Finite Math Examples

\begin{matrix} x & y 0 & 2 1 & 2 3 & 2 4 & 2 5 & 2 \end{matrix}

$\begin{array}{c} x & y \\ 0 & 2 \\ 1 & 2 \\ 3 & 2 \\ 4 & 2 \\ 5 & 2 \end{array}$

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 + 3 + 4 + 5

$\sum_{}^{} x = 0 + 1 + 3 + 4 + 5$

Step 3

Simplify the expression.

\sum x = 13

$\sum_{}^{} x = 13$

Step 4

Sum up the

y

$y$ values.

\sum y = 2 + 2 + 2 + 2 + 2

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

Step 5

Simplify the expression.

\sum y = 10

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

Step 6

Sum up the values of

x \cdot y

$x \cdot y$ .

\sum x y = 0 \cdot 2 + 1 \cdot 2 + 3 \cdot 2 + 4 \cdot 2 + 5 \cdot 2

$\sum_{}^{} x y = 0 \cdot 2 + 1 \cdot 2 + 3 \cdot 2 + 4 \cdot 2 + 5 \cdot 2$

Step 7

Simplify the expression.

\sum x y = 26

$\sum_{}^{} x y = 26$

Step 8

Sum up the values of

x^{2}

$x^{2}$ .

\sum x^{2} = {(0)}^{2} + {(1)}^{2} + {(3)}^{2} + {(4)}^{2} + {(5)}^{2}

$\sum_{}^{} x^{2} = {(0)}^{2} + {(1)}^{2} + {(3)}^{2} + {(4)}^{2} + {(5)}^{2}$

Step 9

Simplify the expression.

\sum x^{2} = 51

$\sum_{}^{} x^{2} = 51$

Step 10

Sum up the values of

y^{2}

$y^{2}$ .

\sum y^{2} = {(2)}^{2} + {(2)}^{2} + {(2)}^{2} + {(2)}^{2} + {(2)}^{2}

$\sum_{}^{} y^{2} = {(2)}^{2} + {(2)}^{2} + {(2)}^{2} + {(2)}^{2} + {(2)}^{2}$

Step 11

Simplify the expression.

\sum y^{2} = 20

$\sum_{}^{} y^{2} = 20$

Step 12

Fill in the computed values.

r = \frac{5 (26) - 13 \cdot 10}{\sqrt{5 (51) - {(13)}^{2}} \cdot \sqrt{5 (20) - {(10)}^{2}}}

$r = \frac{5 (26) - 13 \cdot 10}{\sqrt{5 (51) - {(13)}^{2}} \cdot \sqrt{5 (20) - {(10)}^{2}}}$

Step 13

Simplify the expression.

r = N a N

$r = N a N$

Step 14

Find the critical value for a confidence level of

0

$0$ and

5

$5$ degrees of freedom.

t = 3.18244628

$t = 3.18244628$