Answer to Question #349821 in Statistics and Probability for Wilma

Question

In the given table on the below, solve for Pearson r and interpret the result

X 80,84,86,87,89,90,91,93,94,96

Y 78,83,80,84,89,90,88,91,93,96

Answer 1

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & X & Y & XY & X^2 & Y^2 \\ \hline & 80 & 78 & 6240 & 6400 & 6084 \\ \hdashline & 84 & 83 & 6972 & 7056 & 6889 \\ \hdashline & 86 & 80 & 6880 & 7396 & 6400 \\ \hdashline & 87 & 84 & 7308 & 7569 & 7056 \\ \hdashline & 89 & 89 & 7921 & 7921 & 7921 \\ \hdashline & 90 & 90 & 8100 & 8100 & 8100 \\ \hdashline & 91 & 88 & 8008 & 8281 & 7744 \\ \hdashline & 93 & 91 & 8463 & 8649 & 8281 \\ \hdashline & 94 & 93 & 8742 & 8836 & 8649 \\ \hdashline & 96 & 96& 9216 & 9216 & 9216 \\ \hdashline Sum= & 890 & 872 & 77850 & 79424 & 76340 \\ \end{array}$$

$$\bar{X}=\dfrac{1}{n}\sum _{i}X_i=\dfrac{890}{10}=89$$

$$\bar{Y}=\dfrac{1}{n}\sum _{i}Y_i=\dfrac{872}{10}=87.2$$

$$SS_{XX}=\sum_iX_i^2-\dfrac{1}{n}(\sum _{i}X_i)^2$$ $$=79424-\dfrac{890^2}{10}=214$$

$$SS_{YY}=\sum_iY_i^2-\dfrac{1}{n}(\sum _{i}Y_i)^2$$ $$=76340-\dfrac{(872)^2}{10}=301.6$$

$$SS_{XY}=\sum_iX_iY_i-\dfrac{1}{n}(\sum _{i}X_i)(\sum _{i}Y_i)$$ $$=77850-\dfrac{890(872)}{10}=242$$

$$r=\dfrac{SS_{XY}}{\sqrt{SS_{XX}SS_{YY}}}=\dfrac{242}{\sqrt{214(301.6)}}$$

$$=0.952561$$

Strong positive correlation