Correlation & regression (3)
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Correlation & regression (3)

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Correlation & regression - Unitedworld School of Business

Correlation & regression - Unitedworld School of Business

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Correlation & regression (3) Correlation & regression (3) Presentation Transcript

  • Correlation-Regression
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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 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
  • 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
  • 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 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
  • 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
  • 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 difference N = number of items
  • 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
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