binary/unza dba program sylvia bwalya mutale-mwansa quantitative data analyses assignment 1
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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
50th Percentile=67.00
25th Percentile=61.00
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
element was picked as part of the sample.
K (medicine) = (sample size)/(school population)=n/N=13/1000=54, hence every 54th
element was picked as part of the sample.
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