errata statistics for management

Statistics for Management

[ERRATA]

STATISTICS FOR MANAGEMENT

MB 0040

Book ID: B1129

Sikkim Manipal University DE

Unit Page No.

Location EXISTING CORRECT

1

7

Section 1.2

Analysis of ata

Analysis of data

1 17 Answers to Self

Assessment Questions

2. Industrial Quality control, Investment policies, to find

Market potential for a product.

3. The four components of Statistics are collection,

presentation, analysis and interpretation of data.

2. The four components of Statistics are collection,

presentation, analysis and interpretation of data.

3. Industrial Quality control, Investment policies, to

find Market potential for a product.

1 17 Answers to

Terminal Questions

probabilitie

probabilities

Unit Page No.


3 48 Table 3.4

Departments Age

20 – 40 40 & above

Accounts 2.564 1.282

Finance 2.564 1.795

Personal 3.846 1.282

Production 2.564 2.051

Marketing 1.282 1.795

Total 12.920 8.205

Departments Age

20 – 40 40 & above

Accounts 24 21.429

Finance 20 23.571

Personal 20 20.714

Production 20 18.571

Marketing 16 15.714

Total 100 100

3 49 Table 3.6

Batch Defects

Major Minor

I 8 7

II 15 5

III 25 15

Total 40 27

Batch Defects

Major Minor

I 8 7

II 15 5

III 25 15

IV 32 18

Total 80 45

3 53 Solved Problem 3 Solved Problem 3: Consider the frequency distribution of marks given in table 3.9. Calculate the less than and more than cumulative frequency distribution.

Solved Problem 3: Consider the frequency distribution of marks given in table 3.9. Calculate the derived frequency distributions, less than and more than cumulative frequency distribution.

3 55 Table 3.14c

Age 40-50 Salary

6 6000-12000

15 12000-15000

10 15000-18000

31 Total

Age 40-50 Salary

6 9000-12000

15 12000-15000

10 15000-18000

31

3 64 Section 3.6.2

Solved Problem 10

Solved Problem 10: Construct a frequency polygon for the data represented in table 3.20.

Solved Problem 10: Construct a frequency polygon for the data represented in table 3.21.

3 65

Section 3.6.3

Solved Problem 11

Solved Problem 11: Construct a frequency curve for the data represented in table 3.19.

Solved Problem 11: Construct a frequency curve for the data represented in table 3.21.

Unit Page No.


3 69 Answers to Self

Assessment Questions

1. i. Grouping, Common, Characteristics.

ii. Bulk,

iii. Attribute

iv. Series

v. Two

vi. Series

1. i. Grouping, common characteristics.

ii. Bulk,

iii. Attribute

iv. Series

v. Two

vi. Chronological classification

3 71 Answers to

Terminal Questions

5. The figure 3.17 is the ogive curve for the data given in

terminal question 5. i.16% ii. 47%

5. The figure 3.17 is the ogive curve for the data given

in terminal question 5. i.16% ii. 57%

4 95

Solved Problem 20

Solved Problem 20: Weekly sales of a product on 8 different shops are as follows. Calculate the quartiles. Sales in units: 309, 312, 305, 307, 310, 308, 308, 306, 308

Solved Problem 20: Weekly sales of a product on 8 different shops are as follows. Calculate the quartiles.

Sales in units: 309, 312, 305, 307, 310, 308, 308, 306

4 95 Solution to Solved

Problem 20 305, 306, 307, 308, 309, 310, 312 305, 306, 307, 308, 308, 309, 310, 312


Problem 20

th

24

)1n(2Q

value = 2.25 x 2 = 4.5th value

= 4th value + 0.5 (5th value – 4th value)

= 308 + 0.5 (30/ - 308) = 308

th

24

)1n(2Q

value = 2.25 x 2 = 4.5th value

= 4th value + 0.5 (5th value – 4th value)

= 308 + 0.5 (308 - 308) = 308


Problem 21

P20 class

Q1 class and Q2 class

D7 class

Q3 class

N=200 valueth504

ValueNthQ1

Highlighted four lines to be removed

N=200 value50thValueth 4

NQ1

4 97 Self Assessment

Questions

3. State whether the following questions are ‘True’ or

‘False’.

i. Quantiles are positional value.

ii. Quantiles help us to find percentage of readings

below or above a certain value.

3. State whether the following questions are ‘True’ or

‘False’.

i. Quartiles are positional value.

ii. Quartiles help us to find percentage of

readings below or above a certain value.

Unit Page No.


4 97 Solved Problem

22

Solved Problem 22: Find the 7th decile for the same data

given in solved problem 22.

Solved Problem 22: Find the 7th decile for the same

data given in solved problem 21.

4 98 Table 4.18

Classification Moni Mani Weight

Assignment 60 40 5

Presentation 80 60 10

First Test 50 100 10

Find Test 100 70 20

45

Classification Moni Mani Weight

Assignment 60 40 5

Presentation 80 60 10

First Test 50 100 10

Final Test 100 70 20

45

4 104 Table 4.24

X From Meanx –

145

From Medianx –

143.5

140 5 3.5

141 4 2.5

142 3 1.5

143 2 0.5

144 1 0.5

145 0 1.5

147 2 3.5

158 13 6.5

1160 30 20.0

X From Meanx –

145

From Medianx –

143.5

140 5 3.5

141 4 2.5

142 3 1.5

143 2 0.5

144 1 0.5

145 0 1.5

147 2 3.5

150 5 6.5

1152 22 20.0


Problem 28

Mean deviation from mean = 75.38

30

Coefficient of MD )X( = 0258.0145

75.3

Mean deviation from mean = 75.28

22

N

Coefficient of MD )X( = 018965.0145

75.2

4 105 Line No. 2

Coefficient of MD )X( = 001742.05.143

5.2

Coefficient of MD )X( = 01742.05.143

5.2


Problem 29

24465

6528)X(

24465

)65(82h

f

fdA)X(

4 114 Table 4.32

Table 4.32. Distribution data for terminal question 4

% Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70

No. of Smokers 4 9 19 20 18 7 80

Table 4.32. Distribution data for terminal question 4

% Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70

No. of Smokers

4 9 19 20 18 7 8

Unit Page No.


4 115 Answers to

Terminal Questions

2. 31.64

2. Mean= 31.64, S.D =13.362, CV =42.28%

4 116

Answers to

Terminal Questions

4. 34 4. Median = 35.25, Mode = 33.3

5 129 Key Statistic

nCr = nCr-1

nCr = nCn-r

5 135 Line No.6

5

2

15

6

5

4

3

1

5

1

3

2

5

2

15

6

5

4

3

1

5

1

3

2

Unit Page No.


6 150 Line No.12 theorectical theoretical

6 159 Last Line

0231.025.075.0!3!5

!85

53

0231.025.075.0!3!5

!85

53

7 184 Line No. 8 2

(3)10

97.502

f

fx

f

2fx2

S

2(3)

10

2

f

fx

f

2fx2

S97.50

7 184 Line No.10

0.8660.7500S

asdpoooooooo

Hence, the standard error of the mean ‘S’ is 0.866.

0.8660.7500S

Hence, the standard error of the mean ‘S’ is 0.866.

Unit Page No.


8 208 8.8.2 -p = p = p

8 209 8.8.2 Therefore, standard error of the mean = n/pq

Therefore, standard error of the mean = n/pq

= = 0.057

8 215 Answers to self

assessment Questions

55.01125

615

n

σx

55.01125

615

n

σ

x


Problem 4

5. Conclusion: Since 2cal (7.5) < 2

tab (9.49), ‘Ho’ is rejected.

Hence, absenteeism and days of week are independent.


tab (9.49), ‘Ho’ is

accepted. Hence, absenteeism and days of week are

independent.


Problem 5


tab (7.81), ‘Ho’ is

rejected. Therefore, the result supports the theory.


tab (7.81), ‘Ho’ is

accepted. Therefore, the result supports the theory.

12 292

Line No. 13

Key Statistic

2/1222/122 )dy(dyN)dx(dxN

dydxdydxNr

––(D)

1/2221/222 dy)(dyNdx)( dxN

dydxdydxNr

––(D)

12 294

Solution to Solved

Problem 1

(Line No. 3)

2/1222/122 )Y(YN)XN(XN

YXXYNr

1/2221/222 Y)(YNX)(XN

YXXYNr

12 294

Solution to Solved

Problem 1

(Line No 4)

22 )30()904(5.)60()880(5

)60)(60()840(5r

22 )60()904(5.)60()880(5

)60)(60()840(5

r

Unit Page No.


12 294

Table 12.2b

(Column 4 and

Column 7)

x2 y

2

16 0

4 9

0 4

9 16

1 9

64 9

J 16

25 121

x2 =

120 y

2 =

194

x2 y

2

16 0

4 9

0 4

9 16

1 9

64 9

1 16

25 121

x2 =

120 y

2 =

184

12 295

(Line No. 2)

Solution to Solved

Problem 2

00.61184120

92

)y()x(

xyr

22

0.619

184120

92

)y()x(

xyr

22

Unit Page No.



Problem 5

1286.0

)138(213610)56(135710

)138)(56(83610r

2/122/12

1286.0

)138(213610)56(135710

)138)(56(-836102/122/12

r

12 299 Table 12.5b

Competitor

R1 (Judge 1)

R2 (Judge 2)

D = R1 – R2

D2

1 5 6 -1 1

2 6 4 -2 4

3 4 5 -1 1

4 3 1 2 4

5 2 2 0 0

6 7 7 0 0

7 1 3 2 4

13

Competitor

R1 (Judge

1)

R2 (Judge 2)

D = R1 – R2

D2

1 5 6 -1 1

2 6 4 2 4

3 4 5 -1 1

4 3 1 2 4

5 2 2 0 0

6 7 7 0 0

7 1 3 -2 4

14

12 299 Below Table

12.5b

= 1 – 768.0487

1361

)17(7

1362

Hence, Spearman’s rank correlation coefficient is 0.768

= 1 – 75.025.01487

1461

)17(7

1462

Hence, Spearman’s rank correlation coefficient is

0.75

12 300

Line 1

(Above Solved

Problem 9)

This heading which is given in the beginning of the

page should be ignored

12 307

(Line No. 6 in

page 307)

Section 12.8.2

22y x)dx(dxN

)dy()dx(dxdyNb

and

22y x)dy(dyN

)dy()dx(dxdyNb

22y x)dx(dxN

)dy()dx(dxdyNb

and

22xydy)(dyN

dy)(dx)(dxdyNb

When ranks are repeated Type iii:

12 307 Section 12.8.3 1. xyyx bb rbb xyyx .

12 308

Table 12.10

(Second row of

the table)

Correlation Coefficient

Regression Coefficient

The correlation coefficients,

rxy = ryx

The regression coefficients,

byx = bxy

Correlation Coefficient

Regression Coefficient

The correlation coefficients,

rxy = ryx

The regression coefficients,

byx ≠ bxy

12 309

Line No. 8

(Solution to

Solved Problem

12)

Regression Equation of X and Y is:

byx = 392.124

43

)5(2410

)0()5(4310

2

Regression Equation of X and Y is:

bxy= 792.124

43

)0(2410

)0()5(43102


Problem 15

Regression equation Y on X is given by:

)XX(b)YY(

Regression equation Y on X is given by:

)X(Xb)Y(Y yx

14 345 Table 14.4

Solution: The table 14.5 displays the calculated values of 3

yearly and 5 yearly averages.

Table 14.5: Calculated values of 3 yearly and 5 yearly averages

Year Production

(Thousand Y Tonnes)

3 –yearly moving totals


Ye

Short term fluctuations

(Y - Yc)

1988 21 - - -

1989 22 66 22.00 0

1990 23 70 23.33 - 0.33

1991 25 72 24.00 1.00

1992 24 71 23.67 0.33

1993 22 71 23.67 - 1.67

1994 25 73 24.33 0.67

1995 27 79 26.33 0.67

1996 26 - - -

Solution: The table 14.5 displays the calculated values

of 3 yearly averages.

Table 14.5: Calculated values of 3 yearly averages

Year Production

(In Lakh Tonnes)



Yc

Short term fluctuations

(Y - Yc)

1988 15 - - -

1989 18 49 16.33 1.67

1990 16 56 18.66 -2.66

1991 22 57 19.00 3.00

1992 19 65 21.66 -2.66

1993 24 63 21.00 3.00

1994 20 72 24.00 -4.00

1995 28 70 23.33 4.67

1996 22 80 26.66 -4.66

1997 30 - - -

15 373

Laspeyre’s price

index

100QP

QPIIndexicePrs'Laspeyre

0101

01

100

Q0

ΣP

QΣPLP:IndexPricesLaspeyre'

001

01

15 373

Laspeyre’s index

(Line No. 15)

The quantity index number using Laspeyre’s formula is

given by:

100PQ

PQQ

00

11

01

The quantity index number using Laspeyre’s formula is

given by:

100PQ

PQQ

00

01

01

15 374 Dorbish and

Bowley’s method

1002

01PPLP

DP01

100

1Q0P

1Q1P

0Q0P

oQ1P

1002

01PPLP

DP01

1002

1Q

0PΣ

1Q

1PΣ

0Q

0PΣ

oQ

1PΣ

15 373 Section 15.3.2

In addition to Laspeyres method, Paasche’s method and Dorbish & Bowleys method, there is one more method

called Fisher’s method .Fisher’s Ideal Index is not given in SLM .

Fisher’s Ideal Index = 100

1Q

0PΣ

1Q

1PΣ

0Q

0PΣ

oQ

1PΣ

15 379

Section 15.6.2

(Line No.5)

Family budget method or the method of weighted

relatives is the method where weights are the value (P0Q0)

in the base year often denoted by V.

100V

RVP

01

1.6.15tionsecsubinequationtheassame100QP

QP

00

01

Family budget method or the method of weighted

relatives is the method where weights are the value

(P0Q0) in the base year often denoted by V.

100V

RVP

01

Highlighted yellow Line to be removed

errata statistics for management

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