Question

Which equation of the least squares regression line most closely matches the data set? x y
2 100
4 83
6 50
8 35
10 23




yˆ=−0.31x+37.3

yˆ=−2.3x+97.9

yˆ=−10.1x+118.8

yˆ=−0.09x+11.5

Answer 1

 x      y        x^2      xy
 
2    100        4      200
4     83        16     332
6     50        36     300
8     35        64     280
10   23      100     230

∑x = 30; ∑y = 291; ∑x^2 = 220; ∑xy = 1,342; n = 5

$b= \frac{n\Sigma(xy)-(\Sigma x)(\Sigma y)}{n(\Sigma x^2) - (\Sigma x)^2} = \frac{5(1,342)-(30)(291)}{5(220)-(30)^2} \\ = \frac{6,710-8,730}{1,100-900} = \frac{-2,020}{200} =-10.1 \\ a = \frac{\Sigma y-b\Sigma x}{n} = \frac{291-(-10.1)(30)}{5} = \frac{291+303}{5}= \frac{594}{5} =118.8$

The least squares regression line equation is given by y = bx + a = -10.1x + 118.9.

Answer 2

The least-squares regression line works by making the square of the error. The equation for the regression line can be written as,  y = -10.1x+118.8.

What is the least-squares regression line?

The least-square regression line is the regression line from which the lines have the least vertical distance between the data points and the regression line.

Given to us

 x    y

02   100

04    83

06    50

08    35

10     23

In order to find the regression line of the given points, we need to find the equation of the regression line, therefore,

y=mx+c

where m is the gradient of the line, and c is the y-intercept of the line.

In order to find the slope and the y-intercept of the regression line, we need to calculate the following data,

       x                y                     x²                  xy

       2                100                 4                   200

       4                83                  16                  332

       6                50                  36                 300

       8                35                  64                 280

      10                23                100                 230

∑x= 30      ∑y=  291        ∑x²=220         ∑xy=1342

Also, the value of n is 5, since there are 5 data points.

The slope of the regression line,

$\text{Slope of the line}=m = \dfrac{N(\sum xy)-\sum x \sum y}{N\sum x^2 - (\sum x)^2}$

Substitute the values,

$\text{Slope of the line}=m = \dfrac{5(1342)-(30 \times 291)}{5(220) - (30^2)}\\\\\\\text{Slope of the line}=m =-10.1$

The y-intercept of the regression line,

$\text{Y-intercept of y}= C = \dfrac{\sum y-m\sum x}{N}$

Substitute the values,

$\text{Y-intercept of y}= C = \dfrac{291-(-10.1)(30)}{5}\\\\\\\text{Y-intercept of y}= C = \dfrac{291+303}{5}\\\\\\\text{Y-intercept of y}= C = 118.8$

Thus, the equation for the regression line can be written as,

y = -10.1x+118.8.

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