How to calculate correlation coefficient by hand fast

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.

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.

Yes · 0 No · 0 Add Comment · 0

Add ·

Report