Question

a mathematics teacher wanted to see the correlation between test scores and homework. the homework grade (x) and test grade (y) are given in the accompanying table. write the linear regression equation that represents this set of data, rounding all coefficients to the nearest tenth. using this equation, find the projected test grade, to the nearest integer, for a student with a homework grade of 74.
homework grade (x) test grade (y)
57 51
84 73
78 66
71 58
79 67
62 49
51 54
78 73

Explanation:

Step1: Calculate sums

Let \(n = 8\) (number of data - points).
Calculate \(\sum x\), \(\sum y\), \(\sum x^{2}\), \(\sum xy\):
\(\sum x=57 + 84+78 + 71+79+62+51+78=560\)
\(\sum y=51 + 73+66+58+67+49+54+73=491\)
\(\sum x^{2}=57^{2}+84^{2}+78^{2}+71^{2}+79^{2}+62^{2}+51^{2}+78^{2}\)
\(=3249+7056+6084+5041+6241+3844+2601+6084 = 40190\)
\(\sum xy=57\times51+84\times73+78\times66+71\times58+79\times67+62\times49+51\times54+78\times73\)
\(=2907+6132+5148+4118+5293+3038+2754+5694 = 35184\)

Step2: Calculate slope \(m\) and y - intercept \(b\)

The formula for the slope \(m\) of the regression line is \(m=\frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}}\)
\[

$$\begin{align*} m&=\frac{8\times35184 - 560\times491}{8\times40190-560^{2}}\\ &=\frac{281472-274960}{321520 - 313600}\\ &=\frac{6512}{7920}\\ &\approx0.8 \end{align*}$$

\]
The formula for the y - intercept \(b\) is \(b=\frac{\sum y - m\sum x}{n}\)
\[

$$\begin{align*} b&=\frac{491-0.8\times560}{8}\\ &=\frac{491 - 448}{8}\\ &=\frac{43}{8}\\ & = 5.4 \end{align*}$$

\]
The regression equation is \(y = 0.8x+5.4\)

Step3: Predict test grade

Substitute \(x = 74\) into the regression equation \(y=0.8x + 5.4\)
\(y=0.8\times74+5.4=59.2+5.4 = 64.6\approx65\)

Answer:

65