Question #282141

Compute the correlation coefficient of the data below.




*number of years vs. selling price



X 1 3 5 7 9



y 27 23 25 20 15





1
Expert's answer
2021-12-23T08:47:54-0500
XYXYX2Y2127271729323699529525125256257201404940091513581225Sum=251104961652508\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & X & Y & XY & X^2 & Y^2 \\ \hline & 1 & 27 & 27 & 1 & 729 \\ \hdashline & 3 & 23 & 69 & 9 & 529 \\ \hdashline & 5 & 25 & 125 & 25 & 625 \\ \hdashline & 7 & 20 & 140 & 49 & 400 \\ \hdashline & 9 & 15 & 135 & 81 & 225 \\ \hdashline Sum= & 25 & 110 & 496 & 165 & 2508 \\ \hdashline \end{array}

Xˉ=1ni=1nXi=255=5\bar{X}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nX_i=\dfrac{25}{5}=5

Yˉ=1ni=1nYi=1105=22\bar{Y}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nY_i=\dfrac{110}{5}=22

SSXX=i=1nXi21n(i=1nXi)2=1652525=40SS_{XX}=\displaystyle\sum_{i=1}^nX_i^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nX_i)^2=165-\dfrac{25^2}{5}=40

SSYY=i=1nYi21n(i=1nYi)2=250811025=88SS_{YY}=\displaystyle\sum_{i=1}^nY_i^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nY_i)^2=2508-\dfrac{110^2}{5}=88

SSXY=i=1nXiYi1n(i=1nXi)(i=1nYi)SS_{XY}=\displaystyle\sum_{i=1}^nX_iY_i-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nX_i)(\displaystyle\sum_{i=1}^nY_i)

=49625(110)5=54=496-\dfrac{25(110)}{5}=-54

Correlation coefficient:


r=SSXYSSXXSSYY=544088r=\dfrac{SS_{XY}}{\sqrt{SS_{XX}}\sqrt{SS_{YY}}}=\dfrac{-54}{\sqrt{40}\sqrt{88}}

0.91017\approx -0.91017




0.7<r<11.0-0.7<r<-11.0

r=0.91017,r=-0.91017, strong negative correlation.


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