The data below are the number of absences and the final grades of 9 randomly selected students from a statistics class. Find the standard error of estimate, $\mathrm { s } _ { \mathrm { e } }$ given that ? = -2.75X + 96.14. $$\begin{array} { l | c | c | c | c | c | c | c | c | c | }
\text { Number of absences, } \mathrm { x } & 0 & 3 & 6 & 4 & 9 & 2 & 15 & 8 & 5 \
\hline \text { Final grade, } \mathrm { y } & 98 & 86 & 80 & 82 & 71 & 92 & 55 & 76 & 82
\end{array}$$
A)1.799
B)4.531
C)3.876
D)2.160

Question

The data below are the number of absences and the final grades of 9 randomly selected students from a statistics class. Find the standard error of estimate, $\mathrm { s } _ { \mathrm { e } }$ given that ? = -2.75X + 96.14. $$\begin{array} { l | c | c | c | c | c | c | c | c | c | } 
\text { Number of absences, } \mathrm { x } & 0 & 3 & 6 & 4 & 9 & 2 & 15 & 8 & 5 \
\hline \text { Final grade, } \mathrm { y } & 98 & 86 & 80 & 82 & 71 & 92 & 55 & 76 & 82
\end{array}$$
A)1.799
B)4.531
C)3.876
D)2.160

Quizplus · Accepted Answer

The standard error of estimate is the standard deviation of the residuals, which is the difference between the actual and predicted values. The predicted values are calculated using the regression equation, which is given as y = -2.75x + 96.14. The residuals are then calculated as y - y^, where y is the actual value and y^ is the predicted value. The standard deviation of the residuals is then calculated as the square root of the variance of the residuals, which is the sum of the squared residuals divided by the degrees of freedom. In this case, the degrees of freedom are 7, since there are 9 data points and 2 parameters in the regression equation. The variance of the residuals is 10.85, so the standard error of estimate is the square root of 10.85, which is 3.29.

The Data Below Are the Number of Absences and the Final