Module V - Statistical Techniques

Definition

Measures the degree of linear relationship between two variables \( X \) and \( Y \). Value ranges from \(-1\) to \(+1\).

\[ r = \frac{\sum(x - \bar{x})(y - \bar{y})}{\sqrt{\sum(x - \bar{x})^2 \cdot \sum(y - \bar{y})^2}} = \frac{\sum xy - n\bar{x}\bar{y}}{\sqrt{(\sum x^2 - n\bar{x}^2)(\sum y^2 - n\bar{y}^2)}} \]

Shortcut Formula: \[ r = \frac{n\sum xy - \sum x \sum y}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}} \]

Value of r	Interpretation
\( r = +1 \)	Perfect positive correlation
\( r = -1 \)	Perfect negative correlation
\( r = 0 \)	No linear correlation
\( 0 < r < 1 \)	Positive correlation
\( -1 < r < 0 \)	Negative correlation

Find Karl Pearson's correlation coefficient for the data

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

Prepare calculation table:

X	Y	XY	X²	Y²
1	2	2	1	4
2	4	8	4	16
3	5	15	9	25
4	6	24	16	36
5	8	40	25	64
Σ=15	Σ=25	Σ=89	Σ=55	Σ=145

\( n = 5 \), \( \sum x = 15 \), \( \sum y = 25 \), \( \sum xy = 89 \), \( \sum x^2 = 55 \), \( \sum y^2 = 145 \)
Apply formula: \[ r = \frac{5(89) - 15 \times 25}{\sqrt{[5(55) - 225][5(145) - 625]}} = \frac{445 - 375}{\sqrt{(275-225)(725-625)}} \] \[ = \frac{70}{\sqrt{50 \times 100}} = \frac{70}{\sqrt{5000}} = \frac{70}{70.71} \]

\( r = 0.99 \) (Strong positive correlation)

2. Spearman's Rank Correlation Coefficient (R)

Case 1: Non-Repeated Ranks

\[ R = 1 - \frac{6\sum d^2}{n(n^2 - 1)} \]

where \( d = \) difference between ranks, \( n = \) number of pairs

Find Spearman's rank correlation coefficient

Judge A	1	2	3	4	5
Judge B	2	1	4	3	5

Calculate differences:

Rank A Rank B d = A - B d²

1 2 -1 1

2 1 1 1

3 4 -1 1

4 3 1 1

5 5 0 0

Σd² = 4
\( n = 5 \), \( \sum d^2 = 4 \)
\[ R = 1 - \frac{6 \times 4}{5(25-1)} = 1 - \frac{24}{120} = 1 - 0.2 \]

Rank A	Rank B	d = A - B	d²
1	2	-1	1
2	1	1	1
3	4	-1	1
4	3	1	1
5	5	0	0
Σd² =	4

\( R = 0.8 \)

Case 2: Repeated Ranks

\[ R = 1 - \frac{6\left[\sum d^2 + \frac{m_1^3 - m_1}{12} + \frac{m_2^3 - m_2}{12} + \cdots\right]}{n(n^2 - 1)} \]

where \( m = \) number of times a rank is repeated

For repeated ranks, assign average rank. E.g., if 3rd and 4th positions are tied, both get rank \( \frac{3+4}{2} = 3.5 \)

Find R with repeated ranks

X	40	50	50	60	70	70
Y	30	40	50	50	60	80

Assign ranks (1 = lowest):

X	Rank X	Y	Rank Y	d	d²
40	1	30	1	0	0
50	2.5	40	2	0.5	0.25
50	2.5	50	3.5	-1	1
60	4	50	3.5	0.5	0.25
70	5.5	60	5	0.5	0.25
70	5.5	80	6	-0.5	0.25
Σd² =					2

X: rank 50 repeated 2 times (\(m_1 = 2\)), rank 70 repeated 2 times (\(m_2 = 2\))
Y: rank 50 repeated 2 times (\(m_3 = 2\))
Correction factors: \( \frac{2^3-2}{12} = \frac{6}{12} = 0.5 \) (three times)
\[ R = 1 - \frac{6[2 + 0.5 + 0.5 + 0.5]}{6(36-1)} = 1 - \frac{6 \times 3.5}{210} = 1 - \frac{21}{210} \]

\( R = 0.9 \)

3. Lines of Regression

Line of Regression of Y on X

\[ y - \bar{y} = b_{yx}(x - \bar{x}) \]

where \( b_{yx} = \frac{\sum xy - n\bar{x}\bar{y}}{\sum x^2 - n\bar{x}^2} = r \cdot \frac{\sigma_y}{\sigma_x} \)

Line of Regression of X on Y

\[ x - \bar{x} = b_{xy}(y - \bar{y}) \]

where \( b_{xy} = \frac{\sum xy - n\bar{x}\bar{y}}{\sum y^2 - n\bar{y}^2} = r \cdot \frac{\sigma_x}{\sigma_y} \)

Key Properties

Both lines pass through \( (\bar{x}, \bar{y}) \)
\( r^2 = b_{yx} \times b_{xy} \), so \( r = \pm\sqrt{b_{yx} \times b_{xy}} \)
Sign of \( r \) = sign of \( b_{yx} \) = sign of \( b_{xy} \)

Find regression lines for the data

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

From earlier: \( n=5 \), \( \sum x=15 \), \( \sum y=25 \), \( \sum xy=89 \), \( \sum x^2=55 \), \( \sum y^2=145 \)
\( \bar{x} = 3 \), \( \bar{y} = 5 \)
\[ b_{yx} = \frac{89 - 5(3)(5)}{55 - 5(9)} = \frac{89-75}{55-45} = \frac{14}{10} = 1.4 \]
Y on X: \( y - 5 = 1.4(x - 3) \) → \( y = 1.4x + 0.8 \)
\[ b_{xy} = \frac{89 - 75}{145 - 5(25)} = \frac{14}{145-125} = \frac{14}{20} = 0.7 \]
X on Y: \( x - 3 = 0.7(y - 5) \) → \( x = 0.7y - 0.5 \)

Y on X: \( y = 1.4x + 0.8 \)
X on Y: \( x = 0.7y - 0.5 \)

4. Curve Fitting

First Degree (Linear): \( y = a + bx \)

Fit \( y = a + bx \) to the data

X	1	2	3	4
Y	2	5	7	10

Prepare table:

X Y XY X²

1 2 2 1

2 5 10 4

3 7 21 9

4 10 40 16

10 24 73 30
Normal equations:
\( 24 = 4a + 10b \) ... (1)
\( 73 = 10a + 30b \) ... (2)
From (1): \( a = \frac{24-10b}{4} = 6 - 2.5b \)
Substitute in (2): \( 73 = 10(6-2.5b) + 30b = 60 - 25b + 30b \)
\( 73 = 60 + 5b \) → \( b = 2.6 \)
\( a = 6 - 2.5(2.6) = -0.5 \)

X	Y	XY	X²
1	2	2	1
2	5	10	4
3	7	21	9
4	10	40	16
10	24	73	30

\( y = -0.5 + 2.6x \)

Second Degree (Parabola): \( y = a + bx + cx^2 \)

Normal Equations: \[ \sum y = na + b\sum x + c\sum x^2 \] \[ \sum xy = a\sum x + b\sum x^2 + c\sum x^3 \] \[ \sum x^2y = a\sum x^2 + b\sum x^3 + c\sum x^4 \]

Fit \( y = a + bx + cx^2 \) to the data

X	0	1	2	3
Y	1	2	9	20

Prepare table:

X Y X² X³ X⁴ XY X²Y

0 1 0 0 0 0 0

1 2 1 1 1 2 2

2 9 4 8 16 18 36

3 20 9 27 81 60 180

6 32 14 36 98 80 218
Normal equations:
\( 32 = 4a + 6b + 14c \) ... (1)
\( 80 = 6a + 14b + 36c \) ... (2)
\( 218 = 14a + 36b + 98c \) ... (3)
Solving these simultaneous equations:
From (1): \( 16 = 2a + 3b + 7c \)
From (2): \( 40 = 3a + 7b + 18c \)
From (3): \( 109 = 7a + 18b + 49c \)
Solving: \( a = 1 \), \( b = -1 \), \( c = 2 \)

X	Y	X²	X³	X⁴	XY	X²Y
0	1	0	0	0	0	0
1	2	1	1	1	2	2
2	9	4	8	16	18	36
3	20	9	27	81	60	180
6	32	14	36	98	80	218

\( y = 1 - x + 2x^2 \)

Concept	Formula
Karl Pearson's r	\( r = \frac{n\sum xy - \sum x\sum y}{\sqrt{[n\sum x^2-(\sum x)^2][n\sum y^2-(\sum y)^2]}} \)
Spearman's R (no repeat)	\( R = 1 - \frac{6\sum d^2}{n(n^2-1)} \)
Spearman's R (repeat)	Add \( \frac{m^3-m}{12} \) for each repeated rank
Regression Y on X	\( y - \bar{y} = b_{yx}(x - \bar{x}) \)
Regression X on Y	\( x - \bar{x} = b_{xy}(y - \bar{y}) \)
Relation	\( r^2 = b_{yx} \times b_{xy} \)
Linear fit	\( \sum y = na + b\sum x \); \( \sum xy = a\sum x + b\sum x^2 \)

Module V: Statistical Techniques

1. Karl Pearson's Coefficient of Correlation (r)