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13.9 Unit End Questions (MCQ and Descriptive) A. Descriptive Type: Short Answer Type Questions z Discuss the methods generally used in sampling. Explain in brief the law of statistical regularity. z Distinguish between population and sample. What do you understand by a random sample and how is it drawn.? z What is sampling? What are the requisites of a good sample? Describe any two methods of sampling. z What is a random sample? Discuss the merits and demerits of census and sample surveys. CU IDOL SELF LEARNING MATERIAL (SLM)

Types of Probability 293 © What is meant by sample method of enquiry? When is it adopted? What are its advantages? What are the essential requisites of a good sample? © Explain the terms ‘sample’ and ‘population’. Distinguish between census and sample survey. © Describe briefly the stratified and systematic sampling methods. © Explain the methods of systematic and stratified sampling with suitable illustrations. © State the merits claimed by the method of random sampling in statistical surveys. © Compare and contrast the merits and demerits of the sample survey and the census methods: © Write short notes on: Random sampling Systematic sampling Purposive sampling Stratified sampling Multistage sampling Law of statistical regularity Law of inertia of large numbers © Describe any two sampling methods with suitable illustrations. © Write notes on: Statistical populations and samples Stratified sampling

© Describe briefly the following Statistical random sampling Stratified sampling Systematic sampling CU IDOL SELF LEARNING MATERIAL (SLM)

294 Business Mathematics and Statistics Describe the term ‘random sampling’ and explain why randomness is important in sample design? Describe briefly probability and Non Probability methods of sampling? B. Multiple Choice/Objective Type Questions If N = 12000, e = 0.05, then n = _______________. (a) 450 (b) 350 (c) 300 (d) None of these 2. If p = 0.20, q = 0.80, e = 0.05, z = 1.96, then N = 246 (a) correct (b) incorrect (c) non-representative (d) none of these 3. Purposive sampling is also called as _______________. (a) Systematic sampling (b) Judgement sampling (c) Random sampling (d) None of these 4. Probability and non-probability are the two categories of ___________________. (a) Sample (b) Probability (c) Sampling method (d) All the above Answers: (1) (d); (2) (a); (3) (b); (4) (c) 13.10 References

References of this unit have been given at the end of the book. ˆˆˆ CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 295 UNIT 14 CORRELATION ANALYSIS Structure 14.0 Learning Objectives 14.1 Introduction 14.2 Measuring of Correlation 14.3 Short-cut Method 14.4 The Probable Error in the Coefficient of Correlation 14.5 Coefficient of Correlation in a Grouped Data 14.6 Rank Correlation 14.7 Additional Problems

14.8 Summary 14.9 Key Words/Abbreviations 14.10 Learning Activity 14.11 Unit End Questions (MCQ and Descriptive) 14.12 References 14.0 Learning Objectives After studying this unit, you will be able to:  The concept of correlation and the methods for measuring correlation.   The method for calculating Karl Pearson’s coefficient of correlation using the direct method and the shortcut method.  Calculation of probable error in the coefficient of correlation CU IDOL SELF LEARNING MATERIAL (SLM)

296 Business Mathematics and Statistics 14.1 Introduction It is very often observed that two statistical series sometimes very together in the sense that a variation in one series is quite often accompanied by a variation in the other. In such a case we can study closely the extent of relationship existing between the two related series. The question is : if a change in one series is followed by a change in the other, that what is the amount of such a change relatively? In short, the numerical relationship between the two series is called correlation. The purpose of this topic is to study the methods of such a relationship between two series and also to explain the significance of the numerical relationship. The important set of figures, that is, the first set of figures is called the Subject and the second set of figures is called the Relative. In other words that set which we use as the standard set for the purpose of comparison is called the Subject. The following are some of the related variables : Height and Weight, Demand and Price, Income and Expenditure. If two variables change in the same direction then correlation is direct or positive; but if the variables change in opposite directions then correlation is inverse or negative. 14.2 Measuring of Correlation We can study the correlation between two sets of figures by means of the following methods: z By drawing the graph of the data. z By constructing a Scatter Diagram. z By calculating the coefficient of correlation. 14.2.1 Graphical Method

We represent each of the two sets of figures by plotting the relevant points on the graph and then join the points so plotted. The movements of the two curves so formed are then studied closely. If we find that the curves move in the same direction then we say that correleation between the two sets of figures is positive. If we observe that the curves move in opposite directions, then it is obvious that correlation is negative. CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 297 Fig. 14.1: Correlation is Positive Fig. 14.2: Correlation is Negative 14.2.2 Scatter Diagram Each of the corresponding pairs of numerical values in the two sets of figures, can be represented by a point on the graph. By taking values for the First Series along the X axis and the values for the Second Series along the Y axis, the entire data can be represented by the plotted points. The figure obtained by plotting the points, is called a Scatter Diagram.

Fig. 14.3: Correlation is Positive Fig. 14.4: Correlation is Negative By drawing a line as closely as possible to the plotted points, we can study the nature of correlation. If the direction of the line is upwards from left to right then correlation is positive; but if the direction of the line is downwards from left to right (see Fig. 14.4) then correleation is negative. If we observe that the points are scattered widely then there is no correlation (see Fig. 14.5) CU IDOL SELF LEARNING MATERIAL (SLM)

298 Business Mathematics and Statistics Correlation Fig. 14.5: Correlation: Absent Coefficient of Correlation The extent of correlation between two variables that are related can be determined by means of a coefficient. Karl Pearson devised a formula for this coefficient, which is as follows :  xy \\emdash = n 1 .2 where r = the coefficient of correlation

n = the number of pairs of items x = deviation of the 1st series (Subject) y = deviation of the 2nd series (Relative) 1 = S.D. of the 1st series 2 = S.D. of the 2nd series Remarks : = If r = + 1 then correlation is perfect positive. = If r = – 1 then correlation is perfect negative. = If r = 0 then there is no correlation. = If r is equal to some number between 0 and + 1 then correlation is limited and positive. CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 299 I If r is equal to some negative number between – 1 and 0 then correlation is negative and limited. Problem 1: Calculate the coefficient of correlation for the following data, by Karl Pearson’s Method. X : 46, 52, 63, 59, 73, 70, 87, 86 Y : 39, 41, 52, 68, 60, 64, 76, 80 X x x2 Y y y2 xy 46 – 21 441 39 52 – 15 225 41 – 21 441 + 441 63 59 – 4 16 52 – 19 361 + 285 73 – 8 64 68 70 + 6 36 60 – 8 64 + 32 87 + 3 9 64 86 + 20 400 76 + 8 64 – 64 + 19 361 80 536 0 00 n= 8 + 4 16 + 12 + 16 256 + 320 + 20 400 + 380 1552 480 1602 1406 536 480 x = 8 = 67 y = 8 = 60

 x2 1552  y2 1602 , y = = 8 x = x = = 8 n n 6xy = 1406 Coefficient of correlation is : xy r= n.x.y xy = x2 y2 nn CU IDOL SELF LEARNING MATERIAL (SLM)

300 Business Mathematics and Statistics xy 7  x2 . y2 1406 = 1552  1602 1 = Antilog [ log 1406 – 2 (log 1552 + log 1602 ) 1 19. Antilog [ log 3.1479 – 2 (3.1907 + 3.2047) 20. Antilog [3.1479 – 3.1977) 21. Antilog (1.9502) 22. 0.8917 Coefficient of correlation is + 0.89 approximately. 14.3 Short-cut Method When the respective arithmetic means of both the sets of numerical items are not whole numbers but involve decimals, then the calculation of the coefficient of correlation by the direct

method, becomes a tedious process. To overcome this difficulty and inconvenienc the following modified short-cut method formula is used. n xy –  x.y r=  2 2  2 2  n x –  x  n y – y  where r = coefficient of correlation n = the number of pairs of items x = deviation of the 1st series from an arbitrary mean y = deviation of the 2nd series from an arbitrary mean Problem 2: Calculate the coefficient of correlation for the following data by using the short- cut CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 301 X: 47, 53, 64, 60, 74, 71, 89, 88 Y: 41, 43, 53, 69, 61, 65, 77, 81 Xx x2 Y y y2 xy 47 – 13 784 + 364 53 – 7 169 41 – 28 676 + 182 64 + 4 256 – 64 49 43 – 26 16 53 – 16 60 0 0 69 0 00 74 + 14 196 61 –8 64 – 112 71 + 11 121 65 –4 16 – 44 89 + 29 64 + 232 88 + 28 841 77 +8 144 + 336 784 81 + 12 66 2176 – 62 2004 894 Coefficient of correlation is : n6 xy – 6 x.6y r= ª n.6x 2 – 6x 2 ºª n.6y 2 – 6y2 º ¼¬ ¼ ¬ u 894 – (66) (– 62)

z ª 66 2 ºª 8 u 2004 – –622 º ¬ 8 u 2176 – ¼¬ ¼ 11244 = 13052 u 12188 1 = Antilog [log 11244 – 2 (log 13052 + log 12188)] 1 ? Antilog [4.0508 – 2 (4.1155 + 4.0856) ? Antilog [ 4.0508 – 4.1005 ] ? Antilog [1.9503 ] ? 0.8919 ? + 0.89 approximately. CU IDOL SELF LEARNING MATERIAL (SLM)

302 Business Mathematics and Statistics 14.4 The Probable Error in the Coefficient of Correlation In order to draw precise and correct conclusions about the coefficient of correlation it is necessary to determine the probable error in the coefficient of correlation. The formula is: 0.6745 1 – r2 P.E.= n where P.E = probable error n = number of pairs of items r = coefficient of correlation Remarks : 9. If r is less than P.E., then there is evidence of correlation. 10. If r is greater than six times the P.E., then there is significant correlation. It the P.E. is comparatively smaller than the coefficient of correlation then the following rules hold good : ? If r is less than 0.3, correlation is insignificant, i.e., there is not much evidence of correlation. ? If r is more than 0.3, then there is good evidence of correlation. 14.5 Coefficient of Correlation in a Grouped Data

We have already considered the method of finding the coefficient of correlatin in the case of two series of individual items. But in certain types of data : items may be classified into certain discrete or continuous groups in such a way that some of the items belonging to one group in one series, also belong to another group in the other series. Such data are said to be grouped data or grouped series. For example the following data are grouped data. CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 303 Age Pay Total 150–155 155–160 160–165 165–170 170–175 10 14 25-30 3 5 2 —— 15 12 30-35 2 5 4 3— 9 60 35-40 2 3 4 4 2 40-45 — 2 3 4 3 45-50 — — 2 4 3 Total 7 15 15 15 8 The formula for calculating the coefficient of correlation in a grouped series is as under : n6f dx dy – 6f dx 6f dy r= ª n 6f d2 – 6 f dx 2 ºª n 6 f d2y – 6 f dy 2 º ¼« ¬ »¼ ¬ x where r = coefficient of correlation f = frequency dx = step deviation of X (1st) series dy = step deviation of Y (2nd) series

n = total of frequencies 14.6 Rank Correlation It is not possible to express attributes such as character, conduct, honesty, beauty, morality, intellectual integrity etc. in numerical terms. For example it is quite an easy job for a class teacher to arrange the students of his class in the ascending or descending order of intelligence. This means that he can rank them according to their intelligence. Hence in problems involving attributes of the type mentioned above the method of finding the coefficient of correlation is entirely based on the rank differences between the corresponding items. The following is the procedure for finding the coefficient of correlation by the method of rank differences : = Firstly, assign ranks to the various items of the two series. = Secondly, find the differences between the ranks of the corresponding items (d). CU IDOL SELF LEARNING MATERIAL (SLM)

304 Business Mathematics and Statistics = Thirdly, find the square of each of these differences. (d2). = Lastly, apply the formula, 6 d2 r = 1 – n n2 – 1 where r = coefficient of correlation d = rank n = the number of pairs of items Problem 3: Calculate the coefficient of correlation by the method of rank difference, from the following data: X: 43 56 29 81 96 34 73 62 48 76 Y: 15 26 34 86 19 29 83 67 51 58 XR Y Rd d2 x y 4 4 43 8 15 10 2 16 82 1 56 6 26 64 64 11 4 29 10 34 98 4 72 81 2 86 22 96 1 19 34 9 29 73 4 83

62 5 67 32 4 48 7 51 52 4 76 3 58 41 1 106 (where Rx , Ry denote ranks in X and Y series respectively) Coefficient of correlation is: 6d2 r= 1– 2 n (n – 1) 6 q 106 = 1– 10 (100 – 1) CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 305 636 ? 1– 990 354 = 990 = + .36 approximately. Problem 4: Calculate the coefficient of correlation from the following data by the method of rank differences : X: 24 29 23 38 46 52 41 36 68 56 Y: 110 126 145 131 163 158 131 129 154 140 XR YR d d2 x y 1 1 24 9 110 10 1 36 0.25 29 8 126 9 1 9 1 23 10 145 4 6 2.25 1 38 6 131 6.5 0.5 4 46 4 163 1 3 52 3 158 2 1 41 5 131 6.5 1.5 36 7 129 8 1 68 1 154 3 2

56 2 140 5 3 9 64.50 In this data we find common ranks in the second series. Therefore the formula for the coefficient or correction by the method of rank differences has to be modified as under : ª1 1 º 6 « 6d2 + m13 – m1 + m2 3 – m 2 2 + .....» ¬ 12 12 ¼ r= 1– n n2–1 where m stands for the number of common ranks in a group. In this given data, there are two common ranks. Therefore m1 = 2. 1 m13 – m1 = 1 8–2 = 1 = 0.5 ; and n = 10. Sd2 = 64.50 12 12 2 CU IDOL SELF LEARNING MATERIAL (SLM)

306 Business Mathematics and Statistics ª1 º 6«d 2 + m13 – m1 + .....» ¬ 12 ¼ r= 1– n n2–1 6 > 64.40 + 0.50 @ = 1– 990 6 u 65 = 1– 990 390 = 1– 990 = Antilog [log 600 – log 990] = Antilog [2.7782 – 2.9956] = Antilog [1.7826] = 0.6061 = + 0.61 approximately. 14.7 Additional Problems Problem 7: For the data given below find out Karl Pearson’s coefficient of correlation :

X : 32 35 41 47 52 55 58 64 Y : 14 26 12 20 31 17 36 28 Deviation x2 Deviation xy X from A.M 48 y from A.M 23 y2 144 x y – 39 77 32 – 16 256 14 – 9 81 35 – 13 169 3 41 – 7 49 26 3 9 32 47 – 1 1 – 42 52 4 16 12 – 11 121 130 55 7 49 80 58 10 100 20 – 3 9 64 16 256 385 31 9 64 17 – 6 36 34 13 169 28 5 25 384 0 896 184 0 514 CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 307 384 For X series : A. M. = 8 = 48 184 For Y series : A. M. = 8 = 23 Karl Pearson’s Coefficient of Correlation xy 385 r = x 2 .y2 = 896  514 1 = Antilog [ log 385 – 2 (log 896 + log 514) ] 1 [Ans. r = + 0.5674] z Antilog [2.5855 – 2 (2.9523 + 2.7110) ] z Antilog [2.5855 – 2.8316] z Antilog [ 1.7539 ] = 0.5674 Problem 8: Psychological tests of intelligence of arithmetical ability were applied to 10 children. Here is a record of ungrouped data showing intelligence and arithmetic ratios. Calculate the coefficient of correlation: Child: ABC D E F G H I J Intelligence Ratio: 105 104 102 101 100 99 98 96 93 92

Arithmetic Ratio: 101 103 100 98 95 96 104 92 97 94 Solution: Deviation Devia- from Child Intelli- Arith. Arith- tion from xy gence Mean Ratio x x2 metic Arith. y2 + 18 (X) + 25 Ratio Mean +6 (Y) y 0 –3 A 105 + 6 36 101 +3 9 B 104 + 5 25 103 +5 25 C 102 + 3 9 100 +2 4 D 101 + 2 4 98 0 E 100 + 1 1 95 0 9 –3 CU IDOL SELF LEARNING MATERIAL (SLM)

308 Business Mathematics and Statistics F 99 0 0 96 – 2 4 0 G 98 – 1 1 104 +6 36 – 6 H 96 – 3 9 92 – 6 36 + 18 I 93 – 6 36 97 – 1 1 +6 J 92 – 7 49 94 – 4 16 + 28 Total 990 0 170 980 0 140 92 990 For X series: Arith. Mean = 10 = 99 980 For Y series: Arith. Mean = 10 = 98 x = deviation of items in X series from A.M. Y = deviation of items in Y series from A.M. Calculations show that: 6x2 = 216, 6y2 = 162 6xy = + 97. The coefficient of correlation is: xy 92

r = x2.y2 = 170  140 1 = Antilog [log 92 – 2 (log 170 + log 140)] 1 [Ans: r = + 0.5965] = Antilog [1.9838 – 2 (2.2304 + 2.1461)] = Antilog [1.9638 – 2.61882] = Antilog [1.7756 ] = 0.5965approx. Problem 9: Calculate Karl Pearson’s coefficient of correlation from the following data: X : 55 59 63 68 56 73 82 76 64 74 Y : 60 62 55 54 63 72 78 79 65 82 CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 309 Solution: Xx x2 Y y y2 xy 49 84 55 – 12 144 60 – 7 25 40 59 – 8 64 62 – 5 144 48 63 – 4 16 55 – 12 169 – 13 68 + 1 1 54 – 13 16 44 56 – 11 63 – 4 25 30 121 121 165 144 108 73 + 6 36 72 + 5 46 225 105 82 + 15 225 78 + 11 922 617 76 + 9 81 79 + 12 64 – 3 9 65 – 2 74 + 7 49 82 + 15 670 0 746 670 0 670 670 For X : A. M. = 10 = 67, For Y : A.M. = 10 = 67. The coefficient of correlation is: xy r = x 2 .y2 617 = 745  922

1 = Antilog [log 617 – 2 {log 746 + log 922}] 1 = Antilog [2.7903 – 2 {2.8727 + 2.9647)} = Antilog [2.7903 – 2.9187] = Antilog (1.8716 ) = 0.7440 = + 0.74 approx. CU IDOL SELF LEARNING MATERIAL (SLM)

310 Business Mathematics and Statistics Problem 10: Find the coefficient of correlation between the Sales and Expenses of the following 10 firms. Interpret your result. Firms: 12 3 4 5 6 7 8 9 10 Sales: 50 50 55 60 65 65 65 60 60 50 Expenses: 11 13 14 16 16 15 15 14 13 13 Solution: 580 For Sales: A.M. = 10 = 58 140 For Expenses: A.M. = 10 = 14 The coefficient of correlation is:  xy r = x2.y2 70 = 360  22 1 = Antilog [log 70 – 2 {log 360 + log 22}]

1 (Ans: r = + 0.79 approx.) = Antilog [1.8451 – 2 {2.5563 + 1.3424}] = Antilog [1.8451 – 1.9493] = Antilog (1.8958) ? r = 0.7866 Deviation Deviation from 58 Sales x2 Expenses from 14 y2 xy X –8 –8 (Y) y 9 24 50 –3 1 8 50 64 11 –3 0 0 55 2 64 13 –1 4 4 60 7 9 14 4 14 65 4 16 0 49 16 2 2 CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 311 65 7 49 15 1 17 60 2 4 14 0 00 60 2 4 13 – 1 1 –2 60 – 8 64 13 – 1 18 580 0 360 0 22 70 Interpretation: Correlation is positive and also the value of r being more than 0.5 there is a good evidence of correlation. Problem 11: Calculate the coefficient of correlation for the data given below: X : 26 29 31 34 42 48 51 54 63 57 Y : 19 21 27 31 26 42 49 51 59 60 Solution: n = 10, 6x = 25, 6x2 = 1937 6x = 25, 6x2 = 2155, 6xy = 2022. Deviation y from y2 xy X from x2 assumed mean 36 assumed mean 42 y x

26 – 16 256 19 – 17 289 272 29 – 13 169 31 – 11 121 21 – 15 225 95 34 – 8 42 0 64 27 – 9 81 99 48 + 6 0 51 + 9 36 31 – 5 25 40 54 + 12 81 63 + 21 144 36 0 0 0 67 + 25 441 625 42 + 6 36 36 49 + 13 169 117 51 + 15 225 180 59 + 23 529 483 60 + 24 576 60 Total + 25 1937 + 35 2155 2022 CU IDOL SELF LEARNING MATERIAL (SLM)

312 Business Mathematics and Statistics The coefficient of correlation n6 xy – 6x .6y r= ª n 6x 2 – 6x 2 ºª n 6y 2 – 6y2 º ¼¬ ¼ ¬ = u 2022 – (25) (35) ? >10 u 1937 – 625 @ > 10 u 2155 – 1225 @ 20220 – 875 = 18745 u 20325 19345 = 18745 u 20325 1 = Antilog [ log 19345 – 2 {log 18745 + log 20325} ] 1 ` Antilog [4.2867 – 2 {4.2729 + 4.3081} ] ` Antilog [4.2867 – 4.2905] ` Antilog [ 1.9962 ]

= 0.9913 [Ans. r = + 0.099] Problem 12: Calculate the coefficient of correlation for the following ages of husband and wife. Age of Husband (X) Age of Wife (Y) 24 19 28 21 29 24 29 26 30 27 31 27 34 29 35 31 36 30 37 31 CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 313 Solution: Deviation x2 Deviation y2 xy X from 30 y from 27 36 64 48 x 4 y 36 12 1 9 3 24 – 6 1 19 – 8 1 1 28 – 2 0 21 – 6 0 0 29 – 1 1 24 – 3 0 0 29 – 1 16 26 – 1 4 8 30 0 25 27 0 16 20 31 + 1 36 27 0 9 18 34 + 4 49 29 2 16 28 35 + 5 31 4 36 + 6 30 3 155 138 37 + 7 31 4 13 169 –5 We have, n = 10 , 6x = 13, 6x2 = 169 6y = – 5, 6y2 = 155, 6y = 138 The coefficient of correlation is : n6xy – 6x .6y

r= ª n 6x 2 – 6x 2 º – ª n 6y 2 – 6y2 º ¼ ¬ ¼ ¬ u 138 – (13) (– 5) n ª u 169 – 132 ºª – 52 º ¬10 ¼¬ 10 u 155 – ¼ 1380 + 65 5 >1690 – 169@ >1550 – 25 @ 1445 = 1521  1525 1 = Antilog [ log 1445 – 2 (log 1521 + log 1525) ] CU IDOL SELF LEARNING MATERIAL (SLM)

314 Business Mathematics and Statistics 1 = Antilog [3.1599 – 2 (3.1821 + 3.1832) ] = Antilog [3.1599 – 3.1826 ] = Antilog [ 1.9773] = 0.9491 = + 0.95 approx. [Ans: r = + 0.95 approx.] Problem 13: A random sample of 5 college students is selected and their grades in high school mathematics course and college algebra course are found College Students: 1 2 3 4 5 40 90 High School grade: 85 60 73 50 80 College grade: 93 75 65 Calculate Spearman’s rank correlation coefficient. High school R (x) College grade d d2 grade Y R (y) X 1 1 85 2 93 1 1 1 60 4 75 3 1 0 73 3 65 4 –1 1 40 5 50 5 0 90 1 80 2 –1 4

Hence n = 5 and 6d2 = 4 Spearman’s Rank correlation coefficient: 6d2 = 1– n2 – 1 n 64 = 1– 5 52 –1 = 1 – 24 5  24 CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 315 = 1– 1 5 4 =5 = 0.8 Problem 14: Find the coefficient of rank correlation between production cost and critical appraisal of the following ten motion pictures: Production Cost Rank 12 34 56 78 9 10 Critical Appraisal Ranking 8 6 10 7 5 2 9 1 4 3 Solution : Production Cost Critical Appraisal Rank Difference (d) Rank Rank –7 d2 –4 49 18 –7 16 –3 49 26 0 9 +4 0 3 10 16 47 55 62

79 –2 4 81 +7 49 94 +5 25 10 3 +7 49 266 Coefficient of rank correlation: 6  266 6d2 = 1 – 10 102 – 1 = 1 –n n2 – 1 266 6  266 = 1 –10  99 = 5  33 266 166 – 266 = 1 –165 = 165 CU IDOL SELF LEARNING MATERIAL (SLM)

316 Business Mathematics and Statistics 101 [(Ans. r = – 0.61] `– 165 ` – Antilog [ log 101 – log 165 ] ` – Antilog [ 2.0043 – 2.2175 ] ` – Antilog [ 1.7868 ] = – 0.6120 Problem 15: The following are the ranks of 12 students in two differents tests: (6,5),(8,8),(5,7),(7,6),(4,1),(1,3),(10,12),(12,11),(9,10),(11,9),(2,4),(3,2). Find the coefficient of correlation by the method of rank differences. Solution: Y Rank d d2 5 +1 1 X Rank 8 0 6 7 0 4 8 6 –2 1 5 1 +1 9 7 3 +3 4 4 12 –2 4 1 –2 10

12 11 + 1 1 9 10 – 1 1 11 9 + 2 4 2 4 –2 4 3 2 +1 1 34 Coefficient of rank correlation: 6d2 6  34 r = 1 – n n2 – 1 = 1 – 12 122 – 1 ? 1 – = = 1 – 17 12  143 143 ? 1 – Antilog [ log 17 – log 143 ] ? 1 – Antilog [ 1.2304 – 2.1553] CU IDOL SELF LEARNING MATERIAL (SLM)

Correlation Analysis 317 = 1 – Antilog [ 1.0751 ] = 1 – 1.1189 = 0.8811 ? r = + 0.88 Problem 16: Ten competition in a beauty contest are ranked by three judges in the following order. First Judge: 16 5 10 32 49 78 Second Judge: 35 84 7 10 21 69 Third Judge: 64 98 12 3 10 57 Use the rank correlation coeffcient to discuss which pair of judges has the nearest approach to common tastes in beauty. Solution: We shall calculate the coefficient of correlatin in each of the following cases. ` Ranks assigned by the First and Second Judges. ` Ranks assigned by the Second and Third Judges. ` Ranks assigned by the First and Third Judges. Case (i): First Judge Second Judge Rank difference 1 3 d d2 –2 4