Correlation Coefficient Formula

Correlation coefficients are used to calculate how vital a connection is between two variables. There are different types of correlation coefficients, one of the most popular is Pearson’s correlation (also known as Pearson’s R)which is commonly used in linear regression. 

Correlation coefficient Formula

The correlation coefficient procedure is used to determine how strong a relationship is between the data. The correlation coefficient procedure yields a value between 1 and -1. In which,

• -1 indicates a strong negative relationship
• 1 indicates strong positive relationships
• And an outcome of zero implies no connection at all

Positive Correlation

Negative Correlation

Zero Correlation

Meaning

• A correlation coefficient of -1 means there is a negative decrease of a fixed proportion, for every positive increase in one variable. Like, the amount of gas in a tank decreases in a perfect correlation with the speed.
• A correlation coefficient of  1 means there is a positive increase of a fixed proportion of others, for every positive increase in one variable. Like, the size of the shoe goes up in perfect correlation with foot length.
• Zero means that for every increase, there is neither a positive nor a negative increase. The two just aren’t related.

Types of Formulae

• Pearson’s correlation coefficient formula

[Tex] R= \frac{n(∑xy) – (∑x)(∑y)}{\sqrt{[n∑x²-(∑x)²][n∑y²-(∑y)²}}        [/Tex] [Tex]       [/Tex]

• Sample correlation coefficient formula

[Tex] r_{xy}= cov(x,y) / s_xs_y    [/Tex] [Tex]       [/Tex]

S_xy is the sample Covariance, and S_x and S_y are the sample standard deviations

• Population correlation coefficient formula

[Tex] pxy=  cov(X,Y) / σxσy     [/Tex] [Tex]       [/Tex]

It uses σ_x and σ_y as the population standard deviation and, σ_xy as the population Covariance.

Pearson’s correlation

It is the most common correlation in statistics. The full name is Pearson’s Product Moment correlation in short PPMC. It displays the Linear relation between the two sets of data. Two letters are used to represent the Pearson correlation: the Greek letter rho (ρ) for a population and the letter “r” for a sample correlation coefficient.

Steps to find Pearson’s correlation coefficient

Step 1: Firstly make a chart with the given data like subject,x, and y and add three more columns in it xy, x² and y².

Step 2: Now multiply the x and y columns to fill the xy column. For example:- in x we have 24 and in y we have 65 so xy will be 24×65=1560.

Step 3: Now, take the square of the numbers in the x column and fill the x² column.

Step 4: Now, take the square of the numbers in the y column and fill the y² column.

Step 5: Now, add up all the values in the columns and put the result at the bottom. Greek letter sigma (Σ) is the short way of saying summation.

Step 6: Now, use the formula for Pearson’s correlation coefficient:

[Tex] R= \frac{n(∑xy) – (∑x)(∑y)}{\sqrt{[n∑x²-(∑x)²][n∑y²-(∑y)²}}        [/Tex]

To know which type of variable we have either positive or negative.

Linear Correlation Coefficient

The Pearson’s correlation coefficient is the linear correlation coefficient which returns the value between the -1 and +1. In this -1 indicates a strong negative correlation and +1 indicates a strong positive correlation. If it lies 0 then there is no correlation. This is also known as zero correlation.

The “crude estimations” for analyzing the stability of correlations using Pearson’s Correlation:

r value	crude estimates
+.70 or higher	A very strong positive relationship
+.40 to +.69	Strong positive relationship
+.30 to +.39.	Moderate positive relationship
+.20 to +.29	weak positive relationship
+.01 to +.19	No or negligible relationship
0	No relationship [zero correlation]
-.01 to -.19	No or negligible relationship
-.20 to -.29	weak negative relationship
-.30 to -.39	Moderate negative relationship
-.40 to -.69	Strong negative relationship
-.70 or higher	The very strong negative relationship

Cramer’s V Correlation

It is as similar as the Pearson correlation coefficient. It is used to calculate the correlation with more than 2×2 rows and columns. Cramer’s V correlation varies between 0 and 1. The value close to zero associates that a very little association is there between the variables and if it’s close to 1 it indicates a very strong association.

The “crude estimates” for interpreting strengths of correlations using Cramer’s V Correlation:

Cramer’s V	crude estimates
.25 or higher	Very strong relationship
.15 to .25	Strong relationship
.11 to .15	Moderate relationship
.06 to .10	weak relationship
.01 to .05	No or negligible relationship

Sample Problems

Problem 1: Calculate the correlation coefficient from the following table:

SUBJECT	AGE X	GLUCOSE LEVEL Y
1	42	98
2	23	68
3	22	73
4	47	79
5	50	88
6	60	82

Solution:

Make a table from the given data and add three more columns of XY, X², and Y².

SUBJECT	AGE X	GLUCOSE LEVEL Y	XY	X²	Y²
1	42	98	4116	1764	9604
2	23	68	1564	529	4624
3	22	73	1606	484	5329
4	47	79	3713	2209	6241
5	50	88	4400	2500	7744
6	60	82	4980	3600	6724
∑	244	488	20379	11086	40266

∑xy = 20379

∑x = 244

∑y = 488

∑x² = 11086

∑y² = 40266

n = 6.

Put all the values in the Pearson’s correlation coefficient formula:

[Tex] R= \frac{n(∑xy) – (∑x)(∑y)}{\sqrt{[n∑x²-(∑x)²][n∑y²-(∑y)²}}        [/Tex]

R = 6(20379) – (244)(488) / √[6(11086)-(244)²][6(40266)-(488)² ]                                                 

R = 3202 / √[6980][3452]        

R = 3202/4972.238

R = 0.6439

It shows that the relationship between the variables of the data is a strong positive relationship.

Problem 2: Calculate the correlation coefficient from the following table:

SUBJECT	AGE X	weight Y
1	40	99
2	25	79
3	22	69
4	54	89

Solution:

Make a table from the given data and add three more columns of XY,  X², and Y².

SUBJECT	AGE X	Weight Y	XY	X²	Y²
1	40	99	3960	1600	9801
2	25	79	1975	625	6241
3	22	69	1518	484	4761
4	54	89	4806	2916	7921
∑	151	336	12259	5625	28724

∑xy = 12258

∑x = 151

∑y = 336

∑x² = 5625

∑y² 28724

n = 4

Put all the values in the Pearson’s correlation coefficient formula:

[Tex] R= \frac{n(∑xy) – (∑x)(∑y)}{\sqrt{[n∑x²-(∑x)²][n∑y²-(∑y)²}}        [/Tex]

R = 4(12258) – (151)(336) / √[4(5625)-(151)²][4(28724)-(336)²]        

R = -1704 / √[-301][-2000]        

R=-1704/775.886

R=-2.1961

It shows that the relationship between the variables of the data is a very strong negative relationship.

Problem 3:  Calculate the correlation coefficient for the following data:

X = 7,9,14 and Y = 17,19,21

Solution:

Given variables are,

X = 7,9,14

and,

Y = 17,19,21

To, find the correlation coefficient of the following variables Firstly a table is to be constructed as follows, to get the values required in the formula.

X	Y	XY	X²	Y²
7	17	119	49	36
9	19	171	81	361
14	21	294	196	441
∑ 30	∑ 57	∑ 584	∑ 326	∑ 838

∑xy = 584

∑x = 30

∑y = 57

∑x² = 326

∑y² = 838

n = 3

Put all the values in the Pearson’s correlation coefficient formula:

[Tex] R= \frac{n(∑xy) – (∑x)(∑y)}{\sqrt{[n∑x²-(∑x)²][n∑y²-(∑y)²}}        [/Tex]

R = 3(584) – (30)(57) / √[3(326)-(30)²][3(838)-(57)²]        

R = 42 / √[78][-735]        

R = 42/-239.43

R = -0.1754

It shows that the relationship between the variables of the data is negligible relationship

Problem 4: Calculate the correlation coefficient for the following data:

X = 21, 31, 25, 40, 47, 38 and Y = 70,55,60,78,66,80

Solution:

Given variables are,

X = 21,31,25,40,47,38

And,

Y = 70,55,60,78,66,80

To, find the correlation coefficient of the following variables Firstly a table is to be constructed as follows, to get the values required in the formula.

X	Y	XY	X²	Y²
21	70	1470	441	4900
31	55	1705	961	3025
25	60	1500	625	3600
40	78	3120	1600	6094
47	66	3102	2209	4356
38	80	3040	1444	6400
∑202	∑409	∑13937	∑7280	∑28265

∑xy = 13937

∑x = 202

∑y = 409

∑x² = 7280

∑y² = 28265

n = 6

Put all the values in the Pearson’s correlation coefficient formula:

[Tex] R= \frac{n(∑xy) – (∑x)(∑y)}{\sqrt{[n∑x²-(∑x)²][n∑y²-(∑y)²}}        [/Tex]

R = 6(13937) – (202)(409) / √[6(7280) – (202)²][6(28265) – (409)²]        

R = 1004 /√[2876][2909]        

R = 1004 / 2892.452938

R = 0.3471

It shows that the relationship between the variables of the data is a moderate positive relationship.

Problem 5: Calculate the correlation coefficient for the following data?

X = 5 ,9 ,14, 16 and Y = 6, 10, 16, 20 .

Solution:

Given variables are,

X = 5 ,9 ,14, 16

And

Y = 6, 10, 16, 20.

To, find the correlation coefficient of the following variables Firstly a table is to be constructed as follows, to get the values required in the formula add all the values in the columns to get the values used in the formula

X	Y	XY	X²	Y²
5	6	30	25	36
9	10	90	81	100
14	16	224	196	256
16	20	320	256	400
∑44	∑52	∑664	∑558	∑792

∑xy = 664

∑x = 44

∑y = 52

∑x² = 558

∑y² = 792

n = 4

Put all the values in the Pearson’s correlation coefficient formula:

[Tex] R= n(∑xy) – (∑x)(∑y) / √[n∑x²-(∑x)²][n∑y²-(∑y)²  [/Tex]

R = 4(664) – (44)(52) / √[4(558) – (44)²][4(792) – (52)²]

R = 368 / √[296][464]

R = 368/370.599

R = 0.9930

It shows that the relationship between the variables of the data is a very strong positive relationship.

Problem 6: Calculate the correlation coefficient for the following data:

X = 10, 13, 15 ,17 ,19 and Y = 5,10,15,20,25.

Solution:

Given variables are,

X = 10, 13, 15 ,17 ,19 and Y = 5, 10, 15, 20, 25.

To, find the correlation coefficient of the following variables Firstly a table is to be constructed as follows, to get the values required in the formula also  add all the values in the columns to get the values used in formula,

X	Y	XY	X²	Y²
10	5	50	100	25
13	10	130	169	100
15	15	225	225	225
17	20	340	340	400
19	25	475	475	625
∑74	∑75	∑1103	∑1144	∑1375

∑xy = 1103

∑x = 74

∑y = 75

∑x² = 1144

∑y² = 1375

n = 5

Put all the values in the Pearson’s correlation coefficient formula:

[Tex] R= \frac{n(∑xy) – (∑x)(∑y)}{\sqrt{[n∑x²-(∑x)²][n∑y²-(∑y)²}}       [/Tex]

R = 5(1103) – (74)(75) / √ [5(1144) – (74)²][5(1375) – (75)²]

R = -35 / √[244][1250]  

R = -35/552.26

R = 0.0633

It shows that the relationship between the variables of the data is a negligible relationship.

Problems 7: Calculate the correlation coefficient for the following data:

X = 12, 10, 42, 27,35,56 and Y = 13, 15, 56, 34,65,26

Solution:

Given variables are,

X = 12, 10, 42, 27, 35, 56 and Y = 13, 15, 56, 34, 65, 26

To, find the correlation coefficient of the following variables Firstly a table is to be constructed as follows, to get the values required in the formula also add all the values in the columns to get the values used in the formula

X	Y	XY	X²	Y²
12	13	156	144	169
10	15	150	100	225
42	56	2352	1764	3136
27	34	918	729	1156
35	65	2275	1225	4225
56	26	1456	3136	676
∑182	∑209	∑7307	∑7098	∑9587

∑xy = 7307

∑x = 182

∑y = 209

∑x² = 7098

∑y² = 9587

n = 6

Put all the values in the Pearson’s correlation coefficient formula:

[Tex] R= \frac{n(∑xy) – (∑x)(∑y)}{\sqrt{[n∑x²-(∑x)²][n∑y²-(∑y)²}}       [/Tex]

R = 6(7307) – (182)(209) / √ {[6(7098) – (182)²][6(9587)-(209)²]}

R = 5804 / √[9464][13841] 

R = 5804/11445.139

R = 0.5071

It shows that the relationship between the variables of the data is a strong positive relationship.