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**Answer**

**How to calculate correlation coefficient by hand fast**

1. The initial step is to examine the data pairs.

This can be achieved by putting them horizontally/ vertically in a table labelled x and y. For instance, let us use the following table.

x | y |

2 | 2 |

1 | 4 |

3 | 6 |

2. Calculate the mean of x

Mean is calculated by adding the values of x and dividing them by the number of values.

In our above example we have the mean of x as;

${u}_{x}=(2+1+3)/3$

${u}_{x}=\frac{6}{3}$

=2

3. Calculate the mean of y

Mean is calculated by adding the values of y and dividing them by the number of values.

In our above example we have the mean of y as;

${u}_{y}=(2+4+6)/3$

${u}_{y}=\frac{12}{3}$

=4

4. Determining the standard deviation of x

We use the formula;

${\delta}_{x}=\sqrt{\frac{1}{n-1}\sum _{}{(x-{u}_{x})}^{2}}$

${\delta}_{x}=\sqrt{\frac{1}{3-1}\times ({(2-2)}^{2}}+{(1-2)}^{2}+{(3-2)}^{2}$

${\delta}_{x}=\sqrt{\frac{1}{2}\times (0+1+1)}$

${\delta}_{x}=\sqrt{1}$

${\delta}_{x}=1$

5. Calculating the Standard deviation of y;

We use the formula;

${\delta}_{y}=\sqrt{\frac{1}{n-1}\sum _{}{(y-{u}_{y})}^{2}}$

${\delta}_{y}=\sqrt{\frac{1}{3-1}\times ({(2-4)}^{2}}+{(4-4)}^{2}+{(6-4)}^{2}$

${\delta}_{y}=\sqrt{\frac{1}{2}\times (0+0+4)}$

${\delta}_{y}=\sqrt{4}$

${\delta}_{y}=2$

6. Using the basic formula for correlation coefficient

$\mathcal{l}=\left(\frac{1}{n-1}\right)\sum _{}\left(\frac{x-{u}_{x}}{{\sigma}_{x}}\right)\times \left(\frac{y-{u}_{y}}{{\sigma}_{y}}\right)$

$\mathcal{l}=\left(\frac{1}{3-1}\right)\times \left(\frac{2-2}{1}\right)\times \left(\frac{2-4}{2}\right)+\left(\frac{1-2}{1}\right)\times \left(\frac{4-4}{2}\right)+\left(\frac{3-2}{1}\right)\times \left(\frac{6-4}{2}\right)$

$\mathcal{l}=\left(\frac{1}{2}\right)\times (0+0+1)$

$\mathcal{l}=0.5$

The value 0.5 means a low positive correlation.

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