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a mathematics teacher wanted to see the correlation between test scores…

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 64.
homework grade (x) test grade (y)
59 45
82 81
74 59
61 58
83 78
83 75
85 73
71 56
88 80
answer
regression equation:
final answer:

Explanation:

Step1: Calculate the means

Let $x$ be the homework - grade and $y$ be the test - grade.
$n = 9$
$\bar{x}=\frac{59 + 82+74 + 61+83+83+85+71+88}{9}=\frac{686}{9}\approx76.2$
$\bar{y}=\frac{45 + 81+59+58+78+75+73+56+80}{9}=\frac{615}{9}\approx68.3$

Step2: Calculate the numerator and denominator for the slope $m$

The formula for the slope $m$ of the regression line is $m=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}$
$\sum_{i = 1}^{9}(x_{i}-\bar{x})(y_{i}-\bar{y})=(59 - 76.2)(45 - 68.3)+(82 - 76.2)(81 - 68.3)+(74 - 76.2)(59 - 68.3)+(61 - 76.2)(58 - 68.3)+(83 - 76.2)(78 - 68.3)+(83 - 76.2)(75 - 68.3)+(85 - 76.2)(73 - 68.3)+(71 - 76.2)(56 - 68.3)+(88 - 76.2)(80 - 68.3)$
$=(- 17.2)(-23.3)+(5.8)(12.7)+(-2.2)(-9.3)+(-15.2)(-10.3)+(6.8)(9.7)+(6.8)(6.7)+(8.8)(4.7)+(-5.2)(-12.3)+(11.8)(11.7)$
$=400.76+73.66 + 20.46+156.56+65.96+45.56+41.36+63.96+138.06$
$=1006.3$
$\sum_{i = 1}^{9}(x_{i}-\bar{x})^{2}=(59 - 76.2)^{2}+(82 - 76.2)^{2}+(74 - 76.2)^{2}+(61 - 76.2)^{2}+(83 - 76.2)^{2}+(83 - 76.2)^{2}+(85 - 76.2)^{2}+(71 - 76.2)^{2}+(88 - 76.2)^{2}$
$=(-17.2)^{2}+(5.8)^{2}+(-2.2)^{2}+(-15.2)^{2}+(6.8)^{2}+(6.8)^{2}+(8.8)^{2}+(-5.2)^{2}+(11.8)^{2}$
$=295.84+33.64+4.84+231.04+46.24+46.24+77.44+27.04+139.24$
$=901.6$
$m=\frac{1006.3}{901.6}\approx1.1$

Step3: Calculate the y - intercept $b$

The formula for the y - intercept $b$ is $b=\bar{y}-m\bar{x}$
$b = 68.3-1.1\times76.2$
$b = 68.3 - 83.82=-15.52\approx - 15.5$
The linear regression equation is $y = 1.1x-15.5$

Step4: Solve for $x$ when $y = 64$

$64=1.1x-15.5$
$1.1x=64 + 15.5$
$1.1x=79.5$
$x=\frac{79.5}{1.1}\approx72$

Answer:

Regression Equation: $y = 1.1x-15.5$
Final Answer: $72$