Question

The total of the multiplication deviation of X and Y = 3044
Number of pairs of observations = 10
Total of the deviation of X = −20
Total of the deviation Y = −170
Total of squares of deviation of X = 2264
Total of the squares of deviation of Y = 8288
Find out Karl Pearson' s coefficient of correlation when assumed mean of X and Y are 82 and 68 respectively.

Answer 1

Given:
n = 10
Σd_xd_y = 3044
Σd_x = −17 assumed (instead of −170 see note below)
Σd_y = −20
Σd_x² = 2264
Σd_y² = 8288

$$\begin{gathered} r=\frac{N\sum d_x{d}_y-\left(\sum d_x\sum d_y\right)}{\sqrt{N\sum d_x^2-{\left(\sum d_x\right)}^2}\sqrt{N\sum d_y^2-{\left(\sum d_y\right)}^2}} \\ r=\frac{10\times 3044-\left[\left(-17\right)\times \left(-20\right)\right]}{\sqrt{10\times 2264-{\left(-17\right)}^2}\sqrt{10\times 8288-{\left(-20\right)}^2}} \\ r=\frac{30440-340}{\sqrt{22351}\sqrt{82480}} \\ r=\frac{30100}{149.502\times 287.193} \\ r=\frac{30100}{42935.927} \\ r=0.70 \end{gathered}$$

Hence, Karl Pearson's coefficient of correlation is 0.7.

Note:

1. If we take Σd_x equal to −170, then the calculation of Karl Pearson's coefficient of correlation is as follows:

$$\begin{gathered} r=\frac{N\sum d_x{d}_y-\left(\sum d_x\sum d_y\right)}{\sqrt{N\sum d_x^2-{\left(\sum d_x\right)}^2}\sqrt{N\sum d_y^2-{\left(\sum d_y\right)}^2}} \\ r=\frac{10\times 3044-\left[\left(-170\right)\times \left(-20\right)\right]}{\sqrt{10\times 2264-{\left(-170\right)}^2}\sqrt{10\times 8288-{\left(-20\right)}^2}} \\ r=\frac{30440-3400}{\sqrt{22640-28900}\sqrt{82880-400}} \\ r=\frac{30100}{\sqrt{\overline{)-6260}}\sqrt{82480}} \end{gathered}$$

Since, the square root of a negative number is not defined among the set of real numbers, thus we have assumed Σd_x as −17.

2. The answer as per the book is 0.78 and the answer as per our calculation is 0.70. The difference in the answer may be due to the assumption (i.e. Σd_x = −17) made by us.