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
\[r = \frac{{\sum \left( {X – \overline X } \right)\left( {Y – \overline Y } \right)}}{{\sqrt {\sum {{\left( {X – \overline X } \right)}^2}\sum {{\left( {Y – \overline Y } \right)}^2}} }} = \frac{{122}}{{\sqrt {\left( {136} \right)\left( {138} \right)} }} = \frac{{122}}{{137}} = 0.89\]
Muhammad Haseeb
December 30 @ 8:22 pm
For correlation variables should have different unit but in above case units are same