What does Karl Pearson’s coefficient of correlation measure?

It measures the strength and direction of a linear relationship between two variables.

What is the range of Karl Pearson’s coefficient of correlation?

It always lies between -1 and +1.

What does it mean if r = 0?

It means there is no linear correlation between the two variables.

Why is Karl Pearson’s correlation widely used?

Because it gives both the magnitude and the direction of the relationship, making it a robust method for correlation analysis.

Frequently Asked Questions

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Karl Pearson’s Coefficient of Correlation

Correlation is a statistical measure that shows the degree of relationship between two variables. Karl Pearson’s coefficient of correlation is one of the most widely used methods for determining the strength and direction of this relationship. It plays an important role in statistics, data science, and also in competitive exams like JEE Main, JEE Advanced, and other entrance tests.

1.0Karl Pearson’s Coefficient of Correlation – Definition and Theory

Karl Pearson’s coefficient of correlation measures the linear relationship between two variables.
It is denoted by ‘r’.
The value of Karl Pearson’s coefficient of correlation lies between -1 and +1.
- If r = +1 → Perfect positive correlation
- If r = -1 → Perfect negative correlation
- If r = 0 → No correlation
Karl Pearson’s coefficient of correlation method uses covariance and standard deviation.

Formula:

$r = \frac{Cov ( X , Y )}{σ _{X} σ _{Y}}$

Where,

Cov(X,Y) = Covariance between variables X and Y
$σ_{X}, σ_{Y}$ = Standard deviations of X and Y

2.0Methods to Calculate Karl Pearson’s Coefficient of Correlation

1. Direct Method

This is the standard method using raw data.

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

n: Number of pairs of observations
x, y: Data values of two variables

Use this method when numbers are small and simple.

2. Assumed Mean Method (Short-cut Method)

When data values are large, calculations can be simplified by taking deviations from assumed means.

Let:

$d_{x} = x - A_{x}, d_{y} = y - A_{y}$

Where $A_{x}, A_{y}$ are assumed means.

Formula:

$r = \frac{\sum d _{x} d _{y}}{\sum d _{x}^{2} \cdot \sum d _{y}^{2}}$

This reduces calculation effort.

3. Step Deviation Method (Coding Method)

When data is very large or has a common difference (like in grouped data), divide deviations by a factor.

$d_{x} = \frac{x - A _{x}}{h _{x}}, d_{y} = \frac{y - A _{y}}{h _{y}}$

Formula becomes:

$r = \frac{\sum d _{x} d _{y}}{\sum d _{x}^{2} \cdot \sum d _{y}^{2}}$

This method is useful for grouped frequency distributions.

4. Using Covariance Formula

Another approach is to use:

$r = \frac{Cov ( X , Y )}{σ _{X} σ _{Y}}$

First, calculate the covariance of X and Y.
Then divide by the product of their standard deviations.

Very common in statistics and probability questions.

So, the main methods are:

Direct Method
Assumed Mean Method
Step Deviation Method
Covariance Method

3.0Examples Based on Methods to Calculate Karl Pearson’s Coefficient of Correlation

Method 1 — Direct (raw-sum) formula

Example 1 (Direct method)

Data: (X,Y) = (1, 2), (2, 3), (3, 5), (4, 4). Compute r.

Solution

Compute sums:

$n = 4, \sum x = 10, \sum y = 14, \sum x y = 39, \sum x^{2} = 30, \sum y^{2} = 54.$

Use raw-sum formula

$r = \frac{n \sum x y - ( \sum x ) ( \sum y )}{( n \sum x ^{2} - ( \sum x ) ^{2} ) ( n \sum y ^{2} - ( \sum y ) ^{2} )} .$

Numerator: $4 \cdot 39 - 10 \cdot 14 = 156 - 140 = 16.$

Denominator: $(120 - 100) (216 - 196) = 20 \cdot 20 = 20$ _.

So

$r = \frac{16}{20} = 0.8$

Method 2 — Assumed-mean (shortcut) method

Example 2 (Assumed mean)

Data: X = {98, 102, 100, 105, 95}, Y = {200, 210, 205, 215, 195}.

Take assumed means $A_{x} = 100, A_{y} = 205$ .

Solution

Compute deviations $d_{x} = x - A_{x}, d_{y} = y - A_{y}$ :

x	$d_{x}$	y	$d_{y}$	$d_{x} d_{y}$	$d_{x}^{2}$	$d_{y}^{2}$
98	-2	200	-5	10	4	25
102	2	210	5	10	4	25
100	0	205	0	0	0	0
105	5	215	10	50	25	100
95	-5	195	-10	50	25	100
Sum				120	58	250

Now

$r = \frac{\sum d _{x} d _{y}}{( \sum d _{x}^{2} ) ( \sum d _{y}^{2} )} = \frac{120}{58 \cdot 250} .$

Compute denominator: $14500 \approx 120.415$ _. Hence

$r \approx \frac{120}{120.415} \approx 0.9965$

(very strong positive linear association).

Note: assumed-mean reduces arithmetic — you only work with small deviations.

Method 3 — Step-deviation / coding method (useful when data have constant step)

Example 3 (Step-deviation)

Data: X = {100, 110, 120, 130}, Y = {45, 50, 60, 65}.

Use h = 10. Let $u = (x - A x) / h, v = (y - A y) / h w i t h A_{x} = 115, A_{y} = 55.$

Solution

Compute coded deviations:

x	u = (x-115)/10	y	v = (y-55)/10	uv	$u^{2}$	$v^{2}$
100	-1.5	45	-1.0	1.5	2.25	1.00
110	-0.5	50	-0.5	0.25	0.25	0.25
120	0.5	60	0.5	0.25	0.25	0.25
130	1.5	65	1.0	1.5	2.25	1.00
Sum	0		0	3.5	5	2.5

Then

$r = \frac{\sum uv}{( \sum u ^{2} ) ( \sum v ^{2} )} = \frac{3.5}{5 \cdot 2.5} = \frac{3.5}{12.5} \approx \frac{3.5}{3.5355} \approx 0.990.$

Because we scaled both variables by constants, r is unchanged — coding just simplifies arithmetic.

Method 4 — Covariance and standard deviations (conceptual form)

Example 4 (Covariance method, same data as Example 1)

Data: (1, 2), (2, 3), (3, 5), (4, 4). We already found r = 0.8. Show via covariance.

Solution

Means: $\overset{x}{ˉ} = 2.5, \overset{y}{ˉ} = 3.5$ _.

Compute sums of deviations:

x	$x - \overset{x}{ˉ}$	y	$y - \overset{y}{ˉ}$	$(x - \overset{x}{ˉ}) (y - \overset{y}{ˉ})$	$(x - \overset{x}{ˉ})^{2}$	$(y - \overset{y}{ˉ})^{2}$
1	-1.5	2	-1.5	2.25	2.25	2.25
2	-0.5	3	-0.5	0.25	0.25	0.25
3	0.5	5	1.5	0.75	0.25	2.25
4	1.5	4	0.5	0.75	2.25	0.25
Sum				4.0	5.0	5.0

Population covariance: $Cov (X, Y) = \frac{\sum ( x - x ˉ ) ( y - y ˉ )}{n} = \frac{4}{4} = 1.$

Population standard deviations:

$σ_{x} = \frac{5}{4} = 1.25, σ_{y} = 1.25 .$
Thus

$r = \frac{Cov ( X , Y )}{σ _{X} σ _{Y}} = \frac{1}{( 1.25 ) ^{2}} = \frac{1}{1.25} = 0.8,$ same result as Example 1.

4.0Solved Examples on Karl Pearson’s Coefficient of Correlation

Example 1: Find Karl Pearson’s coefficient of correlation for the following data:

X	1	2	3	4	5
Y	2	4	6	8	10

Solution:

Here, Y = 2X.

Clearly, the relation is perfectly linear.

Thus, r = +1.

Example 2: Given the following data, calculate Karl Pearson’s coefficient of correlation:

X	10	12	13	16	17	18	19
Y	40	38	43	45	37	43	49

Solution:

Step 1: Compute mean of X and Y.

$\overset{ˉ}{X} = \frac{10 + 12 + 13 + 16 + 17 + 18 + 19}{7} = 15$

$\overset{ˉ}{Y} = \frac{40 + 38 + 43 + 45 + 37 + 43 + 49}{7} = 42.14$

Step 2: Find deviations

$d x = X - \overset{ˉ}{X}, d y = Y - \overset{ˉ}{Y}$ .

Step 3: Apply formula:

$r = \frac{\sum d x \cdot d y}{\sum d x ^{2} \cdot \sum d y ^{2}}$

After calculations:

$r \approx 0.82$

So, there is a strong positive correlation.

Example 3: If the correlation coefficient between X and Y is r = 0.6, standard deviation of X = 3, standard deviation of Y = 4, and covariance of (X, Y) = ?

Solution:
We know,

$r = \frac{Cov ( X , Y )}{σ _{X} σ _{Y}}$ $0.6 = \frac{Cov ( X , Y )}{3 \times 4}$ $Cov (X, Y) = 0.6 \times 12 = 7.2$

Example 4: For the data X = {1, 2, 3}, Y = {6, 4, 2}, find r.

Solution:

Here Y = -2X + 8 (perfect linear with negative slope). Hence correlation is perfect negative:

r=−1.

Example 5: For data (X, Y): (1, 2), (2, 3), (3, 5), (4, 4). Compute Karl Pearson’s r.

Solution:

Compute sums (careful arithmetic):

n = 4

$\sum x = 1 + 2 + 3 + 4 = 10 \sum y = 2 + 3 + 5 + 4 = 14 \sum x y = 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 5 + 4 \cdot 4 = 2 + 6 + 15 + 16 = 39 \sum x^{2} = 1 + 4 + 9 + 16 = 30 \sum y^{2} = 4 + 9 + 25 + 16 = 54$

Use formula

$r = \frac{n \sum x y - ( \sum x ) ( \sum y )}{( n \sum x ^{2} - ( \sum x ) ^{2} ) ( n \sum y ^{2} - ( \sum y ) ^{2} )} .$

Numerator: $4 \cdot 39 - 10 \cdot 14 = 156 - 140 = 16.$

Denominator: $(4 \cdot 30 - 100) (4 \cdot 54 - 196) = (120 - 100) (216 - 196) = 20 \cdot 20 = 20.$

So r = 16/20 = 0.8.

r = 0.8 — strong positive linear correlation.

Example 6: Paired grouped data (midpoints used as scores): $(x_{i}, f_{i}, y_{i}) : (5, 4, 50), (15, 6, 60), (25, 5, 70)$ .

Compute r treating $f_{i}$ as weights.

Solution:

First compute weighted sums (N = total frequency = 4 + 6 + 5 = 15).

= 1000 + 5400 + 8750 = 15150.

$\sum f x = 5 \cdot 4 + 15 \cdot 6 + 25 \cdot 5 = 20 + 90 + 125 = 235.$ $\sum f y = 50 \cdot 4 + 60 \cdot 6 + 70 \cdot 5 = 200 + 360 + 350 = 910.$ $\sum f x^{2} = 5^{2} \cdot 4 + 1 5^{2} \cdot 6 + 2 5^{2} \cdot 5 = 100 + 1350 + 3125 = 4575.$ $\sum f y^{2} = 5 0^{2} \cdot 4 + 6 0^{2} \cdot 6 + 7 0^{2} \cdot 5 = 10000 + 21600 + 24500 = 56100.$ $\sum f x y = (5 \cdot 50) \cdot 4 + (15 \cdot 60) \cdot 6 + (25 \cdot 70) \cdot 5$ = 1000 + 5400 + 8750 = 15150.

Now use weighted form:

$r = \frac{N \sum f x y - ( \sum f x ) ( \sum f y )}{( N \sum f x ^{2} - ( \sum f x ) ^{2} ) ( N \sum f y ^{2} - ( \sum f y ) ^{2} )} .$

Numerator: $15 \cdot 15150 - 235 \cdot 910 = 227250 - 213850 = 13400.$

First variance term: $15 \cdot 4575 - 23 5^{2} = 68625 - 55225 = 13400.$

Second variance term: $15 \cdot 56100 - 91 0^{2} = 841500 - 828100 = 13400.$

Denominator: $13400 \cdot 13400 = 13400$ _.

Hence r = 13400/13400 = 1.

r = 1 — perfect positive linear correlation (weighted).

Example 7: Given regression coefficients $b_{x y} = 0.8$ (regression of Y on X) and $b_{y x} = 0.45$ (regression of X on Y), find r.

Solution:

Property: $r^{2} = b_{x y} b_{y x}$ Sign of r equals sign of both coefficients (if both positive → r > 0).

Compute $r^{2} = 0.8 \times 0.45 = 0.36.$

So $r = 0.36 = 0.6.$

r = 0.6.

Example 8: If r = -0.5, $σ_{X} = 4, σ_{Y} = 3,$ find Cov(X, Y).

Solution:

$C o v (X, Y) = - 6.$

$Cov (X, Y) = r σ_{X} σ_{Y} = - 0.5 \times 4 \times 3 = - 6.$ $Cov (X, Y) = - 6.$

Example 9: For a sample of n = 10 pairs you observe r = 0.8. Test significance at 5% level (two-tailed). Use $t = \frac{r n - 2}{1 - r ^{2}}$ _.

Solution:

Degrees of freedom = n – 2 = 8. Compute

$t = 0.8 \frac{8}{1 - 0.64} = 0.8 \frac{8}{0.36} = 0.8 22.22 \overline{2} \approx 0.8 \times 4.714 = 3.771.$

Critical two-tailed $t_{0.025, 8} \approx 2.306. S in ce 3.771 > 2.306$ , the correlation is statistically significant at 5% level.

$Reject H_{0} : r = 0; significant correlation.$

5.0Practice Questions on Karl Pearson’s Coefficient of Correlation

Calculate Karl Pearson’s coefficient of correlation for the following:

X	1	2	3	4	5
Y	9	8	10	12	11

If the correlation coefficient r = -0.75, $σ_{X} = 5, σ_{Y} = 4$ _, find covariance.
For two variables, the regression coefficients are , .b_{xy} = 0.8, \quad b_{yx} = 0.5 Find the value of Karl Pearson’s coefficient of correlation.
Compute r for (X, Y) = (1, 1), (2, 2), (3, 2), (4, 3).

Answer hint: use raw-sum formula.

For a sample n = 6, you get r = 0.45. Is it significant at 5%? (df = 4, $t_{0.025, 4} \approx 2.776$ ).
Answer hint: compute “t” as in Example 6.
Given $\sum x = 40, \sum y = 50, \sum x^{2} = 350, \sum y^{2} = 520, \sum x y = 430$ for n = 10. Find r.

Answer hint: apply the raw-sum formula.

If r=0.3, $σ_{X} = 5, σ_{Y} = 2$ , compute covariance.
Paired grouped data: (x, f, y) = (2, 3, 10), (6, 2, 20), (10, 5, 30). Compute weighted r.

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Karl Pearson’s Coefficient of Correlation

Correlation is a statistical measure that shows the degree of relationship between two variables. Karl Pearson’s coefficient of correlation is one of the most widely used methods for determining the strength and direction of this relationship. It plays an important role in statistics, data science, and also in competitive exams like JEE Main, JEE Advanced, and other entrance tests.

1.0Karl Pearson’s Coefficient of Correlation – Definition and Theory

Karl Pearson’s coefficient of correlation measures the linear relationship between two variables.
It is denoted by ‘r’.
The value of Karl Pearson’s coefficient of correlation lies between -1 and +1.
- If r = +1 → Perfect positive correlation
- If r = -1 → Perfect negative correlation
- If r = 0 → No correlation
Karl Pearson’s coefficient of correlation method uses covariance and standard deviation.

Formula:

$r = \frac{Cov ( X , Y )}{σ _{X} σ _{Y}}$

Where,

Cov(X,Y) = Covariance between variables X and Y
$σ_{X}, σ_{Y}$ = Standard deviations of X and Y

2.0Methods to Calculate Karl Pearson’s Coefficient of Correlation

1. Direct Method

This is the standard method using raw data.

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

n: Number of pairs of observations
x, y: Data values of two variables

Use this method when numbers are small and simple.

2. Assumed Mean Method (Short-cut Method)

When data values are large, calculations can be simplified by taking deviations from assumed means.

Let:

$d_{x} = x - A_{x}, d_{y} = y - A_{y}$

Where $A_{x}, A_{y}$ are assumed means.

Formula:

$r = \frac{\sum d _{x} d _{y}}{\sum d _{x}^{2} \cdot \sum d _{y}^{2}}$

This reduces calculation effort.

3. Step Deviation Method (Coding Method)

When data is very large or has a common difference (like in grouped data), divide deviations by a factor.

$d_{x} = \frac{x - A _{x}}{h _{x}}, d_{y} = \frac{y - A _{y}}{h _{y}}$

Formula becomes:

$r = \frac{\sum d _{x} d _{y}}{\sum d _{x}^{2} \cdot \sum d _{y}^{2}}$

This method is useful for grouped frequency distributions.

4. Using Covariance Formula

Another approach is to use:

$r = \frac{Cov ( X , Y )}{σ _{X} σ _{Y}}$

First, calculate the covariance of X and Y.
Then divide by the product of their standard deviations.

Very common in statistics and probability questions.

So, the main methods are:

Direct Method
Assumed Mean Method
Step Deviation Method
Covariance Method

3.0Examples Based on Methods to Calculate Karl Pearson’s Coefficient of Correlation

Method 1 — Direct (raw-sum) formula

Example 1 (Direct method)

Data: (X,Y) = (1, 2), (2, 3), (3, 5), (4, 4). Compute r.

Solution

Compute sums:

$n = 4, \sum x = 10, \sum y = 14, \sum x y = 39, \sum x^{2} = 30, \sum y^{2} = 54.$

Use raw-sum formula

$r = \frac{n \sum x y - ( \sum x ) ( \sum y )}{( n \sum x ^{2} - ( \sum x ) ^{2} ) ( n \sum y ^{2} - ( \sum y ) ^{2} )} .$

Numerator: $4 \cdot 39 - 10 \cdot 14 = 156 - 140 = 16.$

Denominator: $(120 - 100) (216 - 196) = 20 \cdot 20 = 20$ _.

So

$r = \frac{16}{20} = 0.8$

Method 2 — Assumed-mean (shortcut) method

Example 2 (Assumed mean)

Data: X = {98, 102, 100, 105, 95}, Y = {200, 210, 205, 215, 195}.

Take assumed means $A_{x} = 100, A_{y} = 205$ .

Solution

Compute deviations $d_{x} = x - A_{x}, d_{y} = y - A_{y}$ :

x	$d_{x}$	y	$d_{y}$	$d_{x} d_{y}$	$d_{x}^{2}$	$d_{y}^{2}$
98	-2	200	-5	10	4	25
102	2	210	5	10	4	25
100	0	205	0	0	0	0
105	5	215	10	50	25	100
95	-5	195	-10	50	25	100
Sum				120	58	250

Now

$r = \frac{\sum d _{x} d _{y}}{( \sum d _{x}^{2} ) ( \sum d _{y}^{2} )} = \frac{120}{58 \cdot 250} .$

Compute denominator: $14500 \approx 120.415$ _. Hence

$r \approx \frac{120}{120.415} \approx 0.9965$

(very strong positive linear association).

Note: assumed-mean reduces arithmetic — you only work with small deviations.

Method 3 — Step-deviation / coding method (useful when data have constant step)

Example 3 (Step-deviation)

Data: X = {100, 110, 120, 130}, Y = {45, 50, 60, 65}.

Use h = 10. Let $u = (x - A x) / h, v = (y - A y) / h w i t h A_{x} = 115, A_{y} = 55.$

Solution

Compute coded deviations:

x	u = (x-115)/10	y	v = (y-55)/10	uv	$u^{2}$	$v^{2}$
100	-1.5	45	-1.0	1.5	2.25	1.00
110	-0.5	50	-0.5	0.25	0.25	0.25
120	0.5	60	0.5	0.25	0.25	0.25
130	1.5	65	1.0	1.5	2.25	1.00
Sum	0		0	3.5	5	2.5

Then

$r = \frac{\sum uv}{( \sum u ^{2} ) ( \sum v ^{2} )} = \frac{3.5}{5 \cdot 2.5} = \frac{3.5}{12.5} \approx \frac{3.5}{3.5355} \approx 0.990.$

Because we scaled both variables by constants, r is unchanged — coding just simplifies arithmetic.

Method 4 — Covariance and standard deviations (conceptual form)

Example 4 (Covariance method, same data as Example 1)

Data: (1, 2), (2, 3), (3, 5), (4, 4). We already found r = 0.8. Show via covariance.

Solution

Means: $\overset{x}{ˉ} = 2.5, \overset{y}{ˉ} = 3.5$ _.

Compute sums of deviations:

x	$x - \overset{x}{ˉ}$	y	$y - \overset{y}{ˉ}$	$(x - \overset{x}{ˉ}) (y - \overset{y}{ˉ})$	$(x - \overset{x}{ˉ})^{2}$	$(y - \overset{y}{ˉ})^{2}$
1	-1.5	2	-1.5	2.25	2.25	2.25
2	-0.5	3	-0.5	0.25	0.25	0.25
3	0.5	5	1.5	0.75	0.25	2.25
4	1.5	4	0.5	0.75	2.25	0.25
Sum				4.0	5.0	5.0

Population covariance: $Cov (X, Y) = \frac{\sum ( x - x ˉ ) ( y - y ˉ )}{n} = \frac{4}{4} = 1.$

Population standard deviations:

$σ_{x} = \frac{5}{4} = 1.25, σ_{y} = 1.25 .$
Thus

$r = \frac{Cov ( X , Y )}{σ _{X} σ _{Y}} = \frac{1}{( 1.25 ) ^{2}} = \frac{1}{1.25} = 0.8,$ same result as Example 1.

4.0Solved Examples on Karl Pearson’s Coefficient of Correlation

Example 1: Find Karl Pearson’s coefficient of correlation for the following data:

X	1	2	3	4	5
Y	2	4	6	8	10

Solution:

Here, Y = 2X.

Clearly, the relation is perfectly linear.

Thus, r = +1.

Example 2: Given the following data, calculate Karl Pearson’s coefficient of correlation:

X	10	12	13	16	17	18	19
Y	40	38	43	45	37	43	49

Solution:

Step 1: Compute mean of X and Y.

$\overset{ˉ}{X} = \frac{10 + 12 + 13 + 16 + 17 + 18 + 19}{7} = 15$

$\overset{ˉ}{Y} = \frac{40 + 38 + 43 + 45 + 37 + 43 + 49}{7} = 42.14$

Step 2: Find deviations

$d x = X - \overset{ˉ}{X}, d y = Y - \overset{ˉ}{Y}$ .

Step 3: Apply formula:

$r = \frac{\sum d x \cdot d y}{\sum d x ^{2} \cdot \sum d y ^{2}}$

After calculations:

$r \approx 0.82$

So, there is a strong positive correlation.

Example 3: If the correlation coefficient between X and Y is r = 0.6, standard deviation of X = 3, standard deviation of Y = 4, and covariance of (X, Y) = ?

Solution:
We know,

$r = \frac{Cov ( X , Y )}{σ _{X} σ _{Y}}$ $0.6 = \frac{Cov ( X , Y )}{3 \times 4}$ $Cov (X, Y) = 0.6 \times 12 = 7.2$

Example 4: For the data X = {1, 2, 3}, Y = {6, 4, 2}, find r.

Solution:

Here Y = -2X + 8 (perfect linear with negative slope). Hence correlation is perfect negative:

r=−1.

Example 5: For data (X, Y): (1, 2), (2, 3), (3, 5), (4, 4). Compute Karl Pearson’s r.

Solution:

Compute sums (careful arithmetic):

n = 4

$\sum x = 1 + 2 + 3 + 4 = 10 \sum y = 2 + 3 + 5 + 4 = 14 \sum x y = 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 5 + 4 \cdot 4 = 2 + 6 + 15 + 16 = 39 \sum x^{2} = 1 + 4 + 9 + 16 = 30 \sum y^{2} = 4 + 9 + 25 + 16 = 54$

Use formula

$r = \frac{n \sum x y - ( \sum x ) ( \sum y )}{( n \sum x ^{2} - ( \sum x ) ^{2} ) ( n \sum y ^{2} - ( \sum y ) ^{2} )} .$

Numerator: $4 \cdot 39 - 10 \cdot 14 = 156 - 140 = 16.$

Denominator: $(4 \cdot 30 - 100) (4 \cdot 54 - 196) = (120 - 100) (216 - 196) = 20 \cdot 20 = 20.$

So r = 16/20 = 0.8.

r = 0.8 — strong positive linear correlation.

Example 6: Paired grouped data (midpoints used as scores): $(x_{i}, f_{i}, y_{i}) : (5, 4, 50), (15, 6, 60), (25, 5, 70)$ .

Compute r treating $f_{i}$ as weights.

Solution:

First compute weighted sums (N = total frequency = 4 + 6 + 5 = 15).

= 1000 + 5400 + 8750 = 15150.

$\sum f x = 5 \cdot 4 + 15 \cdot 6 + 25 \cdot 5 = 20 + 90 + 125 = 235.$ $\sum f y = 50 \cdot 4 + 60 \cdot 6 + 70 \cdot 5 = 200 + 360 + 350 = 910.$ $\sum f x^{2} = 5^{2} \cdot 4 + 1 5^{2} \cdot 6 + 2 5^{2} \cdot 5 = 100 + 1350 + 3125 = 4575.$ $\sum f y^{2} = 5 0^{2} \cdot 4 + 6 0^{2} \cdot 6 + 7 0^{2} \cdot 5 = 10000 + 21600 + 24500 = 56100.$ $\sum f x y = (5 \cdot 50) \cdot 4 + (15 \cdot 60) \cdot 6 + (25 \cdot 70) \cdot 5$ = 1000 + 5400 + 8750 = 15150.

Now use weighted form:

$r = \frac{N \sum f x y - ( \sum f x ) ( \sum f y )}{( N \sum f x ^{2} - ( \sum f x ) ^{2} ) ( N \sum f y ^{2} - ( \sum f y ) ^{2} )} .$

Numerator: $15 \cdot 15150 - 235 \cdot 910 = 227250 - 213850 = 13400.$

First variance term: $15 \cdot 4575 - 23 5^{2} = 68625 - 55225 = 13400.$

Second variance term: $15 \cdot 56100 - 91 0^{2} = 841500 - 828100 = 13400.$

Denominator: $13400 \cdot 13400 = 13400$ _.

Hence r = 13400/13400 = 1.

r = 1 — perfect positive linear correlation (weighted).

Example 7: Given regression coefficients $b_{x y} = 0.8$ (regression of Y on X) and $b_{y x} = 0.45$ (regression of X on Y), find r.

Solution:

Property: $r^{2} = b_{x y} b_{y x}$ Sign of r equals sign of both coefficients (if both positive → r > 0).

Compute $r^{2} = 0.8 \times 0.45 = 0.36.$

So $r = 0.36 = 0.6.$

r = 0.6.

Example 8: If r = -0.5, $σ_{X} = 4, σ_{Y} = 3,$ find Cov(X, Y).

Solution:

$C o v (X, Y) = - 6.$

$Cov (X, Y) = r σ_{X} σ_{Y} = - 0.5 \times 4 \times 3 = - 6.$ $Cov (X, Y) = - 6.$

Example 9: For a sample of n = 10 pairs you observe r = 0.8. Test significance at 5% level (two-tailed). Use $t = \frac{r n - 2}{1 - r ^{2}}$ _.

Solution:

Degrees of freedom = n – 2 = 8. Compute

$t = 0.8 \frac{8}{1 - 0.64} = 0.8 \frac{8}{0.36} = 0.8 22.22 \overline{2} \approx 0.8 \times 4.714 = 3.771.$

Critical two-tailed $t_{0.025, 8} \approx 2.306. S in ce 3.771 > 2.306$ , the correlation is statistically significant at 5% level.

$Reject H_{0} : r = 0; significant correlation.$

5.0Practice Questions on Karl Pearson’s Coefficient of Correlation

Calculate Karl Pearson’s coefficient of correlation for the following:

X	1	2	3	4	5
Y	9	8	10	12	11

If the correlation coefficient r = -0.75, $σ_{X} = 5, σ_{Y} = 4$ _, find covariance.
For two variables, the regression coefficients are , .b_{xy} = 0.8, \quad b_{yx} = 0.5 Find the value of Karl Pearson’s coefficient of correlation.
Compute r for (X, Y) = (1, 1), (2, 2), (3, 2), (4, 3).

Answer hint: use raw-sum formula.

For a sample n = 6, you get r = 0.45. Is it significant at 5%? (df = 4, $t_{0.025, 4} \approx 2.776$ ).
Answer hint: compute “t” as in Example 6.
Given $\sum x = 40, \sum y = 50, \sum x^{2} = 350, \sum y^{2} = 520, \sum x y = 430$ for n = 10. Find r.

Answer hint: apply the raw-sum formula.

If r=0.3, $σ_{X} = 5, σ_{Y} = 2$ , compute covariance.
Paired grouped data: (x, f, y) = (2, 3, 10), (6, 2, 20), (10, 5, 30). Compute weighted r.

Frequently Asked Questions

What does Karl Pearson’s coefficient of correlation measure?

What is the range of Karl Pearson’s coefficient of correlation?

What is Karl Pearson’s coefficient of correlation denoted by?

What does it mean if r = 0?

Why is Karl Pearson’s correlation widely used?

Frequently Asked Questions

What does Karl Pearson’s coefficient of correlation measure?

What is the range of Karl Pearson’s coefficient of correlation?

What is Karl Pearson’s coefficient of correlation denoted by?

What does it mean if r = 0?

Why is Karl Pearson’s correlation widely used?

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Karl Pearson’s Coefficient of Correlation

1.0Karl Pearson’s Coefficient of Correlation – Definition and Theory

2.0Methods to Calculate Karl Pearson’s Coefficient of Correlation

1. Direct Method

2. Assumed Mean Method (Short-cut Method)

3. Step Deviation Method (Coding Method)

4. Using Covariance Formula

3.0Examples Based on Methods to Calculate Karl Pearson’s Coefficient of Correlation

4.0Solved Examples on Karl Pearson’s Coefficient of Correlation

5.0Practice Questions on Karl Pearson’s Coefficient of Correlation

Table of Contents

Join ALLEN!

Karl Pearson’s Coefficient of Correlation

1.0Karl Pearson’s Coefficient of Correlation – Definition and Theory

2.0Methods to Calculate Karl Pearson’s Coefficient of Correlation

1. Direct Method

2. Assumed Mean Method (Short-cut Method)

3. Step Deviation Method (Coding Method)

4. Using Covariance Formula

3.0Examples Based on Methods to Calculate Karl Pearson’s Coefficient of Correlation

4.0Solved Examples on Karl Pearson’s Coefficient of Correlation

5.0Practice Questions on Karl Pearson’s Coefficient of Correlation

Table of Contents