Correlation analysis ppt
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Correlation analysis measures the relationship between two or more variables. The sample correlation coefficient r ranges from -1 to 1, indicating the degree of linear relationship between variables. A value of 0 indicates no linear relationship, while values closer to 1 or -1 indicate a strong positive or negative linear relationship. Excel can be used to calculate r using the CORREL function.
![Direct method
N .∑ XY − ∑ X .∑ Y
r=
[ N ∑ X − ( ∑ X ) ]. [ N.∑ Y − ( ∑ Y ) ]
2 2 2 2
08/01/12 anilmishra5555@rediffmail.com 17](https://image.slidesharecdn.com/correlationanalysisppt-120801035650-phpapp02/85/Correlation-analysis-ppt-17-320.jpg)
Correlation analysis ppt
- 1. CORRELATION ANALYSIS It is a statistical measure which shows relationship between two or more variable moving in the same or in opposite direction 08/01/12 [email protected] 1
- 2. Visual Displays and Correlation Analysis • Correlation Analysis • The sample correlation coefficient (r) measures the degree of linearity in the relationship between X and Y. -1 < r < +1 Strong negative relationship Strong positive relationship • r = 0 indicates no linear relationship • In Excel, use =CORREL(array1,array2), where array1 is the range for X and array2 is the range for Y.
- 3. Types of correlation correlation Simple , positive Linear multiple & negative & non-linear & partial 08/01/12 [email protected] 3
- 4. Methods of correlation • Scatter diagram • Product moment or covariance • Rank correlation • Concurrent deviation 08/01/12 [email protected] 4
- 5. Scatter diagram • Perfectly +ve 08/01/12 [email protected] 5
- 6. Less-degree +ve Weak Positive Correlation 08/01/12 [email protected] 6
- 7. High degree +ve Strong Positive Correlation 08/01/12 [email protected] 7
- 9. High degree -ve Strong Negative Correlation 08/01/12 [email protected] 9
- 10. Less degree -ve Weak Negative Correlation 08/01/12 [email protected] 10
- 12. Karl Pearson correlation coefficient cov( x, y ) r= σ x .σ y r= ∑ x. y ∑ x .∑ y 2 2 08/01/12 [email protected] 12
- 13. Where x = X −X and y =Y −Y 08/01/12 [email protected] 13
- 14. problem From the following data find the coefficient of correlation by Karl Pearson method X:6 2 10 4 8 Y:9 11 5 8 7
- 15. Sol. X Y X-6 Y-8 x2 y2 x. y 6 9 0 1 0 1 0 2 11 -4 3 16 9 -12 10 5 4 -3 16 9 -12 8 8 -2 0 4 0 0 4 7 2 -1 4 1 -2 30 40 0 0 40 20 -26
- 16. Sol.cont. X= ∑X = 30 =6 N 5 Y = ∑ Y = 40 = 8 N 5 r= ∑ x. y = − 26 = − 26 ≈ −0.92 ∑ x 2 .∑ y 2 40.20 800
- 17. Direct method N .∑ XY − ∑ X .∑ Y r= [ N ∑ X − ( ∑ X ) ]. [ N.∑ Y − ( ∑ Y ) ] 2 2 2 2 08/01/12 [email protected] 17
- 18. Short-cut method N ∑ d x .d y − ∑ d x .∑ d y r= N ∑ d x − (∑ dx ) . N ∑ d y − (∑ d y ) 2 2 2 2 08/01/12 [email protected] 18
- 19. Where dx = X − A & dy = Y − A A = assume mean 08/01/12 [email protected] 19
- 20. Product moment method r = bxy .byx where bxy = ∑xy ∑ y 2 byx = ∑xy 08/01/12 ∑x [email protected] 2 20
- 21. spearman’s Rank correlation 6∑ D 2 R = 1− N ( N − 1)2 where D = Rx − R y Rx = rank .of . X R y = rank .of . y 08/01/12 [email protected] 21
- 22. problem Calculate spearman’s rank correlation coefficient between advt.cost & sales from the following data Advt.cost :39 65 62 90 82 75 25 98 36 78 Sales(lakhs): 47 53 58 86 62 68 60 91 51 84
- 23. X Y R-x R-y D D 2 Sol. 39 47 8 10 -2 4 65 53 6 8 -2 4 62 58 7 7 0 0 90 86 2 2 0 0 82 62 3 5 -2 4 75 68 5 4 1 1 25 60 10 6 4 16 98 91 1 1 0 0 36 51 9 9 0 0 78 84 4 3 1 1 30
- 24. Sol.cont. 6∑ D 2 R = 1− N −N3 6.30 ⇒ R = 1− 3 10 − 10 2 ⇒ R = 1− 11 9 ⇒ R = = 0.82 11
- 25. In case of equal rank 6 ∑ D + 2 1 3 12 ( m −m + 1 3 12 ) m − m + ........( ) R = 1− N N −1 2 ( ) where m = no.of repeated items 08/01/12 [email protected] 25
- 26. problem A psychologist wanted to compare two methods A & B of teaching. He selected a random sample of 22 students. He grouped them into 11 pairs so that the students in a pair have approximately equal scores in an intelligence test. In each pair one student was taught by method A and the other by method B and examined after the course. The marks obtained by them as follows Pair:1 2 3 4 5 6 7 8 9 10 11 A: 24 29 19 14 30 19 27 30 20 28 11 B: 37 35 16 26 23 27 19 20 16 11 21
- 27. A B R-A R-B D D2 Sol. 24 37 6 1 5 25 29 35 3 2 1 1 19 16 8.5 9.5 -1 1 14 26 10 4 6 36 30 23 1.5 5 -3.5 12.25 19 27 8.5 3 5.5 30.25 27 19 5 8 -3 9 30 20 1.5 7 -5.5 30.25 20 16 7 9.5 -2.5 6.25 28 11 4 11 -7 49 11 21 11 6 5 25 225
- 28. Sol.cont. in A series the items 19 & 30 are repeated twice and in B series16 is repeated twice ∴ 2( 4 − 1) 2( 4 − 1) 2( 4 − 1) 6 ∑ D + 2 + + 12 12 12 R = 1- 11(121 − 1) ⇒ R = −0.0225
- 29. Properties of correlation coefficient • r always lies between +1 & -1 i.e. -1<r<+1 • Two independent variables are uncorrelated but converse is not true • r is independent of change in origin and scale • r is the G.M. of two regression coefficients • r is symmetrical 08/01/12 [email protected] 29
- 30. Probable error 1 −r 2 SE ( r ) = n PE ( r ) = 0.6745 × SE ( r ) or 1 −r 2 PE ( r ) = 0.6745 × n 08/01/12 [email protected] 30
- 31. The coefficient of determination It is the primary way we can measure the extent or strength of the association that exists between two variables x & y . Because we have used a sample of points to develop regression lines . 2 It is denoted by r