Answer to Question #325407 in Statistics and Probability for skum

Question

Fit a least square linear approximation Y=a+bX for the data X: -1, 0, 1, 2, 3 and Y: 0, 1, 2 , 9, 29  

Answer 1

### Theory ###

The least square linear approximation finds a and b coefficients by minimizing the sum of squares of differences between $a+bx_i$ and $y_i$ :

$S=\sum_{i=1}^n(a+bx_i-y_i)^2=min\Rarr\\ \Rarr\frac{\partial S}{\partial a}=0,\space \frac{\partial S}{\partial b}=0\\ \frac{\partial S}{\partial a}=\sum_{i=1}^n2(a+bx_i-y_i)=\\ =2(an+b\sum_{i=1}^{n}x_i-\sum_{i=1}^{n}y_i)=0\Rarr\\ \Rarr a=\frac{1}{n}(\sum_{i=1}^{n} y_i-b\sum_{i=1}^{n}x_i)\\ \frac{\partial S}{\partial b}=\sum_{i=1}^{n}2(a+bx_i-y_i)x_i=\\ 2(a\sum_{i=1}^{n}x_i+b\sum_{i=1}^{n}x_i^2-\sum_{i=1}^{n}x_iy_i)\Rarr\\ \Rarr [insert\space a=\frac{1}{n}(\sum_{i=1}^{n} y_i-b\sum_{i=1}^{n}x_i)]\Rarr\\ \Rarr \sum_{i=1}^{n}x_i\sum_{i=1}^{n}y_i+\\ +b[n\sum_{i=1}^{n}x_i^2-(\sum_{i=1}^{n}x_i)^2]-n\sum_{i=1}^{n}x_iy_i=0\Rarr\\ b=(n\sum_{i=1}^{n}x_iy_i-\sum_{i=1}^{n}x_i\sum_{i=1}^{n}y_i)/\\ /(n\sum_{i=1}^{n}x_i^2-(\sum_{i=1}^{n}x_i)^2)$

### Solution ###

$\sum_{i=1}^{n}x_i=-1+0+1+2+3=5\\ \sum_{i=1}^{n}x_i^2=1+0+1+4+9=15\\ \sum_{i=1}^{n}y_i=0+1+2+9+29=41\\ \sum_{i=1}^{n}x_iy_i=0+0+2+18+87=107\\ b=(5\cdot107-5\cdot41)/(5\cdot15-25)=6.6\\ a=(41-6.6\cdot5)/5=1.6\\ Answer:y=1.6+6.6x$