Answer to Question #340250 in Statistics and Probability for skum

Question

Fit the parabola is Y=a X^2+bX+c for the data x: 1, 3, 4 , 6 ,8 ,9, 11, 14 and y: 1,2, 4, 4, 5, 7, 8, 9.

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c:c} & X & Y & XY & X^2 & X^3 & X^4 & X^2Y\\ \hline & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \hdashline & 3 & 2 & 6 & 9 & 27 & 81 & 18 \\ \hdashline & 4 & 4 & 16 & 16 & 64 & 256 & 64 \\ \hdashline & 6 & 4 & 24 & 36 & 216 & 1296 & 144 \\ \hdashline & 8 & 5 & 40 & 64 & 512 & 4096 & 320 \\ \hdashline & 9 & 7 & 63 & 81 & 729 & 6561 & 567 \\ \hdashline & 11 & 8 & 88 & 121 & 1331 & 14641 & 968 \\ \hdashline & 14 & 9 & 126 & 196 & 2744 & 38416 & 1764 \\ \hdashline Sum= & 56 & 40 & 364 & 524 & 5624 & 65348 & 3846 \\ \hdashline \end{array}$$

$$\bar{X}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nX_i=\dfrac{56}{8}=7$$

$$\bar{Y}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nY_i=\dfrac{40}{8}=5$$

$$\bar{X^2}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nX_i^2=\dfrac{524}{8}=65.5$$

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

$$=524-\dfrac{56^2}{8}=132$$

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

$$=364-\dfrac{56(40)}{8}=84$$

$$SS_{XX^2}=\displaystyle\sum_{i=1}^nX_i^3-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nX_i)(\displaystyle\sum_{i=1}^nX_i^2)$$

$$=5624-\dfrac{56(524)}{8}=1956$$

$$SS_{X^2X^2}=\displaystyle\sum_{i=1}^nX_i^4-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nX_i^2)^2$$

$$=65348-\dfrac{(524)^2}{8}=31026$$

$$SS_{X^2Y}=\displaystyle\sum_{i=1}^nX_i^2Y_i-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nX_i^2)(\displaystyle\sum_{i=1}^nY_i)$$

$$=3846-\dfrac{524(40)}{8}=1226$$

$$b=\dfrac{84(31026)-1226(1956)}{132(31026)-(1956)^2}=0.772286$$

$$a=\dfrac{1226(132)-84(1956)}{132(31026)-(1956)^2}=-0.009173$$

$$c=5-0.772286(7)+0.009173(65.5)=0.194830$$

$$\hat{y}=-0.009173x^2+0.772286x+0.194830$$