove Upper Y with caret Subscript iYi and construct a frequency distribution and histogram. LOADING... Click the icon to view the Home Market Value data. Question content area bottom Part 1 Construct a frequency distribution of the errors, e Subscript iei. (Type whole numbers.) Error Frequency minus−15 comma 00015,000less than< e Subscript iei less than or equals≤minus−10 comma 00010,000 0 minus−10 comma 00010,000less than< e Subscript iei less than or equals≤minus−50005000 5 minus−50005000less than< e Subscript iei less than or equals≤0 21 0less than< e Subscript iei less than or equals≤50005000 9…

The managing director of a consulting group has the accompanying monthly data on total overhead costs and professional labor hours to bill to clients. Complete parts a through c Overhead Costs Billable Hours345000 3000385000 4000410000 5000462000 6000530000 7000545000 8000

Using the accompanying Home Market Value data and associated regression line, Market ValueMarket Valueequals=$28,416plus+$37.066×Square Feet, compute the errors associated with each observation using the formula e Subscript ieiequals=Upper Y Subscript iYiminus−ModifyingAbove Upper Y with caret Subscript iYi and construct a frequency distribution and histogram. Square Feet Market Value1813 911001916 1043001842 934001814 909001836 1020002030 1085001731 877001852 960001793 893001665 884001852 1009001619 967001690 876002370 1139002373 1131001666 875002122 1161001619 946001729 863001667 871001522 833001484 798001589 814001600 871001484 825001483 787001522 877001703 942001485 820001468 881001519 882001518 885001483 765001522 844001668 909001587 810001782 912001483 812001519 1007001522 872001684 966001581 86200

Answer 2

Question

Chapter 11.3, Problem 16E

a.

To determine

Find the value of $b1$ .

a.

Expert Solution

Answer to Problem 16E

The slope $b1$ is –0.154.

Explanation of Solution

Calculation:

The given information is that the sample data consists of 8 values for x and y.

Slope or $b1$ :

$b1=rsysx$

where,

r represents the correlation coefficient between x and y.

$sy$ represents the standard deviation of y.

$sx$ represents the standard deviation of x.

Software procedure:

Step-by-step procedure to find the mean, standard deviation for x and y values using MINITAB is given below:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns of x and y.
• Choose Options Statistics, and select Mean and Standard deviation.
• Click OK.

Output obtained from MINITAB is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 11.3, Problem 16E , additional homework tip 1

Correlation:

$r=1n−1∑(x−x¯sx)(y−y¯sy)$

The table shows the calculation of correlation:

x	y	$x−x¯$	$y−y¯$	$x−x¯sx$	$y−y¯sy$	$(x−x¯sx)(y−y¯sy)$
12	13	–0.38	–0.88	–0.0795	–0.888	0.0706
17	14	4.62	0.12	0.96653	0.12109	0.117
3	16	–9.38	2.12	–1.9623	2.13925	–4.1979
17	13	4.62	–0.88	0.96653	–0.888	–0.8583
16	14	3.62	0.12	0.75732	0.12109	0.0917
11	14	–1.38	0.12	–0.2887	0.12109	–0.035
14	13	1.62	–0.88	0.33891	–0.888	–0.301
9	14	–3.38	0.12	–0.7071	0.12109	–0.0856
Total						–5.1985

Thus, the correlation is

$r=−5.19858−1=−5.19857=−0.743$

$b1=rsysx$

Substitute r as –0.743, $sy$ as 0.991 and $sx$ as 4.78.

$b1=−0.743(0.9914.78)=−0.743(0.207)=−0.154$

Thus, the slope $b1$ is –0.154.

b.

To determine

Find the residual standard deviation $se$ .

b.

Expert Solution

Answer to Problem 16E

The residual standard deviation $se$ is 0.717.

Explanation of Solution

Calculation:

Finding the value of the intercept term before find the residual standard deviation:

Intercept or $b0$ :

$b0=y¯−b1x¯$

$y¯$ represents the mean of y values.

$x¯$ represents the mean of x values.

$b1$ represents the slope coefficient.

Substitute $y¯$ as 13.88, $x¯$ as 12.38 and $b1$ as –0.154.

$b0=13.88−(−0.154)(12.38)=13.88+(0.154)(12.38)=13.88+1.91=15.79$

Thus, the intercept $b0$ is 15.79.

The residual standard deviation $se$ is calculated using the formula,

$se=∑(y−y^)2n−2$

Where,

$∑(y−y^)2$ represents the sum of squares due to error

n represents the sample size.

Thus, the estimated regression equation is $y^=15.79−0.154x$

Use the estimated regression equation to find the predicted value of y for each value of x.

x	y	$y^$	$y−y^$	$(y−y^)2$
12	13	13.942	–0.942	0.88736
17	14	13.172	0.828	0.68558
3	16	15.328	0.672	0.45158
17	13	13.172	–0.172	0.02958
16	14	13.326	0.674	0.45428
11	14	14.096	–0.096	0.00922
14	13	13.634	–0.634	0.40196
9	14	14.404	–0.404	0.16322
Total				3.083

Substitute $∑(y−y^)2$ as 3.083 and n as 8.

$se=3.0838−2=3.0836=0.514=0.717$

Thus, the residual standard deviation $se$ is 0.717.

c.

To determine

Find the sum of squares for x.

c.

Expert Solution

Answer to Problem 16E

The sum of squares for x is 159.88.

Explanation of Solution

Calculation:

The table shows the calculation of sum of squares for x

x	$x−x¯$	$(x−x¯)2$
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
Total		159.88

Thus, the sum of squares for x is 159.88.

d.

To determine

Find the standard error of $b1,sb$ .

d.

Expert Solution

Answer to Problem 16E

The standard error of $b1$ , $sb$ is 0.057.

Explanation of Solution

Calculation:

The standard error of $b1$ is calculated by using the formula:

$sb=se∑(x−x¯)2$

Where,

$se$ represents the residual standard deviation.

$∑(x−x¯)2$ represents the sum of squares due to x.

Substitute $∑(x−x¯)2$ as 159.88 and $se$ as 0.717.

$sb=0.717159.88=0.71712.64=0.057$

Thus, the standard error of $b1$ , $sb$ is 0.057.

e.

To determine

Find the critical value for a 95% confidence interval of $β1$ .

e.

Expert Solution

Answer to Problem 16E

The critical value for a 95% confidence interval of $β1$ is 2.447.

Explanation of Solution

Calculation:

Critical value:

Software procedure:

Step-by-step procedure to find the critical value using MINITAB is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• In Degrees of freedom, enter 6.
• Click the Shaded Area tab.
• Choose Probability and Two tail for the region of the curve to shade.
• Enter the Probability value as 0.05.
• Click OK.

Output obtained from MINITAB is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 11.3, Problem 16E , additional homework tip 2

Thus, the critical value for a 95% confidence interval of $β1$ is 2.447.

f.

To determine

Find the margin of error for a 95% confidence interval of $β1$ .

f.

Expert Solution

Answer to Problem 16E

The margin of error of $b1$ is 0.139.

Explanation of Solution

Calculation:

The margin of error for a 95% confidence interval of $β1$ is calculated using the formula:

$Margin of error=tα2⋅sb1$

Where,

$tα2$ represents the two tailed critical value for a given level of significance.

$sb1$ represents the standard error of $b1$ .

Substitute $tα2$ as 2.447 and $sb1$ as 0.057.

$Margin of error=(2.447)(0.057)=0.139$

Thus, the margin of error of $b1$ is 0.139.

g.

To determine

Construct the 95% confidence interval for $β1$ .

g.

Expert Solution

Answer to Problem 16E

The 95% confidence interval for $β1$ is $(−0.293,−0.015)$

Explanation of Solution

Calculation:

The confidence interval for $β1$ is calculated by using the formula:

$Confidence interval=b1±tα2⋅sb1$

Where,

$b1$ represents the estimated slope coefficient.

$tα2⋅sb1$ represents the margin of error.

$Confidence interval=−0.154±(2.447)⋅(0.057)=−0.154±0.139=−0.293,−0.015$

Thus, the 95% confidence interval for $β1$ is $(−0.293,−0.015)$

h.

To determine

Test the significance of $β1$ using 5% level of significance.

h.

Expert Solution

Answer to Problem 16E

There is a support of evidence to conclude that there is a linear relationship between x and y at 5% level of significance.

Explanation of Solution

Calculation:

The hypotheses used for testing the significance is given below:

Null hypothesis:

$H0:β1=0$

That is, there is no linear relationship between x and y.

Alternate hypothesis:

$H1:β1≠0$

That is, there is a linear relationship between x and y.

Test statistic:

$t=β^1−β1sb1$

Where,

$β^1$ represents the estimated slope coefficient.

$β1$ represents the hypothesized value of slope coefficient.

$sb1$ represents the standard error of slope coefficient.

Substitute $β^1$ as –0.154, $β1$ as 0 and $sb1$ as 0.057.

$t=−0.154−00.057=−0.1540.057=−2.70$

From part e, the critical value is identified as –2.447.

Decision Rule:

If the positive test statistic value is greater than or equal to the positive critical value or less than the negative critical value, then reject the null hypothesis. Otherwise do not reject the null hypothesis.

Conclusion:

The test statistic value is –2.70 and the critical value is –2.447.

The test statistic value is lesser than the critical value.

That is, $−2.70(=test statistic)<−2.447(=critical value)$

Thus, the null hypothesis is rejected.

Hence, there is a support of evidence to conclude that there is a linear relationship between x and y.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 11.3, Problem 16E

a.

To determine

Find the value of $b1$ .

a.

Expert Solution

Answer to Problem 16E

The slope $b1$ is –0.154.

Explanation of Solution

Calculation:

The given information is that the sample data consists of 8 values for x and y.

Slope or $b1$ :

$b1=rsysx$

where,

r represents the correlation coefficient between x and y.

$sy$ represents the standard deviation of y.

$sx$ represents the standard deviation of x.

Software procedure:

Step-by-step procedure to find the mean, standard deviation for x and y values using MINITAB is given below:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns of x and y.
• Choose Options Statistics, and select Mean and Standard deviation.
• Click OK.

Output obtained from MINITAB is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 11.3, Problem 16E , additional homework tip 1

Correlation:

$r=1n−1∑(x−x¯sx)(y−y¯sy)$

The table shows the calculation of correlation:

x	y	$x−x¯$	$y−y¯$	$x−x¯sx$	$y−y¯sy$	$(x−x¯sx)(y−y¯sy)$
12	13	–0.38	–0.88	–0.0795	–0.888	0.0706
17	14	4.62	0.12	0.96653	0.12109	0.117
3	16	–9.38	2.12	–1.9623	2.13925	–4.1979
17	13	4.62	–0.88	0.96653	–0.888	–0.8583
16	14	3.62	0.12	0.75732	0.12109	0.0917
11	14	–1.38	0.12	–0.2887	0.12109	–0.035
14	13	1.62	–0.88	0.33891	–0.888	–0.301
9	14	–3.38	0.12	–0.7071	0.12109	–0.0856
Total						–5.1985

Thus, the correlation is

$r=−5.19858−1=−5.19857=−0.743$

$b1=rsysx$

Substitute r as –0.743, $sy$ as 0.991 and $sx$ as 4.78.

$b1=−0.743(0.9914.78)=−0.743(0.207)=−0.154$

Thus, the slope $b1$ is –0.154.

b.

To determine

Find the residual standard deviation $se$ .

b.

Expert Solution

Answer to Problem 16E

The residual standard deviation $se$ is 0.717.

Explanation of Solution

Calculation:

Finding the value of the intercept term before find the residual standard deviation:

Intercept or $b0$ :

$b0=y¯−b1x¯$

$y¯$ represents the mean of y values.

$x¯$ represents the mean of x values.

$b1$ represents the slope coefficient.

Substitute $y¯$ as 13.88, $x¯$ as 12.38 and $b1$ as –0.154.

$b0=13.88−(−0.154)(12.38)=13.88+(0.154)(12.38)=13.88+1.91=15.79$

Thus, the intercept $b0$ is 15.79.

The residual standard deviation $se$ is calculated using the formula,

$se=∑(y−y^)2n−2$

Where,

$∑(y−y^)2$ represents the sum of squares due to error

n represents the sample size.

Thus, the estimated regression equation is $y^=15.79−0.154x$

Use the estimated regression equation to find the predicted value of y for each value of x.

x	y	$y^$	$y−y^$	$(y−y^)2$
12	13	13.942	–0.942	0.88736
17	14	13.172	0.828	0.68558
3	16	15.328	0.672	0.45158
17	13	13.172	–0.172	0.02958
16	14	13.326	0.674	0.45428
11	14	14.096	–0.096	0.00922
14	13	13.634	–0.634	0.40196
9	14	14.404	–0.404	0.16322
Total				3.083

Substitute $∑(y−y^)2$ as 3.083 and n as 8.

$se=3.0838−2=3.0836=0.514=0.717$

Thus, the residual standard deviation $se$ is 0.717.

c.

To determine

Find the sum of squares for x.

c.

Expert Solution

Answer to Problem 16E

The sum of squares for x is 159.88.

Explanation of Solution

Calculation:

The table shows the calculation of sum of squares for x

x	$x−x¯$	$(x−x¯)2$
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
Total		159.88

Thus, the sum of squares for x is 159.88.

d.

To determine

Find the standard error of $b1,sb$ .

d.

Expert Solution

Answer to Problem 16E

The standard error of $b1$ , $sb$ is 0.057.

Explanation of Solution

Calculation:

The standard error of $b1$ is calculated by using the formula:

$sb=se∑(x−x¯)2$

Where,

$se$ represents the residual standard deviation.

$∑(x−x¯)2$ represents the sum of squares due to x.

Substitute $∑(x−x¯)2$ as 159.88 and $se$ as 0.717.

$sb=0.717159.88=0.71712.64=0.057$

Thus, the standard error of $b1$ , $sb$ is 0.057.

e.

To determine

Find the critical value for a 95% confidence interval of $β1$ .

e.

Expert Solution

Answer to Problem 16E

The critical value for a 95% confidence interval of $β1$ is 2.447.

Explanation of Solution

Calculation:

Critical value:

Software procedure:

Step-by-step procedure to find the critical value using MINITAB is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• In Degrees of freedom, enter 6.
• Click the Shaded Area tab.
• Choose Probability and Two tail for the region of the curve to shade.
• Enter the Probability value as 0.05.
• Click OK.

Output obtained from MINITAB is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 11.3, Problem 16E , additional homework tip 2

Thus, the critical value for a 95% confidence interval of $β1$ is 2.447.

f.

To determine

Find the margin of error for a 95% confidence interval of $β1$ .

f.

Expert Solution

Answer to Problem 16E

The margin of error of $b1$ is 0.139.

Explanation of Solution

Calculation:

The margin of error for a 95% confidence interval of $β1$ is calculated using the formula:

$Margin of error=tα2⋅sb1$

Where,

$tα2$ represents the two tailed critical value for a given level of significance.

$sb1$ represents the standard error of $b1$ .

Substitute $tα2$ as 2.447 and $sb1$ as 0.057.

$Margin of error=(2.447)(0.057)=0.139$

Thus, the margin of error of $b1$ is 0.139.

g.

To determine

Construct the 95% confidence interval for $β1$ .

g.

Expert Solution

Answer to Problem 16E

The 95% confidence interval for $β1$ is $(−0.293,−0.015)$

Explanation of Solution

Calculation:

The confidence interval for $β1$ is calculated by using the formula:

$Confidence interval=b1±tα2⋅sb1$

Where,

$b1$ represents the estimated slope coefficient.

$tα2⋅sb1$ represents the margin of error.

$Confidence interval=−0.154±(2.447)⋅(0.057)=−0.154±0.139=−0.293,−0.015$

Thus, the 95% confidence interval for $β1$ is $(−0.293,−0.015)$

h.

To determine

Test the significance of $β1$ using 5% level of significance.

h.

Expert Solution

Answer to Problem 16E

There is a support of evidence to conclude that there is a linear relationship between x and y at 5% level of significance.

Explanation of Solution

Calculation:

The hypotheses used for testing the significance is given below:

Null hypothesis:

$H0:β1=0$

That is, there is no linear relationship between x and y.

Alternate hypothesis:

$H1:β1≠0$

That is, there is a linear relationship between x and y.

Test statistic:

$t=β^1−β1sb1$

Where,

$β^1$ represents the estimated slope coefficient.

$β1$ represents the hypothesized value of slope coefficient.

$sb1$ represents the standard error of slope coefficient.

Substitute $β^1$ as –0.154, $β1$ as 0 and $sb1$ as 0.057.

$t=−0.154−00.057=−0.1540.057=−2.70$

From part e, the critical value is identified as –2.447.

Decision Rule:

If the positive test statistic value is greater than or equal to the positive critical value or less than the negative critical value, then reject the null hypothesis. Otherwise do not reject the null hypothesis.

Conclusion:

The test statistic value is –2.70 and the critical value is –2.447.

The test statistic value is lesser than the critical value.

That is, $−2.70(=test statistic)<−2.447(=critical value)$

Thus, the null hypothesis is rejected.

Hence, there is a support of evidence to conclude that there is a linear relationship between x and y.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 11.3, Problem 16E

a.

To determine

Find the value of $b1$ .

a.

Expert Solution

Answer to Problem 16E

The slope $b1$ is –0.154.

Explanation of Solution

Calculation:

The given information is that the sample data consists of 8 values for x and y.

Slope or $b1$ :

$b1=rsysx$

where,

r represents the correlation coefficient between x and y.

$sy$ represents the standard deviation of y.

$sx$ represents the standard deviation of x.

Software procedure:

Step-by-step procedure to find the mean, standard deviation for x and y values using MINITAB is given below:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns of x and y.
• Choose Options Statistics, and select Mean and Standard deviation.
• Click OK.

Output obtained from MINITAB is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 11.3, Problem 16E , additional homework tip 1

Correlation:

$r=1n−1∑(x−x¯sx)(y−y¯sy)$

The table shows the calculation of correlation:

x	y	$x−x¯$	$y−y¯$	$x−x¯sx$	$y−y¯sy$	$(x−x¯sx)(y−y¯sy)$
12	13	–0.38	–0.88	–0.0795	–0.888	0.0706
17	14	4.62	0.12	0.96653	0.12109	0.117
3	16	–9.38	2.12	–1.9623	2.13925	–4.1979
17	13	4.62	–0.88	0.96653	–0.888	–0.8583
16	14	3.62	0.12	0.75732	0.12109	0.0917
11	14	–1.38	0.12	–0.2887	0.12109	–0.035
14	13	1.62	–0.88	0.33891	–0.888	–0.301
9	14	–3.38	0.12	–0.7071	0.12109	–0.0856
Total						–5.1985

Thus, the correlation is

$r=−5.19858−1=−5.19857=−0.743$

$b1=rsysx$

Substitute r as –0.743, $sy$ as 0.991 and $sx$ as 4.78.

$b1=−0.743(0.9914.78)=−0.743(0.207)=−0.154$

Thus, the slope $b1$ is –0.154.

b.

To determine

Find the residual standard deviation $se$ .

b.

Expert Solution

Answer to Problem 16E

The residual standard deviation $se$ is 0.717.

Explanation of Solution

Calculation:

Finding the value of the intercept term before find the residual standard deviation:

Intercept or $b0$ :

$b0=y¯−b1x¯$

$y¯$ represents the mean of y values.

$x¯$ represents the mean of x values.

$b1$ represents the slope coefficient.

Substitute $y¯$ as 13.88, $x¯$ as 12.38 and $b1$ as –0.154.

$b0=13.88−(−0.154)(12.38)=13.88+(0.154)(12.38)=13.88+1.91=15.79$

Thus, the intercept $b0$ is 15.79.

The residual standard deviation $se$ is calculated using the formula,

$se=∑(y−y^)2n−2$

Where,

$∑(y−y^)2$ represents the sum of squares due to error

n represents the sample size.

Thus, the estimated regression equation is $y^=15.79−0.154x$

Use the estimated regression equation to find the predicted value of y for each value of x.

x	y	$y^$	$y−y^$	$(y−y^)2$
12	13	13.942	–0.942	0.88736
17	14	13.172	0.828	0.68558
3	16	15.328	0.672	0.45158
17	13	13.172	–0.172	0.02958
16	14	13.326	0.674	0.45428
11	14	14.096	–0.096	0.00922
14	13	13.634	–0.634	0.40196
9	14	14.404	–0.404	0.16322
Total				3.083

Substitute $∑(y−y^)2$ as 3.083 and n as 8.

$se=3.0838−2=3.0836=0.514=0.717$

Thus, the residual standard deviation $se$ is 0.717.

c.

To determine

Find the sum of squares for x.

c.

Expert Solution

Answer to Problem 16E

The sum of squares for x is 159.88.

Explanation of Solution

Calculation:

The table shows the calculation of sum of squares for x

x	$x−x¯$	$(x−x¯)2$
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
Total		159.88

Thus, the sum of squares for x is 159.88.

d.

To determine

Find the standard error of $b1,sb$ .

d.

Expert Solution

Answer to Problem 16E

The standard error of $b1$ , $sb$ is 0.057.

Explanation of Solution

Calculation:

The standard error of $b1$ is calculated by using the formula:

$sb=se∑(x−x¯)2$

Where,

$se$ represents the residual standard deviation.

$∑(x−x¯)2$ represents the sum of squares due to x.

Substitute $∑(x−x¯)2$ as 159.88 and $se$ as 0.717.

$sb=0.717159.88=0.71712.64=0.057$

Thus, the standard error of $b1$ , $sb$ is 0.057.

e.

To determine

Find the critical value for a 95% confidence interval of $β1$ .

e.

Expert Solution

Answer to Problem 16E

The critical value for a 95% confidence interval of $β1$ is 2.447.

Explanation of Solution

Calculation:

Critical value:

Software procedure:

Step-by-step procedure to find the critical value using MINITAB is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• In Degrees of freedom, enter 6.
• Click the Shaded Area tab.
• Choose Probability and Two tail for the region of the curve to shade.
• Enter the Probability value as 0.05.
• Click OK.

Output obtained from MINITAB is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 11.3, Problem 16E , additional homework tip 2

Thus, the critical value for a 95% confidence interval of $β1$ is 2.447.

f.

To determine

Find the margin of error for a 95% confidence interval of $β1$ .

f.

Expert Solution

Answer to Problem 16E

The margin of error of $b1$ is 0.139.

Explanation of Solution

Calculation:

The margin of error for a 95% confidence interval of $β1$ is calculated using the formula:

$Margin of error=tα2⋅sb1$

Where,

$tα2$ represents the two tailed critical value for a given level of significance.

$sb1$ represents the standard error of $b1$ .

Substitute $tα2$ as 2.447 and $sb1$ as 0.057.

$Margin of error=(2.447)(0.057)=0.139$

Thus, the margin of error of $b1$ is 0.139.

g.

To determine

Construct the 95% confidence interval for $β1$ .

g.

Expert Solution

Answer to Problem 16E

The 95% confidence interval for $β1$ is $(−0.293,−0.015)$

Explanation of Solution

Calculation:

The confidence interval for $β1$ is calculated by using the formula:

$Confidence interval=b1±tα2⋅sb1$

Where,

$b1$ represents the estimated slope coefficient.

$tα2⋅sb1$ represents the margin of error.

$Confidence interval=−0.154±(2.447)⋅(0.057)=−0.154±0.139=−0.293,−0.015$

Thus, the 95% confidence interval for $β1$ is $(−0.293,−0.015)$

h.

To determine

Test the significance of $β1$ using 5% level of significance.

h.

Expert Solution

Answer to Problem 16E

There is a support of evidence to conclude that there is a linear relationship between x and y at 5% level of significance.

Explanation of Solution

Calculation:

The hypotheses used for testing the significance is given below:

Null hypothesis:

$H0:β1=0$

That is, there is no linear relationship between x and y.

Alternate hypothesis:

$H1:β1≠0$

That is, there is a linear relationship between x and y.

Test statistic:

$t=β^1−β1sb1$

Where,

$β^1$ represents the estimated slope coefficient.

$β1$ represents the hypothesized value of slope coefficient.

$sb1$ represents the standard error of slope coefficient.

Substitute $β^1$ as –0.154, $β1$ as 0 and $sb1$ as 0.057.

$t=−0.154−00.057=−0.1540.057=−2.70$

From part e, the critical value is identified as –2.447.

Decision Rule:

If the positive test statistic value is greater than or equal to the positive critical value or less than the negative critical value, then reject the null hypothesis. Otherwise do not reject the null hypothesis.

Conclusion:

The test statistic value is –2.70 and the critical value is –2.447.

The test statistic value is lesser than the critical value.

That is, $−2.70(=test statistic)<−2.447(=critical value)$

Thus, the null hypothesis is rejected.

Hence, there is a support of evidence to conclude that there is a linear relationship between x and y.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 11.3, Problem 16E

a.

To determine

Find the value of $b1$ .

a.

Expert Solution

Answer to Problem 16E

The slope $b1$ is –0.154.

Explanation of Solution

Calculation:

The given information is that the sample data consists of 8 values for x and y.

Slope or $b1$ :

$b1=rsysx$

where,

r represents the correlation coefficient between x and y.

$sy$ represents the standard deviation of y.

$sx$ represents the standard deviation of x.

Software procedure:

Step-by-step procedure to find the mean, standard deviation for x and y values using MINITAB is given below:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns of x and y.
• Choose Options Statistics, and select Mean and Standard deviation.
• Click OK.

Output obtained from MINITAB is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 11.3, Problem 16E , additional homework tip 1

Correlation:

$r=1n−1∑(x−x¯sx)(y−y¯sy)$

The table shows the calculation of correlation:

x	y	$x−x¯$	$y−y¯$	$x−x¯sx$	$y−y¯sy$	$(x−x¯sx)(y−y¯sy)$
12	13	–0.38	–0.88	–0.0795	–0.888	0.0706
17	14	4.62	0.12	0.96653	0.12109	0.117
3	16	–9.38	2.12	–1.9623	2.13925	–4.1979
17	13	4.62	–0.88	0.96653	–0.888	–0.8583
16	14	3.62	0.12	0.75732	0.12109	0.0917
11	14	–1.38	0.12	–0.2887	0.12109	–0.035
14	13	1.62	–0.88	0.33891	–0.888	–0.301
9	14	–3.38	0.12	–0.7071	0.12109	–0.0856
Total						–5.1985

Thus, the correlation is

$r=−5.19858−1=−5.19857=−0.743$

$b1=rsysx$

Substitute r as –0.743, $sy$ as 0.991 and $sx$ as 4.78.

$b1=−0.743(0.9914.78)=−0.743(0.207)=−0.154$

Thus, the slope $b1$ is –0.154.

b.

To determine

Find the residual standard deviation $se$ .

b.

Expert Solution

Answer to Problem 16E

The residual standard deviation $se$ is 0.717.

Explanation of Solution

Calculation:

Finding the value of the intercept term before find the residual standard deviation:

Intercept or $b0$ :

$b0=y¯−b1x¯$

$y¯$ represents the mean of y values.

$x¯$ represents the mean of x values.

$b1$ represents the slope coefficient.

Substitute $y¯$ as 13.88, $x¯$ as 12.38 and $b1$ as –0.154.

$b0=13.88−(−0.154)(12.38)=13.88+(0.154)(12.38)=13.88+1.91=15.79$

Thus, the intercept $b0$ is 15.79.

The residual standard deviation $se$ is calculated using the formula,

$se=∑(y−y^)2n−2$

Where,

$∑(y−y^)2$ represents the sum of squares due to error

n represents the sample size.

Thus, the estimated regression equation is $y^=15.79−0.154x$

Use the estimated regression equation to find the predicted value of y for each value of x.

x	y	$y^$	$y−y^$	$(y−y^)2$
12	13	13.942	–0.942	0.88736
17	14	13.172	0.828	0.68558
3	16	15.328	0.672	0.45158
17	13	13.172	–0.172	0.02958
16	14	13.326	0.674	0.45428
11	14	14.096	–0.096	0.00922
14	13	13.634	–0.634	0.40196
9	14	14.404	–0.404	0.16322
Total				3.083

Substitute $∑(y−y^)2$ as 3.083 and n as 8.

$se=3.0838−2=3.0836=0.514=0.717$

Thus, the residual standard deviation $se$ is 0.717.

c.

To determine

Find the sum of squares for x.

c.

Expert Solution

Answer to Problem 16E

The sum of squares for x is 159.88.

Explanation of Solution

Calculation:

The table shows the calculation of sum of squares for x

x	$x−x¯$	$(x−x¯)2$
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
Total		159.88

Thus, the sum of squares for x is 159.88.

d.

To determine

Find the standard error of $b1,sb$ .

d.

Expert Solution

Answer to Problem 16E

The standard error of $b1$ , $sb$ is 0.057.

Explanation of Solution

Calculation:

The standard error of $b1$ is calculated by using the formula:

$sb=se∑(x−x¯)2$

Where,

$se$ represents the residual standard deviation.

$∑(x−x¯)2$ represents the sum of squares due to x.

Substitute $∑(x−x¯)2$ as 159.88 and $se$ as 0.717.

$sb=0.717159.88=0.71712.64=0.057$

Thus, the standard error of $b1$ , $sb$ is 0.057.

e.

To determine

Find the critical value for a 95% confidence interval of $β1$ .

e.

Expert Solution

Answer to Problem 16E

The critical value for a 95% confidence interval of $β1$ is 2.447.

Explanation of Solution

Calculation:

Critical value:

Software procedure:

Step-by-step procedure to find the critical value using MINITAB is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• In Degrees of freedom, enter 6.
• Click the Shaded Area tab.
• Choose Probability and Two tail for the region of the curve to shade.
• Enter the Probability value as 0.05.
• Click OK.

Output obtained from MINITAB is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 11.3, Problem 16E , additional homework tip 2

Thus, the critical value for a 95% confidence interval of $β1$ is 2.447.

f.

To determine

Find the margin of error for a 95% confidence interval of $β1$ .

f.

Expert Solution

Answer to Problem 16E

The margin of error of $b1$ is 0.139.

Explanation of Solution

Calculation:

The margin of error for a 95% confidence interval of $β1$ is calculated using the formula:

$Margin of error=tα2⋅sb1$

Where,

$tα2$ represents the two tailed critical value for a given level of significance.

$sb1$ represents the standard error of $b1$ .

Substitute $tα2$ as 2.447 and $sb1$ as 0.057.

$Margin of error=(2.447)(0.057)=0.139$

Thus, the margin of error of $b1$ is 0.139.

g.

To determine

Construct the 95% confidence interval for $β1$ .

g.

Expert Solution

Answer to Problem 16E

The 95% confidence interval for $β1$ is $(−0.293,−0.015)$

Explanation of Solution

Calculation:

The confidence interval for $β1$ is calculated by using the formula:

$Confidence interval=b1±tα2⋅sb1$

Where,

$b1$ represents the estimated slope coefficient.

$tα2⋅sb1$ represents the margin of error.

$Confidence interval=−0.154±(2.447)⋅(0.057)=−0.154±0.139=−0.293,−0.015$

Thus, the 95% confidence interval for $β1$ is $(−0.293,−0.015)$

h.

To determine

Test the significance of $β1$ using 5% level of significance.

h.

Expert Solution

Answer to Problem 16E

There is a support of evidence to conclude that there is a linear relationship between x and y at 5% level of significance.

Explanation of Solution

Calculation:

The hypotheses used for testing the significance is given below:

Null hypothesis:

$H0:β1=0$

That is, there is no linear relationship between x and y.

Alternate hypothesis:

$H1:β1≠0$

That is, there is a linear relationship between x and y.

Test statistic:

$t=β^1−β1sb1$

Where,

$β^1$ represents the estimated slope coefficient.

$β1$ represents the hypothesized value of slope coefficient.

$sb1$ represents the standard error of slope coefficient.

Substitute $β^1$ as –0.154, $β1$ as 0 and $sb1$ as 0.057.

$t=−0.154−00.057=−0.1540.057=−2.70$

From part e, the critical value is identified as –2.447.

Decision Rule:

If the positive test statistic value is greater than or equal to the positive critical value or less than the negative critical value, then reject the null hypothesis. Otherwise do not reject the null hypothesis.

Conclusion:

The test statistic value is –2.70 and the critical value is –2.447.

The test statistic value is lesser than the critical value.

That is, $−2.70(=test statistic)<−2.447(=critical value)$

Thus, the null hypothesis is rejected.

Hence, there is a support of evidence to conclude that there is a linear relationship between x and y.

Answer 6

Chapter 11.3, Problem 16E

a.

To determine

Find the value of $b1$ .

a.

Expert Solution

Answer to Problem 16E

The slope $b1$ is –0.154.

Explanation of Solution

Calculation:

The given information is that the sample data consists of 8 values for x and y.

Slope or $b1$ :

$b1=rsysx$

where,

r represents the correlation coefficient between x and y.

$sy$ represents the standard deviation of y.

$sx$ represents the standard deviation of x.

Software procedure:

Step-by-step procedure to find the mean, standard deviation for x and y values using MINITAB is given below:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns of x and y.
• Choose Options Statistics, and select Mean and Standard deviation.
• Click OK.

Output obtained from MINITAB is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 11.3, Problem 16E , additional homework tip 1

Correlation:

$r=1n−1∑(x−x¯sx)(y−y¯sy)$

The table shows the calculation of correlation:

x	y	$x−x¯$	$y−y¯$	$x−x¯sx$	$y−y¯sy$	$(x−x¯sx)(y−y¯sy)$
12	13	–0.38	–0.88	–0.0795	–0.888	0.0706
17	14	4.62	0.12	0.96653	0.12109	0.117
3	16	–9.38	2.12	–1.9623	2.13925	–4.1979
17	13	4.62	–0.88	0.96653	–0.888	–0.8583
16	14	3.62	0.12	0.75732	0.12109	0.0917
11	14	–1.38	0.12	–0.2887	0.12109	–0.035
14	13	1.62	–0.88	0.33891	–0.888	–0.301
9	14	–3.38	0.12	–0.7071	0.12109	–0.0856
Total						–5.1985

Thus, the correlation is

$r=−5.19858−1=−5.19857=−0.743$

$b1=rsysx$

Substitute r as –0.743, $sy$ as 0.991 and $sx$ as 4.78.

$b1=−0.743(0.9914.78)=−0.743(0.207)=−0.154$

Thus, the slope $b1$ is –0.154.

Answer 7

Chapter 11.3, Problem 16E

a.

To determine

Find the value of $b1$ .

Answer 8

a.

Expert Solution

Answer to Problem 16E

The slope $b1$ is –0.154.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

The given information is that the sample data consists of 8 values for x and y.

Slope or $b1$ :

$b1=rsysx$

where,

r represents the correlation coefficient between x and y.

$sy$ represents the standard deviation of y.

$sx$ represents the standard deviation of x.

Software procedure:

Step-by-step procedure to find the mean, standard deviation for x and y values using MINITAB is given below:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns of x and y.
• Choose Options Statistics, and select Mean and Standard deviation.
• Click OK.

Output obtained from MINITAB is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 11.3, Problem 16E , additional homework tip 1

Correlation:

$r=1n−1∑(x−x¯sx)(y−y¯sy)$

The table shows the calculation of correlation:

x	y	$x−x¯$	$y−y¯$	$x−x¯sx$	$y−y¯sy$	$(x−x¯sx)(y−y¯sy)$
12	13	–0.38	–0.88	–0.0795	–0.888	0.0706
17	14	4.62	0.12	0.96653	0.12109	0.117
3	16	–9.38	2.12	–1.9623	2.13925	–4.1979
17	13	4.62	–0.88	0.96653	–0.888	–0.8583
16	14	3.62	0.12	0.75732	0.12109	0.0917
11	14	–1.38	0.12	–0.2887	0.12109	–0.035
14	13	1.62	–0.88	0.33891	–0.888	–0.301
9	14	–3.38	0.12	–0.7071	0.12109	–0.0856
Total						–5.1985

Thus, the correlation is

$r=−5.19858−1=−5.19857=−0.743$

$b1=rsysx$

Substitute r as –0.743, $sy$ as 0.991 and $sx$ as 4.78.

$b1=−0.743(0.9914.78)=−0.743(0.207)=−0.154$

Thus, the slope $b1$ is –0.154.

Answer 13

b.

To determine

Find the residual standard deviation $se$ .

b.

Expert Solution

Answer to Problem 16E

The residual standard deviation $se$ is 0.717.

Explanation of Solution

Calculation:

Finding the value of the intercept term before find the residual standard deviation:

Intercept or $b0$ :

$b0=y¯−b1x¯$

$y¯$ represents the mean of y values.

$x¯$ represents the mean of x values.

$b1$ represents the slope coefficient.

Substitute $y¯$ as 13.88, $x¯$ as 12.38 and $b1$ as –0.154.

$b0=13.88−(−0.154)(12.38)=13.88+(0.154)(12.38)=13.88+1.91=15.79$

Thus, the intercept $b0$ is 15.79.

The residual standard deviation $se$ is calculated using the formula,

$se=∑(y−y^)2n−2$

Where,

$∑(y−y^)2$ represents the sum of squares due to error

n represents the sample size.

Thus, the estimated regression equation is $y^=15.79−0.154x$

Use the estimated regression equation to find the predicted value of y for each value of x.

x	y	$y^$	$y−y^$	$(y−y^)2$
12	13	13.942	–0.942	0.88736
17	14	13.172	0.828	0.68558
3	16	15.328	0.672	0.45158
17	13	13.172	–0.172	0.02958
16	14	13.326	0.674	0.45428
11	14	14.096	–0.096	0.00922
14	13	13.634	–0.634	0.40196
9	14	14.404	–0.404	0.16322
Total				3.083

Substitute $∑(y−y^)2$ as 3.083 and n as 8.

$se=3.0838−2=3.0836=0.514=0.717$

Thus, the residual standard deviation $se$ is 0.717.

Answer 14

b.

To determine

Find the residual standard deviation $se$ .

Answer 15

b.

Expert Solution

Answer to Problem 16E

The residual standard deviation $se$ is 0.717.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

Finding the value of the intercept term before find the residual standard deviation:

Intercept or $b0$ :

$b0=y¯−b1x¯$

$y¯$ represents the mean of y values.

$x¯$ represents the mean of x values.

$b1$ represents the slope coefficient.

Substitute $y¯$ as 13.88, $x¯$ as 12.38 and $b1$ as –0.154.

$b0=13.88−(−0.154)(12.38)=13.88+(0.154)(12.38)=13.88+1.91=15.79$

Thus, the intercept $b0$ is 15.79.

The residual standard deviation $se$ is calculated using the formula,

$se=∑(y−y^)2n−2$

Where,

$∑(y−y^)2$ represents the sum of squares due to error

n represents the sample size.

Thus, the estimated regression equation is $y^=15.79−0.154x$

Use the estimated regression equation to find the predicted value of y for each value of x.

x	y	$y^$	$y−y^$	$(y−y^)2$
12	13	13.942	–0.942	0.88736
17	14	13.172	0.828	0.68558
3	16	15.328	0.672	0.45158
17	13	13.172	–0.172	0.02958
16	14	13.326	0.674	0.45428
11	14	14.096	–0.096	0.00922
14	13	13.634	–0.634	0.40196
9	14	14.404	–0.404	0.16322
Total				3.083

Substitute $∑(y−y^)2$ as 3.083 and n as 8.

$se=3.0838−2=3.0836=0.514=0.717$

Thus, the residual standard deviation $se$ is 0.717.

Answer 20

c.

To determine

Find the sum of squares for x.

c.

Expert Solution

Answer to Problem 16E

The sum of squares for x is 159.88.

Explanation of Solution

Calculation:

The table shows the calculation of sum of squares for x

x	$x−x¯$	$(x−x¯)2$
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
Total		159.88

Thus, the sum of squares for x is 159.88.

Answer 21

c.

To determine

Find the sum of squares for x.

Answer 22

c.

Expert Solution

Answer to Problem 16E

The sum of squares for x is 159.88.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

The table shows the calculation of sum of squares for x

x	$x−x¯$	$(x−x¯)2$
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
Total		159.88

Thus, the sum of squares for x is 159.88.

Answer 27

d.

To determine

Find the standard error of $b1,sb$ .

d.

Expert Solution

Answer to Problem 16E

The standard error of $b1$ , $sb$ is 0.057.

Explanation of Solution

Calculation:

The standard error of $b1$ is calculated by using the formula:

$sb=se∑(x−x¯)2$

Where,

$se$ represents the residual standard deviation.

$∑(x−x¯)2$ represents the sum of squares due to x.

Substitute $∑(x−x¯)2$ as 159.88 and $se$ as 0.717.

$sb=0.717159.88=0.71712.64=0.057$

Thus, the standard error of $b1$ , $sb$ is 0.057.

Answer 28

d.

To determine

Find the standard error of $b1,sb$ .

Answer 29

d.

Expert Solution

Answer to Problem 16E

The standard error of $b1$ , $sb$ is 0.057.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation:

The standard error of $b1$ is calculated by using the formula:

$sb=se∑(x−x¯)2$

Where,

$se$ represents the residual standard deviation.

$∑(x−x¯)2$ represents the sum of squares due to x.

Substitute $∑(x−x¯)2$ as 159.88 and $se$ as 0.717.

$sb=0.717159.88=0.71712.64=0.057$

Thus, the standard error of $b1$ , $sb$ is 0.057.

Answer 34

e.

To determine

Find the critical value for a 95% confidence interval of $β1$ .

e.

Expert Solution

Answer to Problem 16E

The critical value for a 95% confidence interval of $β1$ is 2.447.

Explanation of Solution

Calculation:

Critical value:

Software procedure:

Step-by-step procedure to find the critical value using MINITAB is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• In Degrees of freedom, enter 6.
• Click the Shaded Area tab.
• Choose Probability and Two tail for the region of the curve to shade.
• Enter the Probability value as 0.05.
• Click OK.

Output obtained from MINITAB is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 11.3, Problem 16E , additional homework tip 2

Thus, the critical value for a 95% confidence interval of $β1$ is 2.447.

Answer 35

e.

To determine

Find the critical value for a 95% confidence interval of $β1$ .

Answer 36

e.

Expert Solution

Answer to Problem 16E

The critical value for a 95% confidence interval of $β1$ is 2.447.

Answer 37

e.

Expert Solution

Answer 38

e.

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Calculation:

Critical value:

Software procedure:

Step-by-step procedure to find the critical value using MINITAB is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• In Degrees of freedom, enter 6.
• Click the Shaded Area tab.
• Choose Probability and Two tail for the region of the curve to shade.
• Enter the Probability value as 0.05.
• Click OK.

Output obtained from MINITAB is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 11.3, Problem 16E , additional homework tip 2

Thus, the critical value for a 95% confidence interval of $β1$ is 2.447.

Answer 41

f.

To determine

Find the margin of error for a 95% confidence interval of $β1$ .

f.

Expert Solution

Answer to Problem 16E

The margin of error of $b1$ is 0.139.

Explanation of Solution

Calculation:

The margin of error for a 95% confidence interval of $β1$ is calculated using the formula:

$Margin of error=tα2⋅sb1$

Where,

$tα2$ represents the two tailed critical value for a given level of significance.

$sb1$ represents the standard error of $b1$ .

Substitute $tα2$ as 2.447 and $sb1$ as 0.057.

$Margin of error=(2.447)(0.057)=0.139$

Thus, the margin of error of $b1$ is 0.139.

Answer 42

f.

To determine

Find the margin of error for a 95% confidence interval of $β1$ .

Answer 43

f.

Expert Solution

Answer to Problem 16E

The margin of error of $b1$ is 0.139.

Answer 44

f.

Expert Solution

Answer 45

f.

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

Calculation:

The margin of error for a 95% confidence interval of $β1$ is calculated using the formula:

$Margin of error=tα2⋅sb1$

Where,

$tα2$ represents the two tailed critical value for a given level of significance.

$sb1$ represents the standard error of $b1$ .

Substitute $tα2$ as 2.447 and $sb1$ as 0.057.

$Margin of error=(2.447)(0.057)=0.139$

Thus, the margin of error of $b1$ is 0.139.

Answer 48

g.

To determine

Construct the 95% confidence interval for $β1$ .

g.

Expert Solution

Answer to Problem 16E

The 95% confidence interval for $β1$ is $(−0.293,−0.015)$

Explanation of Solution

Calculation:

The confidence interval for $β1$ is calculated by using the formula:

$Confidence interval=b1±tα2⋅sb1$

Where,

$b1$ represents the estimated slope coefficient.

$tα2⋅sb1$ represents the margin of error.

$Confidence interval=−0.154±(2.447)⋅(0.057)=−0.154±0.139=−0.293,−0.015$

Thus, the 95% confidence interval for $β1$ is $(−0.293,−0.015)$

Answer 49

g.

To determine

Construct the 95% confidence interval for $β1$ .

Answer 50

g.

Expert Solution

Answer to Problem 16E

The 95% confidence interval for $β1$ is $(−0.293,−0.015)$

Answer 51

g.

Expert Solution

Answer 52

g.

Expert Solution

Answer 53

Expert Solution

Answer 54

Explanation of Solution

Calculation:

The confidence interval for $β1$ is calculated by using the formula:

$Confidence interval=b1±tα2⋅sb1$

Where,

$b1$ represents the estimated slope coefficient.

$tα2⋅sb1$ represents the margin of error.

$Confidence interval=−0.154±(2.447)⋅(0.057)=−0.154±0.139=−0.293,−0.015$

Thus, the 95% confidence interval for $β1$ is $(−0.293,−0.015)$

Answer 55

h.

To determine

Test the significance of $β1$ using 5% level of significance.

h.

Expert Solution

Answer to Problem 16E

There is a support of evidence to conclude that there is a linear relationship between x and y at 5% level of significance.

Explanation of Solution

Calculation:

The hypotheses used for testing the significance is given below:

Null hypothesis:

$H0:β1=0$

That is, there is no linear relationship between x and y.

Alternate hypothesis:

$H1:β1≠0$

That is, there is a linear relationship between x and y.

Test statistic:

$t=β^1−β1sb1$

Where,

$β^1$ represents the estimated slope coefficient.

$β1$ represents the hypothesized value of slope coefficient.

$sb1$ represents the standard error of slope coefficient.

Substitute $β^1$ as –0.154, $β1$ as 0 and $sb1$ as 0.057.

$t=−0.154−00.057=−0.1540.057=−2.70$

From part e, the critical value is identified as –2.447.

Decision Rule:

If the positive test statistic value is greater than or equal to the positive critical value or less than the negative critical value, then reject the null hypothesis. Otherwise do not reject the null hypothesis.

Conclusion:

The test statistic value is –2.70 and the critical value is –2.447.

The test statistic value is lesser than the critical value.

That is, $−2.70(=test statistic)<−2.447(=critical value)$

Thus, the null hypothesis is rejected.

Hence, there is a support of evidence to conclude that there is a linear relationship between x and y.

Answer 56

h.

To determine

Test the significance of $β1$ using 5% level of significance.

Answer 57

h.

Expert Solution

Answer to Problem 16E

There is a support of evidence to conclude that there is a linear relationship between x and y at 5% level of significance.

Answer 58

h.

Expert Solution

Answer 59

h.

Expert Solution

Answer 60

Expert Solution

Answer 61

Explanation of Solution

Calculation:

The hypotheses used for testing the significance is given below:

Null hypothesis:

$H0:β1=0$

That is, there is no linear relationship between x and y.

Alternate hypothesis:

$H1:β1≠0$

That is, there is a linear relationship between x and y.

Test statistic:

$t=β^1−β1sb1$

Where,

$β^1$ represents the estimated slope coefficient.

$β1$ represents the hypothesized value of slope coefficient.

$sb1$ represents the standard error of slope coefficient.

Substitute $β^1$ as –0.154, $β1$ as 0 and $sb1$ as 0.057.

$t=−0.154−00.057=−0.1540.057=−2.70$

From part e, the critical value is identified as –2.447.

Decision Rule:

If the positive test statistic value is greater than or equal to the positive critical value or less than the negative critical value, then reject the null hypothesis. Otherwise do not reject the null hypothesis.

Conclusion:

The test statistic value is –2.70 and the critical value is –2.447.

The test statistic value is lesser than the critical value.

That is, $−2.70(=test statistic)<−2.447(=critical value)$

Thus, the null hypothesis is rejected.

Hence, there is a support of evidence to conclude that there is a linear relationship between x and y.

Answer 62

Ch. 11.1 - Prob. 1CYU Ch. 11.1 - Prob. 2CYU Ch. 11.1 - Prob. 3CYU Ch. 11.1 - Prob. 4CYU Ch. 11.1 - Prob. 5CYU Ch. 11.1 - Prob. 6CYU Ch. 11.1 - Prob. 7CYU Ch. 11.1 - Prob. 8CYU Ch. 11.1 - Prob. 9E Ch. 11.1 - Prob. 10E

Find the value of b 1 .

Find the value of b1<m:math><m:mrow><m:msub><m:mi>b</m:mi><m:mn>1</m:mn></m:msub></m:mrow></m:math>.

Answer to Problem 16E

Explanation of Solution

Find the residual standard deviation se<m:math><m:mrow><m:msub><m:mi>s</m:mi><m:mi>e</m:mi></m:msub></m:mrow></m:math>.

Answer to Problem 16E

Explanation of Solution

Find the sum of squares for x.

Answer to Problem 16E

Explanation of Solution

Find the standard error of b1,sb<m:math><m:mrow><m:msub><m:mi>b</m:mi><m:mn>1</m:mn></m:msub><m:mo>,</m:mo><m:msub><m:mi>s</m:mi><m:mi>b</m:mi></m:msub></m:mrow></m:math>.

Answer to Problem 16E

Explanation of Solution

Find the critical value for a 95% confidence interval of β1<m:math><m:mrow><m:msub><m:mi>β</m:mi><m:mn>1</m:mn></m:msub></m:mrow></m:math>.

Answer to Problem 16E

Explanation of Solution

Find the margin of error for a 95% confidence interval of β1<m:math><m:mrow><m:msub><m:mi>β</m:mi><m:mn>1</m:mn></m:msub></m:mrow></m:math>.

Answer to Problem 16E

Explanation of Solution

Construct the 95% confidence interval for β1<m:math><m:mrow><m:msub><m:mi>β</m:mi><m:mn>1</m:mn></m:msub></m:mrow></m:math>.

Answer to Problem 16E

Explanation of Solution

Test the significance of β1<m:math><m:mrow><m:msub><m:mi>β</m:mi><m:mn>1</m:mn></m:msub></m:mrow></m:math> using 5% level of significance.

Answer to Problem 16E

Explanation of Solution

Want to see more full solutions like this?

Chapter 11 Solutions

Find the value of $b1$ .

Find the residual standard deviation $se$ .

Find the standard error of $b1,sb$ .

Find the critical value for a 95% confidence interval of $β1$ .

Find the margin of error for a 95% confidence interval of $β1$ .

Construct the 95% confidence interval for $β1$ .

Test the significance of $β1$ using 5% level of significance.