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

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