The error degrees of freedom for Model 1.

Question

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Answer 1

Question

Chapter 11.2, Problem 17E

(a)

Section 1:

To determine

To find: The error degrees of freedom for Model 1.

(a)

Section 1:

Expert Solution

Answer to Problem 17E

Solution: There are 200 error degrees of freedom in Model 1.

Explanation of Solution

Calculation: The two models have been provided in the problem. Model 1 discusses the effect of Gene expression (β1) on popularity of 202 adolescents (n). There is only one explanatory variable present in this case, thus, the statistical model can be written as

yi=β0+β1xi+εi

Here, the variable β0 is the intercept, the variable β1 is the slope coefficient (Gene expression), and the term εi represents the error term.

The degrees of freedom for the model are calculated as

degrees of freedom=k=1

The total degrees of freedom are calculated as:

degrees of freedom=n−1=202−1=201

The error degrees of freedom are calculated as

Error degrees of freedom=Total Degrees of freedom − Model degrees of freedom=201−1=200

Section 2:

To determine

To find: The error degrees of freedom for Model 2.

Section 2:

Expert Solution

Answer to Problem 17E

Solution: There are 199 error degrees of freedom in Model 2.

Explanation of Solution

Calculation: The two models have been provided in the problem. Model 2 discusses the effect of Gene expression (β1) and rule-breaking composite (β2) on popularity of 202 adolescents (n). There are two explanatory variables present in this case, thus, the statistical model can be written as

yi=β0+β1x1+β2x2+εi

Here, the variable β0 is the intercept, the variable β1 is the first slope coefficient (Gene expression), the variable β2 is the second slope coefficient (rule-breaking composite), and the term εi represents the error term.

The degrees of freedom for the model are calculated as

degrees of freedom=k=2

The total degrees of freedom are calculated as

degrees of freedom=n−1=202−1=201

The error degrees of freedom are calculated as

Error degrees of freedom=Total Degrees of freedom − Model degrees of freedom=201−2=199

(b)

To determine

To test: The null hypothesis that “the serotonin gene regression coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value in Model 1.

(b)

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the regression coefficient for gene expression (β1) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the variable gene expression (β1) is equal to zero is stated as

H0:β1=0

The alternative hypothesis that the variable for gene expression (β1) is not equal to zero is stated as

Ha:β1≠0

The level of significance is 0.05. It is provided that

b1=0.204SEb1=0.066

The test statistic (t) is calculated as

tcal=b1SEb1=0.2040.066=3.09

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|3.09|)=2×(0.00114)=0.00228

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

(c)

Section 1:

To determine

To test: The null hypothesis that “the serotonin gene regression coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value for Model 2.

(c)

Section 1:

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the regression coefficient for gene expression (β1) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the regression coefficient for gene expression (β1) is equal to zero is stated as

H0:β1=0

The alternative hypothesis that the regression coefficient for gene expression (β1) is not equal to zero is stated as

Ha:β1≠0

The level of significance is 0.05. It is provided that

b1=0.161SEb1=0.066

The test statistic (t) is calculated as

tcal=b1SEb1=0.1610.066=2.43

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|2.43|)=2×(0.007989)=0.015978

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

Section 2:

To determine

To test: The null hypothesis that “the rule-breaking composite coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value for Model 2.

Section 2:

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the coefficient for rule-breaking composite (β2) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the coefficient of rule-breaking composite (β2) is equal to zero is stated as

H0:β2=0

The alternative hypothesis that the explanatory variable rule-breaking composite (β2) is not equal to zero is stated as

Ha:β2≠0

The level of significance is 0.05. It is provided that

b2=0.100SEb2=0.030

The test statistic (t) is calculated as

tcal=b2SEb2=0.1000.030=3.33

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|3.33|)=2×(0.000517)=0.001033

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

(d)

Section 1

To determine

Whether there lies a positive relationship between “the serotonin gene receptor expression level” and “popularity” after adjusting for rule-breaking.

(d)

Section 1

Expert Solution

Answer to Problem 17E

Solution: The coefficients of the variables “gene expression” and “rule breaking” are positive. It highlights that there lies a positive relationship between the coefficients of the two variables in both the models.

Explanation of Solution

It is observed from the calculations in part (c) that there lies a relationship between the serotonin gene receptor expression and the popularity after adjusting for rule-breaking. Also, it is seen that the coefficients of gene expression and rule breaking for both the models are positive.

Section 2

To determine

To find: The comparison of upsurge in popularity when the gene expression increases by one unit and the rule breaking composite remains fixed.

Section 2

Expert Solution

Answer to Problem 17E

Solution: When the variable gene expression is increased by one unit, the popularity increases by 0.204 units in Model 1, and there is an increase in popularity by 0.161 units in Model 2.

Explanation of Solution

Calculation: The regression model for Model 1 can be written as

y^=β0+0.204x1

When the gene expression is increased by 1 unit, the increase in the popularity is calculated as follows:

y^=β0+0.204x1=0.204(1)=0.204

Also, the regression model for Model 2 can be written as

y^=β0+0.161x1+0.100x2

When the gene expression is increased by 1 unit, keeping all other variables constant, the increase in popularity is calculated as follows:

y^=β0+0.161x1+0.100x2=0.161(1)+0.100(0)=0.161

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Answer 2

Question

Chapter 11.2, Problem 17E

(a)

Section 1:

To determine

To find: The error degrees of freedom for Model 1.

(a)

Section 1:

Expert Solution

Answer to Problem 17E

Solution: There are 200 error degrees of freedom in Model 1.

Explanation of Solution

Calculation: The two models have been provided in the problem. Model 1 discusses the effect of Gene expression (β1) on popularity of 202 adolescents (n). There is only one explanatory variable present in this case, thus, the statistical model can be written as

yi=β0+β1xi+εi

Here, the variable β0 is the intercept, the variable β1 is the slope coefficient (Gene expression), and the term εi represents the error term.

The degrees of freedom for the model are calculated as

degrees of freedom=k=1

The total degrees of freedom are calculated as:

degrees of freedom=n−1=202−1=201

The error degrees of freedom are calculated as

Error degrees of freedom=Total Degrees of freedom − Model degrees of freedom=201−1=200

Section 2:

To determine

To find: The error degrees of freedom for Model 2.

Section 2:

Expert Solution

Answer to Problem 17E

Solution: There are 199 error degrees of freedom in Model 2.

Explanation of Solution

Calculation: The two models have been provided in the problem. Model 2 discusses the effect of Gene expression (β1) and rule-breaking composite (β2) on popularity of 202 adolescents (n). There are two explanatory variables present in this case, thus, the statistical model can be written as

yi=β0+β1x1+β2x2+εi

Here, the variable β0 is the intercept, the variable β1 is the first slope coefficient (Gene expression), the variable β2 is the second slope coefficient (rule-breaking composite), and the term εi represents the error term.

The degrees of freedom for the model are calculated as

degrees of freedom=k=2

The total degrees of freedom are calculated as

degrees of freedom=n−1=202−1=201

The error degrees of freedom are calculated as

Error degrees of freedom=Total Degrees of freedom − Model degrees of freedom=201−2=199

(b)

To determine

To test: The null hypothesis that “the serotonin gene regression coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value in Model 1.

(b)

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the regression coefficient for gene expression (β1) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the variable gene expression (β1) is equal to zero is stated as

H0:β1=0

The alternative hypothesis that the variable for gene expression (β1) is not equal to zero is stated as

Ha:β1≠0

The level of significance is 0.05. It is provided that

b1=0.204SEb1=0.066

The test statistic (t) is calculated as

tcal=b1SEb1=0.2040.066=3.09

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|3.09|)=2×(0.00114)=0.00228

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

(c)

Section 1:

To determine

To test: The null hypothesis that “the serotonin gene regression coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value for Model 2.

(c)

Section 1:

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the regression coefficient for gene expression (β1) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the regression coefficient for gene expression (β1) is equal to zero is stated as

H0:β1=0

The alternative hypothesis that the regression coefficient for gene expression (β1) is not equal to zero is stated as

Ha:β1≠0

The level of significance is 0.05. It is provided that

b1=0.161SEb1=0.066

The test statistic (t) is calculated as

tcal=b1SEb1=0.1610.066=2.43

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|2.43|)=2×(0.007989)=0.015978

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

Section 2:

To determine

To test: The null hypothesis that “the rule-breaking composite coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value for Model 2.

Section 2:

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the coefficient for rule-breaking composite (β2) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the coefficient of rule-breaking composite (β2) is equal to zero is stated as

H0:β2=0

The alternative hypothesis that the explanatory variable rule-breaking composite (β2) is not equal to zero is stated as

Ha:β2≠0

The level of significance is 0.05. It is provided that

b2=0.100SEb2=0.030

The test statistic (t) is calculated as

tcal=b2SEb2=0.1000.030=3.33

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|3.33|)=2×(0.000517)=0.001033

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

(d)

Section 1

To determine

Whether there lies a positive relationship between “the serotonin gene receptor expression level” and “popularity” after adjusting for rule-breaking.

(d)

Section 1

Expert Solution

Answer to Problem 17E

Solution: The coefficients of the variables “gene expression” and “rule breaking” are positive. It highlights that there lies a positive relationship between the coefficients of the two variables in both the models.

Explanation of Solution

It is observed from the calculations in part (c) that there lies a relationship between the serotonin gene receptor expression and the popularity after adjusting for rule-breaking. Also, it is seen that the coefficients of gene expression and rule breaking for both the models are positive.

Section 2

To determine

To find: The comparison of upsurge in popularity when the gene expression increases by one unit and the rule breaking composite remains fixed.

Section 2

Expert Solution

Answer to Problem 17E

Solution: When the variable gene expression is increased by one unit, the popularity increases by 0.204 units in Model 1, and there is an increase in popularity by 0.161 units in Model 2.

Explanation of Solution

Calculation: The regression model for Model 1 can be written as

y^=β0+0.204x1

When the gene expression is increased by 1 unit, the increase in the popularity is calculated as follows:

y^=β0+0.204x1=0.204(1)=0.204

Also, the regression model for Model 2 can be written as

y^=β0+0.161x1+0.100x2

When the gene expression is increased by 1 unit, keeping all other variables constant, the increase in popularity is calculated as follows:

y^=β0+0.161x1+0.100x2=0.161(1)+0.100(0)=0.161

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Answer 3

Question

Chapter 11.2, Problem 17E

(a)

Section 1:

To determine

To find: The error degrees of freedom for Model 1.

(a)

Section 1:

Expert Solution

Answer to Problem 17E

Solution: There are 200 error degrees of freedom in Model 1.

Explanation of Solution

Calculation: The two models have been provided in the problem. Model 1 discusses the effect of Gene expression (β1) on popularity of 202 adolescents (n). There is only one explanatory variable present in this case, thus, the statistical model can be written as

yi=β0+β1xi+εi

Here, the variable β0 is the intercept, the variable β1 is the slope coefficient (Gene expression), and the term εi represents the error term.

The degrees of freedom for the model are calculated as

degrees of freedom=k=1

The total degrees of freedom are calculated as:

degrees of freedom=n−1=202−1=201

The error degrees of freedom are calculated as

Error degrees of freedom=Total Degrees of freedom − Model degrees of freedom=201−1=200

Section 2:

To determine

To find: The error degrees of freedom for Model 2.

Section 2:

Expert Solution

Answer to Problem 17E

Solution: There are 199 error degrees of freedom in Model 2.

Explanation of Solution

Calculation: The two models have been provided in the problem. Model 2 discusses the effect of Gene expression (β1) and rule-breaking composite (β2) on popularity of 202 adolescents (n). There are two explanatory variables present in this case, thus, the statistical model can be written as

yi=β0+β1x1+β2x2+εi

Here, the variable β0 is the intercept, the variable β1 is the first slope coefficient (Gene expression), the variable β2 is the second slope coefficient (rule-breaking composite), and the term εi represents the error term.

The degrees of freedom for the model are calculated as

degrees of freedom=k=2

The total degrees of freedom are calculated as

degrees of freedom=n−1=202−1=201

The error degrees of freedom are calculated as

Error degrees of freedom=Total Degrees of freedom − Model degrees of freedom=201−2=199

(b)

To determine

To test: The null hypothesis that “the serotonin gene regression coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value in Model 1.

(b)

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the regression coefficient for gene expression (β1) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the variable gene expression (β1) is equal to zero is stated as

H0:β1=0

The alternative hypothesis that the variable for gene expression (β1) is not equal to zero is stated as

Ha:β1≠0

The level of significance is 0.05. It is provided that

b1=0.204SEb1=0.066

The test statistic (t) is calculated as

tcal=b1SEb1=0.2040.066=3.09

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|3.09|)=2×(0.00114)=0.00228

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

(c)

Section 1:

To determine

To test: The null hypothesis that “the serotonin gene regression coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value for Model 2.

(c)

Section 1:

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the regression coefficient for gene expression (β1) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the regression coefficient for gene expression (β1) is equal to zero is stated as

H0:β1=0

The alternative hypothesis that the regression coefficient for gene expression (β1) is not equal to zero is stated as

Ha:β1≠0

The level of significance is 0.05. It is provided that

b1=0.161SEb1=0.066

The test statistic (t) is calculated as

tcal=b1SEb1=0.1610.066=2.43

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|2.43|)=2×(0.007989)=0.015978

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

Section 2:

To determine

To test: The null hypothesis that “the rule-breaking composite coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value for Model 2.

Section 2:

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the coefficient for rule-breaking composite (β2) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the coefficient of rule-breaking composite (β2) is equal to zero is stated as

H0:β2=0

The alternative hypothesis that the explanatory variable rule-breaking composite (β2) is not equal to zero is stated as

Ha:β2≠0

The level of significance is 0.05. It is provided that

b2=0.100SEb2=0.030

The test statistic (t) is calculated as

tcal=b2SEb2=0.1000.030=3.33

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|3.33|)=2×(0.000517)=0.001033

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

(d)

Section 1

To determine

Whether there lies a positive relationship between “the serotonin gene receptor expression level” and “popularity” after adjusting for rule-breaking.

(d)

Section 1

Expert Solution

Answer to Problem 17E

Solution: The coefficients of the variables “gene expression” and “rule breaking” are positive. It highlights that there lies a positive relationship between the coefficients of the two variables in both the models.

Explanation of Solution

It is observed from the calculations in part (c) that there lies a relationship between the serotonin gene receptor expression and the popularity after adjusting for rule-breaking. Also, it is seen that the coefficients of gene expression and rule breaking for both the models are positive.

Section 2

To determine

To find: The comparison of upsurge in popularity when the gene expression increases by one unit and the rule breaking composite remains fixed.

Section 2

Expert Solution

Answer to Problem 17E

Solution: When the variable gene expression is increased by one unit, the popularity increases by 0.204 units in Model 1, and there is an increase in popularity by 0.161 units in Model 2.

Explanation of Solution

Calculation: The regression model for Model 1 can be written as

y^=β0+0.204x1

When the gene expression is increased by 1 unit, the increase in the popularity is calculated as follows:

y^=β0+0.204x1=0.204(1)=0.204

Also, the regression model for Model 2 can be written as

y^=β0+0.161x1+0.100x2

When the gene expression is increased by 1 unit, keeping all other variables constant, the increase in popularity is calculated as follows:

y^=β0+0.161x1+0.100x2=0.161(1)+0.100(0)=0.161

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Answer 4

Question

Chapter 11.2, Problem 17E

(a)

Section 1:

To determine

To find: The error degrees of freedom for Model 1.

(a)

Section 1:

Expert Solution

Answer to Problem 17E

Solution: There are 200 error degrees of freedom in Model 1.

Explanation of Solution

Calculation: The two models have been provided in the problem. Model 1 discusses the effect of Gene expression (β1) on popularity of 202 adolescents (n). There is only one explanatory variable present in this case, thus, the statistical model can be written as

yi=β0+β1xi+εi

Here, the variable β0 is the intercept, the variable β1 is the slope coefficient (Gene expression), and the term εi represents the error term.

The degrees of freedom for the model are calculated as

degrees of freedom=k=1

The total degrees of freedom are calculated as:

degrees of freedom=n−1=202−1=201

The error degrees of freedom are calculated as

Error degrees of freedom=Total Degrees of freedom − Model degrees of freedom=201−1=200

Section 2:

To determine

To find: The error degrees of freedom for Model 2.

Section 2:

Expert Solution

Answer to Problem 17E

Solution: There are 199 error degrees of freedom in Model 2.

Explanation of Solution

Calculation: The two models have been provided in the problem. Model 2 discusses the effect of Gene expression (β1) and rule-breaking composite (β2) on popularity of 202 adolescents (n). There are two explanatory variables present in this case, thus, the statistical model can be written as

yi=β0+β1x1+β2x2+εi

Here, the variable β0 is the intercept, the variable β1 is the first slope coefficient (Gene expression), the variable β2 is the second slope coefficient (rule-breaking composite), and the term εi represents the error term.

The degrees of freedom for the model are calculated as

degrees of freedom=k=2

The total degrees of freedom are calculated as

degrees of freedom=n−1=202−1=201

The error degrees of freedom are calculated as

Error degrees of freedom=Total Degrees of freedom − Model degrees of freedom=201−2=199

(b)

To determine

To test: The null hypothesis that “the serotonin gene regression coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value in Model 1.

(b)

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the regression coefficient for gene expression (β1) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the variable gene expression (β1) is equal to zero is stated as

H0:β1=0

The alternative hypothesis that the variable for gene expression (β1) is not equal to zero is stated as

Ha:β1≠0

The level of significance is 0.05. It is provided that

b1=0.204SEb1=0.066

The test statistic (t) is calculated as

tcal=b1SEb1=0.2040.066=3.09

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|3.09|)=2×(0.00114)=0.00228

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

(c)

Section 1:

To determine

To test: The null hypothesis that “the serotonin gene regression coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value for Model 2.

(c)

Section 1:

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the regression coefficient for gene expression (β1) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the regression coefficient for gene expression (β1) is equal to zero is stated as

H0:β1=0

The alternative hypothesis that the regression coefficient for gene expression (β1) is not equal to zero is stated as

Ha:β1≠0

The level of significance is 0.05. It is provided that

b1=0.161SEb1=0.066

The test statistic (t) is calculated as

tcal=b1SEb1=0.1610.066=2.43

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|2.43|)=2×(0.007989)=0.015978

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

Section 2:

To determine

To test: The null hypothesis that “the rule-breaking composite coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value for Model 2.

Section 2:

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the coefficient for rule-breaking composite (β2) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the coefficient of rule-breaking composite (β2) is equal to zero is stated as

H0:β2=0

The alternative hypothesis that the explanatory variable rule-breaking composite (β2) is not equal to zero is stated as

Ha:β2≠0

The level of significance is 0.05. It is provided that

b2=0.100SEb2=0.030

The test statistic (t) is calculated as

tcal=b2SEb2=0.1000.030=3.33

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|3.33|)=2×(0.000517)=0.001033

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

(d)

Section 1

To determine

Whether there lies a positive relationship between “the serotonin gene receptor expression level” and “popularity” after adjusting for rule-breaking.

(d)

Section 1

Expert Solution

Answer to Problem 17E

Solution: The coefficients of the variables “gene expression” and “rule breaking” are positive. It highlights that there lies a positive relationship between the coefficients of the two variables in both the models.

Explanation of Solution

It is observed from the calculations in part (c) that there lies a relationship between the serotonin gene receptor expression and the popularity after adjusting for rule-breaking. Also, it is seen that the coefficients of gene expression and rule breaking for both the models are positive.

Section 2

To determine

To find: The comparison of upsurge in popularity when the gene expression increases by one unit and the rule breaking composite remains fixed.

Section 2

Expert Solution

Answer to Problem 17E

Solution: When the variable gene expression is increased by one unit, the popularity increases by 0.204 units in Model 1, and there is an increase in popularity by 0.161 units in Model 2.

Explanation of Solution

Calculation: The regression model for Model 1 can be written as

y^=β0+0.204x1

When the gene expression is increased by 1 unit, the increase in the popularity is calculated as follows:

y^=β0+0.204x1=0.204(1)=0.204

Also, the regression model for Model 2 can be written as

y^=β0+0.161x1+0.100x2

When the gene expression is increased by 1 unit, keeping all other variables constant, the increase in popularity is calculated as follows:

y^=β0+0.161x1+0.100x2=0.161(1)+0.100(0)=0.161

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Answer 5

Chapter 11.2, Problem 17E

(a)

Section 1:

To determine

To find: The error degrees of freedom for Model 1.

(a)

Section 1:

Expert Solution

Answer to Problem 17E

Solution: There are 200 error degrees of freedom in Model 1.

Explanation of Solution

Calculation: The two models have been provided in the problem. Model 1 discusses the effect of Gene expression (β1) on popularity of 202 adolescents (n). There is only one explanatory variable present in this case, thus, the statistical model can be written as

yi=β0+β1xi+εi

Here, the variable β0 is the intercept, the variable β1 is the slope coefficient (Gene expression), and the term εi represents the error term.

The degrees of freedom for the model are calculated as

degrees of freedom=k=1

The total degrees of freedom are calculated as:

degrees of freedom=n−1=202−1=201

The error degrees of freedom are calculated as

Error degrees of freedom=Total Degrees of freedom − Model degrees of freedom=201−1=200

Section 2:

To determine

To find: The error degrees of freedom for Model 2.

Section 2:

Expert Solution

Answer to Problem 17E

Solution: There are 199 error degrees of freedom in Model 2.

Explanation of Solution

Calculation: The two models have been provided in the problem. Model 2 discusses the effect of Gene expression (β1) and rule-breaking composite (β2) on popularity of 202 adolescents (n). There are two explanatory variables present in this case, thus, the statistical model can be written as

yi=β0+β1x1+β2x2+εi

Here, the variable β0 is the intercept, the variable β1 is the first slope coefficient (Gene expression), the variable β2 is the second slope coefficient (rule-breaking composite), and the term εi represents the error term.

The degrees of freedom for the model are calculated as

degrees of freedom=k=2

The total degrees of freedom are calculated as

degrees of freedom=n−1=202−1=201

The error degrees of freedom are calculated as

Error degrees of freedom=Total Degrees of freedom − Model degrees of freedom=201−2=199

(b)

To determine

To test: The null hypothesis that “the serotonin gene regression coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value in Model 1.

(b)

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the regression coefficient for gene expression (β1) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the variable gene expression (β1) is equal to zero is stated as

H0:β1=0

The alternative hypothesis that the variable for gene expression (β1) is not equal to zero is stated as

Ha:β1≠0

The level of significance is 0.05. It is provided that

b1=0.204SEb1=0.066

The test statistic (t) is calculated as

tcal=b1SEb1=0.2040.066=3.09

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|3.09|)=2×(0.00114)=0.00228

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

(c)

Section 1:

To determine

To test: The null hypothesis that “the serotonin gene regression coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value for Model 2.

(c)

Section 1:

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the regression coefficient for gene expression (β1) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the regression coefficient for gene expression (β1) is equal to zero is stated as

H0:β1=0

The alternative hypothesis that the regression coefficient for gene expression (β1) is not equal to zero is stated as

Ha:β1≠0

The level of significance is 0.05. It is provided that

b1=0.161SEb1=0.066

The test statistic (t) is calculated as

tcal=b1SEb1=0.1610.066=2.43

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|2.43|)=2×(0.007989)=0.015978

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

Section 2:

To determine

To test: The null hypothesis that “the rule-breaking composite coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value for Model 2.

Section 2:

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the coefficient for rule-breaking composite (β2) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the coefficient of rule-breaking composite (β2) is equal to zero is stated as

H0:β2=0

The alternative hypothesis that the explanatory variable rule-breaking composite (β2) is not equal to zero is stated as

Ha:β2≠0

The level of significance is 0.05. It is provided that

b2=0.100SEb2=0.030

The test statistic (t) is calculated as

tcal=b2SEb2=0.1000.030=3.33

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|3.33|)=2×(0.000517)=0.001033

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

(d)

Section 1

To determine

Whether there lies a positive relationship between “the serotonin gene receptor expression level” and “popularity” after adjusting for rule-breaking.

(d)

Section 1

Expert Solution

Answer to Problem 17E

Solution: The coefficients of the variables “gene expression” and “rule breaking” are positive. It highlights that there lies a positive relationship between the coefficients of the two variables in both the models.

Explanation of Solution

It is observed from the calculations in part (c) that there lies a relationship between the serotonin gene receptor expression and the popularity after adjusting for rule-breaking. Also, it is seen that the coefficients of gene expression and rule breaking for both the models are positive.

Section 2

To determine

To find: The comparison of upsurge in popularity when the gene expression increases by one unit and the rule breaking composite remains fixed.

Section 2

Expert Solution

Answer to Problem 17E

Solution: When the variable gene expression is increased by one unit, the popularity increases by 0.204 units in Model 1, and there is an increase in popularity by 0.161 units in Model 2.

Explanation of Solution

Calculation: The regression model for Model 1 can be written as

y^=β0+0.204x1

When the gene expression is increased by 1 unit, the increase in the popularity is calculated as follows:

y^=β0+0.204x1=0.204(1)=0.204

Also, the regression model for Model 2 can be written as

y^=β0+0.161x1+0.100x2

When the gene expression is increased by 1 unit, keeping all other variables constant, the increase in popularity is calculated as follows:

y^=β0+0.161x1+0.100x2=0.161(1)+0.100(0)=0.161

Answer 6

Chapter 11.2, Problem 17E

(a)

Section 1:

To determine

To find: The error degrees of freedom for Model 1.

(a)

Section 1:

Expert Solution

Answer to Problem 17E

Solution: There are 200 error degrees of freedom in Model 1.

Explanation of Solution

Calculation: The two models have been provided in the problem. Model 1 discusses the effect of Gene expression (β1) on popularity of 202 adolescents (n). There is only one explanatory variable present in this case, thus, the statistical model can be written as

yi=β0+β1xi+εi

Here, the variable β0 is the intercept, the variable β1 is the slope coefficient (Gene expression), and the term εi represents the error term.

The degrees of freedom for the model are calculated as

degrees of freedom=k=1

The total degrees of freedom are calculated as:

degrees of freedom=n−1=202−1=201

The error degrees of freedom are calculated as

Error degrees of freedom=Total Degrees of freedom − Model degrees of freedom=201−1=200

Answer 7

Chapter 11.2, Problem 17E

(a)

Section 1:

To determine

To find: The error degrees of freedom for Model 1.

Answer 8

(a)

Section 1:

Expert Solution

Answer to Problem 17E

Solution: There are 200 error degrees of freedom in Model 1.

Answer 9

(a)

Section 1:

Expert Solution

Answer 10

(a)

Section 1:

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation: The two models have been provided in the problem. Model 1 discusses the effect of Gene expression (β1) on popularity of 202 adolescents (n). There is only one explanatory variable present in this case, thus, the statistical model can be written as

yi=β0+β1xi+εi

Here, the variable β0 is the intercept, the variable β1 is the slope coefficient (Gene expression), and the term εi represents the error term.

The degrees of freedom for the model are calculated as

degrees of freedom=k=1

The total degrees of freedom are calculated as:

degrees of freedom=n−1=202−1=201

The error degrees of freedom are calculated as

Error degrees of freedom=Total Degrees of freedom − Model degrees of freedom=201−1=200

Answer 13

Section 2:

To determine

To find: The error degrees of freedom for Model 2.

Section 2:

Expert Solution

Answer to Problem 17E

Solution: There are 199 error degrees of freedom in Model 2.

Explanation of Solution

Calculation: The two models have been provided in the problem. Model 2 discusses the effect of Gene expression (β1) and rule-breaking composite (β2) on popularity of 202 adolescents (n). There are two explanatory variables present in this case, thus, the statistical model can be written as

yi=β0+β1x1+β2x2+εi

Here, the variable β0 is the intercept, the variable β1 is the first slope coefficient (Gene expression), the variable β2 is the second slope coefficient (rule-breaking composite), and the term εi represents the error term.

The degrees of freedom for the model are calculated as

degrees of freedom=k=2

The total degrees of freedom are calculated as

degrees of freedom=n−1=202−1=201

The error degrees of freedom are calculated as

Error degrees of freedom=Total Degrees of freedom − Model degrees of freedom=201−2=199

Answer 14

Section 2:

To determine

To find: The error degrees of freedom for Model 2.

Answer 15

Section 2:

Expert Solution

Answer to Problem 17E

Solution: There are 199 error degrees of freedom in Model 2.

Answer 16

Section 2:

Expert Solution

Answer 17

Section 2:

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation: The two models have been provided in the problem. Model 2 discusses the effect of Gene expression (β1) and rule-breaking composite (β2) on popularity of 202 adolescents (n). There are two explanatory variables present in this case, thus, the statistical model can be written as

yi=β0+β1x1+β2x2+εi

Here, the variable β0 is the intercept, the variable β1 is the first slope coefficient (Gene expression), the variable β2 is the second slope coefficient (rule-breaking composite), and the term εi represents the error term.

The degrees of freedom for the model are calculated as

degrees of freedom=k=2

The total degrees of freedom are calculated as

degrees of freedom=n−1=202−1=201

The error degrees of freedom are calculated as

Error degrees of freedom=Total Degrees of freedom − Model degrees of freedom=201−2=199

Answer 20

(b)

To determine

To test: The null hypothesis that “the serotonin gene regression coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value in Model 1.

(b)

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the regression coefficient for gene expression (β1) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the variable gene expression (β1) is equal to zero is stated as

H0:β1=0

The alternative hypothesis that the variable for gene expression (β1) is not equal to zero is stated as

Ha:β1≠0

The level of significance is 0.05. It is provided that

b1=0.204SEb1=0.066

The test statistic (t) is calculated as

tcal=b1SEb1=0.2040.066=3.09

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|3.09|)=2×(0.00114)=0.00228

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

Answer 21

(b)

To determine

To test: The null hypothesis that “the serotonin gene regression coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value in Model 1.

Answer 22

(b)

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the regression coefficient for gene expression (β1) is not equal to zero.

Answer 23

(b)

Expert Solution

Answer 24

(b)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation: The null hypothesis to test whether the variable gene expression (β1) is equal to zero is stated as

H0:β1=0

The alternative hypothesis that the variable for gene expression (β1) is not equal to zero is stated as

Ha:β1≠0

The level of significance is 0.05. It is provided that

b1=0.204SEb1=0.066

The test statistic (t) is calculated as

tcal=b1SEb1=0.2040.066=3.09

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|3.09|)=2×(0.00114)=0.00228

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

Answer 27

(c)

Section 1:

To determine

To test: The null hypothesis that “the serotonin gene regression coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value for Model 2.

(c)

Section 1:

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the regression coefficient for gene expression (β1) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the regression coefficient for gene expression (β1) is equal to zero is stated as

H0:β1=0

The alternative hypothesis that the regression coefficient for gene expression (β1) is not equal to zero is stated as

Ha:β1≠0

The level of significance is 0.05. It is provided that

b1=0.161SEb1=0.066

The test statistic (t) is calculated as

tcal=b1SEb1=0.1610.066=2.43

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|2.43|)=2×(0.007989)=0.015978

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

Answer 28

(c)

Section 1:

To determine

To test: The null hypothesis that “the serotonin gene regression coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value for Model 2.

Answer 29

(c)

Section 1:

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the regression coefficient for gene expression (β1) is not equal to zero.

Answer 30

(c)

Section 1:

Expert Solution

Answer 31

(c)

Section 1:

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation: The null hypothesis to test whether the regression coefficient for gene expression (β1) is equal to zero is stated as

H0:β1=0

The alternative hypothesis that the regression coefficient for gene expression (β1) is not equal to zero is stated as

Ha:β1≠0

The level of significance is 0.05. It is provided that

b1=0.161SEb1=0.066

The test statistic (t) is calculated as

tcal=b1SEb1=0.1610.066=2.43

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|2.43|)=2×(0.007989)=0.015978

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

Answer 34

Section 2:

To determine

To test: The null hypothesis that “the rule-breaking composite coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value for Model 2.

Section 2:

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the coefficient for rule-breaking composite (β2) is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the coefficient of rule-breaking composite (β2) is equal to zero is stated as

H0:β2=0

The alternative hypothesis that the explanatory variable rule-breaking composite (β2) is not equal to zero is stated as

Ha:β2≠0

The level of significance is 0.05. It is provided that

b2=0.100SEb2=0.030

The test statistic (t) is calculated as

tcal=b2SEb2=0.1000.030=3.33

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|3.33|)=2×(0.000517)=0.001033

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

Answer 35

Section 2:

To determine

To test: The null hypothesis that “the rule-breaking composite coefficient is equal to zero” against “it is not equal to zero,” and the corresponding test statistic and its P-value for Model 2.

Answer 36

Section 2:

Expert Solution

Answer to Problem 17E

Solution: The result is significant, that is, the coefficient for rule-breaking composite (β2) is not equal to zero.

Answer 37

Section 2:

Expert Solution

Answer 38

Section 2:

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Calculation: The null hypothesis to test whether the coefficient of rule-breaking composite (β2) is equal to zero is stated as

H0:β2=0

The alternative hypothesis that the explanatory variable rule-breaking composite (β2) is not equal to zero is stated as

Ha:β2≠0

The level of significance is 0.05. It is provided that

b2=0.100SEb2=0.030

The test statistic (t) is calculated as

tcal=b2SEb2=0.1000.030=3.33

The critical value of test statistic (tcrit), at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

p=2p(t>|tcrit|)=2p(t>|3.33|)=2×(0.000517)=0.001033

Conclusion: The P-value is less than the level of significance. Also, the test statistic results are

tcal>tcrit

Hence, the null hypothesis gets rejected, that is, the result is significant.

Answer 41

(d)

Section 1

To determine

Whether there lies a positive relationship between “the serotonin gene receptor expression level” and “popularity” after adjusting for rule-breaking.

(d)

Section 1

Expert Solution

Answer to Problem 17E

Solution: The coefficients of the variables “gene expression” and “rule breaking” are positive. It highlights that there lies a positive relationship between the coefficients of the two variables in both the models.

Explanation of Solution

It is observed from the calculations in part (c) that there lies a relationship between the serotonin gene receptor expression and the popularity after adjusting for rule-breaking. Also, it is seen that the coefficients of gene expression and rule breaking for both the models are positive.

Answer 42

(d)

Section 1

To determine

Whether there lies a positive relationship between “the serotonin gene receptor expression level” and “popularity” after adjusting for rule-breaking.

Answer 43

(d)

Section 1

Expert Solution

Answer to Problem 17E

Solution: The coefficients of the variables “gene expression” and “rule breaking” are positive. It highlights that there lies a positive relationship between the coefficients of the two variables in both the models.

Answer 44

(d)

Section 1

Expert Solution

Answer 45

(d)

Section 1

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

It is observed from the calculations in part (c) that there lies a relationship between the serotonin gene receptor expression and the popularity after adjusting for rule-breaking. Also, it is seen that the coefficients of gene expression and rule breaking for both the models are positive.

Answer 48

Section 2

To determine

To find: The comparison of upsurge in popularity when the gene expression increases by one unit and the rule breaking composite remains fixed.

Section 2

Expert Solution

Answer to Problem 17E

Solution: When the variable gene expression is increased by one unit, the popularity increases by 0.204 units in Model 1, and there is an increase in popularity by 0.161 units in Model 2.

Explanation of Solution

Calculation: The regression model for Model 1 can be written as

y^=β0+0.204x1

When the gene expression is increased by 1 unit, the increase in the popularity is calculated as follows:

y^=β0+0.204x1=0.204(1)=0.204

Also, the regression model for Model 2 can be written as

y^=β0+0.161x1+0.100x2

When the gene expression is increased by 1 unit, keeping all other variables constant, the increase in popularity is calculated as follows:

y^=β0+0.161x1+0.100x2=0.161(1)+0.100(0)=0.161

Answer 49

Section 2

To determine

To find: The comparison of upsurge in popularity when the gene expression increases by one unit and the rule breaking composite remains fixed.

Answer 50

Section 2

Expert Solution

Answer to Problem 17E

Solution: When the variable gene expression is increased by one unit, the popularity increases by 0.204 units in Model 1, and there is an increase in popularity by 0.161 units in Model 2.