Question

How do you find the slope of the regression line for the following set of data?

X values –3 0 3 5 3
Y values 7 4 –2 2 –3

Answer 1

The slope is `-97/98approx -0.989796`.

Explanation:

You can use the formula that the slope is `b_{1}=(SS_{xy})/(SS_{x x})`, where `SS_{x x}=sum_{i=1}^{n}(x_{i}-bar{x})^{2}=sum_{i=1}^{n}x_{i}^{2}-((sum_{i=1}^{n}x_{i})^{2})/n` and `SS_{x y}=sum_{i=1}^{n}(x_{i}-bar{x})*(y_{i}-\bar{y})`

`=sum_{i=1}^{n}x_{i}y_{i}-((sum_{i=1}^{n}x_{i})*(sum_{i=1}^{n}y_{i}))/n`

In the given situation `n=5` and we have

`sum_{i=1}^{n}x_{i}=-3+0+3+5+3=8`, `sum_{i=1}^{n}y_{i}=7+4-2+2-3=8`, `sum_{i=1}^{n}x_{i}^{2}=9+0+9+25+9=52`, `sum_{i=1}^{n}y_{i}^{2}=49+16+4+4+9=82`, and `sum_{i=1}^{n}x_{i}y_{i}=-21+0-6+10-9=-26`.

Therefore, `SS_{x x}=52-(8^2)/5=(260-64)/5=196/5=39.2` and `SS_{x y}=-26-(8*8)/5=(-130-64)/5=-194/5=-38.8`.

This leads to a slope of `b_{1}=(-38.8)/39.2=-97/98approx -0.989796`.