Question

5 in a mathematics class of ten students, the teacher wanted to determine how a homework grade influenced a student’s performance on the subsequent test. the homework grade and subsequent test grade for each student are given in the accompanying table. the length of the rod decreased by approximately 10.5cm.

homework grade (x)	test grade (y)
95	94
92	95
87	89
82	85
80	78
75	73
65	67
50	45
20	40

Explanation:

1. First, we need to find the correlation coefficient or the regression - line to understand the relationship between homework grades ($x$) and test grades ($y$). The formula for the slope ($b$) of the regression line $y = a+bx$ is:

• The mean of $x$ values, $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$.
• $\sum_{i = 1}^{10}x_{i}=94 + 95+92+87+82+80+75+65+50+20=740$.
• $n = 10$, so $\bar{x}=\frac{740}{10}=74$.
• The mean of $y$ values, $\bar{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}$.
• $\sum_{i = 1}^{10}y_{i}=98 + 94+95+89+85+78+73+67+45+40=774$.
• So $\bar{y}=\frac{774}{10}=77.4$.
• The formula for the slope $b=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}$.
• Calculate $(x_{i}-\bar{x})(y_{i}-\bar{y})$ for each $i$:
• For $x_1 = 94$, $y_1 = 98$: $(94 - 74)(98 - 77.4)=20\times20.6 = 412$.
• For $x_2 = 95$, $y_2 = 94$: $(95 - 74)(94 - 77.4)=21\times16.6 = 348.6$.
• For $x_3 = 92$, $y_3 = 95$: $(92 - 74)(95 - 77.4)=18\times17.6 = 316.8$.
• For $x_4 = 87$, $y_4 = 89$: $(87 - 74)(89 - 77.4)=13\times11.6 = 150.8$.
• For $x_5 = 82$, $y_5 = 85$: $(82 - 74)(85 - 77.4)=8\times7.6 = 60.8$.
• For $x_6 = 80$, $y_6 = 78$: $(80 - 74)(78 - 77.4)=6\times0.6 = 3.6$.
• For $x_7 = 75$, $y_7 = 73$: $(75 - 74)(73 - 77.4)=1\times(- 4.4)=-4.4$.
• For $x_8 = 65$, $y_8 = 67$: $(65 - 74)(67 - 77.4)=(-9)\times(-10.4)=93.6$.
• For $x_9 = 50$, $y_9 = 45$: $(50 - 74)(45 - 77.4)=(-24)\times(-32.4)=777.6$.
• For $x_{10}=20$, $y_{10}=40$: $(20 - 74)(40 - 77.4)=(-54)\times(-37.4)=2019.6$.
• $\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})=412+348.6+316.8+150.8+60.8+3.6 - 4.4+93.6+777.6+2019.6=4189.6$.
• Calculate $(x_{i}-\bar{x})^{2}$ for each $i$:
• For $x_1 = 94$: $(94 - 74)^{2}=400$.
• For $x_2 = 95$: $(95 - 74)^{2}=441$.
• For $x_3 = 92$: $(92 - 74)^{2}=324$.
• For $x_4 = 87$: $(87 - 74)^{2}=169$.
• For $x_5 = 82$: $(82 - 74)^{2}=64$.
• For $x_6 = 80$: $(80 - 74)^{2}=36$.
• For $x_7 = 75$: $(75 - 74)^{2}=1$.
• For $x_8 = 65$: $(65 - 74)^{2}=81$.
• For $x_9 = 50$: $(50 - 74)^{2}=576$.
• For $x_{10}=20$: $(20 - 74)^{2}=2916$.
• $\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=400+441+324+169+64+36+1+81+576+2916=4608$.
• Then $b=\frac{4189.6}{4608}\approx0.91$.
• The formula for the $y$ - intercept $a=\bar{y}-b\bar{x}$.
• $a = 77.4-0.91\times74=77.4 - 67.34 = 10.06$.
• The regression line is $y = 10.06+0.91x$.

1. Interpretation:

• The slope $b = 0.91$ indicates that for every one - unit increase in the homework grade, the test grade is expected to increase by approximately $0.91$ units.

Answer:

The regression line is $y = 10.06+0.91x$, and for every one - unit increase in the homework grade, the test grade is expected to increase by approximately $0.91$ units.