Answer to Question #211617 in Statistics and Probability for Vivek

Question

Following is the distribution of marks (out of 25) obtained by 10 students in Physics and Mathematics. XY Y X No. 1 2 3 4 Mathematics (Y) 21 23 14 23 17 18 16 19 20 23 194 Physics (X) 18 20 11 20 14 15 13 16 17 20 164 5 6 324 400 121 400 196 225 169 256 289 400 2780 441 529 196 529 289 324 256 361 400 529 3854 378 460 154 460 238 270 208 304 480 460 3412 7 8 9 10 Total Draw a scatter diagram for all the 10 students and calculate the correlation between marks of Physics and Mathematics.

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} & X & Y & XY & X^2 & Y^2 \\ \hline & 18 & 21 & 378 & 324 & 441 \\ & 20 & 23 & 460 & 400 & 529 \\ & 11 & 14 & 154 & 121 & 196 \\ & 20 & 23 & 460 & 400 & 529 \\ & 14 & 17 & 238 & 196 & 289 \\ & 15 & 18 & 270 & 225 & 324 \\ & 13 & 16 & 208 & 169 & 256 \\ & 16 & 19 & 304 & 256 & 361 \\ & 17 & 20 & 340 & 289 &400 \\ & 20 & 23 & 460 & 400 & 529 \\ Total= & 164 & 194 & 3272 & 2780 & 3854 \\ \end{array}$$

$$\bar{X}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nX_i=16.4$$

$$\bar{Y}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nY_i=19.4$$

$$SS_{XX}=\displaystyle\sum_{i=1}^nX_i^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nX_i)^2=90.4$$

$$SS_{XY}=\displaystyle\sum_{i=1}^nX_iY_i-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nX_i)(\displaystyle\sum_{i=1}^nY_i) =90.4$$

$$SS_{YY}=\displaystyle\sum_{i=1}^nY_i^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nY_i)^2=90.4$$

$$m=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{90.4}{90.4}=1$$

$$n=\bar{Y}-m\bar{X}=19.4-1(16.4)=3$$

The regression equation is:

$$Y=3+X$$

Correlation coefficient: $r=\dfrac{SS_{XY}}{\sqrt{SS_{XX}}\sqrt{SS_{YY}}}=\dfrac{90.4}{\sqrt{90.4}\sqrt{90.4}}=1$

Strong positive correlation.