Table 14-16
the Superintendent of a School District Wanted to Predict

Table 14-16
The superintendent of a school district wanted to predict the percentage of students passing a sixth-grade proficiency test. She obtained the data on percentage of students passing the proficiency test (% Passing) , daily average of the percentage of students attending class (% Attendance) , average teacher salary in dollars (Salaries) , and instructional spending per pupil in dollars (Spending) of 47 schools in the state.
Following is the multiple regression output with Y = % Passing as the dependent variable, X₁ = % Attendance, X₂ = Salaries and
X₃ = Spending:
$\begin{array}{ll}\text { Model } 1 \\\text { Regression Statistics } \\\hline \text { R Square }& 0.8080 \\\text { AdjustedR S quare }& 0.7568 \\\text { Observations } &20 \end{array}$

ANOVA
$\begin{array}{lrrccc} & \text { df } &{S S} & M S & F & \text { Signuficance F } \\\hline \text { Regression } & 4 & 169503.4241 & 42375.86 & 15.7874 & 2.96869 E-05 \\\text { Residual } & 15 & 40262.3259 & 2684.155 & & \\\text { Total } & 19 & 209765.75 & & & \\\hline\end{array}$

$\begin{array}{lrrrrrrr} && \text { Standard } & & \text {Lower} & \text {Upper }\\& \text {Coefficients} & \text {Error} & \text { t Stat } & \text {p -value} & 90.0 \% & 90.0 \% \\ \hline \text { Intercept }& 421.4277 & 77.8614 & 5.4125 & 7.2 \mathrm{E}-05& 284.9327 & 557.9227 \\ \text { X{1} (Temperature) } & -4.5098 & 0.8129 & -5.5476 &5.58 \mathrm{E}-05 & -5.9349 & -3.0847 \\ \text {X{2} (Insulation) }& -14.9029 & 5.0508 & -2.9505 & 0.0099 & -23.7573 & -6.0485 \\ \text { X{3} (Windows) }& 0.2151 & 4.8675 & 0.0442 & 0.9653 & -8.3181 & 8.7484 \\ \text { X{4} (Furnace Age) } & 6.3780 & 4.1026 & 1.5546 & 0.1408 & -0.8140 & 13.5702 \\\hline\end{array}$

$\begin{array}{ll} \text { Model 2} \\\hline \text {Regression Statistics} \\\hline \text {R Square }& 0.7768\\ \text {Adjusted R Square }&0.7506 \\ \text {Observations }& 20 \\\hline\end{array}$
ANOVA
$\begin{array}{lrrrcc}\hline & \text { d f} & \text { SS } & \text { MS } & \text {SS } & \text {Significance F } \\\hline \text {Regression} & 2 & 162958.2277 & 81479.11 & 29.5923 & 2.9036 \mathrm{E}-06 \\ \text {Residual }& 17 & 46807.5222 & 2753.384 & & \\ \text {Total }& 19 & 209765.75 & & & \\\hline\end{array}$

$\begin{array}{lcccccc} && \text { Standard } & &\text { Lower} &\text { Upper} \\& \text {Coefficients }&\text { Error }&\text { t Stat }& \text { p -value} &95 \%& 95 \% \\\hline \text {Intercept }& 489.3227 & 43.982611 .1253 &3.17 \mathrm{E}-09& 396.5273 & 582.1180 \\\text { X{1} (Temperature) }& -5.1103 & 0.6951-7.3515 & 1.13 \mathrm{E}-06 & -6.5769 & -3.6437 \\\text { X{2} (Insulation) }& -14.7195 & 4.8864-3.0123 & 0.0078 & -25.0290 & -4.4099 \\\hline\end{array}$


-Referring to Table 14-16, which of the following is a correct statement?

A) 60.29% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class holding constant the effect of average teacher salary, and instructional spending per pupil.
B) 60.29% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class, average teacher salary, and instructional spending per pupil after adjusting for the number of predictors and sample size.
C) 60.29% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class after adjusting for the effect of average teacher salary, and instructional spending per pupil.
D) 60.29% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class, average teacher salary, and instructional spending per pupil.