SLM_(1)

6 5 +1 1 5 8 –3 9 10 4 + 6 36 3 7 –4 16 2 10 – 8 64 4 2 +2 4 9 1 +8 64 7 6 +1 1 8 9 –1 1 n = 10, 6d2 = 200 200 Coefficient of correlation: 64d2 r = 1 – n n2 – 1 CU IDOL SELF LEARNING MATERIAL (SLM)

318 Business Mathematics and Statistics 6  200 = 1– 10102 – 1 1200 = 1– 990 = 1 – Antilog [ log 120 – log 99 ] = 1 – Antilog [ 2.0792 – 1.9956 ] = 1 – Antilog [ 0.0836 ] = 1 – 1.213 = – 0.213 In case (i) r = – 0.213 Case (ii): Second Judge Third Judge Rank difference 3 6 5 4 d d2 8 9 –3 9 4 8 +1 1 7 1 –1 1 10 2 – 4 16 2 3 + 6 36 + 8 64 –1 1

1 10 – 9 81 6 5 +1 1 9 7 +2 4 214 n = 10, 6d2 = 214 Coefficient of correlation: 6d2 r= 1– n2 –1 n 6  214 = 1– 10102 – 1 CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 319 1284 ` 1– 990 ` 1 – Antilog [ log 1284 – log 990 ] ` 1 – Antilog [ 3.1086 – 2.9956 ] ` 1 – Antilog [ 0.1130 ] ` 1 – 1.297 = – 0.297 In case (ii) r = – 0.297 Case (iii): Second Judge Third Judge Rank difference 1 6 6 4 d d2 5 9 – 5 25 10 8 +2 4 3 1 – 4 16 2 2 +2 4 4 3 +2 4 9 10 7 5 00 8 7 +1 1 –1 1 +2 4 +1 1 60

n = 10, 6d2 = 60 Coefficient of correlation: 6d2 r= 1– n2 – 1 n 6  60 = 1– 10 102 –1 360 = 1– 990 CU IDOL SELF LEARNING MATERIAL (SLM)

320 Business Mathematics and Statistics ? 1 – Antilog [log 36 – log 99] ? 1 – Antilog [ 1.5563 – 1.9956 ] ? 1 – Antilog [ 1.5607] ? 1 – 0.3637 = 0.6363 In case (iii) r = + 0.6363 Conclusions: In cases (i) and (ii) the correlation is negative, but in case (iii) the correlation is positive and therefore it is obvious that the First and Third Judges have to a large extent identical approach to common tastes in beauty. Problem 17: The coefficient of rank correlation between marks in Statistics and marks in Accountancy obtained by a certain group of students is 0.8. If the sum of squares of differences in ranks is given to be 33, find the number of students in the group. 6d2 Solution: r = 1 – n n2 – 1 Putting r = 0.8, S d2 = 33 in the above, we get 0.8= 1– 6  33 198 n n2 – 1 = 1 – n n2 – 1 198 =1–0.8 198 = 0.2

? n n2 – 1 n n2 – 1 198 n (n2 – 1) = 0.2 = 990 ?n3 – n2 – 990 = 0 (n – 10) (n2 + 10n + 99) = 0 n = 10 CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 321 Problem 18: Calculate the coefficient of correlation between the ages of 100 husbands and wives from the following data: Age of husbands Ages of Wives in Years in years 10–20 20–30 30–40 40–50 50–60 Total 15–25 6 3 — —— 9 25–35 3 16 35–45 — 10 10 —— 29 45–55 — — 55–65 — — 15 7 — 32 7 10 4 21 — 45 9 Total 9 29 32 21 9 100 Solution: We have n = 100, 6f dx dy = 98, 6f dx = – 8 6fd2x = 122, 6fdy = – 8, 6fd2y = 122 Coefficient of correlation: n  f dx dy –  f dx .  f dy r = nfd2x –f dx2nf d2y –f dy2  98 – (– 8) (– 8)

`  100  122 – 64  100  122 – 64  = 9800 – 64 12136  12136 9736 12136 Antilog [ log 9736 – log 12136 ] Antilog [ 3.9884 – 4.0842] Antilog [ 1.9042] = 0.8021 = r = + 0.8 approx. CU IDOL SELF LEARNING MATERIAL (SLM)

Age of wives (in yrs.) (X) 10-20 20-30 30-40 40-50 50-60 15 55 25 35 45 + 20 – 20 – 10 0 + 10 dx +2 Ages of husbands dy – 2 (in yrs.) (Y) –1 0 +1 f CU 25–35 30 – 10 – 1 3 16 10 — — 29 15–25 20 – 20 – 2 6 3 ——— 9 SELFIDO L LEARNING MATERIAL 35–45 40 0 0 — 10 15 7 — 32 50 + 10 + 1 — — 45-55 60 + 20 + 2 — — 7 10 4 21 55–65 —4 59 (SLM) f 9 29 32 21 9 100 fd2x 36 29 0 21 36 122 fdx – 18 – 29 0 21 18 – 8 fd2dy 30 22 0 18 28 98

Correlation Analysis 323 14.8 Summary Two statistical series vary in tandem: a variation in one series is accompanied by a variation in the other. In such case we can study closely the relationship between the two related series. The question is: if a change in one series is followed by a change in the other, what is the magnitude of change in a relative sense? In short, the numerical relationship between two series is called correlation. The following are some of the related variables height and weight, demand and price, income and expenditure. Correlation is direct or positive if two variables change in the same direction, it is inverse or negative if the variables change in opposite directions. We can study the correlation between two sets of figures through the following methods = By drawing the graph of the data = By constructing a scatter diagram. = By calculating the coefficient of correlation. The extent of correlation between two variables that are related can be determined by a coefficient. Karl Pearson devised a formula for this coefficient, which is as follows:  xy z= n 1 .2 Where r =the coefficient of correlation, n = the number of pairs of items, x = deviation of the first series (subject) from AM. Y = deviation of the second series (relative) from AM.

 1 = S D of the 1st series 2 = S D of the 2nd series. Interpretation of the value of r If r = +1, correlation is a perfect positive If r = –1, Correlation is a perfect negative. If r = 0, there is no correlation.

CU IDOL SELF LEARNING MATERIAL (SLM)

324 Business Mathematics and Statistics If r is equal to some positive number between 0 and +1 then correlation is limited and positive. If r is equal to some negative number between -1 and 0 then correlation is negative and limited. Short-cut method: When the arithmetic means of both sets of numerical items are not whole numbers and involve decimals, calculating the coefficient or correlation by the direct method becomes tedious. To overcome this difficulty the following modified short-cut method formula is used: where r = coefficient of correlation. N = the number of pairs of items X = deviation of the first series from an arbitrary mean, Y = deviation of the second series from an arbitrary mean The formula for calculating the coefficient or correlation in a grouped series is where r = coefficient of correlation, f = frequency dx = step deviation of X(1st) series, dy = step deviation of Y (2nd) series, n = total frequencies. Here is the procedure for finding the coefficient of correlation through the method of rank differences where, r = coefficient of correlation D = rank difference between the corresponding ranks N = the number of pairs of items. 14.9 Key Words/Abbreviations

Scatter diagram, correlation groups, Karl Pearson’s coefficient of correlation, Rank correlation, Probable error, concurrent deviation. CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 325 14.10 Learning Activity 12. From the data given below, find out whether there is any significant relationship between age and intelligence. Age in years Test Marks 16 17 18 19 20 150–200 32213 200–250 53422 250–300 41651 300–350 23312 350–400 11231 z The coefficient of rank correlation of the marks obtained by 10 students in English and Economics was found to be 0.5. It was later discovered that the difference in ranks in the two subjects obtained by one of the student’s was wrongly taken as 3 instead of 7. Find the correct coefficient of rank correlation. z Ten recruits were subjected to a selection test to ascertain their suitability for a certain course of training. At the time of the training they were given a proficiency test. The marks secured by the recruits in the selection test (X) and in the proficiency test (Y) are given below : X : 44 49 52 52 47 76 65 60 63 58 Y : 48 55 45 60 43 80 58 50 77 46

Calculate the coefficient of correlation and comment on its value. 4. Compute the Pearson’s coefficient of correlation for the following values of x and y and interpret your result. X : 78 89 97 69 59 79 68 61 Y : 125 137 156 112 107 136 123 108 5. Ten competitors in a music contest were ranked by three different judges as follows : Candidates: I II III IV V VI VII VIII IX X Ranking A: 1 10 9 8 3 5 2 4 6 7 by B: 3 4 1 9 7 8 10 2 5 6 judges C: 5 7 9 8 2 10 1 4 3 6 Find out which two judges differ most in their liking for music. CU IDOL SELF LEARNING MATERIAL (SLM)

326 Business Mathematics and Statistics 14.11 Unit End Questions (MCQ and Descriptive) A. Descriptive Type: Short Answer Type Questions © What is meant by the term correlation coefficient ? Explain the reasons for its calculation. © What is a ‘scatter diagram’? Explain its use in interpreting the correlation between two variables. © Write an explanatory note on the coefficient of correlation and its utility. © Define Karl Pearson’s coefficient of correlation. Interpret about the correlation when the correlation coefficients are r = 1, (ii) r = 0 and (iii) r = – 1 © What is Spearman’s rank correlation coefficient ? Explain its usefulness. 6. Interpret the formula: 0.6745 l – r2 State its practical utility in statistics. n z Define ‘Lag’ and explain its use. z Calculate the coefficient of correlation from the following data and interpret it: Yield of Cotton Price per in lakhs of bales bale 25 347 27 367

26 341 24 232 24 184 22 201 24 198 24 208 26 225 25 210 9. Find the coefficient of correlation from the following data: X : 78 36 98 25 75 82 90 62 65 39 Y : 84 51 91 60 68 62 86 58, 53 47 CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 327 10. Find Karl Pearson’s coefficient of correlation between X and Y series: X series: 17 18 19 19 20 20 21 21 22 23 Y series: 12 16 14 11 15 19 22 16 15 20 22. Find the correlation between height of father (X) and Height of son (Y) from the following data and comment upon its value: X: 65 66 67 67 68 69 70 72 64 61 Y: 67 68 65 56 72 69 71 68 65 60 10. Calculate the correlation coefficient for the following data concerning marks in Statistics and Accountancy of 12 students: Statistics: 52 74 93 55 41 23 92 64 40 71 33 71 Accountancy: 45 80 63 60 35 40 70 58 43 64 51 75 z Calculate coefficient of correlation between the heights and weights of 10 students and by the test of probable error, show whether or not the relationship is significant: S. N. of Students Height (inches) Weight (lb) 1 57 113 2 59 117 3 62 126 4 63 125 5 64 130

6 65 128 7 58 110 8 66 132 9 70 140 10 72 149 CU IDOL SELF LEARNING MATERIAL (SLM)

328 Business Mathematics and Statistics 11. Calculate the coefficient of correlation from the following table relating to the marks obtained by 12 students in two subjects. Interpret your result Students 1 2 3 4 5 6 7 8 9 10 11 12 Seat No. Marks in Subject A 65 40 35 75 63 80 35 20 85 65 55 33 Marks in Subject B 30 55 68 28 76 25 80 85 20 30 45 65 \\emdash Compute the coefficient of correlation by Pearson’s Method and also by the method of ranks, between the variables X and Y from the values given below: X : 78 36 98 25 75 82 90 62 65 39 Y : 84 51 91 60 68 62 86 58 53 47 16. Draw a scatter diagram to represent the following data: X: 7 10 17 16 12 13 9 Y : 15 18 30 27 25 23 30 Calculate the coefficient of correlation between x and y from the above data: 17. Calculate the correlation coefficient from the following data : X : 23 27 28 29 30 31 33 35 36 39

Y : 18 22 23 24 25 26 28 29 30 32 18. Calculate the coefficients of correlation between the values of x and y given below : X : 78 89 97 69 59 79 68 61 Y : 125 137 156 112 107 136 123 138 You may use 69 as working mean for x and 112 as that for y. = State the limits between which correlation coefficient lies. Compute its value for the following data: Age of 22 23 24 25 26 27 28 29 30 31 Bridegroom 18 18 29 18 21 23 25 24 25 26 CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 329 20. Find the coefficient of correlation ‘r’ using the shortcut method. X : 97 34 68 56 71 43 39 52 69 Y : 110 113 124 101 142 115 131 120 103 21. Find if there is any significant correlation between the following heights and weights. Height: 59 62 60 64 61 67 66 63 65 68 (in ins.) Weight: 112 118 114 117 110 125 122 116 124 128 (in lb) 22. Find out the coefficient of correlation for the data given below: X : 68 56 72 80 89 93 94 99 Y : 97 110 112 118 116 122 125 103 Also calculate its Probable Error. J Eight participants in a beauty contest were given the following ranks by three different judges: Judge A: 3 5 4 2 1 67 8 Judge B: 5 2 1 6 3 48 7 Judge C: 1 3 2 5 6 78 4

By using the method of rank differences for finding the coefficient or correlation, find the pair of judges whose tastes of beauty are nearly identical. 24. Calculate the coefficient of concurrent deviations from the following data : X : 40 44 46 50 53 57 61 66 72 78 Y : 42 45 48 52 55 59 64 70 78 86 25. Calculate the coefficient of correlation from the followng table: X 21 22 23 24 25 26 27 Total Y —— — — 3— 58 0–10 10–20 —— — 6 4 7 2 19 20–30 6 — — 21 —— 8 7 CU IDOL SELF LEARNING MATERIAL (SLM)

330 Business Mathematics and Statistics 30–40 — 11 9 5 — — — 25 40–50 — 8 7 2— — — 17 50–60 5 4 6 —— — — 15 60–70 9 4 — —— — — 13 Total 14 27 30 20 13 7 7 118 B. Multiple Choice/Objective Type Questions 1. The following is not a related variables. (a) Height and weight (b) Demand and Price (c) Income and expenditure (d) Weight and price 2. r = coefficient of correlation, If r = +1 the correlation is __________. (a) Perfect positive (b) Perfect negative (c) Imperfect positive (d) None if these 8 If r is equal to some negative number between –1 and 0 then correlation is __________ and limited. (a) Positive (b) Negative (c) Multiple (d) All of these 4. P E is meant for _______________. (a) Public Error (b) Private error (c) Probable error (d) None of these

23. If the plotted points in a scatter diagram move downwards from left top to right bottom, then the condition is __________. (a) Zero (b) Negative (c) Positive (d) None of these Answers: (1) (d); (2) (a); (3) (b); (4) (c); (5) (b) 14.12 References References of this unit have been given at the end of the book. ˆˆˆ CU IDOL SELF LEARNING MATERIAL (SLM)

Regression Analysis 331 UNIT 15 REGRESSION ANALYSIS Structure 15.0 Learning Objectives 15.1 Introduction 15.2 Regression Equation of Y on X 15.3 Regression Equation of X on Y 15.4 Regression Coefficients 15.5 Summary 15.6 Key Words/Abbreviations 15.7 Learning Activity

15.8 Unit End Questions (MCQ and Descriptive) 15.9 References 15.0 Learning Objectives After studying this unit, you will be able to:  Analyse the method of calculating coefficient of correlation in a grouped data.   Elaborate Spearman’s method of calculating rank correlation and as well as the method for finding the coefficient of concurrent deviations   Draw the Galtons graph for ratio of variation and the meaning of lag and lead.   Explain the method of constructing the two regression equation X on Y and Y on X.   Calculate the two equation coefficient and using these value you can find the value of  the correlation coefficient.  Calculate the regression coefficient of the value of r, x and y . CU IDOL SELF LEARNING MATERIAL (SLM)

332 Business Mathematics and Statistics 15.1 Introduction The term “regression” was used by Sir Francis Galton to describe a hereditary phenomenon which he observed in his studies relating to the heights of sons and fathers. His main observation was that though tall fathers have usually, tall sons, the average height of the sons of tall fathers is less than the average height of the fathers. In short, the average height of the sons of tall fathers will regress or go back towards the general average height. This backward or downward tendency in the average height was described by Sir Francis Galton as regression. At present the term regression is being used very widely for describing many other types of economic, business and social phenomena. A line that is drawn as close as possible to the plotted points of the scatter diagram shows the average tendency of the plotted points. This line is known as the regression line and its equation is called the regression equation. The coefficient of correlation indicates the extent of relationship between two sets of figures whereas a regerssion equation enables us to calculate the amount of change in one variable corresponding to a change in the other. 15.2 Regression Equation of Y on X The regression equation that enables us to find out the amount of change in Y corresponding to a change in X, is called the regression equation of Y on X. If this regression equation is repersented by Y = a + bX, then the constants a and b are determined from the two normal equations: 6Y = Na + b6X

6XY = a6X + b6X2 15.3 Regression Equation of X on Y The regression equation that enables us to find out the amount of change in X corresponding to a change in Y is called the regression of X on Y. The regression equation is represented by X = a' + b'Y, where the constant a' and b' are determined from two normal equations: CU IDOL SELF LEARNING MATERIAL (SLM)

Regression Analysis 333 6Y = Na' + b'6X 6XY = a'6X + b'6X2 Conclusions: From the graph of the two regression lines the following conclusions can be drawn: ? The two regression lines intersect at the point x , y where x = Arithmetic mean of the items in X series = Arithmetic mean of the items in Y series 11. If the two regression liens are close to each other then the correlation between X series and Y series is very high. 12. If the two regression lines coincide then there is perfect correlation. 13. If the two regression liens are at a distance from each other then the correlation between X series and Y series is of less degree. 14. If the two regression lines cut at right angles then there is no correlation between X series and Y series. Problem 1: From the data given below, find out: ? Karl Pearson’s coefficient of correlation ? The two regression equations ? The two regression coefficients ? The most likely value of X when Y = 41 ? The most likely value of Y when X = 45

X: 52 60 58 39 41 53 47 34 Y: 40 46 43 54 49 55 48 57 Xx x2 Yy y2 xy 52 + 4 40 – 9 60 + 12 16 46 – 3 81 – 36 58 + 10 144 43 – 6 9 – 36 39 – 9 100 54 + 5 36 – 60 41 – 5 49 0 25 – 45 53 + 5 81 55 + 6 00 49 36 + 30 25 CU IDOL SELF LEARNING MATERIAL (SLM)

334 Business Mathematics and Statistics 47 – 1 1 48 –1 1 +1 34 – 14 196 57 + 8 64 – 112 384 612 392 252 – 258 384 For X: A.M = x = 8 = 48 2 612 x S.D. = x = = = 76.5 = 8.746 n8 392 For Y: A.M = y = 8 = 49 252 y2 S.D. = y = = = 31.5 = 5.612 n 8 xy Coefficient of correlation: r = n x  . y =–.66 258 = 8 (8.746) (5.612) Regression Equations: The regression equation of Y on X is

 y y – y = r x (x – x ) 5.61 y – 49 = – .66x 8.75 (x – 48) y – 49 = – .42 x + 20.16 y = – .42 x + 69.16 The regression equation of X on Y is x x – x = r  y (y – y ) 8.75 x – 48 = – .66 × 5.61 (y – 49) CU IDOL SELF LEARNING MATERIAL (SLM)

Regression Analysis 335 xy 258 x – 48 = – 1.02y + 49.98 r =  x2 y2 = – 612  252 =–.66 x = – 1.02y + 97.98 15.4 Regression Coefficients The regression coefficient of y on x is  y 5.61 ? x = .66 × 8.75 = – .42 The regression coefficient of x on y is  x 8.75 r y = – .66 × 561. = – 1.02 The most likely value of X when Y = 41 = = – 1.02 Y + 97.98 (– 1.02) × 41 + 97.98 – 41.82 + 97.98 56.16

The most likely value of Y when X = 45 Y = – .42X + 69.16 = – 41 × 45 + 69.16 = – 18.90 + 69.16 = 50.26. Problem 2: It is given that two regression coefficients are + .8 and + .6 find out the coefficient of correlation.  y r= r x r  yx = ( .8) (+ .6) CU IDOL SELF LEARNING MATERIAL (SLM)

336 Business Mathematics and Statistics z .48 z + .69 Problem 3: Following data refers to years of service in a factory of seven persons in a specialised field and to their monthly incomes: Years of Service 11 7 9 5 8 6 10 Income in 1000s of ` 7532648 Find the regression equation of income on years of service. Using it, what initial start would you recommend for a person applying for a job after having served in another factory in a similar field for twelve years? Solution: Years Deviation Income Deviation y2 xy X x from y from x2 Y* 4 +6 11 Arithmean Arithmean 00 7 4 –2 9 +3 97 +2 9 +9 5 –1 15 0 10 8 +1 13 1 +2 6 –3 92 –2 06 –3 3 44 +1 –2 –1

10 +2 4 8 +3 9 +6 28 21 56 0 28 35 0 * Income in thousands of Rupees 56 For series X: Arith-Mean = 7 = 8, i.e., x = 8 35 For series Y: Arith-Mean = = 5, i.e., y = 5 7 For X series: S. D. = x 2 28 = 4=2 x = 7 = n For Y series: S. D. = y 2 28 4=2 y == 7 = n CU IDOL SELF LEARNING MATERIAL (SLM)

Regression Analysis 337 The coefficient of correlation is: 3 xy 21 r = n  x y =7  2  2 = 4 = 0.75 Regression Equation y on x is: y x–x y– y =r x 2 i.e., y – 5 = 0.75 ×2 (x – 8) 3 i.e., y – 5 = 4 (x – 8) i.e., 4y – 20 = 3x – 24 3 4y = 3x – 4 ; or y = 4 x – 1 when x = 12, the value of y is obtained from the above equation: 3 i.e., y = 4 × 12 – 1 = 9 – 1, i.e., y = 8.

Remark: For a person who has served in another factory in a similar field for 12 years, the initial start that could be recommended is ` 8,000. Problem 4: For a group of children, mean age is 10 years with standard deviation 2.5 years. The average height of the group is 125 cm. with standard deviation of 13 cm. The coefficient of correlation between age and height is 0.6. Write the equations of the two regression lines and explain their use. Solution: Regression equation of y on x is: y r y – y = x x – x and regression equation of x on y is:  x r x – x = y y – y CU IDOL SELF LEARNING MATERIAL (SLM)

338 Business Mathematics and Statistics = The given values are: Mean age: x = 10 S.D.: x = 2.5 Mean height: y = 12.5 S.D.: y = 13 r = + 0.6 Regression equation of y on x is x y– y = r x– x x 13 i.e., y – 125 = 0.6 × 2.5 (x – 10) 7.8 y – 125 = 2.5 (x – 10) ? y = 3.12 (x – 10) + 125 = 3.12x – 31.2 + 125 ? y = 3.12x + 93.8 Regression equation of x on y is: x y– y r x–x= y

2.5 i.e., x – 10 = 0.6 × 13 (y – 125) 1.50 x – 10 = 13 (y – 125) = x = 0.11 (y – 125) + 10 x = 0.11y – 13.75 + 10 x = 0.11y – 3.75 By using the regression equation of y on x it is possible to estimate the value of y given the value of x. Similarly using the regression equation of x on y we can estimate the value of x on the CU IDOL SELF LEARNING MATERIAL (SLM)

Regression Analysis 339 basis of the given value of y. Corresponding to these two regression equations we have two regression lines. These enable us to find the relationship between the two variables. Problem 5: Given that the regression equation of Y on X is 5y – 3x = 5 and that of X on Y is 3y – 5x – 2 = 0, find the coefficient of correlation between X and Y. Solution: Regression equation of y on x is: 3 5y – 3x = 5 ?y = 5 x – 1 The regression coefficient of y on x is: == 3  r x 5 Regression equation of x on y is: =2 3x – 5x – 2 = 0 ? x = 5 y – 5 Regression coefficient of x on y is: ?3 r= Vx 5 V We have

V x Vy [Ans. r + 0.6] r= r ur Vy Vx 3 33 = 5 u 5 = 5 = 0.6 Problem 6: For two variables x and y the regression of x on y is x = 4y – 3 and the regression equation of y on x is 9y = x + 13. Find the mean of x and y and the coefficient of correlation between x and y. CU IDOL SELF LEARNING MATERIAL (SLM)

340 Business Mathematics and Statistics Solution: Solving the two given regression equations we get the mean values of x and y 4y = x + 3 … (1) 9y = x + 13 … (2) Subtracting equation (1) from (2) 5y = 10 ?y=2 8= x+3 ?x = 8 – 3 = 5 ` The mean values are 5 and 2. Regression coefficient of y on x is byx = r y 1 = x 9 Regression coefficient of x on y is 5xy = r x = 4 y 1 4 9 r = byx  bxy =  4 = 9 2 = 3 = 0.67 [Ans: r = + 0.67] Problem 7: For the data given below obtain the two regression equations. X: 2 314 Y: 4 512 Solution:

x x2 y y2 xy 2 4 4 16 8 3 9 5 25 15 1 1111 4 16 2 4 8 10 30 12 46 32 Let the regression equation of y on x be: y = a + bx, then a and b are determined from the two normal equations. CU IDOL SELF LEARNING MATERIAL (SLM)

Regression Analysis 341 6y = na + b6x … (1) 6xy = a6x + b6x2 … (2) 12 = 4a + 10b 32 = 0a + 30b Multiply equation (1) by 3 and subtract equation (2) from it, then 36 = 12a + 30b 32 = 10a + 30b 4 = 2a ?a=2 From equation (1), we get 12 = 4 × 2 + 10b = 12 = 8 + 10b ? 12 – 8 = 10b 4 ?b = 10 = 0.4 The regression equation of y on x is: y = a + bx i.e., y = 2 + 0.4x Let the regression equation of x on y be: x = a' + b'y then the constants a' and b' are determiend from the two normal equations 6x = na' + b'6x 6xy = a'6x + b'6x2

Therefore, … (1) 10 = 4a' + 12b' … (2) 32 = 12a' + 46b' Multiply equation (1) by 3 and subtract it from (2) 30 = 12a' + 36b' 32 = 12a' – 46b' 2 = 10b' 2 ? b' = 10 = 0.2 CU IDOL SELF LEARNING MATERIAL (SLM)

342 Business Mathematics and Statistics From equation (1) 10 = 4a' + 12 (0.2) 10 = 4a' + 2.4 ?10–2.4 = 4a' = 4a' = 7.6 7.6 a' = 4 = 1.9 The regression equation of x on y is: x = a' + b’y i.e., x = 1.9 + 0.2y Problem 8: Given the following data, estimate the marks in Mathematics obtained by a student who has scored 60 marks in English. : 80 Arithmetic average of marks in Mathematics (all students) : 50 Arithmetic average of marks in English (all students) : 15 S. D. of marks in Mathematics : 10 S. D. of marks in English Coefficient of correlation between marks in Mathematics and marks in English is 0.4 Solution: Maths : x = 80, x = 15 r = 0.4