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Question
the effect of regularly playing video games on grades
a high school biology class conducted a study of whether playing video games had any effect on grades. ten volunteers told the class how many hours per week they spent playing video games and what their grade point average was. the results are tabulated below.
hours spent playing per week grade point average
1 3.49
0 3.05
0 3.24
2 2.82
3 3.19
3 2.78
5 2.31
8 2.54
8 2.03
10 2.51
12
what is the y - intercept, b, of the line of best fit?
Step1: Recall the formula for the line of best - fit in linear regression $y = mx + b$.
The least - squares method is often used to find the coefficients $m$ and $b$. The formula for $b=\bar{y}-m\bar{x}$, where $\bar{x}$ and $\bar{y}$ are the means of the $x$ and $y$ data respectively, and $m$ is the slope of the line of best - fit. First, calculate the means of $x$ and $y$.
Let $x$ be the hours spent playing video games per week and $y$ be the grade - point average.
$n = 10$ (number of data points).
$\sum_{i = 1}^{n}x_{i}=1 + 0+0 + 2+3+3+5+8+8+10+12=52$.
$\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}=\frac{52}{10}=5.2$.
$\sum_{i = 1}^{n}y_{i}=3.49+3.05 + 3.24+2.82+3.19+2.78+2.31+2.54+2.03+2.51=27.96$.
$\bar{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}=\frac{27.96}{10}=2.796$.
Step2: Calculate the slope $m$.
The formula for the slope $m=\frac{n\sum_{i = 1}^{n}x_{i}y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}y_{i}}{n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2}}$.
$\sum_{i = 1}^{n}x_{i}y_{i}=1\times3.49+0\times3.05+0\times3.24 + 2\times2.82+3\times3.19+3\times2.78+5\times2.31+8\times2.54+8\times2.03+10\times2.51+12\times2.51$
$=3.49+0 + 0+5.64+9.57+8.34+11.55+20.32+16.24+25.1+30.12=130.37$.
$\sum_{i = 1}^{n}x_{i}^{2}=1^{2}+0^{2}+0^{2}+2^{2}+3^{2}+3^{2}+5^{2}+8^{2}+8^{2}+10^{2}+12^{2}$
$=1+0+0 + 4+9+9+25+64+64+100+144=420$.
$m=\frac{10\times130.37-52\times27.96}{10\times420 - 52^{2}}$
$=\frac{1303.7-1453.92}{4200 - 2704}=\frac{- 150.22}{1496}\approx - 0.1$.
Step3: Calculate the y - intercept $b$.
Using the formula $b=\bar{y}-m\bar{x}$, substitute $\bar{x}=5.2$, $\bar{y}=2.796$ and $m\approx - 0.1$.
$b=2.796-(-0.1)\times5.2=2.796 + 0.52=3.316$.
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$3.316$