Correlation

Correlation is a fundamental concept in statistical method used to assess the relationship between two variables. It quantifies the degree to which a change in one variable can be associated with a change in another. This blog will explore various aspects of correlation, including calculating Pearson’s correlation coefficient, understanding its meaning, and its relationship with regression and covariance.

1.0What is Correlation?

Fundamentally, correlation measures the strength and direction of the linear relationship between two variables. This relationship is often represented by a correlation coefficient, which ranges from –1 to 1. A correlation coefficient of 1 indicates a perfect positive correlation, while –1 indicates a perfect negative correlation. A coefficient of 0 signifies no correlation at all.

Correlation Coefficient Definition

The correlation coefficient, often denoted as r, is a statistical measure that describes the direction and strength of a relationship between two variables. The formula for calculating Pearson’s correlation coefficient is given by:

$r=\frac{n\left(\sum xy\right)-\left(\sum x\right)\left(\sum y\right)}{\sqrt{\left[n\sum x^2-{\left(\sum x\right)}^2\right]\left[n\sum y^2-{\left(\sum y\right)}^2\right]}}$

where:

• n is the number of paired scores,
• x and y are the two variables being compared.

This formula provides a clear and concise way to quantify the relationship, making it an essential tool in statistical analysis.

2.0Calculating Pearson’s Correlation Coefficient

Let’s consider an example where we want to calculate the correlation coefficient between the hours studied and the scores achieved by students in an exam. The data set is as follows:

1. Calculate the sums and products:

• $\sum x=2+3+4+5+6=20$
• $\sum y=55+65+70+75+80=345$
• $\sum xy=(2\times 55)+(3\times 65)+(4\times 70)+(5\times 75)+(6\times 80)=800$
• $\sum x^2=2^2+3^2+4^2+5^2+6^2=70$
• $\sum y^2=55^2+65^2+70^2+75^2+80^2=23725$

2. Apply the formula:

Substitute these values into the formula for Pearson's correlation coefficient.

Calculating r gives us a value that helps us understand the relationship between hours studied and exam scores.

3.0Coefficient of Correlation Explains

The coefficient of correlation explains how well one variable can predict another. For instance, in our example, a high positive r indicates that students who study more hours tend to score higher on the exam. 

4.0Correlation and Covariance

It is essential to differentiate between correlation and covariance. While both concepts deal with relationships between variables, covariance evaluates the extent to which two random variables vary together. In contrast, correlation standardizes this measure to produce a value between -1 and 1, making it easier to interpret.

5.0Correlation and Regression

Correlation is often confused with regression, but they serve different purposes. While correlation quantifies the strength and direction of a relationship, regression aims to model the relationship between variables to predict outcomes. For example, if we found a strong correlation between hours studied and exam scores, we could create a regression model to predict a student's score based on their study hours.

6.0Correlation Coefficient Meaning

The correlation coefficient meaning can be summarized as follows:

• 1: Perfect positive correlation
• 0: No correlation
• –1: Perfect negative correlation

Understanding this meaning is crucial for interpreting data across multiple domains, such as finance, healthcare, and social sciences.

7.0Solved Example on Correlation

Example 1: Let's consider the following data for five students, showing their study hours and the corresponding test scores they achieved.

Step 1: Calculate the necessary sums

We need to calculate the following values:

• $\sum X:\text{ Total of hours studied }$
• $\sum Y\text{ : Total of test scores }$
• $\sum XY\text{ : Sum of the product of }\mathrm{X}\text{ and }\mathrm{Y}$
• $\sum X^2:\text{ Sum of squared }X\text{ values }$
• $\sum Y^2:\text{ Sum of squared }Y\text{ values }$

Calculating these step by step:

• $\sum X=2+3+4+5+6=20$
• $\sum Y=50+60+70+80+90=350$
• $\sum XY=(2\times 50)+(3\times 60)+(4\times 70)+(5\times 80)+(6\times 90)$
• $=100+180+280+400+540=1500$
• $\sum X^2=2^2+3^2+4^2+5^2+6^2=4+9+16+25+36=90$
• $\sum Y^2=50^2+60^2+70^2+80^2+90^2=2500+3600+4900+6400+8100=25500$

Step 2: Calculate the Pearson Correlation Coefficient

Now, we can plug these values into the Pearson correlation coefficient formula:

$r=\frac{n\left(\sum XY\right)-\left(\sum X\right)\left(\sum Y\right)}{\sqrt{\left[n\sum X^2-{\left(\sum X\right)}^2\right]\left[n\sum Y^2-{\left(\sum Y\right)}^2\right]}}$

Where: n = 5 (number of students)

Substituting the calculated sums into the formula:

1. Calculate the numerator:

$n\left(\sum XY\right)-\left(\sum X\right)\left(\sum Y\right)=5(1500)-(20)(350)$

 = 7500 - 7000 = 500

2. Calculate the denominator:

First, calculate the components:

$n\sum X^2-{\left(\sum X\right)}^2=5(90)-(20{)}^2=450-400=50$

$n\sum Y^2-{\left(\sum Y\right)}^2=5(25500)-(350{)}^2=127500-122500=5000$

Now calculate the denominator:

$\sqrt{\left[n\sum X^2-{\left(\sum X\right)}^2\right]\left[n\sum Y^2-{\left(\sum Y\right)}^2\right]}=\sqrt{50\cdot 5000}=\sqrt{250000}=500$

3. Now substitute into the formula:

$r=\frac{500}{500}=1$

The correlation coefficient r = 1 indicates a perfect positive correlation between the number of hours studied and the test scores. This means that as the hours studied increase, the test scores also increase linearly. 

Example 2: Suppose we want to analyze the relationship between the number of hours students spend preparing for a test (Variable X) and their corresponding test scores (Variable Y). Here is the dataset of five students:

Find the correlation coefficient.

Solution: 

We will calculate the Pearson correlation coefficient to understand the relationship between these two variables.

Step 1: Calculate the Necessary Sums

We need to find the following values:

• $\sum X:\text{ Sum of hours studied }$
• $\sum Y:\text{ Sum of test scores }$
• $\sum XY\text{ : Sum of the product of }\mathrm{X}\text{ and }\mathrm{Y}$
• $\sum X^2:\text{ Sum of squares of }X\text{ values }$
• $\sum Y^2:\text{ Sum of squares of }\mathrm{Y}\text{ values }$

Calculations:

• $\sum X=2+4+6+8+10=30$
• $\sum Y=45+50+65+70+85=315$
• $\sum XY=(2\times 45)+(4\times 50)+(6\times 65)+(8\times 70)+(10\times 85)$

= 90 + 200 + 390 + 560 + 850 

= 2090

• $\sum X^2=2^2+4^2+6^2+8^2+10^2=4+16+36+64+100=220$
• $\sum Y^2=45^2+50^2+65^2+70^2+85^2=2025+2500+4225+4900+7225=20875$

Step 2: Use the Pearson Correlation Coefficient Formula

The formula for Pearson correlation coefficient is given by:

$r=\frac{n\left(\sum XY\right)-\left(\sum X\right)\left(\sum Y\right)}{\sqrt{\left[n\sum X^2-{\left(\sum X\right)}^2\right]\left[n\sum Y^2-{\left(\sum Y\right)}^2\right]}}$

Where: n = 5 (number of students)

Substitute the calculated values:

1. Calculate the Numerator:

$n\left(\sum XY\right)-\left(\sum X\right)\left(\sum Y\right)=5(2090)-(30)(315)$

= 10450 - 9450 = 1000

2. Calculate the Denominator:

First, calculate the components:

$n\sum X^2-{\left(\sum X\right)}^2=5(220)-(30{)}^2=1100-900=200$

$n\sum Y^2-{\left(\sum Y\right)}^2=5(20875)-(315{)}^2=104375-99225=5140$

Now calculate the denominator:

$\sqrt{[200\cdot 5140]}=\sqrt{1028000}=1013.90$

3. Calculate r:

$r=\frac{1000}{1013.90}\approx 0.986$

The correlation coefficient $r\approx 0.986$ indicates a strong positive correlation between the hours studied and the test scores. This means that as the number of hours spent studying increases, test scores also tend to increase significantly.

8.0Practice Questions on Correlation

1. The following table shows the relationship between the number of hours studied and the marks obtained by 7 students. Calculate the correlation coefficient to determine if there is a positive correlation between hours studied and marks obtained.

2. Given the data below, find the correlation coefficient between the monthly temperature (°C) and ice cream sales (in units).