Post on 28-Nov-2014
description
Correlation-Regression
It deals with association between two or more variables
Correlation analysis deals with covariation between two or more variables
Types1. Positive or negativeSimple or multipleLinear or non-linear
Methods of Measuring correlation1. Graphic Method2. Diagramatic Method- Scatter Diagram3. Algebraic methoda. Karl Pearson’s Coefficient of correlationb. Spearman’s Rank Co-efficient Correlationc. Coefficient of Concurrent deviationsd. Least Squares Method
Karl Pearson’s Coefficient of Correlation
Σ dx dy γ ( Gamma) = ------------------------- √ Σ dx2 Σ dy2
Σ dx dy = ------------------------- N σxσy dx = x-xbardy = y- ybardx dy = sum of products of deviations from respective
arithmetic means of both series
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 seriesΣ dx = total of deviations of x seriesΣ dy = total of deviations of y seriesΣ dx2 = total of squared deviations of x seriesΣ dy2 = total of squared deviations of y seriesN= No. of items ( no. of paired items
Karl Pearson’s Coefficient of Correlation After calculating assumed or working mean Ax &
Ay Σ dx x Σ dy Σ dx dy - ---------------- Nγ ( Gamma) = ------------------------- (Σ dx)2 (Σ dy)2
√ [ Σ dx2 - --------- ] x [ Σ dy2 - ------------] N N
Assumptions of Karl Pearson’s Coefficient of Correlation 1. Linear relationship exists between the variablesProperties of Karl Pearson’s Coefficient of Correlation 1.value lies between +1 & - 12.Zero means no correlation3.γ ( Gamma) = √ bxy X byxWhere bxy X byx are regression coefficicentMerit Convenient for accurate interpretation as it gives degree &
direction of relationship between two variables
Limitations 1. Assumes linear relationship , even though it
may not be2. Method & process of calculation is difficult &
time consuming3. Affected by extreme values in distribution
Probable Error of Karl Pearson’s Coefficient of Correlation
1- γ2
Probable Error of γ ( Gamma) = 0.6745 -------- √ N
Q7.Calculate coefficient of correlation for following data
X65 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
1 2 3 4 5 6 7 8 9 10 11 12SumX 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 18.7884.67
dx.dy -0.69 5.81 0.14 6.89 1.89 7.39 1.39 1.72 4.56 -0.19 0.97 10.4740.33
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 5.8438.92
Σ dx dy sum dx2* sumdy2
3294.9
√ Σ dx2 Σ dy2 57.40
coeff of correlation = 0.70
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
1 2 3 4 5 6 7 8 9 10SumX 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 62.41202.
9
dx.dy 62.37 15.17 8.37 3.57 0.77 -0.03 4.37 12.87 21.07 49.77178.
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 39.69158.
1
Σ dx dy sum dx2* sumdy232078.4
9
√ Σ dx2 Σ dy2 179.10
coeff of correlation = 1.00
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 approachAns P&Q= -0.21 , Q &R=--0.3 P &R = +0.64
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
Rank coefficient of correlation 6Σ d2 ρ (rho) = 1 - ------------------- N3-N Σ d2 = total of squared differenceN = number of items
P Q RRp-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
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