a mathematics teacher wanted to see the correlation between test scores and homework. the homework grade (x)…
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.\nhomework grade (x) test grade (y)\n57 51\n84 73\n78 66\n71 58\n79 67\n62 49\n51 54\n78 73
Answer
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