This document discusses different methods of calculating correlation coefficients and regression equations between two variables X and Y, including: - Direct and short cut methods to calculate the Pearson correlation coefficient r - Spearman's rank correlation coefficient rs - Regression equations to predict X from Y and Y from X using the least squares method - The relationship between the correlation coefficient r and the regression coefficients

1. CORRELATION
DIRECT METHOD
r=
∑𝑑𝑥.𝑑𝑦
√∑𝑑𝑥2.∑𝑑𝑦2
where, dx=X-𝑋
̅(ACTUAL MEAN)
dY=Y-𝑌
̅(ACTUAL MEAN)
X 11 10 9 8 7 6 5
Y 20 18 12 8 10 5 4
X Dx dx2
Y dy dy2
dx.dy
11 3 9 20 9 81 27
10 2 4 18 7 49 14
9 1 1 12 1 1 1
8 0 0 8 -3 9 0
7 -1 1 10 -1 1 1
6 -2 4 5 -6 36 12
5 -3 9 4 -7 49 21
Dx=0 ∑dx2
=28 ∑dy2=
226 ∑dx.dy=76
r=
∑𝑑𝑥.𝑑𝑦
√∑𝑑𝑥2.∑𝑑𝑦2
r=
76
√28∗226
r=.96

2. CORRELATION
SHORT CUT METHOD
r= 𝑁∑𝑑𝑥.𝑑𝑦−∑𝑑𝑥.∑𝑑𝑦
{𝑁∑𝑑𝑥2−(∑𝑑𝑥)2}{𝑁∑𝑑𝑦2−(∑𝑑𝑦)2}
dx=X-A(Assumed Mean)
dy=Y-A
N= no.of pairs
X Dx=X-8 dx2
Y Dy dy2
dx.dy
11 3 9 20 10 100 30
10 2 4 18 8 64 16
9 1 1 12 2 4 2
8 0 0 8 -2 4 0
7 -1 1 10 0 0 0
6 -2 4 5 -5 25 10
5 -3 9 4 -6 36 18
=0 ∑dx2
=28 ∑dy=7 ∑dy2=
233 ∑dx.dy=76
r=
𝑁∑𝑑𝑥.𝑑𝑦−∑𝑑𝑥.∑𝑑𝑦
√{𝑁∑𝑑𝑥2−(∑𝑑𝑥)^2}{𝑁∑𝑑𝑦2−(∑𝑑𝑦)^2}
r=
7∗76−0∗7
√{7∗28−(0)}{7∗233−72}
r=+.94
COMMENT-there exists a very high degree of correlation
PROBABLE ERRORS:
PE= .6745
(1−𝑟2)
√𝑁
STANDARD ERROR
SE=
(1−𝑟2)
√𝑁

3. CORRELATION
Relationship between SE & PE
PE= 2/3 S.E.
SPEARMAN’S RANK CORRELATION COEFFICIENT
rs=1-
6 ∑𝐷2
𝑁(𝑁2−1)
∑𝐷2 = 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑜𝑓 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠 𝑖𝑛 𝑟𝑎𝑛𝑘𝑠
N=NO. of pairs
Rx Ry Dxy=
Rx-
Ry
D2
Dyz=Ry-
Rz
D2
Dzx=Rx-
Rz
D2
5 3 2 4
3 7 -4 16
4 5 -1 1
8 9 -1 1
2 2 0 0
1 4 -3 9
7 1 6 36
10 10 0 0
6 8 -2 4
9 6 3 9
∑D2
=80
rs=1-
6 ∑𝐷2
𝑁(𝑁2−1)
rs=1-
6∗80
10(100−1)
=.52

4. CORRELATION
Question 2:
x 22 24 27 35 21 20 27 25 27 23
y 30 38 40 50 38 25 38 36 41 32
Solution:
X Rx Y Ry D=Rx-Ry D2
22 8 30 9 -1 1
24 6 38 5 1 1
27 3 40 3 0 0
35 1 50 1 0 0
21 9 38 5 4 16
20 10 25 10 0 0
27 3 38 5 -2 4
25 5 36 7 -2 4
27 3 41 2 1 1
23 7 32 8 -1 1
=28
=2+3+4/3=3
=4+5+6/3=5
rs=1-
6{∑𝐷2+
1
12
(𝑚3−𝑚)+
1
12
(𝑚3−𝑚)}
𝑁(𝑁2−1)
where, m= No. OF TIMES AN ITEM has been repeated (e.g., 27 & 38 has
been repeated thrice)
=1-
6{28+
1
12
(33−3)+
1
12
(33−3)}
10(102−1)
=+.81

5. CORRELATION
REGRESSION EQUATIONS
SHORT CUT METHOD
X 0N Y
X-𝑋
̅=r*
σx
σy
(y-𝑦
̅)
OR
X-𝑋
̅=bxy(y-𝑦
̅)
y 0N X
Y-𝑌
̅=r* σY/ σX(X-𝑋
̅)
OR
y-𝑦
̅=byx(X-𝑋
̅)
X Dx dx2
Y Dy dy2
dx.dy
1 -2 4 6 -1 1 2
2 -1 1 8 1 1 -1
3 0 0 7 0 0 0
4 1 1 6 -1 1 -1
5 2 4 8 1 1 2
10 4 2
𝑥̅=3, 𝑦
̅=7
X 1 2 3 4 5
Y 6 8 7 6 8

6. CORRELATION
r=
∑𝑑𝑥.𝑑𝑦
√∑𝑑𝑥2.∑𝑑𝑦2
=2/√10 ∗ 4
=.32
σX =√∑𝑑𝑥2/𝑁
=√10/5
=1.41
σY =√∑𝑑𝑌2/𝑁
=√4/5
=.89
X on Y
X-𝑋
̅=r*
σx
σy
(y-𝑦
̅)
X-3=.32*1.41/.89(Y-7)-----------(1)
X=0.51Y-.57
Y on X
Y-𝑌
̅=r* σY/ σX(X-𝑋
̅)
Y-7=.32*.89/1.41(X-3)
Y=.2X+6.4

7. CORRELATION
Least squares method/ Direct Method
X on Y
X=a+by------------(1)
∑x= Na+b ∑y
∑xy= a ∑y+b ∑y2
Y on X
Y=a+bX------------(2)
∑Y= Na+b ∑X
∑xy= a ∑X+b ∑X2
QUESTION2:
Y=?,X=30
X Y XY X2
Y2
1 6 6 1 36
2 8 16 4 64
3 7 21 9 49
4 6 24 16 36
5 8 40 25 64
∑X=15 ∑Y=35 ∑XY=107 55 249
X=a+by------------(1)
∑x= Na+b ∑y
X(population) 1 2 3 4 5
Y(no. of TV
demanded)
6 8 7 6 8

8. CORRELATION
∑xy= a ∑y+b ∑y2
15= 5a+35b
107= a35+249b
a=-0.5
b=0.5
lets substitute the values in the equation-(1)
X=.5y-.5
Y on X
Y=a+bX------------(2)
∑Y= Na+b ∑X
∑xy= a ∑X+b ∑X2
(35= 5a+b 15)*3
107= a15+55b---B
-105=-15a-45b---A
----------------------------
2=10b
b=.2
put the value of b in eq B
a=6.4
b=0.2
lets substitute the values in the equation-(2)

9. CORRELATION
Y=6.4+0.2X
Relationship b/w Correlation coefficient & regression
coefficients
Correlation coefficient
r=√𝑏𝑦𝑥 ∗ 𝑏𝑥𝑦
r=√. 2 ∗ .5
r=.32
find the value of y when x =7?
Y=6.4+0.2*7
Y=7.8
Exercise question:
X 1 2 3 4 5 6 7 8 9
Y 9 8 10 12 11 13 14 16 15
Regression equations, correlations coefficients_?
Y=?, x=6.2