# Examples of Correlation

Calculate and analyze the correlation coefficient between the number of study hours and the number of sleeping hours of different students.

 Number of Study Hours 2 4 6 8 10 Number of Sleeping Hours 10 9 8 7 6

Solution:

The necessary calculations are given below:

 $X$ $Y$ $\left( {X – \overline X } \right)$ $\left( {Y – \overline Y } \right)$ $\left( {X – \overline X } \right)\left( {Y – \overline Y } \right)$ ${\left( {X – \overline X } \right)^2}$ ${\left( {Y – \overline Y } \right)^2}$ 2 10 -4 +2 -8 16 4 4 9 -2 +1 -2 4 1 6 8 0 0 0 0 0 8 7 +2 -1 -2 4 1 10 6 +4 -2 -8 16 1 $\begin{gathered} \sum X \\ = 30 \\ \end{gathered}$ $\begin{gathered} \sum Y \\ = 40 \\ \end{gathered}$ $\begin{gathered} \sum \left( {X – \overline X } \right) \\ = 0 \\ \end{gathered}$ $\begin{gathered} \sum \left( {Y – \overline Y } \right) \\ = 0 \\ \end{gathered}$ $\begin{gathered} \sum \left( {X – \overline X } \right)\left( {Y – \overline Y } \right) \\ = – 20 \\ \end{gathered}$ $\begin{gathered} \sum {\left( {X – \overline X } \right)^2} \\ = 40 \\ \end{gathered}$ $\begin{gathered} \sum {\left( {Y – \overline Y } \right)^2} \\ = 10 \\ \end{gathered}$

$\overline X = \frac{{\sum X}}{n} = \frac{{30}}{5} = 6$   and   $\overline Y = \frac{{\sum Y}}{n} = \frac{{40}}{5} = 8$
${r_{XY}} = \frac{{\sum \left( {X – \overline X } \right)\left( {Y – \overline Y } \right)}}{{\sqrt {\sum {{\left( {X – \overline X } \right)}^2}\sum {{\left( {Y – \overline Y } \right)}^2}} }} = \frac{{ – 20}}{{20}} = – 1$

There is a perfect negative correlation between the number of study hours and the number of sleeping hours.

Example:

From the following data, compute the coefficient of correlation between $X$ and $Y$:

 $X$ Series $X$ Series Number of Items 15 15 Arithmetic Mean 25 18 Sum of Square Deviations 136 138

The summation of the products of the deviations of $X$ and $X$ series from their arithmetic means = 122.

Solution:

Here $n = 15,{\text{ }}\overline X = 25,{\text{ }}\overline Y = 18,{\text{ }}\sum {\left( {X – \overline X } \right)^2} = \sum {\left( {Y – \overline Y } \right)^2} = 138$
$\sum {\left( {X – \overline X } \right)^2}{\left( {Y – \overline Y } \right)^2} = 122$ and hence

$r = \frac{{\sum \left( {X – \overline X } \right)\left( {Y – \overline Y } \right)}}{{\sqrt {\sum {{\left( {X – \overline X } \right)}^2}\sum {{\left( {Y – \overline Y } \right)}^2}} }} = \frac{{122}}{{\sqrt {\left( {136} \right)\left( {138} \right)} }} = \frac{{122}}{{137}} = 0.89$