Answer to Question #349784 in Statistics and Probability for Monay

Question #349784

THE FOLLOWING ARE THE HEIGHT IN CENTIMETER AND HEIGHTS IN KILOGRAM OF 5 TEACHERS IN A CERTAIN SCHOOL.DETERMINE THE RELATIONSHIP BETWEEN THE HEIGHT (CM) AND WEIGHT (KG) OF THE TEACHERS


TEACHER A,B,C,D,E,F,G


HEIGHT (CM) X 163,160,168,159,170


WEIGHT (KG) Y 52,50,64,51,69

1
Expert's answer
2022-06-13T11:19:27-0400

In order to compute the regression coefficients, the following table needs to be used:


XYXYX2Y216352847626559270416050800025600250016864107522822440961595181092528126011706911730289004761Sum=8202864706713457416662\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & X & Y & XY & X^2 & Y^2 \\ \hline & 163 & 52 & 8476 & 26559 & 2704 \\ \hdashline & 160 & 50 & 8000 & 25600 & 2500 \\ \hdashline & 168 & 64 & 10752 & 28224 & 4096 \\ \hdashline & 159 & 51 & 8109 & 25281 & 2601 \\ \hdashline & 170 & 69 & 11730 & 28900 & 4761 \\ \hdashline Sum= & 820 & 286 & 47067 & 134574 & 16662 \\ \hdashline \end{array}




Xˉ=1niXi=8205=164\bar{X}=\dfrac{1}{n}\sum _{i}X_i=\dfrac{820}{5}=164




Yˉ=1niYi=2865=57.2\bar{Y}=\dfrac{1}{n}\sum _{i}Y_i=\dfrac{286}{5}=57.2




SSXX=iXi21n(iXi)2SS_{XX}=\sum_iX_i^2-\dfrac{1}{n}(\sum _{i}X_i)^2=13457482025=94=134574-\dfrac{820^2}{5}=94




SSYY=iYi21n(iYi)2SS_{YY}=\sum_iY_i^2-\dfrac{1}{n}(\sum _{i}Y_i)^2=16662(286)25=302.8=16662-\dfrac{(286)^2}{5}=302.8




SSXY=iXiYi1n(iXi)(iYi)SS_{XY}=\sum_iX_iY_i-\dfrac{1}{n}(\sum _{i}X_i)(\sum _{i}Y_i)=47067820(286)5=163=47067-\dfrac{820(286)}{5}=163




r=SSXYSSXXSSYY=16394(302.8)r=\dfrac{SS_{XY}}{\sqrt{SS_{XX}SS_{YY}}}=\dfrac{163}{\sqrt{94(302.8)}}




=0.9662=0.9662


Strong positive correlation



m=slope=SSXYSSXX=16394=1.7340m=slope=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{163}{94}=1.7340n=YˉmXˉ=57.216394(164)=227.1830n=\bar{Y}-m\bar{X}=57.2-\dfrac{163}{94}(164)=-227.1830



The regression equation is:


y=227.1830+1.7340xy=-227.1830+1.7340x

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