A mathematics teacher wanted to see the correlation between test scores and homework. The homework grade [tex]\(( x )\)[/tex] and test grade [tex]\(( y )\)[/tex] are given in the accompanying table.

Write the linear regression equation that represents this set of data, rounding all coefficients to the nearest hundredth. Using this equation, find the projected test grade, to the nearest integer, for a student with a homework grade of 61.

[tex]\[
\begin{tabular}{|c|c|}
\hline Homework Grade (x) & Test Grade (y) \\
\hline 72 & 75 \\
\hline 53 & 39 \\
\hline 90 & 81 \\
\hline 77 & 70 \\
\hline 82 & 83 \\
\hline 74 & 78 \\
\hline 78 & 75 \\
\hline 74 & 70 \\
\hline
\end{tabular}
\][/tex]



Answer :

To solve this problem, we will use the method of linear regression which involves finding the line of best fit for a set of data points. The linear regression equation has the form:

[tex]\[ y = mx + b \][/tex]

where:
- [tex]\( m \)[/tex] is the slope of the line,
- [tex]\( b \)[/tex] is the y-intercept.

Given the data points:

[tex]\[ \begin{array}{|c|c|} \hline \text{Homework Grade (x)} & \text{Test Grade (y)} \\ \hline 72 & 75 \\ 53 & 39 \\ 90 & 81 \\ 77 & 70 \\ 82 & 83 \\ 74 & 78 \\ 78 & 75 \\ 74 & 70 \\ \hline \end{array} \][/tex]

We will use the following formulas to find the slope [tex]\( m \)[/tex] and y-intercept [tex]\( b \)[/tex]:

[tex]\[ m = \frac{N (\sum xy) - (\sum x)(\sum y)}{N (\sum x^2) - (\sum x)^2} \][/tex]
[tex]\[ b = \frac{(\sum y)(\sum x^2) - (\sum x)(\sum xy)}{N (\sum x^2) - (\sum x)^2} \][/tex]

where [tex]\( N \)[/tex] is the number of data points.

First, we calculate the required sums:
[tex]\[ \sum x = 72 + 53 + 90 + 77 + 82 + 74 + 78 + 74 = 600 \][/tex]
[tex]\[ \sum y = 75 + 39 + 81 + 70 + 83 + 78 + 75 + 70 = 591 \][/tex]
[tex]\[ \sum x^2 = 72^2 + 53^2 + 90^2 + 77^2 + 82^2 + 74^2 + 78^2 + 74^2 = 44919 \][/tex]
[tex]\[ \sum y^2 = 75^2 + 39^2 + 81^2 + 70^2 + 83^2 + 78^2 + 75^2 + 70^2 = 47165 \][/tex]
[tex]\[ \sum xy = 72 \cdot 75 + 53 \cdot 39 + 90 \cdot 81 + 77 \cdot 70 + 82 \cdot 83 + 74 \cdot 78 + 78 \cdot 75 + 74 \cdot 70 = 45687 \][/tex]

Now, substituting these values into the slope and intercept formulas:

[tex]\[ m = \frac{8 \cdot 45687 - 600 \cdot 591}{8 \cdot 44919 - 600^2} = \frac{365496 - 354600}{359352 - 360000} = \frac{10896}{-648} = -16.81 \][/tex]

[tex]\[ b = \frac{591 \cdot 44919 - 600 \cdot 45687}{8 \cdot 44919 - 600^2} = \frac{26524929 - 27412200}{359352 - 360000} = \frac{-887271}{-648} = 1369.22 \][/tex]

The linear regression equation:

[tex]\[ y = -16.81x + 1369.22 \][/tex]

To predict the test grade for a homework grade of 61, we substitute [tex]\( x = 61 \)[/tex]:

[tex]\[ y = -16.81(61) + 1369.22 \][/tex]
[tex]\[ y = -1025.41 + 1369.22 \][/tex]
[tex]\[ y = 343.81 \][/tex]

Rounded to the nearest integer, the predicted test grade is [tex]\( 344 \)[/tex].

So the predicted test grade for a student with a homework grade of 61 is [tex]\( 344 \)[/tex].