Question

find the regression analysis, what is the value of a,b, and what is expected y if x=18

X: 17 12 10 20

Y: 10 10 8 12

Answer 1

$$\begin{matrix} & X & Y & XY & X^2 & Y^2 \\ & 17 & 10 & 170 & 289 & 100 \\ & 12 & 10 & 120 & 144 & 100 \\ & 10 & 8 & 80 & 100 & 64 \\ & 20 & 12 & 240 & 400 & 144 \\ Sum =& 59 & 40 & 610 & 933 & 408 \\ \end{matrix}$$

$$\bar{X}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nX_i=\dfrac{59}{4}=14.75$$

$$\bar{Y}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nY_i=\dfrac{40}{4}=10$$

$$SS_{XX}=\displaystyle\sum_{i=1}^nX_i^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nX_i)^2$$

$$=933-\dfrac{(59)^2}{4}=62.75$$

$$SS_{YY}=\displaystyle\sum_{i=1}^nY_i^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nY_i)^2$$

$$=408-\dfrac{(40)^2}{4}=8$$

$$SS_{XY}=\displaystyle\sum_{i=1}^nX_iY_i-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nX_i)(\displaystyle\sum_{i=1}^nY_i)$$

$$=610-\dfrac{59(40)}{4}=20$$

The regression coefficients (the slope $m,$ and the y-intercept  $n$) are obtained as follows:

$$m=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{20}{62.75}=0.318725$$

$$n=\bar{Y}-m\bar{X}=10-\dfrac{20}{62.75}(14.75)=5.298805$$

We find that the regression equation is:

$$Y=5.298805+0.318725X$$

Correlation coefficient

$$r=\dfrac{SS_{XY}}{\sqrt{SS_{XX}}\sqrt{SS_{YY}}}=\dfrac{20}{\sqrt{62.75}\sqrt{8}}=0.892644$$

Strong positive correlation.

$$r^2=\dfrac{(20)^2}{62.75(8)}=0.796813$$

$$Y(18)=5.298805+0.318725(18)=11.035855$$