Chapter 11.1, Problem 9E

(a)

To determine

To explain: The expected signs of coefficients.

Answer to Problem 9E

Solution: The signs of the coefficients are apt according to their relation with the response variable. The signs for the variables Math course anxiety, Math test anxiety and the Numerical task anxiety should be negative and hence, are negative in the provided table. The signs for the variables: Enjoyment, Self-confidence, Motivation and Feedback usefulness should be positive, and hence, are positive in the provided problem.

Explanation of Solution

The explanatory variables in the provided table are: Math course anxiety, Math test anxiety, Numerical task anxiety, Enjoyment, Self-confidence, Motivation, Feedback usefulness. The signs for the variables Math course anxiety, Math test anxiety and the Numerical task anxiety should be negative as they are inversely related to the response variable ‘student’s final exam score’. These variables lower the student’s scores, and as expected, these have negative signs in the table. The signs for the variables: Enjoyment, Self-confidence, Motivation and Feedback usefulness should be positive as they are directly related to the response variable ‘student’s final exam score’. These variables will be the reasons of an increase in the student’s scores, and as expected, these have positive signs in the table.

(b)

To determine

The degrees of freedom for the model and error.

Answer to Problem 9E

Solution: The degrees of freedom of the Model (DFM) are 7 and the degrees of freedom of Error (DFE) are 158.

Explanation of Solution

The total number of observations (denoted as $n$ ) are provided in the referred problem 11.9 as:

$n=166$

There are 7 explanatory variables (Math course anxiety, Math test anxiety, Numerical task anxiety, Enjoyment, Self-confidence, Motivation, Feedback usefulness) in the problem. Hence, the number of independent variables is denoted by $p$ equal to:

$p=7$

The number of explanatory variables are the degrees of freedom of the Model (DFM) and are equal to $p$, that is, 7. The degrees of freedom of Error (DFE) are obtained by the formula:

$\begin{array}{c}n-p-1=166-7-1\\ =158\end{array}$

(c)

To determine

To test: The hypothesis that the coefficient of variable ‘Math course anxiety’ is zero against the hypothesis that it is not zero.

Answer to Problem 9E

Solution: The null hypothesis that the coefficient of variable ‘Math course anxiety’ is zero is true. Hence, it proves that the result of insignificant.

Explanation of Solution

Calculation: The multiple linear regression equation for the provided problem with the response variable ‘student’s final exam score’ (denoted as $y$ ) with explanatory variables Math course anxiety (denoted as $x_1$ ), Math test anxiety (denoted as $x_2$ ), Numerical task anxiety (denoted as $x_3$ ), Enjoyment (denoted as $x_4$ ), Self-confidence (denoted as $x_5$ ), Motivation (denoted as $x_6$ ) and Feedback usefulness (denoted as $x_7$ ). The coefficients for these variables are denoted as ${\beta }_1,{\beta }_2,{\beta }_3,{\beta }_4,{\beta }_5,{\beta }_6$ and ${\beta }_7$ respectively. The deviations for each individual independent variable is denoted as ${\epsilon }_i$. The multiple-linear regression model for the problem can be written as:

$y={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+{\beta }_3{x}_3+{\beta }_4{x}_4+{\beta }_5{x}_5+{\beta }_6{x}_6+{\beta }_7{x}_7+{\epsilon }_i$

From the regression model, the coefficient of $x_1$ is denoted as ${\beta }_1$. According to the provided problem, the null hypothesis to test the significance of the coefficient ${\beta }_1$ will be formulated as follows:

$H_0:{\beta }_1=0$

And the alternative hypothesis is a two-sided alternative. Hence, it will be formulated as follows:

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

The total number of observations has been provided in the problem as:

$n=166$

Since, there are 7 explanatory variables in the regression equation, the degrees of freedom for the provided problem can be calculated as follows:

$\begin{array}{c}n-p-1=166-7-1\\ =158\end{array}$

The value of coefficient of $x_1$ is $b_1=-0.212$ and the standard error is ${\text{SE}}_{b_1}=0.114$. Since, the test is two-tailed, the value of $t^{*}$ for $95\%$ confidence level for 158 degrees of freedom is obtained from the table as: 1.96. The $t-\text{statistic}$ is calculated as:

$\begin{array}{c}{t}_{b_1}=\frac{b_1}{{\text{SE}}_{b_1}}\\ =\frac{-0.212}{0.114}\\ =-1.85\end{array}$

Conclusion: Since, the calculated value of $t$ is less than the observed value of $t$, thus, the null hypothesis does not get rejected. Hence, the null hypothesis that the coefficient of variable $x_1$ is zero is true. This implies that the test is insignificant.

To determine

To test: The hypothesis that the coefficient of variable ‘Math test anxiety’ is zero against the hypothesis that it is not zero.

Answer to Problem 9E

Solution: The null hypothesis that the coefficient of variable ‘Math test anxiety’ is zero is true. Hence, it proves that the result of insignificant.

Explanation of Solution

Calculation: The coefficient of $x_2$ is denoted as ${\beta }_2$. According to the provided problem, the null hypothesis to test the significance of the coefficient ${\beta }_2$ will be formulated as follows:

$H_0:{\beta }_2=0$

And the alternative hypothesis is a two-sided alternative. Hence, it will be formulated as follows:

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

The total number of observations have been provided in the problem as:

$n=166$

Since, there are 7 explanatory variables in the regression equation, the degrees of freedom for the provided problem can be calculated as follows:

$\begin{array}{c}n-p-1=166-7-1\\ =158\end{array}$

The value of coefficient of $x_2$ is $b_2=-0.155$ and the standard error is $S{E}_{b_2}=0.119$. Since, the test is two-tailed, the value of $t^{*}$ for $95\%$ confidence interval for 158 degrees of freedom is obtained from the table as: 1.96. The $t-\text{statistic}$ is calculated as:

$\begin{array}{c}{t}_{b_2}=\frac{b_2}{S{E}_{b_2}}\\ =\frac{-0.155}{0.119}\\ =-1.3\end{array}$

Conclusion: Since, the calculated value of $t$ is less than the observed value of $t$, thus, the null hypothesis does not get rejected. Hence, the null hypothesis that the coefficient of variable $x_2$ is zero is true. This implies that the test is insignificant.

To determine

To test: The hypothesis that the coefficient of variable ‘Numerical task anxiety’ is zero against the hypothesis that it is not zero.

Answer to Problem 9E

Solution: The null hypothesis that the coefficient of variable ‘Numerical task anxiety’ is zero is true. Hence, it proves that the result of insignificant.

Explanation of Solution

Calculation: The coefficient of $x_3$ is denoted as ${\beta }_3$. According to the provided problem, the null hypothesis to test the significance of the coefficient ${\beta }_3$ will be formulated as follows:

$H_0:{\beta }_3=0$

And the alternative hypothesis is a two-sided alternative. Hence, it will be formulated as follows:

$H_a:{\beta }_3\ne 0$

The total number of observations has been provided in the problem as:

$n=166$

Since, there are 7 explanatory variables in the regression equation, the degrees of freedom for the provided problem can be calculated as follows:

$\begin{array}{c}n-p-1=166-7-1\\ =158\end{array}$

The value of coefficient of $x_3$ is $b_3=-0.094$ and the standard error is ${\text{SE}}_{b_3}=0.116$. Since, the test is two-tailed, the value of $t^{*}$ for $95\%$ confidence interval for 158 degrees of freedom is obtained from the table as: 1.96. The $t-\text{statistic}$ is calculated as:

$\begin{array}{c}{t}_{b_3}=\frac{b_3}{{\text{SE}}_{b_3}}\\ =\frac{-0.094}{0.116}\\ =-0.81\end{array}$

Conclusion: Since, the calculated value of $t$ is less than the observed value of $t$, thus, the null hypothesis does not get rejected. Hence, the null hypothesis that the coefficient of variable $x_3$ is zero is true. This implies that the test is insignificant.

To determine

To test: The hypothesis that the coefficient of variable ‘Enjoyment’ is zero against the hypothesis that it is not zero.

Answer to Problem 9E

Solution: The null hypothesis that the coefficient of variable ‘Enjoyment’ is zero is true. Hence, it proves that the result of insignificant.

Explanation of Solution

Calculation: The coefficient of $x_4$ is denoted as ${\beta }_4$. According to the provided problem, the null hypothesis to test the significance of the coefficient ${\beta }_4$ will be formulated as follows:

$H_0:{\beta }_4=0$

And the alternative hypothesis is a two-sided alternative. Hence, it will be formulated as follows:

$H_a:{\beta }_4\ne 0$

The total number of observations has been provided in the problem as:

$n=166$

Since, there are 7 explanatory variables in the regression equation, the degrees of freedom for the provided problem can be calculated as follows:

$\begin{array}{c}n-p-1=166-7-1\\ =158\end{array}$

The value of coefficient of $x_4$ is $b_4=0.176$ and the standard error is ${\text{SE}}_{b_4}=0.114$. Since, the test is two-tailed, the value of $t^{*}$ for $95\%$ confidence interval for 158 degrees of freedom is obtained from the table as: 1.96. The $t-\text{statistic}$ is calculated as:

$\begin{array}{c}{t}_{b_4}=\frac{b_4}{{\text{SE}}_{b_4}}\\ =\frac{0.176}{0.114}\\ =1.54\end{array}$

Conclusion: Since, the calculated value of $t$ is less than the observed value of $t$, thus, the null hypothesis does not get rejected. Hence, the null hypothesis that the coefficient of variable $x_4$ is zero is true. This implies that the test is insignificant.

To determine

To test: The hypothesis that the coefficient of variable ‘Self-confidence’ is zero against the hypothesis that it is not zero.

Answer to Problem 9E

Solution: The null hypothesis that the coefficient of variable ‘Self-confidence’ is zero is true. Hence, it proves that the result of insignificant.

Explanation of Solution

Calculation: The coefficient of $x_5$ is denoted as ${\beta }_5$. According to the provided problem, the null hypothesis to test the significance of the coefficient ${\beta }_5$ will be formulated as follows:

$H_0:{\beta }_5=0$

And the alternative hypothesis is a two-sided alternative. Hence, it will be formulated as follows:

$H_a:{\beta }_5\ne 0$

The total number of observations has been provided in the problem as:

$n=166$

Since, there are 3 explanatory variables in the regression equation, the degrees of freedom for the provided problem can be calculated as follows:

$\begin{array}{c}n-p-1=166-7-1\\ =158\end{array}$

The value of coefficient of $x_5$ is $b_5=0.118$ and the standard error is $S{E}_{b_5}=0.114$. Since, the test is two-tailed, the value of $t^{*}$ for $95\%$ confidence interval for 158 degrees of freedom is obtained from the table as: 1.96. The $t-\text{statistic}$ is calculated as:

$\begin{array}{c}{t}_{b_5}=\frac{b_5}{{\text{SE}}_{b_5}}\\ =\frac{0.118}{0.114}\\ =1.035\end{array}$

Conclusion: Since, the calculated value of $t$ is less than the observed value of $t$, thus, the null hypothesis does not get rejected. Hence, the null hypothesis that the coefficient of variable $x_5$ is zero is true. This implies that the test is insignificant.

To determine

To test: The hypothesis that the coefficient of variable ‘Motivation’ is zero against the hypothesis that it is not zero.

Answer to Problem 9E

Solution: The null hypothesis that the coefficient of variable ‘Motivation’ is zero is true. Hence, it proves that the result of insignificant.

Explanation of Solution

Calculation: The coefficient of $x_6$ is denoted as ${\beta }_6$. According to the provided problem, the null hypothesis to test the significance of the coefficient ${\beta }_6$ will be formulated as follows:

$H_0:{\beta }_6=0$

And the alternative hypothesis is a two-sided alternative. Hence, it will be formulated as follows:

$H_a:{\beta }_6\ne 0$

The total number of observations has been provided in the problem as:

$n=166$

Since, there are 3 explanatory variables in the regression equation, the degrees of freedom for the provided problem can be calculated as follows:

$\begin{array}{c}n-p-1=166-7-1\\ =158\end{array}$

The value of coefficient of $x_6$ is $b_6=0.097$ and the standard error is $S{E}_{b_6}=0.115$. Since, the test is two-tailed, the value of $t^{*}$ for $95\%$ confidence interval for 158 degrees of freedom is obtained from the table as: 1.96. The $t-\text{statistic}$ is calculated as:

$\begin{array}{c}{t}_{b_6}=\frac{b_6}{{\text{SE}}_{b_6}}\\ =\frac{0.097}{0.115}\\ =0.8435\end{array}$

Conclusion: Since, the calculated value of $t$ is less than the observed value of $t$, thus, the null hypothesis does not get rejected. Hence, the null hypothesis that the coefficient of variable $x_6$ is zero is true. This implies that the test is insignificant.

To determine

To test: The hypothesis that the coefficient of variable ‘Feedback usefulness’ is zero against the hypothesis that it is not zero.

Answer to Problem 9E

Solution: The null hypothesis that the coefficient of variable ‘Feedback usefulness’ is zero is not true. Hence, it proves that the result is significant.

Explanation of Solution

Calculation: The coefficient of $x_7$ is denoted as ${\beta }_7$. According to the provided problem, the null hypothesis to test the significance of the coefficient ${\beta }_7$ will be formulated as follows:

$H_0:{\beta }_7=0$

And the alternative hypothesis is a two-sided alternative. Hence, it will be formulated as follows:

$H_a:{\beta }_7\ne 0$

The total number of observations has been provided in the problem as:

$n=166$

Since, there are 3 explanatory variables in the regression equation, the degrees of freedom for the provided problem can be calculated as follows:

$\begin{array}{c}n-p-1=166-7-1\\ =158\end{array}$

The value of coefficient of $x_7$ is $b_7=0.644$ and the standard error is ${\text{SE}}_{b_7}=0.194$. Since, the test is two-tailed, the value of $t^{*}$ for $95\%$ confidence interval for 158 degrees of freedom is obtained from the table as: 1.96. The $t-\text{statistic}$ is calculated as:

$\begin{array}{c}{t}_{b_7}=\frac{b_7}{{\text{SE}}_{b_7}}\\ =\frac{0.644}{0.194}\\ =3.319\end{array}$

Conclusion: Since, the calculated value of $t$ is more than the observed value of $t$, therefore, the null hypothesis gets rejected. Hence, the null hypothesis that the coefficient of variable $x_7$ is zero is not true. This implies that the test is significant.