Simple Tabulation 7and.ppt

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Simple Tabulation

Cross Tabulation

Chi-square



1. In a questionnaire-based marketing research project, each question usually

represents a variable under study. The basic form of analysis of onevariable in a questionnaire is Simple Tabulation of the answers. This could

be in the form of simple counting of the frequencies (how many people

answered Yes, and No, for example), and percentages.

This consists of every question being treated separately and tabulated. For

every question, the number of responses in each category of answers iscounted. Assuming the sample size is 500, and all 500 have answered the

question, the simple tabulation of the respondents' gender may look like

the following

Frequency

Male 300

Female 200

Total 500

Table 1: Gender



Graduation Frequency Percent

B.Com 82 35.5

B.Sc 67 29.0

B.A 19 8.2

B.E 35 15.2

Others 28 12.1

Total 231 100.0

Table 2: Education



Table 2 analysis

• The analysis of Table 2 furnishes the details of

respondents and education. It is evident that less

than half (82) 35.5% of the respondents are from

B.Com (Bachelor of Commerce) background; whereas 62 (29%) are from B.Sc (Bachelor of Science)

background. On an average the B.A (Bachelor of Arts)

students seems to be less with only 19 students

coming from this background.

• Irrespective of their graduation (B.Com, B.Sc, B.A and

B.E ) the graduates preferred post graduation course

in management



Table 3: Choice of Industry

Frequency Percent

Banking 44 19.0

Finance 43 18.6

IT&ITES 43 18.6

FMCG 41 17.7

Pharma 18 7.8

Retailing 30 13.0

Others 12 5.2

Total 231 100.0

Write the data analysis for Table 3



Table 3: Analysis

• From the table it is evident that majority

which is off course less than one-fourths of

the students (44, 19%) preferring Banking

sector closely followed by Finance and IT &

ITES (43, 18.6%). FMCG is the preference of

41 (17.7%) students, while 30 students opted

for retailing.



Table: 4

Delivery Time

Frequency Percent

Strongly Agree 0 0

Agree 0 0

Neutral 46 39.3

Disagree 52 44.4

Strongly

Disagree

19 16.2

Total 117 100.0



Graph 1

Strongly Agree

0%

Agree

0%

Neutral

39.3%

Disagree

44.4%

Strongly Disagree16.2%

Delivery Time



2. Two different questions in a questionnaire may represent two variables, and

if we count these two together, this is called a cross-tabulation. An example

could be “10 people from Income Group 1 said they liked Brand A”. Here, the

two variables are “INCOME GROUP” and “LIKING FOR BRANDS A TO E”, measured separately in two different questions on the questionnaire.

3. Simple and Cross tabulation is a very useful form of analysis for all nominally

and ordinally scaled variables. For these two scales, calculations such as

average (mean) and standard deviation are not permitted. Therefore, frequencyand percentages are used to analyse such variables.



Table 5: Graduation * Choice of

Industry Cross tabulation

CHOICE OF INDUSTRY

Graduati

on Banking Finance

IT&ITES

FMCG

Pharmacy

Retailing

Other s Total

B.Com 18 23 7 14 3 10 7 82

B.Sc 12 10 6 15 9 13 2 67

B.A1 0 7 7 1 3 0 19

B.E 5 7 14 1 2 4 2 35

Others 8 3 9 4 3 0 1 28

Total 44 43 43 41 18 30 12 231



Dependent and Independent Variables

1. If two or more variables are analysed together, it may be necessary to spell out

the relationship between the two variables. The concept of dependent and

independent variables is useful in spelling out the relationship. Two variables arecalled independent variables if a change in one does not influence or cause a

change in the other. But if a change in one variable causes a change in the other,

the first one is called an independent variable, and the second one is called a

dependent variable (dependent on the first).

2. A common example of a dependent variable in marketing is “Sales”. Annual

sales of a brand usually depend on several factors or variables. One of the

independent variables on which annual sales depend could be the quantum of

advertising (in rupees) done for the brand. A second variable on which sales may

depend could be the number of retailers stocking the brand.

3. In a consumer research questionnaire, the dependent variable could be

satisfaction with the brand, which may depend on taste (if it is a food brand), and

easy availability. Another example is the quantity of a product bought, a

dependent variable, which depends on family size and household income.



Demographic Variables

1. Many demographic variables such as age, location, income, occupation,

gender, education are generally independent variables for the purposes of

most marketing studies. This is because other variables “depend” on them.

2. Attitude towards a brand, or the brand purchased, or intention to buy, are

usually treated as dependent variables in many marketing studies. For a

marketing researcher, these variables or similar ones, are the real variables

of interest, as they help in arriving at strategies for increasing sales or

market share.

3. The other major types of independent variables are the elements of the

four ‘P’s of marketing. The marketing effort of a company can be measured

in terms of its promotional efforts, price variations and distribution changes.It can also be gauged from new product launches, or repositioning or

repackaging of existing brands.

4. Therefore, we could measure sales as the dependent variable with any of

the marketing ‘P’s as independent variables.



Manual Calculation of Chi-square

• The sales manager of a showroom feels that

the daily demand of a product follows uniform

distribution. The observed frequencies of

demand values are given in the table below.Check whether the given data follows uniform

distribution at a significance level of .05

Demand 30 31 32 33 34 35 36 37 38 39

Observed

Frequency

13 10 7 10 6 9 12 10 14 9



Calculation with hypothesis

• H0 : The given data follows uniform distribution or there is no difference inthe data distribution

• H1 : The given data do not follow uniform distribution or there isdifference in the data distribution

• The total observed frequencies = 100

• Number of demand values = 10

• Expected frequency for each demand value = 100/10 = 10

• Formula χ 2 = (Oij – Eij )2

Σ ____________________________

Ei j

Where O is observed frequency of the cell in the ith row and jth columnand E is expected frequency of the cell in the ith row and jth column



Intrepretation of Chi-square value

• if calculated value is less than table value

accept the H0 and reject the H1

• if calculated value is more than table value

accept the H1 and reject the H0



Degree of Freedom Determination

• If there are 10 frequency classes and there is

one independent constraint, then there are

(10-1) = 9 Degree of Freedom.







Hypothesis

• H0 : The given data follows uniform distribution

• H1 : The given data does not follow uniform

distribution



S.no Demand Observed

Frequency

Expected

Frequency

(O – E ) (O – E )2 (O – E)2

____________________________

E

1 30 13

2 31 10

3 32 7

4 33 10

5 34 6

6 35 9

7 36 12

8 37 10

9 38 14

10 39 9




Frequency

Expected

Frequency

(O – E ) (O – E )2 (O – E)2

____________________________

E

1 30 13 10

2 31 10 10

3 32 7 10

4 33 10 10

5 34 6 10

6 35 9 10

7 36 12 10

8 37 10 10

9 38 14 10

10 39 9 10




Frequency

Expected

Frequency

(O – E ) (O – E )2 (O – E)2

____________________________

E

1 30 13 10 3

2 31 10 10 0

3 32 7 10 -3

4 33 10 10 0

5 34 6 10 -4

6 35 9 10 1

7 36 12 10 2

8 37 10 10 0

9 38 14 10 4

10 39 9 10 1




Frequency

Expected

Frequency

(O – E ) (O – E )2 (O – E)2

____________________________

E

1 30 13 10 3 9

2 31 10 10 0 0

3 32 7 10 -3 9

4 33 10 10 0 0

5 34 6 10 -4 16

6 35 9 10 1 1

7 36 12 10 2 4

8 37 10 10 0 0

9 38 14 10 4 16

10 39 9 10 1 1




Frequency

Expected

Frequency

(O – E ) (O – E )2 (O – E)2

____________________________

E

1 30 13 10 3 9 .9

2 31 10 10 0 0 0

3 32 7 10 -3 9 .9

4 33 10 10 0 0 0

5 34 6 10 -4 16 1.6

6 35 9 10 1 1 .1

7 36 12 10 2 4 .4

8 37 10 10 0 0 0

9 38 14 10 4 16 1.6

10 39 9 10 1 1 .1



Result and interpretation of problem 1

Total χ 2= 5.6The tabulated chi-square value with (n-1) (10-

1)=9 at 5% significance level is 16.919.

since the calculated value 5.6 is less than

tabulated value 16.919 accept H0 which

means H1 is not accept.

Hence, it was found that the daily demand of a

product does follow uniform distribution



Sum Two

• Is there is any association between the

demand and price?

Demand 28 33 38 47 52 56 61

Price 8 7 6 5 4 3 2



• H0 : Demand is independent of price

• H1 : Demand is dependent on price

•The total observed frequency (Price) = 35

• Number of demand values = 7

• Expected frequency for each demand value =

35/7= 5




Frequency

(Price)

Expected

Frequency(Oi – Ei)

2

____________________________

Ei

1 28 8 5 1.8

2 33 7 5 .8

3 38 6 5 .2

4 47 5 5 0

5 52 4 5 .2

6 56 3 5 .8

7 61 2 5 1.8

5.6



Interpretation of sum 2 result

• The table value with (7-1) = 6 DF at 5%

significance level is 12.591.

• Since the calculated value 5.6 is less than

tabulated value 12.591 accept H0, reject H1 ,

which means that demand is independent of

price.



Sum 3 Calculation of Chi-square• A total of 1600 families were selected at random in a city to

test the belief that high income families usually send theirchildren to private schools and low income families often send

their children to government schools.

Income Private

schools

Government Total

Low 494 506 1000

High 162 438 600

656 944 1600

Test whether income and type of school are independent



• H0 : Income and choice of school are not related or (areindependent)

• H1 : income and choice of school are related or (aredependent) (Choice of school is dependent on income)

• To calculate the expected value we should apply theformula as

• Row total multiplied by column total divided by grandtotal

•656 X 1000/1600 = 410 for row 1 public schools

• Degree of Freedom shall be (r-1) x (c-1) = (2-1) x(2-1) =1



S.no Observat

ion

Observed

Frequency

Expected

Frequency

(O – E ) (O – E )2 (O – E)2

____________________________

E

1 Row 1

(low)

494

2 Row 2

(Govt)

506

3 Row 1

(Public)

162

4 Row 2

(Govt)

438

1600



S.no Observat

ion

Observed

Frequency

Expected

Frequency

(O – E ) (O – E )2 (O – E)2

____________________________

E

1 Row 1

(low)

494 410

2 Row 2

(Govt)

506 590

3 Row 1

(Public)

162 246

4 Row 2

(Govt)

438 354

1600



S.no Observat

ion

Observed

Frequency

Expected

Frequency

(O – E ) (O – E )2 (O – E)2

____________________________

E

1 Row 1

(low)

494 410 84

2 Row 2

(Govt)

506 590 -84

3 Row 1

(Public)

162 246 -84

4 Row 2

(Govt)

438 354 84

1600



S.no Observat

ion

Observed

Frequency

Expected

Frequency

(O – E ) (O – E )2 (O – E)2

____________________________

E

1 Row 1

(low)

494 410 84 7056/410

2 Row 2

(Govt)

506 590 -84 7056/590

3 Row 1

(Public)

162 246 -84 7056/246

4 Row 2

(Govt)

438 354 84 7056/354

1600



S.no Observat

ion

Observed

Frequency

Expected

Frequency

(O – E ) (O – E )2 (O – E)2

____________________________

E

1 Row 1

(low)

494 410 84 7056/410 17.21

2 Row 2

(Govt)

506 590 -84 7056/590 11.96

3 Row 1

(Public)

162 246 -84 7056/246 28.68

4 Row 2

(Govt)

438 354 84 7056/354 19.93

1600 77.78

The critical of χ 2 for 1 Df at 5% level is 3.841



Interpretation of sum 3

• As the calculated value 77.78 of χ 2 is greater

than table value 3.841, H0 is rejected. H1 is

accepted.

• Hence, There is relationship between income

and choice of school or it means that choice of

school is dependent on income.



Sums for Practice

• On the basis of information given below aboutthe treatment of 200 patients suffering from adisease, state whether the new treatment is

comparatively superior to the conventionaltreatment.

No of Patients

Treatment Favorable

Response

No Response

New 60 20

Conventiona

l

70 50



H0= The new treatment is equal to conventional treatment

H1= The new treatment is superior to conventional treatment

No of Patients Total

Treatment Favorable Response No Response

New 60 20 80

Conventional 70 50 120

130 70 200

2



S.no Observat

ion

Observed

Frequency

Expected

Frequency

(O – E ) (O – E )2 (O – E)2

____________________________

E

1 N.F.R 60 52 8 64 1.230

2 C.F.R 70 78 -8 64 .8205

3 N.N.R 20 28 8 64 2.285

4 C.N.R 50 42 -8 64 1.523

5.859

The critical of χ 2 for 1 Df at 5% level is 3.841



Interpretation of practice sum

• Since the calculated value 5.859 is more than

the table value of 3.84146, it was found that

H1 is accepted therefore, that the new

treatment is comparatively superior to theconventional treatment.



Sums for Practice

• 200 digits were chosen at random from a set

of tables. The frequencies of the digits were:

Digit 0 1 2 3 4 5 6 7 8 9

Frequency 18 19 23 21 16 25 22 20 21 15

Calculate the chi-square and test whether data is uniformly distributed



Sums for Practice

Age

Beauty Soap Brand

Total ‘ p’ Value Popular

Brands of

Soaps

Premium

Brands of

Soaps

Economy

Brands of

Soaps

< 25 years (46) (47) (10) (103)

0.028 26-50 years (138) (71) (9) (218)

> 51 years (73) (42) (11) (126)

Total (257) (160) (30) (447)



• 447 respondents from three age groups opted

for three categories of beauty soaps. The

results of the data is as below.

• Find how is age associated with the choice of

beauty soap categories?



Hypothesis

• H0= Age has no influence on the choice of

different categories of beauty soap brands

• H1= There is influence of age on the choice on

different categories of beauty soap brands



S.no Observati

on

Observed

Frequenc

y

Expected

Frequency

(O – E ) (O – E )2 (O – E)2

___________________

E

1 46 59.22 -13.22 174.75 2.95

2 138 125.34 12.66 160.275 1.2783 73 72.44 0.56 0.313 0.004

4 47 36.87 10.13 102.61 2.78

5 71 78.03 7.03 49.42 0.633

6 42 45.10 3.1 9.61 0.213

7 10 6.91 6.91 9.54 1.38

8 9 14.63 14.63 31.69 2.166

9 11 8.46 8.46 55.65 0.667

Total = 12.071

A l i



• The calculated value 12.071 is more than 9.49

Df at 5% level is 9.49, therefore H1 is

accepted. Hence, it is concluded that age

influences the choice of different categories of beauty soap brands.

Analysis



To Obtain a Chi-Square Test

• From the menus choose:

• AnalyzeNonparametric Tests

Chi-Square...• Select one or more test variables. Each

variable produces a separate test.

• Optionally, you can click Options fordescriptive statistics, quartiles, and control of the treatment of missing data.



Analysis

• In a chi-square test, for a 95% confidence

level, if the significance level is greater than or

equal to .05, it signifies that there is no

association between the two variables in thecross-tabulation and if significance level is less

than .05, then it signifies that there is a

significant relationship between the selectedvariables.



• Social Factors

• In this chapter the social factors influencing purchase decision

of personal care products is analyzed with the help of chi-square test.

Frequenc

y Percent

Don't use 5 .5

Friends 208 22.8

Family 455 49.8

Neighbors 5 .5

colleagues 32 3.5

Not Influenced by

anyone 207 22.6

All influences 2 .2

Total 914 100.0

Table: 5.1.1

Social Factors and Beauty Soap

Data Analysis: Out of all the social

factors like family, friends, neighbors

and colleagues the largest

influencer was found to be family

(49.8%) (455). Friends were found

to be (22.8%) (208) the secondlargest source of group influence.

Further, it was also found that less

than one-fourth (22.6%) (207) of

the respondents opinioned that

there were not influenced by any of

the social factor.

Chi l i i SPSS



Chi-square analysis using SPSS

Value df

Asymp.

Sig. (2-

sided)

Pearson Chi-

Square 466.042(a) 180 .000

Likelihood Ratio 387.650 180 .000

Linear-by-Linear

Association 20.258 1 .000

N of Valid Cases 913

Value

Asym

p. Std.

Error(

a)

Appro

x. T(b)

Approx

. Sig.

Nominal by

Nominal

Contingency

Coefficient

.581 .000

N of Valid Cases 913

Table: 5.1.2


.

Table: 5.1.3


(In the chi-square test, for a 95 percent confidence

level, if the significance level is greater than or

equal to .05, it signifies that there is no associationbetween the two variables and the if significance

level is less than .05, then it signifies that there is a

significant relationship between the two variables.)

At 95 percent confidence level, with 180 degree of

freedom the above analysis demonstrates

significant association between social factors and

beauty soap. The calculated value .000 is less

than the standard value .05. From the obtained

contingency coefficient value of .581, it may be

inferred that the association between the

dependent and independent variable is significant,

as the value .581 is closer to 1 than to 0.



Data Analysis of previous table

• (In the chi-square test, for a 95 percent confidence level, if the

significance level is greater than or equal to .05, it signifies that

there is no association between the two variables and the if

significance level is less than .05, then it signifies that there is a

significant relationship between the two variables.)• At 95 percent confidence level, with 180 degree of freedom the

above analysis demonstrates significant association between

social factors and beauty soap. The calculated value .000 is less

than the standard value .05. From the obtained contingencycoefficient value of .581, it may be inferred that the association

between the dependent and independent variable is significant,

as the value .581 is closer to 1 than to 0.

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Simple Tabulation 7and.ppt