Ho Chi Minh City University of Technology Faculty of Civil Engineering Department of Water Resources Engineering & Management

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1 Lecturer: Associ. Prof. Dr. NGUYỄN Thống or Web: 4/5/ Tél. (08) CONTENTS Chapter 1: Orientation. Evaluation of mathematical skill. Chapter 2: Taylor series (1). Partial derivatives. Chapter 3: Taylor series (2). Directional derivatives. Chapter 4: Gradient vector. Engineering application. Chapter 5: Mean, variance and standard deviation. Normal distribution. Chapter 6: Least square method. Chapter 7: Correlation coefficient. Chapter 8: Engineering applications. 4/5/

2 Course Goals Understand and describe the physical concepts and mathematical treatment of Taylor series, partial derivatives, directional derivatives, and gradient vector. Understand and describe the physical concepts and mathematical treatment of mean, variance, standard deviation, least square method and correlation coefficient. Apply the obtained skill to fundamental engineering problems. 4/5/ Lecture Notes: 4/5/ Course Materials [1 ] M.YUHI AND Y. MAENO. Physical mathematics as an engineering tool. Nakanishiya Pub. Co., 2004 Reference books: [1] RAYMOND A. BERNETT et al. Applied Mathematics. DELLEN PUBLISHING COMPANY, [2] ROBERT WREDE, Ph.D et al. Theory and problems of advanced calculus. Schaum ouline. McGRAW-HILL, [3] JAMES T. McCLAVE et al. Statistics for Businessand Economics. MAXWELL MACMILLAN INTERNATIONAL EDITIONS

3 Chapter 7: Correlation coefficient 4/5/ A correlation coefficient measures the strength and direction of a linear association between two variables. It ranges from 1 to -1. The closer the absolute value is to 1, the stronger the relationship. A correlation of zero indicates that there is no linear relationship between the variables. 4/5/

4 The coefficient can be either negative or positive. The scatterplots below indicate two linear associations of the same strength but opposite directions. 4/5/ positive relation! negative relation! 4/5/

5 4/5/ All the correlation coefficients are negative. Figure (g) represents a perfect correlation of -1, all the points fall on a straight line with a negative slope, an unlikely occurence in any social science data set. Figure (h) represents a strong negative correlation. In 4/5/ Figure (i) the correlation is moderately strong. 5

6 COMPUTING THE CORRELATION COEFFICIENT (CC.) Correlation cofficient It knows as Pearson s r N x i X y i Y i 1 2 x X y Y 2 i i 4/5/ Correlation is an effect size and so we can verbally describe the strength of the correlation using the guide that Evans (1996) suggests for the absolute value of r: very weak weak moderate strong very strong 4/5/

7 Example: Consider the literacy (%) and life expectency of six selected nation. Computing the correlation cofficient r: Solution r=12.5/ =0.92 4/5/ The Kendall Rank Correlation Coefficient The Kendall (1955) rank correlation coefficient evaluates the degree of similarity between two sets of ranks given to a same set of objects. This coefficient depends upon the number of inversions of pairs of objects which would be needed to transform one rank order into the other. 4/6/

8 For example with the following set of N = 4 objects S = {a, b, c, d} the ordered set O1 = [a,c,b,d] gives the ranks R1 = [1,3,2,4]. An ordered set on N objects can be decomposed into N(N 1)/2 ordered pairs. 4/6/ For example, O1 is composed of the following 6 ordered pairs: P1 = {[a,c], [a,b], [a,d], [c,b], [c,d], [b,d]} In order to compare two ordered sets (on the same set of objects), the approach of Kendall is to count the number of different pairs between these two ordered sets. 4/6/

9 This number gives a distance between sets (see entry on distance) called the symmetric difference distance. The symmetric difference distance between two sets of ordered pairs P1 and P2 is denoted d (P1,P2) 4/6/ The formulat for Kendall rank correlation coefficient: 1 2 N(N 1) d 1 N(N 1) 2 (P1, P2) 2 d (P1, P2) 1 N(N 1) 4/6/

10 4/6/ EXAMPLE Suppose that two experts order four wines called {a, b, c, d}. The first expert gives the following order: O1 = [a,c,b,d], which corresponds to the following ranks: R1 = [1,3,2,4]; and the second expert orders the wines as O2 = [a,c,d,b] which corresponds to the following ranks: R2 = [1,4,2,3]. The order given by the first expert is composed of the following 6 ordered pairs P1 = {[a,c], [a,b], [a,d], [c,b], [c,d], [b,d]} (1) The order given by the second expert is composed of the following 6 ordered pairs P2 = {[a,c], [a,b], [a,d], [c,b], [c,d], [d,b]} (2) 4/6/

11 The set of pairs which are in only one set of ordered pairs is {[b,d] [d,b]}, which gives a value of d (P1,P2) = 2. With this value of the symmetric difference distance we compute the value of the Kendall rank correlation coefficient between the order given by these two experts as: 4/6/ N(N 1) d (P1, P2) 2x N(N 1) 4x3 2 This large value of τ indicates that the two experts strongly agree on their evaluation of the wines (in fact their agree about everything but one pair). 4/6/

12 Exercise: Consider now the first expert gives the order R1 = [1,4,2,3]; and the seconds gives the order R2 = [1,3,4,2]. Caculation the Kendall Rank Correlation Coefficient. Conclusion? 4/6/ END OF CHAPTER 7 4/5/