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, estimate the homework grade, to the nearest integer, for a student with a test grade of 42. homework grade (x) test grade (y) 61 49 81 67 86 86 70 56 74 66 74 77 55 52

Explanation:

Step1: Calculate means

Let \(n = 7\).
\(\bar{x}=\frac{61 + 81+86+70+74+74+55}{7}=\frac{491}{7}\approx70.1\)
\(\bar{y}=\frac{49+67+86+56+66+77+52}{7}=\frac{453}{7}\approx64.7\)

Step2: Calculate \(S_{xx}\) and \(S_{xy}\)

\(S_{xx}=\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=(61 - 70.1)^{2}+(81 - 70.1)^{2}+(86 - 70.1)^{2}+(70 - 70.1)^{2}+(74 - 70.1)^{2}+(74 - 70.1)^{2}+(55 - 70.1)^{2}\)
\(=(-9.1)^{2}+(10.9)^{2}+(15.9)^{2}+(-0.1)^{2}+(3.9)^{2}+(3.9)^{2}+(-15.1)^{2}\)
\(=82.81+118.81+252.81+0.01+15.21+15.21+228.01 = 712.87\)

\(S_{xy}=\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})\)
\(=(61 - 70.1)(49 - 64.7)+(81 - 70.1)(67 - 64.7)+(86 - 70.1)(86 - 64.7)+(70 - 70.1)(56 - 64.7)+(74 - 70.1)(66 - 64.7)+(74 - 70.1)(77 - 64.7)+(55 - 70.1)(52 - 64.7)\)
\(=(-9.1)(-15.7)+(10.9)(2.3)+(15.9)(21.3)+(-0.1)(-8.7)+(3.9)(1.3)+(3.9)(12.3)+(-15.1)(-12.7)\)
\(=142.87+25.07+338.67+0.87+5.07+47.97+191.77 = 752.3\)

Step3: Calculate slope \(b\)

\(b=\frac{S_{xy}}{S_{xx}}=\frac{752.3}{712.87}\approx1.2\)

Step4: Calculate intercept \(a\)

\(a=\bar{y}-b\bar{x}=64.7-1.2\times70.1=64.7 - 84.12=-19.42\approx - 22.8\)

So the linear regression equation is \(y = 1.2x-22.8\)

Step5: Estimate \(x\) when \(y = 42\)

\(42=1.2x-22.8\)
\(1.2x=42 + 22.8=64.8\)
\(x=\frac{64.8}{1.2}=54\)

Answer:

Linear regression equation: \(y = 1.2x - 22.8\). Estimated homework grade for \(y = 42\) is \(54\).