Linear Algebra Examples

$\begin{array}{c} x & y \\ 0 & 4 \\ 1 & 12 \\ 2 & 36 \\ 3 & 108 \end{array}$

Step 1

The linear correlation coefficient measures the relationship between the paired values in a sample.

$r = \frac{n (\sum_{}^{} x y) - \sum_{}^{} x \sum_{}^{} y}{\sqrt{n (\sum_{}^{} x^{2}) - {(\sum_{}^{} x)}^{2}} \cdot \sqrt{n (\sum_{}^{} y^{2}) - {(\sum_{}^{} y)}^{2}}}$

Step 2

Sum up the $x$ values.

$\sum_{}^{} x = 0 + 1 + 2 + 3$

Step 3

Simplify the expression.

$\sum_{}^{} x = 6$

Step 4

Sum up the $y$ values.

$\sum_{}^{} y = 4 + 12 + 36 + 108$

Step 5

Simplify the expression.

$\sum_{}^{} y = 160$

Step 6

Sum up the values of $x \cdot y$ .

$\sum_{}^{} x y = 0 \cdot 4 + 1 \cdot 12 + 2 \cdot 36 + 3 \cdot 108$

Step 7

Simplify the expression.

$\sum_{}^{} x y = 408$

Step 8

Sum up the values of $x^{2}$ .

$\sum_{}^{} x^{2} = {(0)}^{2} + {(1)}^{2} + {(2)}^{2} + {(3)}^{2}$

Step 9

Simplify the expression.

$\sum_{}^{} x^{2} = 14$

Step 10

Sum up the values of $y^{2}$ .

$\sum_{}^{} y^{2} = {(4)}^{2} + {(12)}^{2} + {(36)}^{2} + {(108)}^{2}$

Step 11

Simplify the expression.

$\sum_{}^{} y^{2} = 13120$

Step 12

Fill in the computed values.

$r = \frac{4 (408) - 6 \cdot 160}{\sqrt{4 (14) - {(6)}^{2}} \cdot \sqrt{4 (13120) - {(160)}^{2}}}$

Step 13

Simplify the expression.

$r = 0.91651513$