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Research in business studies
Department of Business Administration
SPRING 2015-16
Quantitative and Qualitative Data Analysis
Chapter 6
by
Prof. Dr. Sami Fethi
Research Methods in Business Studies
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Quantitative data analysis
 Examining differences
 Relationship between variables
 Explaining and predicting relationship between variables
 Data reduction, structure and dimension
 Additional methods
 Characteristic of qualitative research
 Qualitative data
 Analytical procedure
 Interpretation
 Strategies for qualitative analysis
 Quantify qualitative data
 Validity in qualitative research
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Testing Hypothesis
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Hypotheses about one mean
 In research we often have to make statements
about the mean. When the population variance
is unknown, the stadard error of the mean is
also unknown. The standard error of the mean
must be estimated from sample data.
e.g. SDX= SD‘/ N
where
SDX= standard error of mean
SD‘= estimated standard deviation
N= sample size


2
SD‘=
N-1 is degrees of freedom
N
i 1
( xi  X )
N 1
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Testing Hypothesis/t-test n<30
Example 1: For a supermarket chain to add a
new product, at least 100 units must be sold
per week. The new product is tested in ten
randomly selected stores for a limited time.
Apply a test such as one-tailed t test and
answer the question that will the new product
sell more than 100 unit per week?
a) construct hypothesis
b) calculate mean and standard deviation if they
are not given.
c) calculate standart error of mean
d) find t- value
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Testing Hypothesis/t-test n<30
a)
H0: X<=100
H1: X>100
b) X and SD are given 109.4 and 14.90
respectively.
c) SDX = 14.90/ 10  1 =4.55
d) t= (X-µ)/SDX=(109.4-100)/4.55=2.07
Where t-table is 1.83 at 5% significant
level. We reject the null.
will the new product sell more than 100
5
unit per week? Yes...
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t-table
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10  1
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Testing Hypothesis/t-test n<30
Example 2: A type of bulb has mean operating
life at least 4200 hr with std.dev of 200 hr. The
mean operating life for a random sample is 10
whereas the mean is 4000. Assume that this is
under the normal distribution.
Apply a test such as one-tailed t test and
answer the question that will the mean operating
life of the type of bulb be at least 4200 hr ?
a) construct hypothesis
b) calculate mean and standard deviation if they
are not given.
c) calculate standart error of mean
d) find t- value
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Testing Hypothesis/t-test n<30
a)
H0: X<=4200
H1: X>4200
b) SDX = 200/
=63.3 hr
10
d) t= (X-µ)/SDX=(4000-4200)/63.3=-3.16
Where t-table is -1.83 at 5% significant
level. We reject the null.
will the mean operating life of the type of
bulb be at least 4200 hr ? No, It should
be greater than 4200.
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Testing Hypothesis/t-test n<30 or z-test n≥30
Examining differences
Hypotheses about two means
 This is usually associated with such a question: Are
the tastes in region A different from the tastes in
region B?
 e.g.
Where
X1= sample mean for the first sample
X2= sample mean for the second sample
( X 1  X 2 )  ( 1   2 )
z /t 
SD X1  X 2
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Testing Hypothesis
Examining differences
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SDX 1  X 2
= the standard eror of differences in means
µ1 and µ2 are the unknown population means and
the general estimate of:
SDX1  X 2  SD  SD
2
X1
2
X2

SD 21 SD 2 2

N1
N2
 In assuming the two population variances to be equal,
the common population variance can be generated by
pooling the samples. When the variances are
unknonw and the standard errors of means must be
estimated, then the t represents an adequate test
statistics, distributed with v= N1+ N2-2- degrees of
freedom.
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Testing Hypothesis/t-test n<30
Examining differences
Example3:
A manufacturer has developed a new
product and wonders whether the label of the
package should be red or blue. The new
products with two different labels are tested
in ten randomly selected stores. The means
sales obtained for the red package are 403.0
and for the blue package 390.3. The
standard error of estimate for the difference
means is 8.15.
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Testing Hypothesis/t-test n<30
Examining differences
a) construct hypothesis
b) find t- value
a) H0: (µ1- µ2 )=0
H1: (µ1- µ2 )≠0
or
H0: (µ1- µ2 )<=0
H1: (µ1- µ2 )>0
b) t  ( X 1  X 2 )  (1   2 )
SDX1  X 2
=((403.0-390.3)-0)/8.15=1.56
V=10+10-2=18 degrees of freedom...5% and df 18 so
critical value from the table is 2.101. This means that null
hypothesis is accepted.. H0: (µ1- µ2 )=0. This means that the two
unknown population means are assumed to be same.
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Testing Hypothesis/z-test n≥30
Examining differences
 Example 4: The mean weekly wage for a
sample of n1=30, its mean is $280000 with std
dev $14. In other large firm, a sample of
n2=40, its mean is $270000 with std dev $10.
 Test the hypothesis that there is no difference
between the mean weekly wage amounts in the
two firms at 5% level of significance.
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Testing Hypothesis/z-test n≥30
Examining differences
a) construct hypothesis
b) find t- value
a) H0: (µ1- µ2 )=0
H1: (µ1- µ2 )≠0
or
H0: (µ1- µ2 )<=0
H1: (µ1- µ2 )>0
b)
z
( X 1  X 2 )  ( 1  2 )
=((280-270)-0)/3.01=3.32
SDX1  X 2
5% and so critical value from z table is 1.96. This means
that null hypothesis is rejected.. H0: (µ1- µ2 )=0. This means that the
two unknown population means are assumed to be different.
SDX1 = 14/ sqrt of 30
SDX2 = 10/ sqrt of 40
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=2.56
=1.58
SDX1  X 2  2.562X1 1.582X 2  3.01
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Z-table
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Testing Hypothesis/z-test n≥30
Examining differences
 Example 5: The mean weekly income for a
sample of household is n=30, its mean is
$24000 with std dev $2000. In the area,
average income mean is.
 Test the hypothesis that the mean household
income in the population is at least $25000 at
5% level of significance.
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Testing Hypothesis/z-test n≥30
Examining differences
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a) construct hypothesis
b) find z- value
a) H0: µ<=0
H1: µ>0
b)
( x1  2 ) (24000  25000)
z

 2.74
SDX
364.96
So critical value at 5% from z table is -1.64. This means that
null hypothesis is rejected.
This means that the mean of household income is greater
than population.
SDX1 = 2000/ sqrt of 30
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=$ 364.96
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Chi-squared
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o A chi-squared test, also referred to as test (or chi-square test), is any
statistical hypothesis test in which the sampling distribution of the test
statistic is a chi-square distribution when the null hypothesis is true. Chisquared tests are often constructed from a sum of squared errors, or through
the sample variance.
Chi-Square Formula
Degrees of freedom (df) = n-1 where n is the number of classes
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Chi-squared
Example 6-Answer
o
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The chi-squared test is a hypothesis test so the general approach is the
same as with parametric test. But the chi-squared test looks at the frequency
of observations and sees if these match the expected frequencies
Defect
Days
1
26
2
42
3
18
4
9
5
5
Required:
a) State the null and alternative hypotheses
b) Calculate the actual value of Chi-squared
c) Compare the actual value of Chi-squared with the critical value of Chi-squared
d) Decide whether to accept or reject the null hypothesis and state the conclusion
reached.
e) Briefly explain the following statement ‘‘does the evidence suggest that some sites
have statistically more accidents than others?
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Chi-squared
Example 6-Answer
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H0
each defect expects the same number of days
HI
each defect does not expect the same number of days
Check
5%d.f=n-k-1
Freq
O
4
E
O-E
9.48-critical value
(O-E)*2
(O-E)*2/E
1
26
20
6
36
1.8
2
42
20
22
484
24.2
3
18
20
-2
4
0.2
4
9
20
-11
121
6.05
5
5
20
-15
225
11.25
100
100
43.5
Reject null and accept alternative since 43.5 is greater than 9.48.
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Chi-squared-table
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Research in business studies
Useful alternative tests




o




In problems involving one or two population means, t-methods are
usually appropriate, but often non-parametric methods are good
alternatives.
e.g. Non-parametric methods have advantage of requiring less in
terms of assumptions and less powerful than t-methods (see siegel
and Castella; 1998).
e.g. The main difference between them is that t-method associates
with means while non-parametric methods are concerned with
medians.
ANOVA- analysis of variance measures comparisons of more than
two groups simultaneously. This method rests on comparing the ratio
of systematic variance to unsystematic variance.
In ANOVA, the following is computed:
Total variation by comparing each observation with the grand mean.
The between-group variation by comparing the treatment means with
the grand mean.
The within-group variation by comparing each score in the group with
the group mean.
Recall-MANOVA-multivariate analysis of variance. This has more
than one dependent variable compared to ANOVA:
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F-TEST
Comparison of more than two group
Example 3: In the
following table,
three advertising
campaigns tested
in 24 randomly
selected cities
comparable in size
and demographics.
The following
output is an anova
analysis results:
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Source Sum Degree Mean F-ratio
of
of
sq.
freedom
sq.
Between 49.0
2
24.1 5.88
group
Within
group
87.5
21
total
136.5
23
4.17
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Example 7
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a) construct hypothesis
b) find F- value whether significant or not
c) Comment on the F-values
a) H0: G1= G2= G3
H1: G1≠ G2 ≠ G3
d.f= 24-1=23, between group 3-1=2 within group 232=21.
b) Fcalculated=24.1/4.17=5.88
Fcritical=n-k,k-1=24-3,3-1=(21,2). From F-distribution,
Fcritical is 3.47.
c) Since 5.88 is greater than 3.47, we reject the null
hypothesis, that is, the group means are equal and
accept the alternative hypothesis that the advertising
campaigns vary in effectiveness.
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F-Table
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Relationship between variables
 In research, we are often preoccupied with whether
there is a relationship or two or more variables covary.
o Correlation coefficient
Based on the Pearson criterion, it examines the strength
of linear relationship between two variables, for example
x and y.
o Theoretically, the Correlation coefficient can take the
values from -1 to 1. A correlation coefficient of 1 tells us
that two variables perfectly covary positively whereas -1
shows that two variables perfectly inversely related.
Close to 0 indicates that the variables are unrelated.
The formula of the Correlation coefficient as fololw:
Where X and Y represent the sample means of X and Y.
rXY 
 ( x  X )( y  Y )
 (x  X )  ( y  Y )
i
i
2
i
26
2
i
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Relationship between variables
o Correlation coefficient
A Correlation coefficient shows covariation between two variables,
and not that the variables are causally related.
The square of the Correlation coefficient is the coefficient of
determination.
R2=Explained variation/Total variation
o Example 4- partial correlation
Using the following table (Table 1) and calculate the relationship
between advertisement recognition, appeal and sex. In other words,
Is the relationship between advertisement recognition (1) and appeal
influenced (2) by controling for sex (3)?
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Example 8
o This is partial correlation and can be formulated as follow
based on partial Correlation coefficient r123 as such ad.roc,
appeal, sex
r123 
r12  (r13 ) (r23 )
1  r13
2
1  r23
2
rAd .roc, appeal, sex 
0.24  (0.33) (0.09)
1  (0.33)
2
1  (0.09)
2
 0.29
o This shows that controlling for sex the observed
relationship between ad.roc, and appeal positive and
strengthened.
o Note: a correlation between two variables when the
effects of one or more related variables are removed.
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Explaining and predicting relationship between variables
o Explaining and predicting relationship between variables are
important tasks in business research. One of the most applied and
useful approaches to examining relationships between variables is
regression analysis. In regression analysis, we want to fit a model
that best describes the data which is done in regression analysis by
applying the method of least squares. More precisely, this is done by
fitting a straight line that minimizes the squared vertical deviations
from that line as shown in following figure.
o Single Linear Regression
Y= a0+a1xi+ei
Where Y=the outcome variable, X=predictor variable, a1=slope of the
straight line fitted to the data and a0=intercept of the line and
ei=difference between the score predicted and the score actually
obtained. This is called residual.
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Single Linear Regression
Explaining and Predicting Relationship between Variables
Figure 1 The linear model
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Single Linear Regression
Example 9
o Assume that a car dealer collects data for six months on four
variables; Tv advertising, printing advertising, competitors’
advertising and sales. Y is sales. The car dealer expects carsales to
be positively correlated with TV-ads and Print-ads and negatively
correlated with competitors’ ads.
Table 2 Data matrix
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Simple Mean Regression-output
Example 10
o Assume that a car dealer collects data for six months on four
variables; Tv advertising, printing advertising, competitors’
advertising and sales. Y is sales. The car dealer expects carsales to
be positively correlated with TV-ads and Print-ads and negatively
correlated with competitors’ ads. Based on the information below,
comment on the estimated coefficinent and T-ratio as well as R2
on Tv-Ads.
Table 3 Simple mean regression-output
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Simple Mean Regression-output
Example 10-Answer
o The estimated constant term 0.7 shows that If the dealer does not
use Tv-ads at all (Tv-ads=0), the estimated expected value of
carsale is 0.7 unit that is 7 car. The estimated regression coefficient
of sales on Tv-Ads is 0.9. This coefficient shows that if the variable
Tv-ads is increased by 1 unit, the estimated expected value of
carsales increases by 0.9 units, that is nine car. The result, Rsquare, R2 that is 85.3 percent shows that the sample determination of
coefficient is equal to 0.853. Practically speaking, this means that the
variation in the variable Tv-ads has explained 85.3 percent of the
variations in the dependent variable carsales. Estimated t-value on Tvads is 4.81 which is greater than 2 (tabular value from t-distribution) or
rule of thumb so it is signficant 5% and 1% levels. This means that we
can reject the null hypothesis that is the corresponding population
regression coefficient is equal to zore. The conclusion then is that Tvads and sales are significantly related to each other or Tv-ads has
positive impact on sales.
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Assumptions in Regression analysis
o The expected value of the error term is zero
o The variance for the error term for each X is constant.
This term homoscedasticity. If the variance to e varies with
X, this is termed heteroscedasticity.
o The error for the observations are uncorrelated.
o e should be normally distributed for each X.
o The error term should not be correlated with x-corr(e,
x)=0
o It is also a common assumption that the regression
model should be linear in its parameters.
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Correlation Coefficients-output
Example 11
o Assume that a car dealer collects data for six months
on four variables; Tv advertising, printing advertising,
competitors’ advertising and sales. Y is sales. The car
dealer expects carsales to be positively correlated with
TV-ads and Print-ads and negatively correlated with
competitors’ ads. Use the concept of correlation
coefficient and explain the relationships between the
variable under inspection based on the information given
in table 4.
Table 4 Correlation coefficients-output
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Correlation Coefficients-output
Example 11 -Answer
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o The relationship between carsales (dependent) and Tv
advertising, printing advertising, competitors’ advertising
(explanatory) are expected to be high. The relationship
between the explanatory variables as such Tv
advertising, printing advertising, competitors’ advertising
are expected to be low. So high correlation coefficient
between for example Tv advertising and printing
advertising shows a high degree of multicollinearity.
This influences the estimates results badly. To remedy
this situation, the relevant variable can be dropped from
the regression equation. For example between sales
and Tv-ads is 0.92 which is highly reasonable score or
between sales and Comp-ads is 0.155 which is very low
score .
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Multiple Regression
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o In multiple regression, at least two or more independent or
explanatory variables are applied to explain/predict the dependent
variable. The purpose is to make the model more realistic, control for
other variables, and explain more of the variance in the dependent
variable as well as reduce the residuals. The following is a typical
example output for a multiple regression.
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Table 5 Multiple regression – output
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Example 12 Multiple Regression
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o In the following table, Dependent variable
is consumer’s
willingness (ACC) and the explanatory variables; REP (brand
reputation), SIM (similarity), PKNOW (consumer knowledge) CONC
(consumer’perception) and UNCERT (uncertainty).
o Interpret and make comments on the results estimated.
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Table 5 Multiple regression – output
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Example 12- Answers
o For example; REP coefficient shows that if the variable is
increased by 1 unit, the estimated expected value of consumer’s
willingness (ACC) is increased by 0.35. The result, R- square, R2 that
is 38.4 percent shows that the sample determination of coefficient is
equal to 0.384. Practically speaking, this means that the variation in the
variable Brand reputation (REP) has explained 38.4 percent of the
variations in the dependent variable consumer’s willingness.
Estimated t-value on (REP) is 6.49 which is greater than 2 (tabular
value from t-distribution) or rule of thumb so it is signficant 5% and 1%
levels. This means that we can reject the null hypothesis that is the
corresponding population regression coefficient is equal to zore. The
conclusion then is that (REP) and consumer’s willingness (ACC) are
significantly related to each other or (REP) has positive impact on
consumer’s willingness (ACC) .
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Dummy Variables
o In a multiple regression, dummy variable can be used in two ways.
As a dependent variables where its values take 1 or 0 that is also
called dichotomous. The other type can be used as independent
variable which takes the value 0 or 1. The dummy variable used in
an analysis when there does not exist as numerical values. For
example, in the following table that is a nominal scaled variable that
can not be ranked so to be applied in a regression analysis, the
seasons need to be assigned numbers
Table 6 Coding of dummy variable
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Dummy variables
Example 13
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o In the following table, there three new variables A, B and C and indicates
that the four seasons are different combinations of zeros and ones. Assume
that the following regression model for sales of women’s clothing where the
price (P) is also included, has been estimated:
Sale=1000 - 0.5P+100A - 20B - 50C
a) Calculate the sales in the summer by considering dummy variables as
well (i.e. p=$200 ).
b) Calculate the sales in the autumn by considering dummy variables as well
(i.e. p=$200 ).
c) Compare the sales in winter and spring by keeping the same price.
Table 6 Coding of dummy variable
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Dummy variables
Example 13-Answer
o
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In the following table, there three new variables A, B and C and
indicates that the four seasons are different combinations of zeros and
ones. Assume that the following regression model for sales of women’s
clothing where the price (P) is also included, has been estimated:
Sale=1000 - 0.5P+100A - 20B - 50C
a) Calculate the sales in the summer by considering dummy variables as
well (i.e. p=$200 ).
Sale=1000 - 0.5 (200)+100(1) – 20(0) – 50(0)=$1000
b) Calculate the sales in the autumn by considering dummy variables as
well (i.e. p=$200 ).
Sale=1000 - 0.5 (200)+100(0) – 20(1) – 50(0)= $880
c) Compare the sales in winter and spring by keeping the same price.
Winter- Sale=1000 - 0.5 (200)+100(0) – 20(0) – 50(1)= $950
spring- Sale=1000 - 0.5 (200)+100(0) – 20(0) – 50(0)= $900
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Discriminant Analysis
If the dependent variable is categorical by taking value as
0 and 1, this analysis is called discriminant analysis.
o e.g. Succesfull firms and unsuccesful firms
o The objective of discriminant analysis is to derive a variate
o e.g the linear combination of two independent variable can
best discriminate two a priori defined groups.
o
The discriminant function can be shown in the following form:
Z=a+WX1+WX2+...
Z is discriminant z scores
W is discriminant weight
X is independent variable
Excluded from Final exam subjects
Research Methods in Business Studies
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Research in business studies
Example 14: Discriminant Analysis
o CASE: A manufacturer has conduct a study to
examine whether it is possible to find out what
discriminates buyers and non buyers. The groups
are
age
and
income
whether
posses
discriminating power or not (Buy=1, not buy=0).
o Calculate G0 and G1 using the figures estimated
in the following table where age and income are
54 year old and $ 250 respectively.
Excluded from Final exam subjects
Research Methods in Business Studies
44
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Example 14: Discriminant Analysis
45
Research Methods in Business Studies
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Example 14: Discriminant Analysis
o G0= -18.127+0.225 Age+ 0.120 Income
o G1= -31.758+0.052 Age + 0.203 Income
o When we use age as 54 and income 250, we can
estimate the following results:
o G0= 24.02 and G1= 21.8.
o Since G0>G1, the person belongs to G0. In other
words, the person will not buy the product.
Excluded from Final exam subjects
Research Methods in Business Studies
46
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Thanks
47
Research Methods in Business Studies
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.