binary/unza dba program sylvia bwalya mutale-mwansa quantitative data analyses assignment 1

BINARY/UNZA DBA PROGRAM

SYLVIA BWALYA MUTALE-MWANSA

QUANTITATIVE DATA ANALYSES

ASSIGNMENT 1

Sylvia B Mwansa

9/28/2015


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QUANTITATIVE DATA ANALYSES

ASSIGNMENT 1

SECTION A: ANALYSIS ON UNIVARIATE DATA (25 MARKS)

MARKS OBTAINED BY STUDENTS IN AN EXAMINATION

(a) Obtain the mean, median and standard deviation for the data and interpret the results.

According to Zikmund, Babin, Carr & Griffin et al (2013), the Mean is the arithmetic average

and is perhaps the most common measure of Central tendency. The Median is the midpoint of

the distribution, or the fiftieth percentile. They contend that in other words, the median is the

value below which half the values in the sample fall and above which half of the values fall.

Standard deviation is a quantitative index of a distribution’s spread, or variability; the square root

of the variance for a distribution.

Table 1: Mean, Median and Standard Deviation for Marks obtained in an examination

Statistics Scores

Mean 67.25

Median 67.00

Standard Deviation 11.538

MEAN:

The mean therefore in this table, is the arithmetic average which is obtained by dividing the sum

of all the scores and the total number of the marks. The mean score from the marks obtained by

students in an examination is 67.25. This implies that on average each student mark was

67.25. However, the mean is sensitive to extreme cases. By this it is implied that this mean may

have been affected by values which are extremely large or small.

MEDIAN:

The median therefore in tTable 1 of a distribution being the middle score after arranging the

cases in either ascending or descending order. The median of an even number of cases is

obtained by the average of the two middle most marks. For the marks obtained by students in an

examination as the case understudy, the median was found to be 67.00.


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STANDARD DEVIATION:

The standard deviation therefore in table 1 being the average variation of each score in the

examination from the mean which in this case is 67.25. The standard deviation was found to be

11.538, hence it can be noted that on average each mark obtained from the examination was

deviated from the mean (67.25) by 11.538.

(b) What comments can you give based on the box plot for the data?

Coakes & Ong (2011) confirm that the box plot provide more information about the actual

value in the distribution. The boxplot, illustrated below summarizes information about the

distribution of marks obtained in an exam.

Table 2: Box Plot Descriptive for Marks Obtained in an Exam

Statistic Std.

Error

Marks obtained by

students in an

examination

Mean 67.25 1.739

95% Confidence

Interval for Mean

Lower

Bound 63.74

Upper

Bound 70.76

5% Trimmed Mean 67.68

Median 67.00

Variance 133.122

Std. Deviation 11.538

Minimum 41

Maximum 88

Range 47

Interquartile Range 17

Skewness -.544 .357

Kurtosis -.094 .702


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Figure 1: Box Plot for Marks obtained in an Exam

From the boxplot above, it was observed that the maximum marks obtained were 88 and

the minimum obtained was 41.

The 25th

Percentile as indicated in figure 1was 61.00. This mark is the highest mark

possible at the 25th

position and below when the marks are sorted in ascending order.

The 50th

percentile represents the Median score which is 67.00. This means that when

the marks are arranged in ascending order, at 50th

percent position, the highest mark was

67.00 and all those below that position got less than 67. The 75th

Percentile was 77.75

which meant that at the 75th

position when marks are sorted in ascending order the

highest mark was 77.75, the rest of the marks below this position got less than 77.75.

The position of the Median is in the bottom half of the box plot which indicates positive

skewedness. This therefore implies a slight departure from normal distribution.

However, the spread or variability of the scores indicated medium variability because the

length of the box plot was not too small nor too large which means it was medium as

observed in Figure 1 as indicated on page 41 Sheridan J Coaks & Clara Ong (2011).

We noted that there were no outliers in the box plot because the image does not have any

asterisk (*). This can further be proved using the formula:

Top Outliers: Q3+1.5 x (Q3 – Q1) = 102.875, which is not among the marks recorded as

the highest was 88.

Bottom Outliers: Q1-1.5 x (Q3 – Q1) = 35.875, which is not among the marks obtained as

the lowest was 41.

(c) Test whether the marks are normally distributed.

Zikmund, Babin, Carr & Griffin et al (2013), contend that one of the most common

probability distributions in statistics is the normal distribution, commonly represented by

Maximum = 88

75th Percentile=77.75



Minimum=41


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the normal curve. They add that, this mathematical and theoretical distribution describes

the expected distribution of sample means and many other chance occurrences. The

normal curve is bell shaped and almost all (99%) of its values are within ±3 standard

deviations from its mean. To test whether the Marks were normally distributed, we

follow the following process:

Step 1: Stating the Hypothesis

Ho: The marks are normally distributed.

Ha: The marks are not normally distributed.

Step 2: Stating the Significance level

P-Value = 0.05

Step 3: Stating the rejection level

Reject Ho, when α < 0.05

Step 4: Stating the analysis of results

Table 3: Test for Normality Distribution for Marks Obtained in an Exam

Tests of Normality

Kolmogorov-Smirnova Shapiro-Wilk

Statistic df Sig. Statistic df Sig.

Marks obtained by students

in an examination .094 44 .200

* .960 44 .135

*. This is a lower bound of the true significance.

a. Lilliefors Significance Correction

Since the sample size = 44, which is less than 2000 confirms Maths Statistics Tutor (2015),

Shapiro-Wilk test was used to determine normality. The Shapiro-Wilk statistic value was 0.960,

degree of freedom was 44 and significance value was 0.135.

Step 5: Stating the Conclusion

Since the α = 0.135 > 0.05, we fail to reject Ho. Therefore, we conclude that the Marks were

normally distributed.

(d) (i) Recode the data to form the following grades: A:≥80, B:70-79, C60-69,

D:50-59, E:<50.


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Table 4: Distribution of grades in an Exam

Mark Grade Mark Grade Mark Grade Mark Grade

84 A 60 C 78 B 54 D

73 B 54 D 66 C 67 C

75 B 62 C 57 D 64 C

67 C 65 C 77 B 49 E

66 C 51 D 80 A 78 B

70 B 43 E 76 B 41 E

61 C 64 C 79 B 82 A

80 A 70 B 68 C 68 C

81 A 59 D 72 B 65 C

72 B 41 E 80 A 67 C

88 A 79 B 61 C 65 C

The table above is a result of recoding the marks into grades based on the grading criteria given

in the question.


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(ii) Construct a pie chart showing the distribution of grades for the class and comment

on the distribution.

Figure 2: Pie Chart of the Distribution of Grades

Commenting on the distribution, it was observed that 4 Students representing 9.09% obtained

grade E, 5 students representing 11.36% obtained grade D, 16 students representing 36.36%

obtained grade C, 12 students representing 27.27% obtained grade B and 7 students representing

15.91% obtained grade A. From this distribution it can be noted that the highest number of

students obtained grade C and the lowest number of students obtained grade E.


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SECTION B – DATE ANALYSIS & REPORT (50 MARKS)

Conduct analyses on the employee data set (given) and prepare a report for the attention of the

management of Delco Ltd. Your report should include the following sections:

(i) Introduction-background of the company, main activities and purpose of study (about

half a page).

INTRODUCTION

Data Analysis is the process of systematically applying statistical and/or logical techniques to

describe and illustrate, condense and recap, and evaluate data. According to Shamoo and Resnik

(2003) various analytic procedures “provide a way of drawing inductive inferences from data

and distinguishing the signal (the phenomenon of interest) from the noise (statistical fluctuations)

present in the data”. While data analysis in qualitative research can include statistical

procedures, many times analysis becomes an ongoing iterative process where data is

continuously collected and analyzed almost simultaneously.

Delco was established exactly 30 years ago, situated in Lusaka’s Makeni Area. The Company is

a corporate governed family business. It has a total of 130 employees. Delco’s main activities

are in Fashion and Printing & Advertising Solutions. Delco has four branches for the clothing

outlets. The outlets are categorized in Formal and Functional Office clothing, the sporty and

maternity fashion for young adults, the weddings and gifts outlet and the hospital and hospitality

outlet. The printing and advertisement outlet is situated together with the head office for ease of

supervision.

The purpose of the study was conduct analyses on the employee data set to find out the actual

status of the company. The report will enable the company participate and take advantage of the

opportunity that arose in the financing of women ran organisations. Delco having different

branches needed to interconnect the activities of all branches to centrally manage all activities

from head office using information technology. Among the criterial for selecting the recipients

for funding was knowledge of the organisation in terms of its educational distribution taking

gender into consideration. The other consideration was employee remuneration according to

employee productivity as well as years served in the company. In addition, the study looked at

differences in terms of salaries and allowances of female and male employees. One of the

requirements of funder was to establish if the female employees annual allowance was

significantly greater than their specified $10,000.


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(ii) Sample composition – cross-tab (sex versus education level) supported with a suitable

diagram and comments.

Table 5: Gender Sample Composition Level of Education Cross Tabulation

Gender * Level of education Cross tabulation

Level of education Total

Primary Secondary Tertiary

Gender

Male

Count 3 7 4 14

% within Gender 21.4% 50.0% 28.6% 100.0%

% within Level of

education 37.5% 50.0% 50.0% 46.7%

% of Total 10.0% 23.3% 13.3% 46.7%

Female

Count 5 7 4 16

% within Gender 31.2% 43.8% 25.0% 100.0%

% within Level of

education 62.5% 50.0% 50.0% 53.3%

% of Total 16.7% 23.3% 13.3% 53.3%

Total

Count 8 14 8 30

% within Gender 26.7% 46.7% 26.7% 100.0%

% within Level of

education 100.0% 100.0% 100.0% 100.0%

% of Total 26.7% 46.7% 26.7% 100.0%

It was observed from the cross tab of gender versus education level that 3 out of 14 male

representing 21.4% of males had primary education level. This represented 37.5% of the total

number of employees who had primary education level. 7 males out of 14 representing 50.0%

had Secondary education level. This represented 50% of all the employees with secondary

education level. 4 males out of 14 employees representing 28.6% had tertiary education level.

This represented 50% of the total number of employees with tertiary level of education.

It was further observed that 5 out of 16 female employees representing 31.2% of all the females

had primary education level. This number represented 62.5% of the total number of employees

with primary education level. 7 out of 16 female employees representing 43.8% had secondary

education level. This number also represented 50.0% of all the employees with Secondary

education level. 4 out of 16 female employees representing 25.0% of the female employees had

tertiary education level. This also represented 50% of the total number of employees with

tertiary education level.


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Figure 3: Clustered Bar Chart Gender Verses Education Level

From the clustered bar chart showing educational level and gender, it was observed that 3 male

employees representing 10.00% had primary education level, 7 male employees representing

23.33% had secondary education level and 4 male employees representing 13.33% had tertiary

education level.

Further the clustered bar chart showing educational level and gender, it was observed that 5

female employees representing 16.67% had primary education level, 7 female employees

representing 23.33 had secondary education level and 4 female employees representing 13.33%

had tertiary education level.


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(iii) Correction Analysis-Annual Salary versus years of service. (The analysis should be

presided with a scatter plot with comments).

Figure 4: Scatter Plot Annual Salary versus years of Service Section B (iii)

It was observed from the scatter plot of annual salary and years of service had a relationship

between the two variables because the dots from the plots formed a pattern like a straight line

moving from the bottom to the top right. The pattern depicted signified a positive relationship

which implied that as years of service increased, annual salaries increased accordingly.

Table 6: Pearson Correlation of years of service verses annual income

Correlations

Annual Salary

($000s)

Years of Service

Annual Salary

($000s)

Pearson Correlation 1 .811**

Sig. (2-tailed) .000

N 30 30

Years of Service

Pearson Correlation .811**

1

Sig. (2-tailed) .000

N 30 30

**. Correlation is significant at the 0.01 level (2-tailed).


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The Pearson correlation coefficient for the relationship between years of service and annual

salary r=.811. This value indicates a strong correlation between the two variables because it is

in the region of 0.7 and 0.9 according to Andrews (2015) The value of r is positive implying a

positive relationship between annual salary and years of service. Hence as years of service

increase, the annual salary increases accordingly.

(iv) Test to determine whether there is a significant difference in the mean annual income

(annual salary + annual allowances) of male and female employees.

Table 7: Group Statistics for the independent sample T-Test

Group Statistics

Gender N Mean Std. Deviation Std. Error Mean

Total_annual_income male 14 49.43 12.835 3.430

female 16 45.00 9.230 2.308

This table has the descriptive statistics for both groups of the variables being tested.

Table 8: Independent Sample T-Test

Independent Samples Test

Levene's Test for

Equality of

Variances

t-test for Equality of Means

F Sig. t df Sig. (2-

tailed)

Mean

Differen

ce

Std. Error

Differenc

e

95% Confidence

Interval of the

Difference

Lower Upper

Total_a

nnual_in

come

Equal

variances

assumed

1.377 .250 1.095 28 .283 4.429 4.044 -3.856 12.713

Equal

variances not

assumed

1.071 23.294 .295 4.429 4.134 -4.118 12.975

This table represents the results of the independent sample t-test. The Levene’s results had an F-

statistics of 1.377 with a significance value of 0.250. Because 0.250 is greater than 0.05, the two

variables had statistically same variance distributions. Therefore, we use the first row of T-Test

information to determine if the 2 means are statistically and significantly different from each

other.


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The T-Statistics value from the first row was 1.095. The degree of freedom was 28. The 2-tailed

significance value was 0.283. The difference between the means was $4.429. The standard error

of this difference was $4.044.

The 95% confidence interval of the difference ranged from $-3.856 to $12.713.

Conclusion:

Because 0.283 > 0.05, we conclude that the average total annual income for male and female is

not significantly different at $4.429. It can therefore be concluded that the average total annual

income for male employees was not significantly higher than that of female employees.

(v) Test to whether the mean annual allowance of the female employees is significantly

greater than $10,000

Table 9: One-Sample Statistics Descriptive

One-Sample Statistics

N Mean Std.

Deviation

Std. Error

Mean

Annual Allowance

($000s) 16 10.31 2.651 .663

The one-sample statistics table shows the total number of female employees equal to 16, average

annual allowance was $10.31 ($000s), the standard deviation from the mean $2.651 ($000s) and

the standard error of the mean was $0.663 ($000s).

Table 10: One Sample T-Test

One-Sample Test

Test Value = 10

t df Sig. (2-tailed) Mean

Difference

95% Confidence Interval of

the Difference

Lower Upper

Annual Allowance

($000s) .471 15 .644 .313 -1.10 1.73

The table above is a One-sample Test, the T-value was 0.471, and degrees of freedom were 15,

the 2-tailed significant value was 0.644. The mean difference was $0.313 ($000s) and 95%

confidence interval of the difference was from $-1.10 ($000s) to $1.73 ($000s).

Conclusion: Since 0.644>0.05, we conclude that the average annual allowance of the female

employees was not statistically and significantly greater than $10.00 ($000s) by $0.313 ($000s).


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From the analyses it can be concluded that education distribution over gender is almost the same

with a small difference in the primary educational level where we have more female than male

employees. Otherwise, the study revealed the same number of female and male employees with

Secondary and tertiary education. It was further concluded that the company has got an

increment policy for employee annual remuneration. In addition, the study revealed that there

was no significant difference between the mean total annual income (annual allowance + annual

salary) of male and female employees. It can furthermore be concluded that female employees

mean annual allowance was not significantly greater than $10,000.

(vi) Conclusions – a brief write-up (about half a page) based on the results of the analyses.

In conclusion of Delco, in its educational distribution taking Gender into consideration, it

was observed that there was no difference in the education distribution of male and female

employees. However, the company had more female than male employees in the Primary

education category.

It was further observed that the company awarded annual salary increments proportionate to

years of service. It can also be concluded that the average annual income for male and

female does not differ significantly. The average annual allowance of female employees was

found statistically and significantly not greater than $10,000.

Delco being a woman ran company and the results of the research conducted has proven that

the organisation qualifies for funding to enable it expand its operations by creating a

centralized head office to help coordinate the branches. The funding will mainly be

channeled to Information Communication Technology (ICT) Dash-Board to observe and to

strategically manage the branches. The system where is has been installed and effectively

used has proved to have added value in the monitoring and evaluating of projects and

effectiveness of the companies’ staff productivity.


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SECTION C – SAMPLING (25 MARKS)

(a) Suppose there are 3500 students in a college. How large a sample is required to estimate

the proportion of students who smoke cigarettes within a margin of error of ± 10%.

Formula for Sample size used:

n = (NZ2S

2) / (Z

2S

2+NE

2)

n=sample size=?

N=population size=3500

Z=Z score at 90% confidence interval=1.645

S=Standard deviation=0.5

E=margin of error=0.1

Substitute in the formula:

n= (3500*1.6452)/(1.645

2*0.5

2+3500*0.1

2)

n=66.

Non-Response Bias

According to greenbook.org in data collection, there are two types of non-response: item

and unit non-response. Item non-response occurs when certain questions in a survey are

not answered by a respondent. Unit non-response takes place when a randomly sampled

individual cannot be contacted or refuses to participate in a survey. To take care of this

we apply as follows:

We add 30%

130% sample size

60 x 1.3 = 86

n=86

(b) Suggest an appropriate probability sampling, you should outline clearly matters

pertaining to sampling frame, stratification, sample size and sample selection for the

specific sampling scheme that you are proposing.

Sheldon M Rosse (2010) contend that, one of the key concerns of statistics is the drawing

of conclusions from a set of observed data. These data will usually consists of a sample

of certain elements of a population, and the objective will be to use the sample to draw

conclusions about the entire population.

According to lecture notes Marianpan (2015), Probability sample is a sample in which

every unit in the population has a chance of being selected in the sample, and this

probability can be accurately determined. Some of the sampling methods include simple

random sampling, systematic sampling and stratified sampling.


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Non probability sampling is any sampling method where some elements of the population

have no chance of selection or where the probability of selection can’t be accurately

determined. It involves the selection of elements based on assumptions regarding the

population of interests, which forms the criteria for selection. Some of the Non

probabilities sampling methods are; judgment sampling, quota sampling and cluster

sampling.

With a student population of 3500 and a margin of error of 10%, the probability sampling

that is going to be applied to the above study is stratified sampling. According to

Zikmund, Babin, Carr & Griffin et al (2013) stratified sampling is a probability sampling

procedure in which simple random subsamples that are more or less equal on some

characteristic are drawn from within each stratum of the population. In this sampling, the

data obtained from the sampling frame (the student record book) will be used.

School of business= 1000

School of humanities= 1800

School medicine =700

Total = 3500

With a margin of error of 10%, the sample size is 66 stratified according to the above

groupings/stratum.

The first step is to find the total number of students (3500) and calculate the percentage in

each stratum.

Percentage of Business Student=1000/3500=29%

Percentage of Humanities Student=1800/3500=51%

Percentage of Medicine Students=700/3500=20%

This means that with a sample size of 66 students:

29% of Business Students=(29/100)*66=19

51% of Humanities students=(51/100)*66=34

20% Medicine Students=(20/100)*66=13

Having identified the sample size of each stratum in the above computation, the next step

is to conduct systematic random sampling. Zikmund, Babin, Carr & Griffin et al

(2013) contend that systematic sampling is a sampling procedure in which a starting point

is selected by a random process and then every Kth number on the list is selected.


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Following method will be followed; the population in each school was ordered and given

an identification number. Then the required sample was divided into the school

population to find the selection interval. N/n=k(selection interval), this meant that every

kth

element was picked to be part of the sample.

K (business) = (sample size)/(school population)=n/N=19/1000=53, hence every 53rd

element was picked as part of the sample.

K (humanities) = (sample size)/(school population)=n/N=34/1800=53, hence every 53rd


K (medicine) = (sample size)/(school population)=n/N=13/1000=54, hence every 54th


For non-probability sampling according to Zikmund, Babin, Carr & Griffin et al (2013) is a

sampling technique in which units of the sample are selected on the basis of personal judgment

or convenience, the probability of any particular member of the population being chosen in

unknown. In this case, quota sampling was suggested. Quota sampling is a nonprobability

sampling procedure that ensures that various subgroups of a population will be represented on

pertinent characteristics to the exact extent that the investigator desires contends Zikmund,

Babin, Carr & Griffin et al (2013).

With a student population of 3500 and a margin of error of 10%, the non-probability sampling

that is going to be applied to the above study is quota sampling. Using the data obtained from the

quota frame (the student record book):

School of business= 1000

School of humanities= 1800

School medicine =700

Total = 3500

With a margin of error of 10%, the sample size is 66 grouped according to quota frame.

The first step is to find the total number of students (3500) and calculate the percentage in each

quota.

Percentage of Business Student=1000/3500=29%

Percentage of Humanities Student=1800/3500=51%

Percentage of Medicine Students=700/3500=20%

This means that with a sample size of 66 students:

29% of Business Students=(29/100)*66=19

51% of Humanities students=(51/100)*66=34


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20% Medicine Students=(20/100)*66=13

In order to pick the students from each quota to make up the sample, simple random sampling

was used. This was done by going to each quota and randomly picking 19 for School of

Business, 34 for School of Humanities and 13 for School of Medicine respectively.

According to Black, T. R. (1999) the advantages and disadvantages of non-probability sampling

methods are stated in the illustration below. In a study where a researcher wants to know how

many students out of the 3500 graduated in the field of Humanities, Business and Medical

schools in the university, the only people who can give the researcher first hand advise are the

individuals who graduated from the University. With this very specific and very limited pool of

individuals that can be considered, the researcher must use the non-probability sampling in the

chosen quota.

Table 11: Sampling techniques: Advantages and disadvantages

Technique Descriptions Advantages Disadvantages

Convenience

sampling

Refers to the collection

of information from

members of the

population which are

conveniently available

Best way of getting some

basic information quickly

and efficiently.

Convenience samples are

cheap.

Convenience samples do

not produce

representative results.

Judgment

Sampling

The researcher selects

the units to be observed

on the basis of his own

judgment.

The approach is

understood as well and has

been refined through

experience over many

years.

It is unscientific and

usually too large samples

are selected and It is

wasteful;

Personal bias in, to

choice the sample is

unavoidable;

Quota Select individuals as

they come to fill a

quota by characteristics

proportional to

populations

Ensures selection of

adequate numbers of

subjects with appropriate

characteristics

Not possible to prove that

the sample is

representative of

designated population


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REFERENCES:

Andrews Education (2015): Viewed on 14th August 2015:

http://www.andrews.edu/̴calkins/math/edrm611/edrm05.html).

Black, T. R. (1999). doing quantitative research in the social sciences: An integrated approach

to research design, measurement, and statistics. Thousand Oaks, CA: SAGE Publications, Inc.

(p. 118)

Maths Statistics Tutor.com viewed 14th

August 2015: http://www.maths-statistics-

tutor.com/normality_test_pasw_spss.php

Non-responsive bias, viewed on 27th

September 2015: http://www.greenbook.org/marketing-

research/non-response-bias

Shamoo, A.E., Resnik, B.R. (2003). Responsible Conduct of Research. Oxford University Press.

Sheldon M Rosse (2010). Introductory to Statistics, Third Edition, University of Southern

California Academic Press, Elsevier

Sheridan J Coakes and Clara Ong SPSS Version 18.0, Analysis without Anguish 2011

William G. Zikmund, Barry J. Babin, Jon C. Carr and Mitch Griffin Business Research Methods,

Ninth Edition International Edition South-Western Cengage Learning 2013

http://www.greenbook.org/marketing-

http://www.greenbook.org/marketing-

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