Given the table below representing the number of hours of television nine Math II class students watched the night before a big test on
triangles
along with the grades they each earned on that test.
[tex]\begin{center}
\begin{tabular}
{|c|c|}
Hours Spent Watching TV & Grade on Test (out of 100) \\ [1ex]
4 & 71 \\
2 & 81 \\
4 & 62 \\
1 & 86 \\
3 & 77 \\
1 & 93 \\
2 & 84 \\
3 & 80 \\
2 & 85
\end{tabular}
\end{center}[/tex]
Let the number the number of hours of television each of the students watched the night before the test be x while the grades they each earned on that test be y.
We use the following table to find the equation of the line of best fit of the regression analysis of the data.
[tex]\begin{center} \begin{tabular} {|c|c|c|c|} x & y & x^2 & xy \\ [1ex] 4 & 71 & 16 & 284 \\ 2 & 81 & 4 & 162 \\ 4 & 62 & 16 & 248 \\ 1 & 86 & 1 & 86 \\ 3 & 77 & 9 & 231 \\ 1 & 93 & 1 & 93 \\ 2 & 84 & 4 & 168 \\ 3 & 80 & 9 & 240 \\ 2 & 85 & 4 & 170 \\ [1ex]\Sigma x=22 & \Sigma y=719 & \Sigma x^2=64 & \Sigma xy=1,682 \end{tabular} \end{center}[/tex]
Recall that the equation of the line of best fit of a regression analysis is given by
[tex]y=a+bx[/tex]
where:
[tex]a= \frac{(\Sigma y)(\Sigma x^2)-(\Sigma x)(\Sigma xy)}{n(\Sigma x^2)-(\Sigma x)^2} [/tex]
and
[tex]b= \frac{n(\Sigma xy)-(\Sigma x)(\Sigma y)}{n(\Sigma x^2)-(\Sigma x)^2} [/tex]
[tex]y=\frac{(719)(64)-(22)(1,682)}{9(64)-(22)^2}+\frac{9(1,682)-(22)(719)}{9(64)-(22)^2}x \\ \\ = \frac{46,016-37,004}{576-484} + \frac{15,138-15,818}{576-484} x \\ \\ = \frac{9,012}{92} + \frac{-680}{92} x \\ \\ =97.95-7.391x[/tex]
Thus, the equation of the line of best fit is given by y = 97.95 - 7.391x
A student that watched 1.5 hours of TV will have a score given by
y = 97.95 - 7.391(1.5) = 97.95 - 11.0865 = 86.8635
Therefore, a student’s score if he/she watched 1.5 hours of TV to the nearest whole number is 87.
