Nm Shah 2018 Solutions for Class 11 Commerce Economics Chapter 8 Measures Of Correlation are provided here with simple step-by-step explanations. These solutions for Measures Of Correlation are extremely popular among Class 11 Commerce students for Economics Measures Of Correlation Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Nm Shah 2018 Book of Class 11 Commerce Economics Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Nm Shah 2018 Solutions. All Nm Shah 2018 Solutions for class Class 11 Commerce Economics are prepared by experts and are 100% accurate.

Page No 323:

Question 1:

The following pairs give the value of variables of capital employed and profit.

Capital employed (in crores of â‚¹ ) (X) : 2 3 5 6 8 9
Profit (in lacs of â‚¹ ) (Y) : 6 5 7 8 12 11
(a) Make a scatter diagram.
(b) Do you think that there is any correlation between profit and capital employed? Is it positive or negative ? Is it high or low?
(c) By graphic inspection , draw an estimating line.
 

Answer:

a) Scatter Diagram





b) The points obtained on the scatter diagram lie close to each other and reflect an upward trend. Thus, there exists a high degree of positive correlation between capital employed and profits earned.

c)

Page No 323:

Question 2:

Plot the following data as a scatter diagram and comment on the result obtained:

X : 11 10 15 13 10 16 13 8 17 14
Y : 6 7 9 9 7 11 9 6 12 11

Answer:

Scatter Diagram


Thus, there exists a positive correlation of moderate degree between X and Y.

Page No 323:

Question 3:

Following are the heights and weights of 10 students in a class. Draw a scatter diagram and indicate whether the correlation is positive or negative.

Height (in inches) : 72 60 63 66 70 75 58 78 72 62
Weight (in kg)  : 65 54 55 61 60 54 50 63 65 50
 

Answer:




Thus, there exists a very low degree of positive correlation between height and weight of students.

Page No 323:

Question 4:

Draw a scatter diagram for the data given below and interpret it:

X : 10 20 30 40 50 60 70 80
Y : 32 20 24 36 40 28 48 44
 

Answer:




Thus, there exists a moderate degree of correlation between X and Y.

Page No 323:

Question 5:

Draw a scatter diagram of the following data:

X : 15 18 30 27 25 23 30
Y : 7 10 17 16 12 13 9
 

Answer:




Thus, there exists a moderate degree of positive correlation between X and Y.

Page No 323:

Question 6:

From the following data compute the product moment correlation between X and Y.

  X series Y series
Arithmetic Mean 25 18
Sum of Square of deviations from Arithmetic Mean 136 138
Summation of products of deviations of X and Y series from their respective means = 122
Number of points of values = 15 
 

Answer:

Given:X=25  Σx2=136  Σxy=122Y=18  Σy2=138   n=15Now, r=ΣxyΣx2 ×Σy2r=122136 ×138=122136×138=12211.66×11.74r=122136.996=0.89
Thus, the product moment correlation between X and Y is 0.89.

Page No 323:

Question 7:

Calculate Karl Pearson's Coefficient of Correlation on the following data:

X : 15 18 21 24 27 30 36 39 42 48
Y : 25 25 27 27 31 33 35 41 41 45
 

Answer:

X Y X-X
(x)
x2 Y-Y
(y)
y2 xy
15 25 −15 225 −8 64 120
18 25 −12 144 −8 64 96
21 27 −9 81 −6 36 54
24 27 −6 36 −6 36 36
27 31 −3 9 −2 4 6
30 33 0 0   0 0 0
36 35 6 36   2 4 12
39 41 9 81   8 64 72
42 41 12 144   8 64 96
48 45 18 324   12 144 216
Æ©X = 300 Æ©Y =330   Æ©x2 =1080   Æ©y2 =480 Æ©xy =708

N = 10

X=ΣXN=30010=30Y=ΣYN=33010=33r=ΣxyΣx2 ×Σy2 =7081080× 480=70832.86×21.91 =708720  =0.983
Thus, the value of correlation coefficient is 0.983.

Page No 323:

Question 8:

Calculate Karl Pearson's Coefficient of Correlation between the sales and expenses of the following 10 firms:

Firms : 1 2 3 4 5 6 7 8 9 10
Sales(in â‚¹ '000) : 50 50 55 60 65 65 65 60 60 50
Expenses (in â‚¹ '000) : 11 13 14 16 16 15 15 14 13 13
 

Answer:

Sales
(X)
Expenses
(Y)
X-X
x
x2 Y-Y
y
y2 xy
50 11 −8 64 −3 9 24
50 13 −8 64 −1 1   8
55 14 −3 9   0 0   0
60 16    2 4   2 4   4
65 16    7 49   2 4  14
65 15    7 49   1 1   7
65 15    7 49   1 1   7
60 14    2 4   0 0   0
60 13    2 4 −1 1 −2
50 13 −8 64 −1 1   8
580 140   360   22 70

N = 10

X=ΣXN=58010=58Y=ΣYN=14010=14r=ΣxyΣx2 Σy2=70360 22=7018.973×4.690=7088.9833=0.786

Note: As per the textbook, coefficient of correlation is 0.67. However, as per the above solution coefficient of correlation should be 0.786.



Page No 324:

Question 9:

Calculate correlation coefficient between X, the number of rainy days per month and Y, the number of rain coats sold in that month in a certain shop for 12 months. Interpret the results.

X : 14 8 18 10 22 9 3 5 6 11 13 13
Y : 15 11 20 12 15 7 3 4 7 10 11 29
 

Answer:

X Y X-X
(x)
x2 Y-Y
(y)
y2 xy
14 15   3 9   3 9 9
8 11 −3 9 −1 1 3
18 20   7 49   8 64 56
10 12 −1 1   0 0 0
22 15  11 121   3 9 33
9 7  −2 4 −5 25 10
3 3  −8 64 −9 81 72
5 4  −6 36 −8 64 48
6 7  −5 25 −5 25 25
11 10    0 0 −2 4 0
13 11    2 4 −1 1 −2
13 29    2 4 17 289 34
Æ©X =132 Æ©Y =144   Æ©x2 =326   Æ©y2 =572 Æ©xy =288

N = 12

X=ΣXN=13212=11Y=ΣYN=14412=12r=ΣxyΣx2× Σy2 =288326×572=28818.05×23.91 =288431.82   =0.6669 =0.67(approx.)

There is a moderate degree of (+) correlation between the number of rainy days and the number of rain coats sold. In other words, as the number of rainy days increases in a month, the number of rain coats sold in that month increases moderately.

Note: As per the textbook, coefficient of correlation is -0.67. However, as per the above solution coefficient of correlation should be +0.67.

Page No 324:

Question 10:

The deviations from their means of two series (X and Y) are given below:

X : −4 −3 −2 −1 0 +1 +2 +3 +4
Y : +3 −3 −4 0 +4 +1 +2 −2 −1

Calculate the Karl Pearson's coefficient of correlation and interpret the result .
 

Answer:

x y x2 y2 xy
−4 3 16 9 −12
−3 −3 9 9 9
−2 −4 4 16 8
−1 0 1 0 0
  0 4 0 16 0
  1 1 1 1 1
  2  2 4 4 4
  3 −2 9 4 −6
  4 −1 16 1 −4
    Æ©x2 =60 Æ©y2 =60 Æ©xy =0

r=ΣxyΣx2 Σy2 =060×60=0
Thus, there is no correlation between series X and series Y.

Page No 324:

Question 11:

Find the product moment correlation of the following data:

X : 1 2 3 4 5
Y : 9 8 10 12 11
 

Answer:

X Y X-X
x
Y-Y
y
x2 y2 xy
1 9 −2 −1 4 1 2
2 8 −1 −2 1 4 2
3 10 0 0 0 0 0
4 12 1  2 1 4 2
5 11 2 1 4 1 2
Æ©X =15 Æ©Y =50     Æ©x2 = 10 Æ©y2 = 10 Æ©xy = 8

N = 5

X=ΣXN=155=3Y=ΣYN=505=10r=ΣxyΣx2× Σy2=810 ×10=810=0.80
Thus, the product moment correlation is 0.80.

Page No 324:

Question 12:

Calculate the correlation coefficient of the marks obtained by 12 students in Mathematics and Statistics and interpret it.

Students  : A B C D E F G H I J K L
Marks (in Maths) : 50 54 56 59 60 62 61 65 67 71 71 74
Marks ( in Statis.) : 22 25 34 28 26 30 32 30 28 34 36 40
 

Answer:

Maths
(X)
Statis.
(Y)
X-X
x
x2 Y-Y
(y)
y2 xy
50 22 −12.5 156.25 −8.41 70.72 105.12
54 25   −8.5   72.25 −5.41 29.26   45.98
56 34   −6.5   42.25   3.59 12.88 −23.33
59 28   −3.5   12.25 −2.41   5.80     8.43
60 26   −2.5     6.25 −4.41 19.44   11.02
62 30     −0.5      0.25   −0.41   0.168   0.205
61 32   −1.5    2.25   1.59   2.52  −2.38
65 30     2.5    6.25   −.41   0.168  −1.02
67 28     4.5  20.25 −2.41   5.80 −10.84
71 34     8.5  72.25   3.59 12.88    30.51
71 36     8.5  72.25   5.59 31.24    47.51
74 40   11.5 132.25   9.59 91.96 110.28
Æ©X = 750 Æ©Y= 365   Æ©x2 = 595   Æ©y2 = 282.836 Æ©xy = 321.485

N = 12

X=ΣXN=75012=62.5Y=ΣYN=36512=30.41r=ΣxyΣx2× Σy2  =321.485595 ×282.836 =321.48524.392×16.80 =321.485410.22 =0.783
Thus, there exists sufficiently high degree of positive correlation marks in Mathematics and marks in Statistics.

Page No 324:

Question 13:

The height of fathers and sons are given below:

Height of fathers (in inches)  : 65 66 67 67 68 69 71 73
Height of sons (in inches) : 67 68 64 68 72 70 69 70
Calculate Karl Pearson's  coefficient of correlation .
 

Answer:

Father's Height
(X)
Son's Height
(Y)
X-X
x
x2 Y-Y
y
y2 xy
65 67 −3.25 10.56 −1.5   2.25   4.875
66 68 −2.25   5.06   −0.5    0.25   1.125
67 64 −1.25   1.56 −4.5 20.25   5.625
67 68 −1.25   1.56   −0.5    0.25     0.625
68 72   −0.25   0.06    3.5 12.25 −0.875
69 70    0 .75     0.56    1.5   2.25   1.125
74 69   2.75   7.56    0.5     0.25   1.375
73 70   4.75 22.56    1.5   2.25   7.125
Æ©X = 546 Æ©Y = 548   Æ©x2 = 49.48   Æ©y2 = 40 Æ©xy = 21

N = 8

X=ΣXN=5468=68.25Y=ΣYN=5488=68.5r=ΣxyΣx2× Σy2=2149.48× 40r=217.034×6.324=0.472Thus, the correlation coefficient between height of father and height of son is 0.472. 

Page No 324:

Question 14:

Find Karl Pearson's coefficient of correlation from the following index numbers and interpret it.

Wages (₹) : 100 101 103 102 100 99 97 98 96 95
Cost of living : 98 99 99 97 95 92 95 94 90 91
 

Answer:

Wages
(X)
Cost of Living
(Y)
X-X
x
x2 Y-Y
y
y2 xy
100 98    0.9     0.81    3 9   27
101 99   1.9   3.61    4 16  7.6
103 99   3.9 15.21    4 16  5.6
102 97   2.9   8.41    2 4  5.8
100 95     0.9     0.81    0 0     0
99 92   −0.1     0.01  −3 9    0.3
97 95 −2.1   4.41    0 0     0
98 94 −1.1   1.21  −1 1   1.1
96 90 −3.1   9.61  −5 25 15.5
95 91 −4.1 16.81  −4 16 16.6
Æ©X = 991 Æ©Y = 950   Æ©x2 = 60.9   Æ©y2 = 96 Æ©xy = 65.2

N = 10

X=ΣXN=99110=99.1Y=ΣYN=95010=95r=ΣxyΣx2× Σy2 =65.260.9× 96 =65.27.80×9.79  = 0.85

Wages and cost of living has high positive correlation. If wages increase by 1 unit, the cost of living increases by 0.85 units.

Page No 324:

Question 15:

Find the product moment correlation between sales and expenses of the following 10 firms.

Firms : 1 2 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
 

Answer:

Sales
(X)
Expenses
(Y)
X-X
x
x2 Y-Y
y
y2 xy
50 11  −8 64  −3 9   24
50 13  −8 64  −1 1    8
55 14  −3 9    0 0    0
60 16    2 4    2 4    4
65 16   7 49    2 4   14
65 15   7 49    1 1    7
65 15    7 49    1 1    7
60 14    2 4    0 0    0
60 13    2 4  −1 1  −2
50 13  −8 64  −1 1    8
Æ©X = 580 Æ©X = 140   Æ©x2 = 360   Æ©y2 = 22  Æ©xy =  70

N = 10

X=ΣXN=58010=58Y=ΣYN=14010=14r=ΣxyΣx2 ×Σy2 =70360 ×22=7018.97×4.69 =7088.9693= 0.7867 or 0.787(approx.)

Thus, there exists high positive correlation between sales of firm and expenses.

Page No 324:

Question 16:

Calculate the coefficient of correlation for the following ages of husbands and wives in years at the time of their marriage.

Age of Husbands : 23 27 28 28 29 30 31 33 35 36
Age of Wives : 18 20 22 27 21 29 27 29 28 29
 

Answer:

Husband
(X)
Wife
(Y)
X-X
x
x2 Y-Y
y
y2 xy
23 18 −7  49 −7  49  49
27 20 −3   9 −5  25  15
28 22 −2   4 −3   9    6
28 27 −2   4   2   4  −4
29 21 −1   1 −4 16    4
30 29   0   0   4 16    0
31 27   1   1   2   4    2
33 29   3   9   4 16  12
35 28   5 25   3   9  15
36 29   6 36   4 16  24
Æ©X = 300 Æ©Y = 250   Æ©x2 =138   Æ©y2 =164 Æ©xy =123

N = 10

X=ΣXN=30010=30Y=ΣYN=25010=25r=ΣxyΣx2× Σy2 =123138 ×164 =12311.74×12.80 =0.818 or 0.82 (approx.)

Thus, there exists high positive correlation between age of husband and age of wife.

Page No 324:

Question 17:

Find Karl Pearson's coefficient of correlation for the following data:

Fertilizers used ( in tons ) : 15 18 20 24 30 35 40 50
Productivity ( in tons) : 85 90 95 105 120 130 150 160
 

Answer:

Fertilizers
(X)
Productivity
(Y)
X-X
x
x2 Y-Y
y
y2 xy
15   85 −14  196 −32.25 1040.06   451.5
18   93 −11  121 −24.25   588.06 266.75
20   95  −9    81 −22.25   495.06 200.25
24 105  −5    25 −12.25   150.06   61.25
30 120    1      1     2.75      7.56    2.75
35 130    6    36   12.75  162.56    76.5
40 150   11  121   32.75 1072.56 360.25
50 160   21  441   42.75 1827.56 897.75
Æ©X =232 Æ©Y = 938   Æ©x2 =1022   Æ©y2 =5343.48 Æ©xy =2317

N = 8

X=ΣXN=2328=29Y=ΣYN=9388=117.25r=ΣxyΣx2 Σy2=23171022 5343.48=231731.96×73.09=23172335.95=0.99
Thus, there exists high positive correlation between the amount of fertilizers used and the productivity.



Page No 325:

Question 18:

Calculate product moment correlation  between age of cars and annual maintenance cost and comment.

Age of cars (years) : 2 4 6 7 8 10 12
Annual maintenance
​
Cost (in â‚¹)
: 1600 1500 1800 1900 1700 2100 2000
 

Answer:

Age of Cars
(X)
Maintenance
(Y)
Y-Y
x
Y-Y
y
x2 y2 xy
2 1600  −5  −200 25 40000 1000
4 1500  −3    −30   9 90000   900
6 1800  −1        0   1        0       0
7 1900    0    100   0 10000       0
8 1700    1 −100   1 10000 −100
10 2100    3   300   9 90000   900
12 2000    5   200 25 40000 1000
Æ©X = 49 Æ©Y =12600     Æ©x2 =70 Æ©y2 =280000 Æ©xy =3700

N = 7

X=ΣXN=497=7Y=ΣYN=126007=1800r=ΣxyΣx2× Σy2=370070 ×280000=37008.36×529.15=0.836
Thus, there exists high positive correlation between age of car and maintenance cost.

Page No 325:

Question 19:

Calculate coefficient of  correlation by Pearson's method between the density of population and death rate.

Cities : A B C D E F
Density : 200 500 400 700 600 300
Death rate : 10 16 14 20 17 13
 

Answer:

Density
(X)
Death
rate
(Y)
X-X
x
x2 Y-Y
y
y2 xy
200 10  −250 62500  −5 25 1250
500 16      50   2500    1 1    50
400 14    −50   2500  −1 1    50
700 20    250 62500    5 25 1250
600 17    150 22500    2 4   300
300 13 −150 22500  −2 4   300
Æ©X = 2700 Æ©Y =90   Æ©x2 =175000   Æ©y2 =60 Æ©xy =3200

N = 6

X=ΣXN=27006=450Y=ΣYN=906=15r=ΣxyΣx2 Σy2=3200175000 60=3200418.33×7.74=0.988

Thus, there exists high positive correlation between density of population and death rate.

Page No 325:

Question 20:

The total of the multiplication deviation of X and Y = 3044
Number of pairs of observations = 10
Total of the deviation of X = −20
Total of the deviation Y =  −170
Total of squares of deviation of X = 2264
Total of the squares of deviation of Y = 8288
Find out Karl Pearson' s coefficient of correlation when assumed mean of X and Y are 82 and 68 respectively.

Answer:

Given:
n = 10
Σdxdy = 3044
Σdx = −17 assumed (instead of −170 see note below)
Σdy = −20
Σdx2 = 2264
Σdy2 = 8288

r=Ndxdy-dxdyNdx2-dx2 Ndy2-dy2r=10×3044--17×-2010×2264--172 10×8288--202r=30440-34022351 82480r=30100149.502×287.193r=3010042935.927r=0.70

Hence, Karl Pearson's coefficient of correlation is 0.7.

Note:

1. If we take Σdx equal to −170, then the calculation of Karl Pearson's coefficient of correlation is as follows:

r=Ndxdy-dxdyNdx2-dx2 Ndy2-dy2r=10×3044--170×-2010×2264--1702 10×8288--202r=30440-340022640-28900 82880-400r=30100-6260 82480

Since, the square root of a negative number is not defined among the set of real numbers, thus we have assumed Σdx as −17.

2. The answer as per the book is 0.78 and the answer as per our calculation is 0.70. The difference in the answer may be due to the assumption (i.e. Σdx = −17) made by us.

Page No 325:

Question 21:

Number of pairs of observations of X and Y series = 10 .
X series   : Arithmetic Average = 65
                : Standard Deviation = 23.33
Y series   : Arithmetic Average = 66
                   Standard deviation = 14.9
Summation  of products of corresponding deviations of X and Y series from their respective means = 2704.
Calculate product moment correlation of X and series.
      

Answer:

Given:X=65Y=66 σx=23.33σy=14.9 

Now,
r=ΣxyN·σx σy =270410×23.33×14.9=27043476.17=0.7778 or 0.78 (approx.)
Thus, product moment correlation of X and Y series is 0.78.

Page No 325:

Question 22:

Calculate Spearman's rank correlation from the following data:

X : 10 12 8 15 20 25 40
Y : 15 10 6 25 16 12 8
 

Answer:

X R1 Y R2 (R1R2)
D
D2
10 2 15 5 −3 9
12 3 10 3   0 0
8 1 6 1   0 0
15 4 25 7 −3 9
20 5 16 6 −1 1
25 6 12 4   2 4
40 7 8 2   5 25
          D2=48

Given:
N = 7
Now,
rk=1-6ΣD2N3-N=1-6×48343-7   =1-0.857   =0.143
Thus, the rank coefficient of correlation is 0.143.

Page No 325:

Question 23:

The following are the marks obtained (out of 100) by a group of candidates in an employment interview held by two independent judges separately. Calculate the rank coefficient of correlation.

Candidates : A B C D E F G H I J
Judge X : 20 25 18 15 12 16 11 13 14 10
Judge Y : 22 20 15 14 10 8 11 12 13 9
 

Answer:

Candidate Judge
X
R1 Judge
Y
R2 (R1R2)
D
D2
A 20 9 22 10  −1 1
B 25 10 20 9    1 1
C 18 8 15 8    0 0
D 15 6 14 7  −1 1
E 12 3 10 3    0 0
F 16 7 8 1    6 36
G 11 2 11 4  −2 4
H 13 4 12 5  −1 1
I 14 5 13 6  −1 1
J 10 1 9 2  −1 1
            D2=46

Given:
N = 10
Now,
rk=1-6ΣD2N3-N=1-6×461000-10=1-276990=1-0.278rk=0.721
Thus, the rank coefficient of correlation is 0.721.

Page No 325:

Question 24:

Two judges in a beautry competition rank the 12 entries as follows :

X : 1 2 3 4 5 6 7 8 9 10 11 12
Y : 12 9 6 10 3 5 4 7 8 2 11 1
Calculate rank coefficient of correlation.
 

Answer:

X
(R1)
Y
(R2)
(R1R2)
D
D2
1 12 −11 121
2 9  −7 49
3 6  −3 9
4 10  −6 36
5 3    2 4
6 5    1 1
7 4    3 9
8 7    1 1
9 8    1 1
10 2    8 64
11 11    0 0
12 1   11 121
      D2=416

Given,
N = 12
Now,
rk=1-6ΣD2N3-N=1-6×4161728-12=1-24961716=1-1.4545=-0.454
Thus, the rank coefficient of correlation is -0.454.



Page No 326:

Question 25:

 Calculate rank coefficient of correlation of the following data.

X : 80 78 75 75 68 67 60 59
Y : 12 13 14 14 14 16 15 17
 

Answer:

 

X R1 Y R2 (R1R2)
D
D2
80
78
75
75
68
67
60
59
8
7
5.5
5.5
4
3
2
1
12
13
14
14
14
16
15
17
1
2
4
4
4
7
6
8

7
5
1.5
1.5
0
-4
-4
-7
 
49
25
2.25
2.25
0
16
16
49
          D2=159.5

Given:
N = 8
Now, in the given case, where a rank is repeated more than once, the following formula will be used,
rk=1-6ΣD2+112 M13-M1+112M23-M2N3-N   =1-6 159.5+112 23-2+112 33-3512-8   =1-6 159.5+0.5+2504   =1-972504   =1-1.928  =-0.928 or -0.93 (approx.)
Thus, the rank coefficient of correlation is -0.93.

Page No 326:

Question 26:

Twelve entries were submitted in a flower show  competition. They were ranked by two judges as under:

Entries : 1 2 3 4 5 6 7 8 9 10 11 12
Judge A : 7 8 2 1 9 3 12 11 4 10 6 5
Judge B : 6 4 1 3 11 2 12 10 5 9 7 8
 

Answer:

Entries Judge A
(R1)
Judge B
(R2)
(R1R2)
D
D2
1 7 6 −1 1
2 8 4   4 16
3 2 1   1 1
4 1 3 −2 4
5 9 11 −2 4
6 3 2   1 1
7 12 12   0 0
8 11 10   1 1
9 4 5 −1 1
10 10 9   1 1
11 6 7 −1 1
12 5 8 −3 9
        D2=40

Given,
N = 12
To calculate Spearman's Rank correlation, the following formula is used,
rk=1-6ΣD2N3-N   =1-6×401728-12   =1-2401716   =1-0.139   =0.860
Thus, the Spearman's rank correlation is 0.860. 

Page No 326:

Question 27:

 Calculate the coefficient of rank correlation from the following data.

X : 48 33 40 9 16 16 65 25 15 57
Y : 13 13 24 6 15 4 20 9 6 19
 

Answer:

X R1 Y R2 (R1R2)
D
D2
48    8 13 5.5   2.5    6.25
33    6 13 5.5     0.5      0.25
40    7 24  10    −3        9
  9    1   6 2.5 −1.5    2.25
16 3.5 15    7 −3.5  12.25
16 3.5   4    1    2.5    6.25
65  10 20    9       1         1
25    5   9    4       1         1
15    2   6  2.5    −0.5      0.25
57    9 19     8       1         1
          D2=39.5

Given:
N = 10
When, ranks are repeated more than once, the following formula is used to calculate coefficient of rank correlation.
rk=1-6ΣD2+112 m3-m+112 m3-m+112 m3-mN3-N  =1-6 39.5+112 23-2+112 23-2+112 23-21000-10  =1-6×41990=1-246990=1-0.248  =0.751
Thus, the coefficient of rank correlation is 0.751.

Page No 326:

Question 28:

Calculate rank coefficient of correlation between years of service and efficiency rating.

Persons : A B C D E F G H I J
Years of Service : 24 30 12 25 29 19 16 10 11 7
Efficiency rating : 66 51 84 66 45 81 72 97 92 70
 

Answer:

Persons Years of Service R1 Efficiency Rating R2 (R1R2)
D
D2
A 24 7 66 3.5 3.5 12.25
B 30 10 51    2    8      64
C 12 4 84    8  −4      16
D 25 8 66 3.5 4.5 20.25
E 29 9 45    1    8      64
F 19 6 81    7  −1        1
G 16 5 72    6  −1        1
H 10 2 97  10  −8      64
I 11 3 92    9  −6      36
J 7 1 70    5  −4      16
            D2=294.5

Given:
N = 10
When ranks are repeated more than once, the following formula is used to calculate rank coefficient of correlation.
rk=1-6 ΣD2+112 m3-mN3-N    =1-6 294.5+112 23-21000-10    =1-6 294.5+0.5990    =1-1770990    =1-1.787    =-0.787
Thus, the rank coefficient of correlation is -0.787.

Page No 326:

Question 29:

From the following data calculate coefficient of correlation by the method of rank differences.

X : 75 68 95 70 60 80 81 50
Y : 120 134 150 115 110 140 142 100
 

Answer:

X R1 Y R2 (R1R2)
D
D2
75 5 120 4    1 1
68 3 134 5 −2 4
95 8 150 8   0 0
70 4 115 3   1 1
60 2 110 2   0 0
80 6 140 6   0 0
81 7 142 7   0 0
50 1 100 1   0 0
          D2=6

Given:
N = 8
Now,
rk=1-6ΣD2N3-N   =1-6×6512-8   =1-36504=1-0.0714   =0.928 or 0.93 (approx.)
Thus, the coefficient of rank correlation is 0.93.



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