Chapter 11, Problem 19E

(a)

Section 1:

To determine

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

Answer to Problem 19E

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 $\left({\beta }_1\right)$ on popularity of 202 adolescents $\left(n\right)$. There is only one explanatory variable present in this case, thus, the statistical model can be written as

$y_i={\beta }_0+{\beta }_1{x}_i+{\epsilon }_i$

Here, the variable ${\beta }_0$ is the intercept, the variable ${\beta }_1$ is the slope coefficient (Gene expression), and the term ${\epsilon }_i$ represents the error term.

The degrees of freedom for the model are calculated as

$\begin{array}{c}\text{degrees of freedom}=k\\ =1\end{array}$

The total degrees of freedom are calculated as:

$\begin{array}{c}\text{degrees of freedom}=n-1\\ =202-1\\ =201\end{array}$

The error degrees of freedom are calculated as

$\begin{array}{c}\text{Error degrees of freedom}=\text{Total Degrees of freedom }-\text{ Model degrees of freedom}\\ =201-1\\ =200\end{array}$

Section 2:

To determine

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

Answer to Problem 19E

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 $\left({\beta }_1\right)$ and rule-breaking composite $\left({\beta }_2\right)$ on popularity of 202 adolescents $\left(n\right)$. There are two explanatory variables present in this case, thus, the statistical model can be written as

$y_i={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+{\epsilon }_i$

Here, the variable ${\beta }_0$ is the intercept, the variable ${\beta }_1$ is the first slope coefficient (Gene expression), the variable ${\beta }_2$ is the second slope coefficient (rule-breaking composite), and the term ${\epsilon }_i$ represents the error term.

The degrees of freedom for the model are calculated as

$\begin{array}{c}\text{degrees of freedom}=k\\ =2\end{array}$

The total degrees of freedom are calculated as

$\begin{array}{c}\text{degrees of freedom}=n-1\\ =202-1\\ =201\end{array}$

The error degrees of freedom are calculated as

$\begin{array}{c}\text{Error degrees of freedom}=\text{Total Degrees of freedom }-\text{ Model degrees of freedom}\\ =201-2\\ =199\end{array}$

(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 to Problem 19E

Solution: The result is significant, that is, the regression coefficient for gene expression $\left({\beta }_1\right)$ is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the variable gene expression $\left({\beta }_1\right)$ is equal to zero is stated as

$H_0:{\beta }_1=0$

The alternative hypothesis that the variable for gene expression $\left({\beta }_1\right)$ is not equal to zero is stated as

$H_a:{\beta }_1\ne 0$

The level of significance is 0.05. It is provided that

$\begin{array}{c}{b}_1=0.204\\ S{E}_{b_1}=0.066\end{array}$

The test statistic $\left(t\right)$ is calculated as

$\begin{array}{c}{t}_{cal}=\frac{b_1}{S{E}_{b_1}}\\ =\frac{0.204}{0.066}\\ =3.09\end{array}$

The critical value of test statistic $\left(t_{crit}\right)$, at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

$\begin{array}{c}p=2p\left(t>\left|t_{crit}\right|\right)\\ =2p\left(t>\left|3.09\right|\right)\\ =2\times \left(0.00114\right)\\ =0.00228\end{array}$

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

$t_{cal}>t_{crit}$

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.

Answer to Problem 19E

Solution: The result is significant, that is, the regression coefficient for gene expression $\left({\beta }_1\right)$ is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the regression coefficient for gene expression $\left({\beta }_1\right)$ is equal to zero is stated as

$H_0:{\beta }_1=0$

The alternative hypothesis that the regression coefficient for gene expression $\left({\beta }_1\right)$ is not equal to zero is stated as

$H_a:{\beta }_1\ne 0$

The level of significance is 0.05. It is provided that

$\begin{array}{c}{b}_1=0.161\\ S{E}_{b_1}=0.066\end{array}$

The test statistic $\left(t\right)$ is calculated as

$\begin{array}{c}{t}_{cal}=\frac{b_1}{S{E}_{b_1}}\\ =\frac{0.161}{0.066}\\ =2.43\end{array}$

The critical value of test statistic $\left(t_{crit}\right)$, at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

$\begin{array}{c}p=2p\left(t>\left|t_{crit}\right|\right)\\ =2p\left(t>\left|2.43\right|\right)\\ =2\times \left(0.007989\right)\\ =0.015978\end{array}$

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

$t_{cal}>t_{crit}$

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.

Answer to Problem 19E

Solution: The result is significant, that is, the coefficient for rule-breaking composite $\left({\beta }_2\right)$ is not equal to zero.

Explanation of Solution

Calculation: The null hypothesis to test whether the coefficient of rule-breaking composite $\left({\beta }_2\right)$ is equal to zero is stated as

$H_0:{\beta }_2=0$

The alternative hypothesis that the explanatory variable rule-breaking composite $\left({\beta }_2\right)$ is not equal to zero is stated as

$H_a:{\beta }_2\ne 0$

The level of significance is 0.05. It is provided that

$\begin{array}{c}{b}_2=0.100\\ S{E}_{b_2}=0.030\end{array}$

The test statistic $\left(t\right)$ is calculated as

$\begin{array}{c}{t}_{cal}=\frac{b_2}{S{E}_{b_2}}\\ =\frac{0.100}{0.030}\\ =3.33\end{array}$

The critical value of test statistic $\left(t_{crit}\right)$, at 201 degrees of freedom at 0.05 level of significance, is 1.972.

The P-value for the test is calculated as

$\begin{array}{c}p=2p\left(t>\left|t_{crit}\right|\right)\\ =2p\left(t>\left|3.33\right|\right)\\ =2\times \left(0.000517\right)\\ =0.001033\end{array}$

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

$t_{cal}>t_{crit}$

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.

Answer to Problem 19E

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.

Answer to Problem 19E

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

$\widehat{y}={\beta }_0+0.204x_1$

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

$\begin{array}{c}\widehat{y}={\beta }_0+0.204x_1\\ =0.204\left(1\right)\\ =0.204\end{array}$

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

$\widehat{y}={\beta }_0+0.161x_1+0.100x_2$

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

$\begin{array}{c}\widehat{y}={\beta }_0+0.161x_1+0.100x_2\\ =0.161\left(1\right)+0.100\left(0\right)\\ =0.161\end{array}$