TABLE 14-7
the Department Head of the Accounting Department $\text {SUMMARY OUTPUT}$

TABLE 14-7
The department head of the accounting department wanted to see if she could predict the GPA of students using the number of course units (credits) and total SAT scores of each. She takes a sample of students and generates the following Microsoft Excel output:
$\text {SUMMARY OUTPUT}$
$\begin{array}{ll}\hline \text { Regression Statistics } \\\hline \text { Multiple R }& 0.916 \\ \text { R Square} & 0.839 \\ \text { Adjusted R Square} & 0.732 \\ \text { Standard Error }& 0.24685 \\ \text { Observations} & 6 \\\hline\end{array}$

ANOVA
$\begin{array}{lccclc}\hline & \text {d f }& \text { SS }& \text { M S } & \text { F } & \text {Significance F } \\\hline \text { Regression} & 2 & 0.95219 & 0.47610 & 7.813 & 0.0646 \\ \text {Residual} & 3 & 0.18281 & 0.06094 & & \\ \text {Total} & 5 & 1.13500 & & & \\\hline\end{array}$

$\begin{array}{lrcrr}\hline & \text {Coefficients }& \text {Standard Error} & \text {t Stat } & \text { p -value} \\\hline \text { Intercept }& 4.593897 & 1.13374542 & 4.052 & 0.0271 \\ \text {Units }& -0.247270 & 0.06268485 & -3.945 & 0.0290 \\ \text {SAT Total }& 0.001443 & 0.00101241 & 1.425 & 0.2494 \\\hline\end{array}$

-Referring to Table 14-7, the department head wants to test H0 : þ1 = þ2 = 0. At a level of significance of 0.05, the null hypothesis is rejected.