Business Research Methodology by SRINIVAS R RAO - HTML preview

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2

4

14

Sum of squares of deviations for B:

B

B B = B − 7

( − )2

B B

7

0

0

6

-1

1

6

-1

1

9

2

4

6

Sum of squares of deviations for C:

C

C C = C − 6

( − )2

C C

6

0

0

6

0

0

7

-1

1

5

-1

1

2

170

Sum of squares of deviations within

Varieties = ∑( − )2 + ∑( − )2

A A

B B + ∑( C C)2

= 14 + 6 + 2

= 22

Sum of squares of deviations for total variance:

Sales -

Sales person

sales

X = sales

– 7

( Sales − )2

7

A

8

1

1

A

9

2

4

A

5

-2

4

A

10

3

9

B

7

0

0

B

6

-1

1

B

6

-1

1

B

9

2

4

C

6

-1

1

C

6

-1

1

C

7

0

0

C

5

2

4

30

171

ANOVA Table

Source of

Degrees of

Sum of squares

Variance

variation

freedom

of deviations

Between varieties

3 – 1 = 2

8

8 = 4

2

Within varieties

12 – 3 = 9

22

22 = 2.44

9

Total

12 – 1 = 11

30

Calculation of F value:

F

=

Greater Variance = 4.00 =1.6393

Smaller Variance

2.44

Degrees of freedom for greater variance ( df 1) = 2Degrees of freedom for

smaller variance ( df 1) = 9

Let us take the level of significance as 5%

The table value of F = 4.26

Inference:

The calculated value of F is less than the table value ofF. Therefore,

the null hypothesis is accepted. It is concluded that there is no significant

difference in the performance of the sales persons, at 5% level of

significance.

Method II (Short cut Method):

∑ A = 32, ∑ B = 28, ∑ C = 24.

T= Sum of all the sample items

172

= ∑ A+ ∑ B + ∑ C

= 32 + 28 + 24

= 84

N = Total number of items in all the samples = 4 + 4 + 4 =12

2

2

Correction Factor = T

84

=

= 588

N

12

Calculate the sum of squares of the observed values as follows:

Sales Person

X

X2

A

8

64

A

9

81

A

5

25

A

10

100

B

7

49

B

6

36

B

6

36

B

9

81

C

6

36

C

6

36

C

7

49

C

5

25

618

Sum of squares of deviations for total variance =

2

X - correction

factor

= 618 – 588 = 30.

173

Sum of squares of deviations for variance between samples

(∑ A)2 (∑ B)2 (∑ C)2

=

+

+

CF

N

N

N

1

2

3

2

2

2

32

28

24

=

+

+

− 588

4

4

4

1024 784 576

=

+

+

− 588

4

4

4

= 256 +196 +144 − 588

= 8

ANOVA

Table

Source of

Degrees of

Sum of squares

Variance

variation

freedom

of deviations

Bet

3-1 = 2

8

ween varieties

8 = 4

2

Within varieties

12 – 3 = 9

22

22 = 2.44

9

Total

12 – 1 = 11

30

It is to be noted that the ANOVA tables in the methods I and II

are one and the same. For the further steps of calculation of F value and

drawing inference, refer to method I.

174

Problem 2

The following are the details of plinth areas of ownership apartment

flats offered by 3 housing companies A,B,C. Use analysis of variance to

determine whether there is any significant difference in the plinth areas of

the apartment flats.

H o u s i n g

Plinth area of apartment flats

Company

A

1500

1430

1550

1450

B

1450

1550

1600

1480

C

1550

1420

1450

1430

Use analysis of variance to determine whether there is any

significant difference in the plinth areas of the apartment’s flats.

Note:

As the given figures are large, working with them will be difficult.

Therefore, we use the following facts:

i).

Variance ratio is independent of the change of origin.

ii.)

Variance ratio is independent of the change of scale.

In the problem under consideration, the numbers vary from 1420

to 1600. So we follow a method called the coding method. First, let us

subtract 1400 from each item. We get the following transformed data:

Company

Transformed measurement

A

100

30

150

50

B

50

150

100

80

C

150

20

50

30

Next, divide each entry by 10.

175

The transformed data are given below.

Company

Transformed measurement

A

10

3

15

5

B

5

15

10

8

C

15

2

5

3

We work with these transformed data. We have

∑ =

A 10+3+15+5=33

B 5+

= 15+10+8=38

C=15+2+5+3=25

T = ∑ A+∑ B +∑ C

= 33+ 38 + 25

= 96

N = Total number of items in all the samples = 4 + 4 + 4 = 12

2

2

Correction factor = T

96

=

= 768

N

12

Calculate the sum of squares of the observed values as follows:

Company

X

X2

A

10

100

A

3

9

A

15

225

A

5

25

B

5

25

B

15

225

B

10

100

B

8

64

C

15

225

176

C

2

4

C

5

25

C

3

9

1036

Sum of squares of deviations for total variance =

2

X - correction

factor

= 1036 – 768 = 268

Sum of squares of deviations for variance between samples

(∑ A)2 (∑ B)2 (∑ C)2

=

+

+

CF

N

N

N

1

2

3

2

2

2

33

38

25

=

+

+

− 768

4

4

4

1089 1444 625

=

+

+

− 768

4

4

4

= 272.25 + 361+156.25 − 768

= 789.5 − 768

=

21.5

ANOVA Table

Source of

Degrees of

Sum of squares

Variance

variation

freedom

of deviation

Between varieties

3-1 = 2

21.5

21.5 =10.75

2

Within varieties

12 – 3 = 9

264.5

24.65 = 27.38

9

Total

12 – 1 = 11

268

177

Calculation of F value:

F = Greater Variance = 27.38 = 2.5470

Smaller Variance 10.75

Degrees of freedom for greater variance ( df 1) = 9

Degrees of freedom for smaller variance ( df 2 ) = 2

The table value of f at 5% level of significance = 19.38

Inference:

Since the calculated value of F is less than the table value of F, the

null hypothesis is accepted and it is concluded that there is no significant

difference in the plinth areas of ownership apartment flats offered by the

three companies, at 5% level of significance.

Problem 3

A finance manager has collected the following information on the

performance of three financial schemes.

Sum of squares of

Source of variation

Degrees of freedom

deviations

Treatments

5

15

Residual

2

25

Total (corrected)

7

40

Interpret the information obtained by him.

Note: ‘Treatments’ means ‘Between varieties’.

‘Residual’ means ‘Within varieties’ or ‘Error’.

178

Solution:

Number of schemes = 3 (since 3 – 1 = 2)

Total number of sample items = 8 (since 8 – 1 = 7)

Let us calculate the variance.

Variance between varieties =

15 = 7.5

2

Variance between varieties =

25 = 5

5

F

=

Greater Variance = 7.5 =1.5

Smaller Variance

5

Degrees of freedom for greater variance

( df 1) = 2

Degrees of freedom for smaller variance

( df 1) = 5

The total value of F at 5% level of significance

= 5.79

Inference:

Since the calculated value of F is less than the table value of F

we accept the null-hypothesis and conclude that there is no significant

difference in the performance of the three financial schemes.

179

SELF ASSESSMENT QUESTIONS---END OF CHAPTER

QUESTIONS.[IMPORTANT FOR EXAMS]

1. Define analysis of variance.

2. State the assumptions in analysis of variance.

3. Explain the classification of linear models for the sample

data.

4. Explain anova table.

5. Explain how inference is drawn from anova table.

6. Explain the managerial applications of analysis of variance.

180

CHAPTER IV

3. Partial And Multiple Correlation

The Concept Of Partial Correlation

The Concept Of Multiple Correlation

Determine Partial Correlation Coefficient

Determine Multiple Correlation Coefficient

I. PARTIAL CORRELATION

Simple correlation is a measure of the relationship between a

dependent variable and another independent variable. For example, if

the performance of a sales person depends only on the training that he

has received, then the relationship between the training and the sales

performance is measured by the simple correlation coefficient r. However,

a dependent variable may depend on several variables. For example, the

yarn produced in a factory may depend on the efficiency of the machine,

the quality of cotton, the efficiency of workers, etc. It becomes necessary

to have a measure of relationship in such complex situations. Partial

correlation is used for this purpose. The technique of partial correlation

proves useful when one has to develop a model with 3 to 5 variables.

Suppose Y is a dependent variable, depending on n other variables

X , X , …, X .. Partial correlation is a measure of the relationship between

1

2

n

Y and any one of the variables X , X ,…,X , as if the other variables have

1

2

n

been eliminated from the situation.

181

The partial correlation coefficient is defined in terms of simple correlation

coefficients as follows:

Let r denote the correlation of X and X by eliminating the

12. 3

1

2

effect of X .3

Let r be the simple correlation coefficient between X and X .

12

1

2

Let r be the simple correlation coefficient between X and X .

13

1

3

Let r be the simple correlation coefficient between X and X .

23

2

3

Then we have

r 12 r 13 r 23

r 12.3 =

2

2

(1− r 13) (1− r 23)

Similarly,

r 13 r 12 r 32

r 13.2 =

2

2

(1− r 12) (1− r 32)

and

r 23 r 21 r 13

r 32.1 =

2

2

(1− r 21) (1− r 13)

182

Problem 1

Given that r = 0.6, r = 0.58, r = 0.70 determine the partial

12

13

23

correlation coefficient r12.3

Solution:

We have

0.6−0.58 x 0.70

=

2

2

(1−(0.58) ) (1−(0.70) )

0.6−0.406

=

(1−0.3364) (1−0.49)

0.194

= 0.6636 x 0.51

0.194

=

0.8146 x 0.7141

0.194

= 0.5817

= 0.3335

183

Problem 2

If r = 0.75, r = 0.80, r = 0.70, find the partial correlation

12

13

23

coefficient r

13.2

Solution:

We have

r r r

13

12

32

r 13.2 =

2

2

(1− r 12) (1− r 32)

0.8 0.75 X 0.70

=

2

2

(1− (0.75) ) (1− (0.70) )

0.8 − 0.525

= (1−0.5625) (1−0.49)

0.275

= (0.4375) (0.51)

0.275

=

0.6614 X 0.7141

0.275

=

0.4723

= 0.5823

II. MULTIPLE CORRELATION

When the value of a variable is influenced by another variable, the

relationship between them is a simple correlation. In a real life situation, a

variable may be influenced by many other variables. For example, the sales

achieved for a product may depend on the income of the consumers, the

price, the quality of the product, sales promotion techniques, the channels

of distribution, etc. In this case, we have to consider the joint influence

184

of several independent variables on the dependent variable. Multiple

correlations arise in this context.

Suppose Y is a dependent variable, which is influenced by n other

variables X , X , …,X . The multiple correlation is a measure of the

1

2

n