Finite Math Examples

\begin{matrix} x & y 0 & 1 1 & 4 2 & 16 3 & 64 4 & 256 \end{matrix}

$\begin{array}{c} x & y \\ 0 & 1 \\ 1 & 4 \\ 2 & 16 \\ 3 & 64 \\ 4 & 256 \end{array}$

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}}}

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

Step 2

Sum up the

x

$x$ values.

\sum x = 0 + 1 + 2 + 3 + 4

$\sum_{}^{} x = 0 + 1 + 2 + 3 + 4$

Step 3

Simplify the expression.

\sum x = 10

$\sum_{}^{} x = 10$

Step 4

Sum up the

y

$y$ values.

\sum y = 1 + 4 + 16 + 64 + 256

$\sum_{}^{} y = 1 + 4 + 16 + 64 + 256$

Step 5

Simplify the expression.

\sum y = 341

$\sum_{}^{} y = 341$

Step 6

Sum up the values of

x \cdot y

$x \cdot y$ .

\sum x y = 0 \cdot 1 + 1 \cdot 4 + 2 \cdot 16 + 3 \cdot 64 + 4 \cdot 256

$\sum_{}^{} x y = 0 \cdot 1 + 1 \cdot 4 + 2 \cdot 16 + 3 \cdot 64 + 4 \cdot 256$

Step 7

Simplify the expression.

\sum x y = 1252

$\sum_{}^{} x y = 1252$

Step 8

Sum up the values of

x^{2}

$x^{2}$ .

\sum x^{2} = {(0)}^{2} + {(1)}^{2} + {(2)}^{2} + {(3)}^{2} + {(4)}^{2}

$\sum_{}^{} x^{2} = {(0)}^{2} + {(1)}^{2} + {(2)}^{2} + {(3)}^{2} + {(4)}^{2}$

Step 9

Simplify the expression.

\sum x^{2} = 30

$\sum_{}^{} x^{2} = 30$

Step 10

Sum up the values of

y^{2}

$y^{2}$ .

\sum y^{2} = {(1)}^{2} + {(4)}^{2} + {(16)}^{2} + {(64)}^{2} + {(256)}^{2}

$\sum_{}^{} y^{2} = {(1)}^{2} + {(4)}^{2} + {(16)}^{2} + {(64)}^{2} + {(256)}^{2}$

Step 11

Simplify the expression.

\sum y^{2} = 69905

$\sum_{}^{} y^{2} = 69905$

Step 12

Fill in the computed values.

r = \frac{5 (1252) - 10 \cdot 341}{\sqrt{5 (30) - {(10)}^{2}} \cdot \sqrt{5 (69905) - {(341)}^{2}}}

$r = \frac{5 (1252) - 10 \cdot 341}{\sqrt{5 (30) - {(10)}^{2}} \cdot \sqrt{5 (69905) - {(341)}^{2}}}$

Step 13

Simplify the expression.

r = 0.83455433

$r = 0.83455433$

Step 14

Find the critical value for a confidence level of

0

$0$ and

5

$5$ degrees of freedom.

t = 3.18244628

$t = 3.18244628$