Instruction 13.36
The Head of the Accounting Department wanted to see if she could predict the average grade of students using the number of course units (credits) and total university entrance exam scores of each. She takes a sample of students and generates the following Microsoft Excel output:
OUTPUT
SUMMARY
$\begin{array} { l l } \text { Regression Statistics } & \\ \text { MultipleR } & 0.916 \\ \text { R Square } & 0.839 \\ \text { Adj. R Square } & 0.732 \\ \text { Std. Error } & 0.24685 \\ \text { Observations } & 6 \end{array}$

ANOVA
$\begin{array} { l l l l l l } & d f & \text { SS } & \text { MS } & F & \text { Signif F } \\ \text { Regression } & 2 & 0.95219 & 0.47610 & 7.813 & 0.0646 \\ \text { Residual } & 3 & 0.18281 & 0.06094 & & \\ \text { Total } & 5 & 1.13500 & & & \\ & & & & & \\ & \text { Coeff } & \text { StdError } & t \text { Stat } & p \text { value } & \\ \text { Intercept } & 4.593897 & 1.13374542 & 4.052 & 0.0271 & \\ \text { GDP } & - 0.247270 & 0.06268485 & - 3.945 & 0.0290 & \\ \text { Price } & 0.001443 & 0.00101241 & 1.425 & 0.2494 & \end{array}$ Note: Adj. R Square = Adjusted R Square; Std. Error = Standard Error
-Referring to Instruction 13.36,the Head of Department wants to use a t test to test for the significance of the coefficient of X₁.At a level of significance of 0.05,the department head would decide that ?₁ ? 0.

Question

Instruction 13.36
The Head of the Accounting Department wanted to see if she could predict the average grade of students using the number of course units (credits) and total university entrance exam scores of each. She takes a sample of students and generates the following Microsoft Excel output:
OUTPUT
SUMMARY

\begin{array} { l l } \text { Regression Statistics } & \\ \text { MultipleR } & 0.916 \\ \text { R Square } & 0.839 \\ \text { Adj. R Square } & 0.732 \\ \text { Std. Error } & 0.24685 \\ \text { Observations } & 6 \end{array}

ANOVA

\begin{array} { l l l l l l } & d f & \text { SS } & \text { MS } & F & \text { Signif F } \\ \text { Regression } & 2 & 0.95219 & 0.47610 & 7.813 & 0.0646 \\ \text { Residual } & 3 & 0.18281 & 0.06094 & & \\ \text { Total } & 5 & 1.13500 & & & \\ & & & & & \\ & \text { Coeff } & \text { StdError } & t \text { Stat } & p \text { value } & \\ \text { Intercept } & 4.593897 & 1.13374542 & 4.052 & 0.0271 & \\ \text { GDP } & - 0.247270 & 0.06268485 & - 3.945 & 0.0290 & \\ \text { Price } & 0.001443 & 0.00101241 & 1.425 & 0.2494 & \end{array}

Note: Adj. R Square = Adjusted R Square; Std. Error = Standard Error
-Referring to Instruction 13.36,the Head of Department wants to use a t test to test for the significance of the coefficient of X₁.At a level of significance of 0.05,the department head would decide that ?₁ ? 0.

Quizplus · Accepted Answer

The output does not provide the t value for X₁, but since the department head wants to use a t test to test for the significance of the coefficient, this implies that the output provides the p-value for the coefficient. Thus, if the p-value is less than the level of significance (0.05), the department head would reject the null hypothesis and conclude that the coefficient is significantly different from zero. Therefore, the statement "the department head would decide that β₁ ≠ 0" is true if the p-value is less than 0.05.

Instruction 13 ANOVA Note: Adj R Square = Adjusted R Square; Std