ࡱ> #% !"~u` ,bjbjss <.Qa.......B8,J|B`vvv$hBR.[@[[..vvP [.v.v[..v u4$0.8v~TDE5[[[[BBBBBBBBB...... Example 18 From the following bi-variate distribution, compute two regression coefficients, coefficient of variation, coefficient of correlation and estimate the value of Y when value of X is 45. X Y10 2020 3030 4040 5010 20 20 30 30 40 40 5020 8 26 14 4  37 18 4 3 6 Solution Let, A = 25, B = 35 u = EMBED Equation.3 and v = EMBED Equation.3  Here, N = Sf = 140 Sf u = 49, Sf v = - 141 Sf u2 = 123, Sf v2 = 253, Sfuv = 27 i. We have correlation coefficient for bi-variate distribution is, =  eq \f(NSfuv - Sfu Sfv, \r(NSfu2 - (Sfu))2 \r(SNfv2 - (Sfv)2) ) = EMBED Equation.3 = EMBED Equation.3 = EMBED Equation.3 = 0.706 ii. For coefficient of variation, first we calculate the means and standard deviation of X and Y. Mean of X, EMBED Equation.3= A + EMBED Equation.3 = 25 + EMBED Equation.3 = 25 + 3.5 = 28.5 Mean of Y, EMBED Equation.3 = B + EMBED Equation.3 = 35 - EMBED Equation.3 = 35 10.071 = 24.929 S.D. of X, EMBED Equation.3 = EMBED Equation.3 =EMBED Equation.3 = EMBED Equation.3 = 8.695 S.D. of Y, sy = EMBED Equation.3 = EMBED Equation.3 = EMBED Equation.3 = 8.904. \ Coefficient of variation of X = EMBED Equation.3 = EMBED Equation.3 = 30.51% Coefficient of variation of Y = EMBED Equation.3 = EMBED Equation.3 = 35.72% iii. The regression coefficient of Y on X is byx = rEMBED Equation.3 = 0.706 EMBED Equation.3= 0.723 And, the regression coefficient of X on Y is bxy = r EMBED Equation.3 = 0.706 EMBED Equation.3= 0.689 iv. To estimate the value of Y when value of X is given, we have to find the regression line of Y on X as, Y - EMBED Equation.3= byx (X- EMBED Equation.3 ) Y 24.929 = 0.723(X 28.5) Y = 24.929 + 0.723X 20.6055  EMBED Equation.3 = 0.723X + 4.3235 Which is required regression line of Y on X. Then, value of Y when, X = 45 is  EMBED Equation.3  = 0.723 EMBED Equation.3 45 + 4.3235 \  EMBED Equation.3  = 36.858 Example 19 In two sets of variables X and Y with 50 observations each the following data were observed.  eq \o(__,X)  = 10,  eq \o(__,Y)  = 6, x = 3, x = 2 Coefficient of correlation between X and Y is 0.3. However on subsequent verification it was found that one pair (X = 10, Y = 6) was inaccurate and hence waived out. With the remaining 49 pairs of values, how is the original value of correlation coefficient affected? Solution We are given, n = 50,  eq \o(,X)  = 10,  eq \o(,Y)  = 6, x = 3, x = 2, rxy = 0.3 We have,  eq \o(,X)  =  eq \f( SX, n) eq \o(,Y)  =  eq \f( SY, n)10 =  eq \f( SX, 50) 6 =  eq \f( SY,50)SX = 500SY = 300 sx2 =  eq \f(1,n)  S(X  eq \o(,X) )2 sY2 =  eq \f(1,n)  S(Y  eq \o(,Y) )2 sx2 =  eq \f(SX2,n)  ( eq \o(,X) )2 sY2 =  eq \f(SY2,n)  ( eq \o(,Y) )2 n sX2 = SX2 - n  eq \o(,X) 2 n sY2 = SY2 - n  eq \o(,Y) 2 SX2 = n (sX2 +  eq \o(,X) 2) SY2 = n (sY2 +  eq \o(,Y) 2) =50 ( 9 + 100) = 50 (4 +36) = 5450 = 2000 Also, r =  eq \f(C ov (X, Y), sX sY)  \ r sX sY = Cov (X, Y) =  eq \f(SXY,n)    eq \o(,X)   eq \o(,Y)  0.3 3 2 =  eq \f(SXY,50)   10 6 \ SXY = 50 61.8 = 3090 One pair of observations (X = 10, Y = 6) is wrong. Omitting this pair of observations we have, n = 50  1= 49 Now, the corresponding correct values are SX = 500  10 = 490 SY = 300  6 = 294 SX2 = 5450  102 = 5350 SY2 = 2000  62 = 1964 SXY= 3090  10 6 = 3030 Now putting the corrected values of SX, SY, SX2, SY2 and SXY in the following formula we get corrected correlation coefficient r =  eq \f(nSXY SX SY, \r(n SX2  (SX)2) \r(n SY2 (SY)2)) =  eq \f(49 3030 490 294, \r(49 5350 (490)2) \r(49 1964 (294)2) ) =  eq \f(148470 144060, \r(262150 240100) \r(96236 86436))  =  eq \f(4410,\r(22050 9800) ) =  eq \f(4410,14700)  = 0.3 Therefore, the corrected correlation coefficient is 0.3. Thus in this case, the original value of correlation coefficient is not affected. Example 20 A computer while calculating correlation coefficient between two variables X and Y from 25 pairs of observations obtained the following results. n = 25, X = 125, X2 = 650, Y = 100, Y2 = 460, XY = 508 It was however discovered at the time of checking that two pairs of observations were not correctly copied. They were taken as (6, 14) and (8, 6) while the correct values were (8, 12) and (6, 8). Prove that the correct value of the correlation coefficient should be  eq \f(2,3) . Solution We have to add the correct values and subtract the wrong values as in all sum values. The corresponding corrected sum values are Correct SX = 125  6 8 + 8 + 6 = 125 Corrected SY = 100  14  6 + 12 + 8 = 100 Corrected SX2 = 650  62  82 + 82 + 62 = 650 Corrected SY2 = 460  142  62 + 122 + 82 = 436 Corrected SXY= 508  6 14  8 6 + 8 12 + 6 8 = 520 Corrected value of r is given by Corrected r =  eq \f(nSXY (SX) (SY), \r(nSX2 (SX)2) \r(nSY2 (SY)2))  =  eq \f(25 520 125 100,\r(25 650 (125)2) \r(25 436 (100)2))  =  eq \f(13000 12500, \r(16250 15625) \r(10900 10000))  =  eq \f(500,\r(625 900))  =  eq \f(500, 25 30)  =  eq \f(500,750)  =  eq \f(2,3)  Thus verified Example 21 A student calculates the value of r as 0.7, when the value of n is 5 and he concludes that r is highly significant. Does he correct? Calculate the limits for population correlation coefficient. If the calculated value of PE (r) = 0.085 for r = 0.7 find the value of n. Solution We have, r = 0.7, n = 5 PE (r) = 0.6745  eq \f(1 r2, \r(n))  = 0.6745  eq \f(1 (0.7)2, \r(5))  = 0.154 and, 6 PE (r) = 6 0.154 = 0.924 Hence, this shows that r is not greater than 6 PE. Thus, we can not make any decision about the significance of correlation coefficient. It is seen that his conclusion becomes wrong. Limits for population correlation coefficients are r PE (r) = 0.7 0.154 \ Upper limit of r = 0.7 + 0.15 = 0.854 Lower limit of r = 0.7 0.154 = 0.546 Now, if PE(r) = 0.085, r = 0.7, n = ? We have, PE(r) = 0.6745  eq \f(1 r2,\r(n))  0.085 = 0.6745  eq \f(1 (0.7)2,\r(n))  0.085  eq \r(n)  = 0.344  eq \r(n)  =  eq \f(0.344, 0.085)   eq \r(n)  = 4 .047 n = 16 (approximately) Example 22 Following figures give the ages in years of newly married husbands and wives. Represent the data by a bivariate frequency distribution. (Age of husband, age of wife): (25, 17) (26,18) (27,19) (25,17) (28,20) (24,18) (27,18) (28,19) (25,18) (26,19) (25,17) (26,18) (27,19) (25,19) (27,20) (26,19) (25,17) (26,20) (26,17) (26,18) Also, find Karl Pearson's correlation coefficient. Test its significance. Solution Let, X and Y be age of husband and wife respectively. We observe that the variable X takes the values from 24 to 28 and Y takes the values from 17 to 20. We obtain the bivariate discrete frequency distribution given as Bivariate frequency distribution X YAge of Husbands 2425262728Row totalAge of wives17131518113161912216201113Column Total 2574220Let, u = X A = X 26 v = Y B = Y 18 Calculation of Karl Pearsons correlation coefficient  X Y v u242526272821012ffvfv2fuv171 2 1 3 3  0 1  5555180 0 1 0 1 0 3 0 1  6000191    1 1 0 2 2 2 2 16662202     0 1 2 1 4 136126f257422072314fu 4 5044 1fu28504825fuv2204614 Here, N = 20, Sfu =  1, Sfv = 7, Sfu2 = 25, Sfv2= 23, Sfuv = 14 \ r =  eq \f(NSfuv  Sfu Sfv,\r(NSfu2  (Sfu)2) \r(NSfv2  (Sfv)2) ) =  EMBED Equation.3  =  eq \f(280 + 7, \r(500  1) \r(460  49))  =  eq \f(287,\r(499) \r(411))  = 0.612 Thus, there is positive relationship between husband s age and wife s age. Now, PE = 0.6745  eq \f(1 - r2, \r(n))  = 0.6745  EMBED Equation.3  = 0.0944 \ 6 PE = 6 0.0944 = 0.5664 Since, r is greater than 6 PE; the correlation coefficient is significant i.e. there is evidence of correlation coefficient. Example 23 For the following information X Y Arithmetic mean(in Rs)68Standard deviation (in Rs)540/3Correlation coefficient between X and Y is 8/15. Find a. The regression coefficient of Y on X and X on Y b. The two regression equations c. The most likely value of Y when X = 100 rupees. Solution We have,  eq \o(,X) = 6,  eq \o(,Y)  = 8, sx = 5, sy = 40/3, r = 8/15 a. Regression coefficient of Y on X is byx = r  eq \f(sy,sx)  =  eq \f(8,15)   eq \f(40/3, 5)  = 1.422 Similarly, regression coefficient of Y or Y is bxy = r  eq \f(sx,sy)  =  eq \f(8,15)   eq \f(5,40/3) = 0.2 b. The regression equation of Y on X is Y -  eq \o(,Y)  = byx (X  eq \o(,X) ) Y - 8 = 1.422 (X 6)  eq \o(^,Y)  = 1.422 X 0.532 Similarly, the regression equation of X on Y is X  eq \o(,X)  = bxy (Y  eq \o(,Y) ) X 6 = 0.2 (Y 8)  EMBED Equation.3  = 0.2Y + 4.4 c.  eq \o(^,Y)  = ? When X = 100  eq \o(^,Y)  = 1.422 100 0.532 = 142.2 0.532 = 141.67 Thus, the most likely value of Y is Rs 141.67. EXERCISE 6 THEORETICAL QUESTIONS 1. What do you mean by correlation? Mention its types. 2. Explain the concept of simple multiple and partial correlation coefficient. 3. What are different methods of finding correlation between two variables? Explain briefly. 4. Define Karl Pearsons correlation coefficient and interpret the result of its coefficient. 5. Define Spearman's rank correlation coefficient. When it is used? 6. Define Probable error of correlation coefficient. Mention it's utilities. 7. Define the term 'regression. Discuss two regression lines. 8. Mention the properties of regression coefficients. PRACTICAL PROBLEMS 9. Draw a scatter diagram from the following data. Height (inch)62727060677064656070Weight (lbs)50656352566059585465 Also, indicate whether correlation is positive or negative. 10. If the covariance between X and Y variable is 18 and the variance of X and Y are 25 and 81 respectively. Find the coefficient of correlation between them. 11. Calculate the correlation coefficient between X and Y series from the following data. X seriesY seriesNo. of observations:1616Standard deviation:3.013.03EMBED Equation.3(X -  eq \o(,X) ) (Y -  eq \o(,Y)) =12212. For 10 observations on Height (X) and Weight (Y), the following data were obtained (in approximate units) SX = 130, SY = 220, SX2 = 2290, SY2 = 5510 and SXY = 3467 Compute the coefficient of correlation. 13. Calculate the coefficient of correlation using product moment formula from the data of price and supply given below: Price (Rs.)160162165161163164166Supply6362646362666814. The following table gives the age and blood pressure in appropriate unit of 10 patients. Age56423647494260726355Pressure147125118128145140155160149150 Compute the coefficient of correlation assuming 49 and 140 as the assumed means of age and pressure respectively. 15. Calculate Karl Pearsons coefficient of correlation between expenditure on advertising (Rs. 000) and seles (lakh Rs.) from the data given below. Adv. Expenses39656290827525983678Sales4753588662686091518416. From the following table calculate the coefficient of correlation by Karl Pearsons method. X621048Y911?87 Arithmetic means of X and Y series are 6 and 8 respectively. 17. Calculate the Karl Pearsons coefficient of correlation from the following data: Sum of deviation of X = 5 Sum of deviation of Y = 4 Sum of squares of deviation of X =40 Sum of squares of deviation of Y =50 Sum of product of deviation of X and Y = 32 and Number of pairs of observation = 10 18. Calculate the coefficient of correlation between the age of students and pass percentage given below: Age (year)% PassAge (year)% Pass13 - 143918 - 193914 - 154019 - 204815 - 164320 - 214916 - 174421 - 225417 - 183619. Find correlation coefficients between the age and playing habit of the people from the following information. Age group (year)15-2020-2525-3030-3535-4040-45No. of people200270340320400300No. of players150162170180180120 Interpret the calculated correlation coefficient. 20. Family income and percentage spent on food in case of 100 families gave the following bi-variate frequency distribution. Find correlation coefficient between them. Food exp. in (%)Family income200-300300-400400-500500-600600-70010---3715-49432076125-25310198-21. Following data represents the bi-variate frequency distribution of 25 students getting marks in Statistics and Economics. Find the coefficient of correlation. Marks in EconomicsMarks in Statistics30-4040-5050-6060-7025-35311-35-45261245-55122155-65-11122. From the data given below, find the coefficient of correlation between the drivers age and the number of accidents made by him. Number of accidentsDriver's age25 - 3030 - 3535 - 4040 - 4545 - 500-33781--941235103-3496--412731-23. The marks obtained by 25 students in Economics and Statistics are given below. The first figure in brackets indicates the marks in Economics and second in Statistics. (13, 11) (14, 17) (10, 10) (11, 7) (15, 15) (6, 10) (4, 1) (11, 14) (8, 3) (19, 15) (19, 18) (11, 7) (10, 13) (13, 16) (16, 14) (2, 8) (12, 18) (9, 11) (5, 3) (17, 14) (4, 12) (0, 2) (1, 5) (7, 3) (15, 9) Prepare a two way table taking the magnitudes of each class interval as 5 marks for Economics and 4 marks for Statistics. Also, find correlation coefficient between them. 24. In order to find the correlation coefficient between two variables X and Y from 12 pairs of observations, the following calculations were made. SX = 30, SY = 5, SX2= 670, SY2= 285 and SXY= 334 On subsequent verification, it was found that the pair (X = 11, Y = 4) was copied wrongly, the correct value being (X = 10, Y = 14). Find the correct value of correlation coefficient. 25. For a sample of 25 observations, the correlation coefficient is found to be 0.7. Find the limits within which correlation coefficient lies for population. 26. If the correlation coefficient is found to be 0.6 for a pair of 64 observations, find the probable error of r and determine the limits of population correlation coefficient. 27. The manager of Machine and Tool Company wants to know the impact of TV advertisement on sales of his products. He sought information regarding the frequency of advertisement per week and the volume of sales per week. The information supplied to him was as follows. Advertisement on TV21282835354242Sales (Lakhs)20353028454042 Taking into consideration the enormous cost that is involved in advertisement, the manager decided that if the relationship between the volume of sales and the frequency of advertisement on TV were significant, he would continue to advertise otherwise not, what will be his decision? 28. A sample of 100 firms was taken and these were classified according to the sales executed by them and profits earned consequently. The results are shown in the table below. Determine the correlation between sales and profits. And also, compute the probable error. Sales (million of Rs) Profits ('00 Rs)7 - 88 - 99 - 1010 -1111 -1212 -1350 - 705370 - 90385490 - 110171122110 - 13045156130 - 1502746Total915193712829. In a beauty contest, two judges rank the 10 competitors in the following order Competitors12345678910Rank by A64321798105Rank by B41675810932 Is there any concordance between the two judges? 30. Ranks provided by two judges for twelve competitors in painting competition are shown below. Competitors123456789101112Judge I523416871091211Judge II452167109111238 Find Spearman's rank correlation coefficient. 31. Ten industries of Nepal have been raked as follows according to profits earned in 2004 and working capital for the year. Industry12345678910R1 (Profit)86723510194R2 (working capital)10872546193 Calculate the Spearman's rank correlation coefficient 32. Ten competitors in an I.Q. test are raked by three judges in the following ordered. Competitors12345678910Rank by I16510324978Rank by II35847102169Rank by III64981231057 Use the method of rank correlation to gauge which pair of judge has the nearest approach to common linking in IQ. 33. Calculate the coefficient of rank correlation between demand and supply of the cement in '000 metric tones. Demand1525182016211012Supply810126914181634. Calculate Spearman's coefficient of rank correlation for the following data. X1152214825183473259270164Y8438520011029215286120321144 35. A company took an examination of eight applicants for an accountant post. From the marks obtained by the applicant in the Accountancy and Statistics papers, compute the rank correlation coefficient. Applicant 12345678Accountancy1520281240602080Statistics 403050302010306036. Quotations of index number of equity share prices of a certain joint stock company and of prices of preference shares are given below. Years1999200020012002200320042005Preference Shares732858789758772812838Equity Shares978992988983983967971 Use the method of rank correlation to determine the relationship between equity share and preference share prices. 37. Calculate Spearman's rank correlation coefficient between the age of person and blood pressure. Age56423647494260726355Pressure145125118128145145155160149150 38. Obtain Spearman's rank correlation coefficient for the following data. X68647550648075405564Y6258684581606848507039. A sample of 12 father and their eldest son gave the following data about their heights in inches: Father656367646862706668676971Son686668656966686571676870 Calculate the correlation coefficient by rank method. 40. The coefficient of rank correlation of the marks obtained by 10 students in English and Business Mathematics in +2 was found to be 0.8. 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Fit regression line of Y on X from the following data. X5758595960616264Y676865687272697142. Find the two regression equations from the following data. X12345Y1357943. From the data given below, estimate the most likely height of a brother whose sister's height is 50 cm. BrotherSisterMean Height 170 cm75 cmS.D. of Heights 6 cm6 cm The coefficient of correlation between the heights of brothers and sister is 0.60. 44. Find the most likely price in market A corresponding to the price of Rs 75 at market B from the following information. Average price in market A = Rs 67. Average price in market B = Rs 65. Coefficient of variation at market A = 5.22 Coefficient of variation at market B = 3.85 Correlation coefficient between them= 0.82 45. Estimate the loss in production in a day when the number of workers on strike is 18000 from the following information. Mean number of workers on strike = 800 Mean loss of daily production in '000 Rs = 35 Standard deviation of number of workers on strike = 100 Standard deviation of daily production in '000 Rs = 2 Coefficient of correlation between number of workers on strike and daily production was = 0.8 46. In a partially destroyed record of the following data available, Variance of X = 25 Two regression equations: 5X - Y - 22 = 0 and 64X - 45Y - 24 = 0 find a. Mean values of X and Y b. Coefficient of correlation between X and Y c. Standard deviation of Y. 47. The equation of two regression lines between two variables are expressed as 3X 4Y + 30 = 0 and 5X - 2Y + 8 = 0. a. Identify which of the two can be called regression equation of Y on X and X on Y. b. Find the mean of X and Y and correlation coefficient. 48. The following table gives the ages and blood pressure of 10 women. Age56423647494260726355Pressure147125118128145140155160149150 Estimate the blood pressure of a woman whose age is 45 years. 49. From the data given below, Marks in Statistics252835323136293834Marks in Economics434649413632313033Find, i. Two regression equations ii. The coefficient of correlation between the marks in Statistics and Economics. iii. The most likely marks in Economics when the mark in Statistics is 30. 50. From the following data between age of husbands and wives, calculate the two regression equations and find the husbands age when wifes age is 20. Wifes age18202223272830Husbands age2325273032313551. Obtain the lines of regression for the following bi-variate frequency distribution. Sales revenue ('00 Rs)Advertisement time on TV (second)5 - 1515 - 2525 - 3535 - 45 75 - 12541125 - 1757621175 - 2251342225 - 275113452. From the given bi-variate frequency distribution, find out if there exists any relationship between the age of wives and husbands and test for the significance of the result and interpret the result. Also determine the age of the wife whose husbands age is 75 years. Age of wives (years)Age of husbands (yrs)20 -3030-4040-5050-6060-7015 25593--25 35-10252-35 45-1122-45 55--416555 65---4253. From the following bi-variate frequency table, compute two regression coefficients, coefficient of variation, coefficient of correlation and estimate the expenditure of a person when his income is Rs. 4,000. Expenditure (Rs.)Income (Rs.)0-500500-10001000-15001500-20002000-25000 - 4001268--400 - 800218451800 - 1200-810241200 - 1600-110211600 - 2000--123Total143333119 ANSWERS 9. Positive 10. r = 0.40 11. r = 0.836 12. r = 0.957 13. r = 0.725 14. r = 0.892, High 15. r = 0.7804 16. r = 0.92 17 r = 0.704 18. r = 0.7225 19. r = 0.918, high and negative 20. r = 0.438 21. r = 0.394 22. r = 0.699, negatively correlated 23. r = 0.58 24. Corrected r = 0.77 25. Lower limit = 0.631 and Upper limit = 0.769 26. PE = 0.054, Lower limit = 0.546 and Upper limit = 0.654 27. r = 0.771, PE = 0.103, r is significant, Continues the advertisement 28. r = 0.6227, PE = 0.042 29. Yes, R= 0.25 30. R = 0.46 31. R = 0.8232 32. R12 = 0.212, R13 = 0.636, R23 = 0.297, 1st and 3rd 33. R= 0.405 34. R = 0.721 35. R = 0 36. R = 0.125 37. R = 0.8606 38. R = 0.545 39. R = 0.722 40. Rc = 0.606 41.  EMBED Equation.3  = 29 + 0.67 X 42.  EMBED Equation.3  = 0.5 + 0.5Y and  EMBED Equation.3  = 1 + 2X 43. 155 cm 44. Rs 78.46 45.  EMBED Equation.3  = 0.016 X + 22.2, Rs. 310200 46. a. 6 and 8 b. 0.533 c.13.33 47. a. Y on X is 3X 4Y + 30 = 0 and X on Y is 5X - 2Y + 8 = 0. b.  eq \o(,X)  = 2 and  eq \o(,Y) = 9 and r = 0.5477 48.  EMBED Equation.3  = 83.756 + 1.11 X, Blood pressure = 133.708 49. i.  EMBED Equation.3 = 40.88 0.2337 Y and  EMBED Equation.3  = 59.146 0.664X ii. r = - 0.394 iii. 39.23 year 50. 25.34 year 51.  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