The document discusses correlation and regression analysis. It describes correlation as dealing with the association between two or more variables. There are different types of correlation including positive, negative, simple, and multiple correlations. Methods for measuring correlation include graphical, diagrammatic, and algebraic methods such as Karl Pearson's coefficient of correlation and Spearman's rank correlation coefficient. Karl Pearson's coefficient formula is provided and its assumptions and properties are described.

1. Correlation-Regression

2. It deals with association between two or
more variables
Correlation analysis deals with
covariation between two or more
variables
Types
1. Positive or negative
Simple or multiple
Linear or non-linear

3. Methods of Measuring correlation
1. Graphic Method
2. Diagramatic Method- Scatter Diagram
3. Algebraic method
a. Karl Pearson’s Coefficient of correlation
b. Spearman’s Rank Co-efficient Correlation
c. Coefficient of Concurrent deviations
d. Least Squares Method

4. Karl Pearson’s Coefficient of Correlation
Σ dx dy
γ ( Gamma) = -------------------------
√ Σ dx2
Σ dy2
Σ dx dy
= -------------------------
N σxσy
dx = x-xbar
dy = y- ybar
dx dy = sum of products of deviations from
respective arithmetic means of both series

5. Karl Pearson’s Coefficient of Correlation
After calculating assumed or working mean Ax &
Ay
Σ dx dy – (Σ dx) *( Σ dy)
γ ( Gamma) = --------------------------------
√ [ NΣ dx2
- (Σ dx)2
* [Σ Ndy2
- (Σ dy)2
]
Σ dx dy = total of products of deviation from
assumed means of x and y series
Σ dx = total of deviations of x series
Σ dy = total of deviations of y series
Σ dx2
= total of squared deviations of x series
Σ dy2
= total of squared deviations of y series
N= No. of items ( no. of paired items

6. Karl Pearson’s Coefficient of Correlation
After calculating assumed or working mean Ax &
Ay
Σ dx x Σ dy
Σ dx dy - ----------------
N
γ ( Gamma) = -------------------------
(Σ dx)2
(Σ dy)2
√ [ Σ dx2
- --------- ] x [ Σ dy2
- ------------]
N N

7. Assumptions of Karl Pearson’s Coefficient of
Correlation
1. Linear relationship exists between the variables
Properties of Karl Pearson’s Coefficient of
Correlation
1.value lies between +1 & - 1
2.Zero means no correlation
3.γ ( Gamma) = √ bxy X byx
Where bxy X byx are regression coefficicent
Merit
Convenient for accurate interpretation as it gives
degree & direction of relationship between two
variables

8. Limitations
1. Assumes linear relationship , even though it
may not be
2. Method & process of calculation is difficult &
time consuming
3. Affected by extreme values in distribution

9. Probable Error of Karl Pearson’s Coefficient of
Correlation
1- γ2
Probable Error of γ ( Gamma) = 0.6745 --------
√ N

10. Q7.Calculate coefficient of correlation for following data
X
65 63 67 64 68 62 70 66 68 67 69 71
Y 68 66 68 65 69 66 68 65 71 67 68 70
Ans Σ dx dy
γ ( Gamma) = -------------------------
√ Σ dx2
Σ dy2
Σ dx dy
= -------------------
N σxσy

11. 1 2 3 4 5 6 7 8 9 10 11 12
Su
mX Xbar
X 65 63 67 64 68 62 70 66 68 67 69 71 800 66.67
Y 68 66 68 65 69 66 68 65 71 67 68 70 811 67.58
dx=x-xbar -1.67 -3.67 0.33 -2.67 1.33 -4.67 3.33 -0.67 1.33 0.33 2.33 4.33
dx2
2.78 13.44 0.11 7.11 1.78 21.78 11.11 0.44 1.78 0.11 5.44 18.78
84.
67
dx.dy -0.69 5.81 0.14 6.89 1.89 7.39 1.39 1.72 4.56 -0.19 0.97 10.47
40.
33
dy=y-ybar 0.42 -1.58 0.42 -2.58 1.42 -1.58 0.42 -2.58 3.42 -0.58 0.42 2.42
dy2
0.17 2.51 0.17 6.67 2.01 2.51 0.17 6.67 11.67 0.34 0.17 5.84
38.
92
Σ dx dy
sum dx2*
sumdy2
3294.
9
√ Σ dx2 Σ dy2 57.40
coeff of
correlation = 0.70

12. Q8. following information about age of husbands
& wives. Find correlation coefficient
Husband
23 27 28 29 30 31 33 35 36 39
Wife 18 22 23 24 25 26 28 29 30 32
γ ( Gamma) =0.99

13. 1 2 3 4 5 6 7 8 9 10
Sum
X Xbar
X 23 27 28 29 30 31 33 35 36 39 311 31.10
Y 18 22 23 24 25 26 28 29 30 32 257 25.70
dx=x-
xbar -8.10 -4.10 -3.10 -2.10 -1.10 -0.10 1.90 3.90 4.90 7.90
dx2
65.61 16.81 9.61 4.41 1.21 0.01 3.61 15.21 24.01 62.41
202.
9
dx.dy 62.37 15.17 8.37 3.57 0.77 -0.03 4.37 12.87 21.07 49.77
178.
3
dy=y-
ybar -7.70 -3.70 -2.70 -1.70 -0.70 0.30 2.30 3.30 4.30 6.30
dy2
59.29 13.69 7.29 2.89 0.49 0.09 5.29 10.89 18.49 39.69
158.
1
Σ dx dy sum dx2* sumdy2
32078.4
9
√ Σ dx2 Σ dy2 179.10
coeff of correlation
= 1.00

14. Q9. ten competitors in a cooking competition are
ranked by three judges in the following way .by
using rank coorelation method find out which
pair of judges have nearest approach
Ans P&Q= -0.21 , Q &R=--0.3 P &R = +0.64

15. Q9. ten competitors in a cooking competition are ranked
by three judges in the following way .by using rank
coorelation method find out which pair of judges have
nearest approach
P Q R
1 1 3 6
2 6 5 4
3 5 8 9
4 10 4 8
5 3 7 1
6 2 10 2
7 4 2 3
8 9 1 10
9 7 6 5
10 8 9 7

16. Rank coefficient of correlation
6Σ d2
ρ (rho) = 1 - -------------------
N3
-N
Σ d2
= total of squared difference
N = number of items

17. P Q R
Rp-
Rq dpq2
Rq-
Rr dqr2
Rp-
Rr dpr2
1 1 3 6 -2 4 -3 9 -5 25
2 6 5 4 1 1 1 1 2 4
3 5 8 9 -3 9 -1 1 -4 16
4 10 4 8 6 36 -4 16 2 4
5 3 7 1 -4 16 6 36 2 4
6 2 10 2 -8 64 8 64 0 0
7 4 2 3 2 4 -1 1 1 1
8 9 1 10 8 64 -9 81 -1 1
9 7 6 5 1 1 1 1 2 4
10 8 9 7 -1 1 2 4 1 1
1000 200 214 0 60
6Sigma d2 1200 1284 360
N3
-N 990 6Sigma d2/N3
-N 1.21 1.297 0.3636
P= -0.21 -0.297 0.636364