Question

(a) Sketch the line that appears to be the best fit for the given points.

11. [-13.22 Points] DETAILS LARLINALG8 2.6.017. Consider the following. 5 (1,5) 4 (2, 4) 3 2 (2, 2) (3, 1) - Х - 1 2 3 4 -14 (a) Sketch the line that appears to be the best fit for the given points.

(b) Find the least squares regression line. y(x)=

(c) Calculate the sum of squared error.

Answer 1

(a)

The figure below illustrates a scatter plot and a line of best fit.

The equation of the line is, , where m is the slope of the line, and c is the y-intercept.

y = mx + c

The line passes through the points, A(1,5), and, B(3,1).

Slope, m=$\frac{5-1}{1-3}$

i.e., m=-2

Therefore the equation of the line becomes, y=-2x+c

The above line passes through the point, A(1,5). Using the co-ordinates of the point A(1,5), in the above, equation of the line, we have,

5=(-2)(1)+c

i.e., c=7

Substituting the value of, c=7 in the above, equation of the line, we have, the equation of the line,

$y=-2x+7$

The equation of the line of best fit, is, $y=-2x+7$ .

7 CO -5 +3 N -4 -3 -2 -1 0 2 3 4 5 6 00 9 10 1

(b)

The least squares regression line is given by the formula,

$\hat{y}=b_0+b_1x$,

Its slope
bi
and y-intercept
bo
are computed using the formulas, $b_1=\frac{SS_{xy}}{SS_{xx}}$ , and, $b_0=\bar{y}-b_1\bar{x}$ , where,

$SS_{xx}=\sum x^2-\frac{1}{n}\left ( \sum x \right )^2$, $SS_{xy}=\sum xy-\frac{1}{n}\left ( \sum x \right )\left ( \sum y \right )$

T
is the mean of all the x-values, $\bar{y}$ is the mean of all the y-values, and n is the number of pairs in the data set.

	x	y	x^2	xy
	1	5	1	5
	2	2	4	4
	2	4	4	8
	3	1	9	3
$\sum$	8	12	18	20

In the last line of the table above, we have the sum of the numbers in each column. Using them we compute:

$SS_{xx}=\sum x^2-\frac{1}{n}\left ( \sum x \right )^2=18-\frac{1}{4}(8)^2=2$

$SS_{xy}=\sum xy-\frac{1}{n}\left ( \sum x \right )\left ( \sum y \right )=20-\frac{1}{4}(8)(12)=-4$

$\bar{x}=\frac{\sum x}{n}=\frac{8}{4}=2$

$\bar{y}=\frac{\sum y}{n}=\frac{12}{4}=3$

and we now have,

$b_1=\frac{SS_{xy}}{SS_{xx}}=\frac{-4}{2}=-2$

Using the value of $b_1=-2$ , $\bar{x}=2$ , and, $\bar{y}=3$ in the equation of the least squares regression line, we have, $b_0=\bar{y}-b_1\bar{x}=3-(-2)(2)=7$

We now have the least squares regression line for given data and it is, $\hat{y}=7-2x$

(c)

We now calculate the sum of squared error.

$\textup{SSE}=\sum_{i=1}^{n}(x_i-\bar{x})^2$

	T		$x_i-\bar{x}=x_i-2$	$(x_i-\bar{x})^2$
	1	5	-1	1
	2	2	0	0
	2	4	0	0
	3	1	1	1
$\sum$	8	12	0	2

$\textup{SSE}=\frac{(x_1-2)^2+(x_2-2)^2+(x_3-2)^2+(x_4-2)^2}{4}$

i.e., $\textup{SSE}=\frac{(1-2)^2+(2-2)^2+(2-2)^2+(3-2)^2}{4}$

i.e., $\textup{SSE}=\frac{1+0+0+1}{4}$

i.e., $\textup{SSE}=\frac{2}{4}=\frac{1}{2}$