Answer to Question #349275 in Statistics and Probability for GAB

Question

The Following are the height in centimeters and weights in kilogram of 5 teachers in a certain schools.Determine they relationship between the height (cm) and weight (kg)

Teacher A B C D E

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

Weight (kg) Y 52, 50, 64, 51, 69

Answer 1

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

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

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

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

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

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

$$SS_{XY}=\sum_iX_iY_i-\dfrac{1}{n}(\sum _{i}X_i)(\sum _{i}Y_i)$$ $$=47067-\dfrac{820(286)}{5}=163$$

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

$$=0.9662$$

Strong positive correlation

$$m=slope=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{163}{94}=1.7340$$

$$n=\bar{Y}-m\bar{X}=57.2-\dfrac{163}{94}(164)=-227.1830$$

The regression equation is:

$$y=-227.1830+1.7340x$$