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:
$\begin{array}{lr}\hline\text {Regression Statistics } \\\hline \text {Multiple R }& 0.7930 \\\text {R Square} & 0.6288 \\\text {Adjusted R Square} & 0.6029 \\\text {Standard Error }& 10.4570 \\\text {Observations }& 47 \\\hline\end{array}$


ANOVA
$\begin{array}{lccccc}\hline &\text { d f } &\text { SS }& \text { MS }& \text { F } & \text { Significance F} \\\hline \text { Regression }& 3 & 7965.08 & 2655.03 & 24.2802 & 2.3853 \mathrm{E}-09 \\\text { Residual} & 43 & 4702.02 & 109.35 & & \\\text { Total }& 46 & 12667.11 & & & \\\hline\end{array}$

$\begin{array}{lrrrrrr}\hline &\text { Coeffs} & \text { Stnd Err} &\text { t Stat} &\text { p -value }&\text { Lower 95\%} \text { Upper 95\%} \\\hline\text { Intercept }& -753.4225 & 101.1149 & -7.4511 & 2.88 \mathrm{E}-09 & -957.3401 & -549.5050 \\\%\text { Attend }& 8.5014 & 1.0771 & 7.8929 & 6.73 \mathrm{E}-10 & 6.3292 & 10.6735 \\\text { Salary} & 6.85 \mathrm{E}-07 & 0.0006 & 0.0011 & 0.9991 & -0.0013 & 0.0013 \\\text { Spending }& 0.0060 & 0.0046 & 1.2879 & 0.2047 & -0.0034 & 0.0153 \\\hline\end{array}$

-Referring to Table 14-16, the alternative hypothesis H1 : At least one of þj × 0 for j = 1, 2, 3 implies that percentage of students passing the proficiency test is related to all of the explanatory variables.