University of the Philippines, Diliman, Quezon City
Department of Mathematics
summer term 2010
Assoc.Prof. Dr.Christian Traweger
University of Innsbruck, Austria
Department of Political Science
Social Science Methode Group
christian.traweger@uibk.ac.at
overview - topics
lecture-content
basics in market research, definitions
frequency distributions, measures of central tendency,
measures of dispersion
boxplots, statistical testing
hypotheses, hypotheses-testing, level of significance
probability, probability tables, contingency tables
Chisquare-test for testing two/more independent samples
for nominal variables
Mann-Whintey U-test for ordinal data
Kruskal Wallis test for ordinal data
analysis of variance (ANOVA), t-test
measures of association – Cramer´s V for nominal data
correlation coefficient for ordinal data – Spearman
correlation coefficient for metric data – Pearson
regression analysis
written exam on May 17th 2007
2
Basics in market research
 formulation of goals; what should be analysed?
 survey methods
 questionnaires
 sampling
 data collection (field work)
 plausibilty and follow up work regarding interviewers
 representativeness
 data analysis and statistical analysis
 written report
 presentation of results
(press conferences, board meetings,.....)
3
formulation of goals:
What should be analysed ?
Is market research the proper tool for solving the
problem? What are the constraints ? What is to consider ?
Survey methods:
To choose the right survey method is determined from the budget, the time frame and
how much information should be retrieved trough the survey.
4 kinds of surveys:
surveys can be conducted
- face to face
- through mailing/post-mail
- by telephone (CATI, CI,......)
- via Internet, text
Discuss the advantages and disadvantages of the above survey methods.
questionnaire:
- questions, where every answer is possible
- questions with specific answers:
- dichotomous scale: only two answers two chose from
- Multiple choice question: three or more alternatives and you can choose
one or more answers
- Likert-Scale: the subject may be asked to indicate his/her degree of agreement
with each of a series of statements relevant to the attitude.
e.g.: Travelling with big and well known airlines is safer.
Strongly agree
agree
undecided disagree strongly disagree





- Semantic differential: scale with a pair of attitudes
e.g.: the hotel xy is: modern ------------- old-fashioned
4
- Itemized rating scale:
e.g.: As a President, Gloria Arroyo has been
 very competent
 somewhat competent
 neither competent nor incompetent
 somewhat incompetent
 very incompetent
- stapel scale:
e.g.: As a President, G.A. has been
-3
-2
-1
competent +1
+2
+3
Always try to avoid batteries of same answers within more than 12 items.
Scales:
In order to chose the correct statistical test and /or statistical measures, it is important
to know what scale is used:
- nominal
- ordinal
- metric (interval, ratio)
Sampling methods
A sample is part of a population (e.g.: Philippine population). In order to relate the
results of the sample to the population, the sample has to be an exact representation of
the population, with regard to various demographic characteristics.
- Quota sampling
- Random sampling
1) Determinatioin of sample-points: area sampling
2) Determination of households (randomized)
3) Determination of the person in the household
„Each element must have the same chance to be selected“.
5
Sample Error: if one has no survey-results (e.g. before starting the survey), you
have to start with the worst case: 50% : 50% answering pattern
Sample error e =
1.96 2  p  (1  p)
n
p = % (e.g. 50% = 0.5)
Sample size:
The sample size is determined through…….
- the exactness of the sample (interval for the most likely result)
- ... how sure someone wants to be, when predicting a result (95%), and
- the budget which is available by the customer
n = 1.962
.
p  (1  p)
e2
n......sample size
p......percentage of respondents, that gave a certain answer
e......sample error
6
Table to determine the sample error:
p........percentage of people giving a certain answer
n........sample size
Sample error in %
10 / 90 15 / 85 20 / 80 25 / 75 30 / 70 35 / 65 40 / 60 45 / 55 50/50
=0.1
=0.15
=0.2
=0.25
=0.3
=0.35
=0.4
=0.45
=0.5
%
n=
100
200
300
400
500
600
700
800
900
1000
1500
2000
2500
3000
5.88%
4.16%
3.39%
2.94%
2.63%
2.40%
2.22%
2.08%
1.96%
1.86%
1.52%
1.31%
1.18%
1.07%
7.00%
4.95%
4.04%
3.50%
3.13%
2.86%
2.65%
2.47%
2.33%
2.21%
1.81%
1.56%
1.40%
1.28%
Sample Error =
7.84%
5.54%
4.53%
3.92%
3.51%
3.20%
2.96%
2.77%
2.61%
2.48%
2.02%
1.75%
1.57%
1.43%
8.49%
6.00%
4.90%
4.24%
3.80%
3.46%
3.21%
3.00%
2.83%
2.68%
2.19%
1.90%
1.70%
1.55%
1,96 2  p  (1  p)
n
8.98%
6.35%
5.19%
4.49%
4.02%
3.67%
3.39%
3.18%
2.99%
2.84%
2.32%
2.01%
1.80%
1.64%
9.35%
6.61%
5.40%
4.67%
4.18%
3.82%
3.53%
3.31%
3.12%
2.96%
2.41%
2.09%
1.87%
1.71%
9.60%
6.79%
5.54%
4.80%
4.29%
3.92%
3.63%
3.39%
3.20%
3.04%
2.48%
2.15%
1.92%
1.75%
9.75%
6.89%
5.63%
4.88%
4.36%
3.98%
3.69%
3.45%
3.25%
3.08%
2.52%
2.18%
1.95%
1.78%
9.80%
6.93%
5.66%
4.90%
4.38%
4.00%
3.70%
3.46%
3.27%
3.10%
2.53%
2.19%
1.96%
1.79%
p = in %
example:
Within a survey 500 travellers on the airport in Manila where asked, how they book
their trips. 30% said, that they book their holidays "last minute". Now one can
maintain, that the real percentage of people, that book their holiday "last minute", is
with a likelihood of 95% between 26% and 34% ( 30% +- 4.02%).
Is the sample more than 10% of the population, the sample error has to be computed
as:
Sample Error SP =
1,96 2  p  (1  p)  ( N  n)
n  ( N  1)
p = %, N = population
SP...corrected for small population !!! (sample is more than 10% of population)
7
data collection:
Actual fieldwork:
- face to face
- by telephone
- or by mailing
by specially trained interviewers
Plausibility and follow up work regarding interviewers
This part is done visually as well as computer-assisted. Special questions in the
questionnaire will help to verify , how serious the respondent was.
Special market-research companies, working with a CATI-system, are able to monitor
every interview.
Representativeness
The objective of a representative survey is an exact "picture" of the respective
population. The sample has to be an exact representation of the population, with regard
to various demographic characteristics. Usually representativeness is based on:
Sex
Age
Education
8
Data analysis and statistical anaylsis:
Within the data analysis, one has always to refer to the scale of each question; so you
know when it makes sense to calculate a mean, median, mode, variance,….. .
The same differentiation has to be made when applying statistical tests.
Statistical measures according to variable-type (scale):
Variable-type
Nominal
Ordinal
Metric
measures
Frequency distribution, Mode
Frequency distribution,
Mode, Median, Quartiles
Min., Max., Median, Mean,
Variance, Std.Dev., Std. Err.;
Graphical representation
bar and pie charts
bar and pie charts
boxplots
histogramm,....
Concerning graphical representation statisticians can apply any graphic that visualizes the results in
an optimal way. The above table gives only a brief overview of graphical representation.
Written report:
The final report should contain tables, graphs and a verbal interpretation of each
question. Interesting hypotheses should be tested and explained.
Presentation of results:
Beside the long written report a short report should be made and discussed within a
board meeting and with the customer. A press conference with a power-point
presentation should be prepared.
9
Descriptive Statistics:
Frequency Distribution:
Shows the number of times each observation occurs when the values of a
variable are arranged in order according to their magnitude.
Example: student-scores on exam
score:
xi (absolute
hi (relative frequency)
frequency)
Hi (cumulative relative
frequency)
1
2
8.3
8.3
2
4
16.7
25.0
3
9
37.5
62.5
4
6
25.0
87.5
5
3
12.5
100.0
total
24
100.0
How to compute relative frequencies:
h2 = (x2/N)*100 = 16.7 %
How to compute cumulative relative frequencies:
% H3 = h1 + h2 + h3 = 8.3 + 16.7 + 37.5 = 62.5 %
A frequency curve is the result of:
a histogram
curve-polygon in the histogram
smoothing of the curve-polygon
density-curve
10
Statistical measures:
Mode: xMode  hi max
Definition: Observation that occurs with the greatest frequency. In the previous
frequency distribution table, the mode is score 3.
Mean :
x
1 N
*  xi
N i 1
Def: The mean is the arithmetic average and is defined as the sum of a set of values
divided by their number; as its computation involves algebraic manipulation of the
individual data values, the mean is an approximate measure of central location for
metric data only.
Excurs : mean of given class-intervals
x
1 k
*  xi * xmi
N i 1
Bsp:
class
1
2
3
4
x
IQ-Interval
80 – 100
101 – 121
122 – 142
143 – 163
Mean-value in class
90
111
132
153
xi
4
9
9
3
Σ= 25
xi xmi
360
999
1188
459
Σ= 3006
1 k
*  xi * xmi = 1/25 * 3006 = 120.24
N i 1
Geometric Mean: xG = n x1 * ......xn
xi > 0 !!!!!
Def: To be used on annual growth-processes or loss-processes.
e.g. the mean company turnover as a time-serie:
turnover-increase from year 1 to year 2: 2%
turnover-increase from year 1 to year 2: 18%
IMPORTANT: Calculation via growth-factors (1.02 bzw. 1.18) and not growth-rates!
xG = 1,02 *1,18 = 1.097 ; that is 9.7% mean turnover growth.
11
Median :
odd number of n: z  x( n 1) / 2
Even number of n: z 
xn / 2  x( n / 2)1
2
Def: Observation or potential observation in a set that devides the set so that the same
number of observations lie on each side of it.
Quantiles:
Divisions of a probability distribution or frequency distribution into equal, ordered
subgroups, for examples: quartiles or percentiles:
Quartiles: four groups 25% each
Perzentiles: the set of divisions that produce exactly 100 equal parts in a series
of
continuous values
Very commonly used quartiles:
Q1 : 1.Quartile = x 0,25
Q2 : 2.Quartile = x 0,5 = z =Median
Q3 : 3.Quartile = x 0,75
With min-value, Q1, z, Q3, max-value a Box-Plot can be drawn to show a better
picture of the distribution of a set of observations.
12
Population Variance: σ2 =
Sample Variance s2 =
1 k
*  ( xi  x ) 2
N i 1
k

1
* ( xi  x ) 2
N  1 i 1
Def: The average of square differences between observations and their mean
Standard Deviation: s = s 2 or
s=
2
Def: Square root of the variance.
Normal distribution/ Normal curve: distribution under the bell shaped curve
x ±s
= ~ 68% of the observations
x ±2s
= ~ 95% of the observations
x ±3s
= ~ 99.7% of the observations
Standard Error of Mean: s x =
s
N
We can be 95% confident, that real mean lies in an interval of x ± 1.96 s x .
Variation Coefficient: VC =
s  100
%
x
Def: This coefficient shows if a distribution is homogeneous. If the coefficient shows
a result of more than 50%, the distribution is more likely not homogeneous and so the
mean would not be a good measure.
13
Graphical representation:
 Boxplot
1600,00
ek
1400,00
1200,00
1000,00
A graphical method of displaying the important characteristics of a set of
observations. The display is based on the five-number summary
- x0,25 = begin of the box
- x0,75 = end of the box
- the median in the box, which covers the inter quartile-range
- two "whiskers"
extending to xmin und xmax, outliers or "outside
observations" are being indicated separately.
In order to compute so called outliers, the inter quartile-range is used:
(dQ = x0.75 - x0.25 )
The lower and upper end of the whiskers are:
zu = x0.25 – 1.5* dQ
zo = x0.75 + 1.5* dQ
All observations greater zo or smaller zu are outliers or "outside
observations".
14
Statistical Testing:
Statistical tests within data analysis are needed to determine differences
between groups or to compare groups. This could be two different surveys
with same questions or in our case one single sample, which is split into two
or more sub-samples on the basis of some characteristic; e.g. creating subsamples of Males and Females, different age-groups, different educationlevels,…
Another point of interest could be to compute relationships between two
variables; this is also called the measure of association, where especially the
magnitude of the relationship is what we are interested in. These measures
of association are usually calibrated so as to range between 0 and +/- 1, with
0
indicating
no
relationship
between
the
variables
(=complete
independence) and 1 a perfect relationship (the + or – sign indicates the
direction of the relationship; (e.g. +1…..the higher someones income, the
more money he/she spends for christmas presents).
As a very rough rule of thumb, a relationship is usually considered
 very strong, if the association measure is larger than 0.8
 strong, if the association measure is between 0.7 and 0.8
 moderate, if the association measure is between 0.4 and 0.7
 weak, if the association measure is below 0.4
15
The table below gives an overview about the different statistical tests
referring to their level of measurement or scale:
Scale
Comparing groups
Nominal
Chisquare test
Computing relationships
Contingency-coefficient or
Cramer´s V
n=2 groups: Mann Whitney U-test
Ordinal
n>2 groups: Kruskal-Wallis test
Metric
One-way ANOVA;
Pre-condition: the metric variable has to
be
Normally distributed and
Homogenity of Variance
Spearman´s rank order
correlation
Pearson´s product moment
correlation
16
Hypothesis-Testing:
One approach to making inferences about the population is via hypothesis
testing. Whereas in estimation the focus was on making some informed
guesses about the values of population parameters using our sample data
and a relevant sampling distribution, in testing hypotheses the aim is to
examine whether a particular proposition concerning the population is likely
to hold or not.
Within a bivariate analysis one presumption concerning for instance men
and women could be, that there is a differnce in their smoking behaviour.
Men are more likely to smoke than women. These presumptions for a
population based on a survey are also called hypotheses.
Basically there are two types of hypotheses:
- Null-hypothesis: H0
- Alternative-hypothesis: H1
In the case of comparing groups, the null-hypothesis indicates that there is
no difference between two or more groups.
Only null-hypotheses can be tested; if they are rejected, this is taken to
signify support for the alternative hypothesis. We can never test an
alternative hypothesis directly and nor can we ever prove a hypothesis.
17
Steps in hypothesis testing:
 Formulate the null and the alternative hypothesis
 Specify the significance level
 Select an appropriate statistical test
 Identify the probability distribution of the test statistic and define the
region of rejection
 Compute the value of the test statistic from the data and decide
whether to reject or not reject the null hypothesis; any statistical
software computes the exact level of significance;again it is your turn
to decide whether to reject or not reject the null hypothesis
Formulating the null and the alternative hypothesis
The null hypothesis should contain a statement of equality or no (=null)
difference between groups or no (=null) relationship between variables. By
convention, a null hypothesis is denoted – as mentioned previously – as H0
(read: H-nought) and is given always the benefit of the doubt, that it is
assumed to be true unless it is rejected as a result of the testing procedure.
Note that inability to reject the null hypothesis does not prove that H 0 is
actually true; it may be true, but our tests are only capable of disproving
(but not confirming) a hypothesis. If, as a result of testing, the null
hypothesis is rejected, this is interpreted as signifying support for the
alternative hypothesis which, again by convention is denoted as H 1 (read Hone). Remember that since H0 and H1 are complementary, the alternative
hypothesis should always include a statement of inequality.
18
Specification of Significance Level
Having formulated the null and the alternative hypothesis, the next step is to
specify the circumstances under which H0 will be rejected and H1 will not
be rejected.
For a better understanding of this decision problem the following example
should help:
As a market researcher you are consulting a big company, which is
inventing a new product on the market. One result of your market study
was, that significantly more men than women will buy this new product
(=you rejected H0 in favour of H1). Based on your study the marketingmanager of the company will develop a new strategy and advertising
campaign in order to have two equal groups of buyers; the product should
be sold equally under men and women. All this costs a lot of money.
The very first question of the marketing researcher concerning your study
could be:
“How sure are you about your advise ?” or “How high is the possibility of
an error?”, or in statistical notation: “How high is the risk of wrongly
rejecting H0?
As you want to minimize the possibility of an error, one possible answer
could be: “The chance for an error is ≤ 5% !!!”
19
There are four possible outcomes whenever you test any hypotheses:
Situation in the population
H0
Decision made
H0 not rejected correct decision
H1
Beta-error
based on a sample…..
H1 / H0 rejected Alpha-error ()
correct decision
We denote  as our significance level and use it to indicate the maximum
risk we are willing to take in rejecting a true null hypothesis; the less risk
we are willing to assume, the lower the . Typical values for  are 0.05 ,
0.01 and 0.001.
Always remember that the significance level is a probability of making a
mistake: rejecting a true null-hypothesis.
Having specified a significance level, the way we use it is simple. If the
result of our statistical test is such that the value obtained has a probability
of occurrence less than or equal to , then we reject H0 in favour of H1 and
we declare the test result as significant. Is the probability, associated with
the test result, greater than , we cannot reject H0 and we denote the test
result as non-significant.
This is the reason why statistical tests are often referred to as
significance tests and hypothesis-testing as significance-testing.
20
Selection of an Appropriate Statistical Test
Basically there are three criteria to look at, when selecting a statistical test.
 First, the type of hypothesis to be tested, will require a different test
each time. For example, different tests are appropriate for hypotheses
concerning differences between groups compared with those for
hypotheses concerning relationships between variables.
 Second, the distributional assumptions made regarding the population
from which the sample was drawn will affect the choice of test. The
most used assumption in this context is, that the sample data have
been drawn from a normally distributed population. Another question
could be “are the samples coming from populations with equal
variances?”; (especially when applying an ANOVA).
 Third, the level of measurement (data scale) of the variables involved
in the analysis is also relevant to consider.
Based on above criteria the following tests are commonly used:
Scale
Comparing groups
Nominal
Chisquare test
Computing relationships
Contingency-coefficient or
Cramer´s V
n=2 groups: Mann Whitney U-test
Ordinal
n>2 groups: Kruskal-Wallis test
Metric
Spearman´s rank order
correlation
One-way ANOVA;
Pre-condition: the metric variable has to be Pearson´s product moment
correlation
Normally distributed and
Homogenity of Variance
21
Identification of the Probability Distribution of the Test Statistic
Each test generates what is known as a test statistic, which is a measure for
expressing the results of the test. Referring to the table on the last page test
statistics are e.g. Chisquare-value, U-value, F-value,…
Each of these test-statistics can be computed from the sample data and has
to be compared with his comparable value drawn from a table (see
appendices in statistics literature); this value – drawn from a table – is also
called the critical value, that separates the rejection and the acceptance
region. In a next step you make a graph from the distribution, apply the
critical value from the table, define the acceptance region and the rejection
region and either reject or don’t reject H0.
Computation of the test statistic
How to compute the test statistic will be shown on the next pages,
depending which test is used. Every test statistic can be transformed in a
standard normal distribution (=bell shape curve). All mentioned test
statistics (their value from the tables) from above are depending from a
statistical measure, called degrees of freedom and of course the applied
level of significance. In order to standardize these test statistics a ztransformation has to be applied.
The critical z-value for the rejection of H0 (|z-value| greater than…) within
the standard normal distribution are at a significance level of
1%, or α=0.01
the z-value +/- 2.575
5%, or α=0.05
the z-value +/- 1.96
10%, or α=0.10
the z-value +/- 1.645
22
Probability – in brief:
The subject view of probability states that the probability is an estimate of
what an individual thinks is the likelihood that an event will happen. This
could also mean that two individuals estimate the prbability differently.
How can we proof, that probabilioty is somehow determined.
 Flip a coin and you will see the probability for head is 50% as well as
the probability for tail.
 Throw a dice: The possibility for a 6 is 1/6=16.66%
The set of all possible results is called the probability space; each possible
result is also called an outcome. An event is a set that consists of a group of
outcomes.
 Probability of an event: First we count the number of outcomes in A.
We call that N(A), which is the number of outcomes in A. s is thetotal
number of outcomes in the probability space. The probability is now
defined as as:
o Probability that event A will occur is P(A)=[N(A)]/s
So probability is sometimes also defined as the long term relative frequency
with which an outcome or event occurs.
As there are many other ways of describing probabilities, e.g.:
 probability of a union
 probability of an intersection, and other probability related topics like
o sampling with replacement
o sampling without replacement
we will focus more on the sort of probability that is related to frequency
tables.
23
CONDITIONAL PROBABILITY:
Widely to be interpreted probabilities within a frequency- or contingencytable are:
- Joint probability
- Marginal probability
- conditional probability
events and probabilities
event E
event F
Total
Count
% within row
% within column
% of Total
Count
% within row
% within column
% of Total
Count
% within row
% within column
event C
52
59,1%
45,6%
23,5%
62
46,6%
54,4%
28,1%
114
51,6%
100,0%
event D
36
40,9%
33,6%
16,3%
71
53,4%
66,4%
32,1%
107
48,4%
100,0%
Total
88
100,0%
39,8%
39,8%
133
100,0%
60,2%
60,2%
221
100,0%
100,0%
- the row % of Total are the joint probabilities:
o joint prob. between C and E =p(C&E)=52/221=0.235=23.5%
o joint prob. between F and D =p(F&D)=71/221=0.321=32.1%
o E and D / F and C accordingly
- the marginal probability is to find under Total within row for event C and
D and under Total within column for event E and F.
o marginal prob. event C = p(C)=114/221 = 0.516 = 51.6%
o marginal prob. event E = p(E)=88/221 = 0.398 = 39.8%
o marginal prob. for D and F accordingly
24
- the conditional probability are:
o E/C = the probability for event E under the condition of C
is 52/114= 0.456=45.6%
o F/C = the probability for event F under the condition of C
is 62/114= 0.544=54.4%
………….
………….
o C/F = the probability for event C under the condition of F
is 62/133= 0.466=46.6%
o D/F = the probability for event D under the condition of F
is 71/133= 0.534=53.4%
25
Tests to compare differences
Chisquaretest:
The Chisquaretest is the most widely-used non-parametric test. Our form of
chisquaretest is the test of homogeneity, to test two samples, if they come
from populations with similar or like distributions. It is applied on nominal
variables, to compute possible differences between two or more groups.
The Pearson Chisquaretest is computed:
(O  E ) 2
 
E
i 1
n
2
O....... observed values
E....... expected values
It is the comparison of observed versus expected values in a twodimensional table, where the expected values are those calculated from the
product of the appropriate row and column totals divided by the number of
observations. The expected values would be the answering pattern, when
there is no difference between e.g. two groups.
expected values
c......column-total/
eij 
ci*r j
N
r.......row-total/
N......TOTAL(sample)
26
Below example shows the application of the chisquare-test:
The qquestion is: Are there differences between men and women regarding
their book-reading behaviour?
The two hypotheses are:
H0: There is no difference between men and women regardin their
book reading behaviour.
H1: Between men and women there are significant differences
regarding their book reading behaviour.
The following table shows the result of an survey within 300 students.
1.step: observed frequencies:
Men
Women
total: absolute
in %
detective novels
Love novels
SciFi
total
65
70
135
45%
25
50
75
25%
50
40
90
30%
140
160
300
100%
detective novels
Love novels
SciFi
63
72
35
40
42
48
2.step: expected frequencies:
Men
Women
2calc. = ((65-63)2/63) + ((25-35)2/35) +......+((40-48)2/48) = 8.333
27
In order to find the critical 2-value in the table of the Chisquaredistribution, we need to know the degrees of freedom of our test-problem
and the proposed level of significance.
For the chisquare-distribution the degrees of freedom are calculated as:
DF= (c-1) * (r-1) = (3-1)*(2-1) = 2
that means 2 degrees of freedom
The level of significance should be 5%; this is α=0.05 !
The critical value then is:
2Table 0.95; 2 = 5.991
As the – from the sample – calculated chisquare-value is greater than the
critical chisquare-value we reject H0.
The chisquare-distribution can be approximized through the normaldistribution. The appropriate z-value can be calculated as:
z=
2

2
 2  DF  1
(this is only one way to calculate z, others may be found in the SPSS-handbook)
for the above example we calculate z as follows:
z=
2  8,333  (2 * 2)  1
= 2.351
28
The exact level of significance (=aerea on the very left and very right side
of our bell shape curve of the standard normal distribution) can be
computed by an Integral with the lower and upper limit of z =  2.4 :
z
2
x
1
1
2

e
2
d x  0.016395
z
The probability of making a type 1-error (=α-error), which means, we
reject H0 in favor of H1, but in the population (~reality) H0 would be true
(=the right decision), is only 1,6%.
According to the result above:
Sig=0.016395  0.05 we can reject H0.
 There statistically significant differences between men and
women regarding their book reading behaviour.
 If we reject H0, the contingency table has to be interpreted
(see output below !).
29
SPSS-Output:
SEX * BOOK Crosstabulation
SEX
male
female
Total
Count
% within SEX
% within BOOK
Count
% within SEX
% within BOOK
Count
% within SEX
% within BOOK
detective s tories
65
46,4%
48,1%
70
43,8%
51,9%
135
45,0%
100,0%
BOOK
love novels
25
17,9%
33,3%
50
31,3%
66,7%
75
25,0%
100,0%
Sc iFi
50
35,7%
55,6%
40
25,0%
44,4%
90
30,0%
100,0%
Total
140
100,0%
46,7%
160
100,0%
53,3%
300
100,0%
100,0%
Chi-Square Te sts
Pearson Chi-Square
Lik elihood Ratio
Linear-by-Linear
As soc iation
N of Valid Cases
Value
8,333a
8,459
,661
2
2
As ymp. Sig.
(2-sided)
,016
,015
1
,416
df
300
a. 0 c ells (,0% ) have expected count less than 5. The
minimum expected count is 35, 00.
Interpretation of the contingency-table !!!
30
Special application of the chisquare-test:
- With a 2x2 table:
- chisquare test with continuity correction or Yates correction for
continuity
When testing for independence in a contingency table, a continuous
probability distribution, namely the chisquare distribution, is used as an
approximation to the discrete probability of observed frequencies, namely
the multinoial distribution. To improve this approximation Yates suggested
a correction that involves subtracting 0.5 from the positive discrepancies
(observed-expected) and adding 0.5 to the neagtive discrepancies beforde
these values are squared in the calculation of the usual chisquare statistics.
If the sample size is very large, the correction will have only little effect on
the value of the test statistic.
Example: Is there are difference between men and women regarding their
religiosity?
n
[ O  E  0.5]2
i 1
E
2  
The hypotheses are:
 H0: Between men an women there is no difference regarding their
religiosity.
H0: M = F
31
 H1: Between men and women there are differences regarding
their religiosity.
H0: M  F
SEX * Are you a religious person ? Crosstabulation
SEX
female
male
Total
Are you a religious
person ?
yes
no
6
2
75,0%
25,0%
Count
% within SEX
% within Are you a
religious person ?
Count
% within SEX
% within Are you a
religious person ?
Count
% within SEX
% within Are you a
religious person ?
Total
8
100,0%
60,0%
20,0%
40,0%
4
33,3%
8
66,7%
12
100,0%
40,0%
80,0%
60,0%
10
50,0%
10
50,0%
20
100,0%
100,0%
100,0%
100,0%
Chi-Square Tests
Pearson Chi-Square
Continuity Correction a
Likelihood Ratio
Fis her's Exact Test
Linear-by-Linear
As sociation
N of Valid Cases
Value
3,333b
1,875
3,452
3,167
df
1
1
1
1
As ymp. Sig.
(2-sided)
,068
,171
,063
Exact Sig.
(2-sided)
Exact Sig.
(1-sided)
,170
,085
,075
20
a. Computed only for a 2x2 table
b. 2 cells (50,0%) have expected count less than 5. The minimum expected count is
4,00.
Above example shows, that the  - based on the continuity
correction - is =0.171; this indicates that there is no difference
between men and women regarding their religiosity, we cannot
rejetc H0. If we would reject H0, the risk for a type 1-error, which
means, rejecting H0 in favor of H1 , while H0 is the right decision,
would H0 be 17%.
32
NONPARAMETRIC TESTS
The following statistical tests Mann Whitney U-Test and Kruskal
Wallis Test are used when one compares two or more groups on an
ordinal variable or one metric Variable that was not normally
distributed. .
 Mann-Whitney U-Test:
The Mann-Whitney U-test (also known as the Wilcoxon rank sum W test) is
very useful when you have two groups to compare on a variable which is
measured at ordinal level or a metric variable which is not normally
distributed. The test focuses on differences in central location and makes
the assumption that any differences in the distributions of the two
populations are due only to differences in locations (rather than, say,
variability). Basically all observations are ranked and divided in the two
observed groups. Where one test-statistic is the sum of ranks in the smaller
group, the U-test statistic is computed via this so called W (= sum of ranks).
The null hypothesis tested by the Mann-Whitney U-Test is, that there is no
difference between the two groups in terms of location, focusing on each
mean rank as a measure of central tendency, only when comparing them.
The value of the mean rank itself does not state anything about how good or
bad something was evaluated or ranked. On an ordinal scale collected data
only allow to interprete informations based on their ranks. Mean and
Variance are meaningless in their interpretation.
33
The following example shows the application of the Mann Whitney
U-test:
The question is:
Is there a difference between male and female students in how they rate the
possibilities of doing at least some sport in the area where they live.
The hypotheses are:
H0: There is no difference between male and female students regarding how
they rate the possibilities of doing at least some sport in the area where they
live.
H1: There is a significant difference between male and female students
regarding how they rate the possibilities of doing at least some sport in the
area where they live.
(Alpha = 5%)
The below table shows the data as collected:
male
female
ti
(ti3-ti)
Very good
6
3
9
720
Good
5
1
6
210
Less good
2
3
5
120
bad
0
4
4
60
total
n1=13
n2=11
N=24
=1110
rating
In a first step all data are ranked by their rating and – in this case – by sex
(=the grouping variable); so we calculate a rank-value for all very good
ratings, good ratings, less good ratings and bad ratings, independent of the
34
grouping variable sex. So we have 24 rankings, starting with the first male
person that said very good until the last female – in our example 24th –
person, that rated with bad.
In a next step the rank values are weighted by the grouping variable (=sex;
number of males and females) and so the ranksums for all male and female
persons can be calculated:
Rating
1=Very good
2=Good
3=Less good
4=Bad
Male
Female
1+2+3+4+5+6+
5 x 6=30
7+8+9=45/9=5
5 x 3=15
10+11+12+13+14+
12.5 x 5=62.5
16+17+
18 x 2=36
+
22.5 x 0=0
15=75/6=12.5
12.5 x 1=12.5
18+19+20=90/5=18
18 x 3=54
21+22+23+24=90/4=22.5
22.5 x 4=90
R1=128.5
R2=171.5
128.5/13=9.88
171.5/11=15.59
Sum of ranks R
Mean rank
R1 + R2 = [N*(N+1)]/2
128.5 + 171.5 = 300
In a next step we have to determine the U-test-statistic:
U1 = R1 – [n1*(n1+1)]/2
U1 = 128.5 – 91 = 37.5
U2 = R2 – [n2*(n2+1)]/2
U2 = 171.5 – 66 = 105.5
35
The U-test-statistic is always the smaller value of U1 and U2:
U = Min! (U1, U2) = 37.5
According to the U-distribution table the critical value for U is:
U/2;n1;n2 = U0.025;13;11 = 42
As the calculated U-value (=37.5) is smaller (remember: U-distribution)
than the critical U-value from the table (Utabulated = 42), we are able to
reject H0 and favor H1; so we can interpret the calculated mean ranks.
SPSS-Output: Mann Whitney U-Test
Ranks
rating s portposs ibilities in your area
sex
male
female
Total
N
13
11
24
Mean Rank
9,88
15,59
Sum of Ranks
128,50
171,50
36
In order to get the exact level of significance, we first have to compute z:
The following formula for computing z has a correction-term for so called
ties (=more people could have given the same answer):
U
z=
n1* n2
2
n1* n2
* (N 3  N 
12 * N * ( N  1)
m
 (t
3
i
 ti )
i 1
z = - 2.054
Excurs:---------------------------------------------------------------------Is each data value of the ordinal or non normally distributed variable
only one of a kind (each value of this variable appears only one time
within this variable) , then we would not need a correction for ties and z
could be computed as:
n1* n2
2
n1* n2 * (n1  n2  1)
12
U
z=
-------------------------------------------------------------------------------Computing the exact level of significance (limits */- 2.054):
z
2
x
1
1
2

2
e d x  0.028307
0.040
z
sig = 0.040 = 4.0%
As 0.040  0.05  H1 ; in other words: we reject H0 !!
37
Conclusion:
There is a significant difference between male and female students
regarding how they rate the possibilities of doing at least some sport in the
area where they live.
How can we interpret the difference ?
Take a look at the mean ranks.
Mean Ranks
Male
Female
9.88
15.59
The mean rank itself does not help us from it´s value. As soon as we
compare both mean ranks we can see the difference. The mean rank of the
female persons is higher than the one from the male persons.
Let´s have a look at the coding values of the variable rating the possibilities
of doing sport in the area, where one lives:
1 = very good
2 = good
3 = less good
4 = bad
A higher value is a worse rating than a low value; this means:
Female persons rated the possibilities worse than men.
We cannot say how much worse!! In order to clarify the result even more,
we could take a closer look at the contingency-table of the two analysed
variables.
38
Output of Mann Whitney U-Test continued:
Mann-Whitney-U-Test
Test Statisticsb
Mann-Whitney U
Wilcoxon W
Z
As ymp. Sig. (2-tailed)
Exact Sig. [2*(1-tailed
Sig.)]
rating
sportpossibilities
in your area
37,500
128,500
-2,054
,040
a
,047
a. Not corrected for ties.
b. Grouping Variable: sex
To determine if H0 or H1 , we look at the „Asymptotic Significance” (0.04),
which is corrected for ties and we decide to reject H0 .
Once again: If we rejct H0 , we interpret the mean ranks or take a look
at the contingency-table.
Ranks
rating s portposs ibilities in your area
sex
male
female
Total
N
13
11
24
Mean Rank
9,88
15,59
Sum of Ranks
128,50
171,50
For interpretation only: contingency-table
rating sportpossibilities in your area * sex Crosstabulation
sex
male
rating s portpos sibilities
in your area
very good
good
les s good
bad
Total
Count
% within sex
Count
% within sex
Count
% within sex
Count
% within sex
Count
% within sex
6
46,2%
5
38,5%
2
15,4%
0
,0%
13
100,0%
female
3
27,3%
1
9,1%
3
27,3%
4
36,4%
11
100,0%
Total
9
37,5%
6
25,0%
5
20,8%
4
16,7%
24
100,0%
39
 Kruskal-Wallis H-Test:
The H-test by Kruskal-Wallis is used to compare more than 2 independet
groups (n>2) regarding an ordinal or not normally distributed metric
variable. Like the Mann-Whitney U-test, the Kruskal-Wallis-H-Test tests
the same null hypothesis, but across k rather than two groups.
Basically all observations are ranked and divided in the k observed groups.
Where one test-statistic is the sum of ranks in the smaller group, the
Chisquare-test statistic will be computed in order to find out if H 0 will be
rejected or not. The test statistic is based on an approximation of the
chisquare distribution with k-1 degrees of freedom. Were k is the number of
groups compared. An adjustment, taking into account tied ranks, is also
provided.
The null hypothesis tested by the Kruskal-Wallis H-Test is, that there is no
difference between the k groups in terms of location, focusing on each mean
rank as a measure of central tendency, only when comparing them. The
value of the mean rank itself does, as within the Mann Whitney U-test, not
state anything about how good or bad something was evaluated or ranked.
On an ordinal scale collected data only allow to interprete informations
based on their ranks. Mean and Variance are meaningless in their
interpretation.
40
Example to demonstrate the application of the Kruskal-Wallis-H- test:
Is there a difference between the four regions of origin regarding how they
rate the possibilities of doing at least some sport in the area where they live?
The hypotheses are:
H0: There is no difference between students originating from the four areas
(Luzon,Visayas, Mindanao, abroad) regarding how they rate the
possibilities of doing at least some sport in the area where they live..
H1: There is a significant difference between students originating from the
four areas (Luzon,Visayas, Mindanao, abroad) regarding how they rate the
possibilities of doing at least some sport in the area where they live.
(Alpha = 5%)
The computation of the sum of ranks and the mean ranks is the same as
within the Mann Whitney U-Test.
We start with the following table:
Luzon Visayas Mindano abroad
ti
(ti3-ti)
1=Very good
6
1
0
2
9
720
2=Good
5
0
0
1
6
210
3=Less good
2
2
1
0
5
120
4=Bad
2
1
1
0
4
60
TOTAL
15
4
2
3
24
1110
41
Luzon
Visayas
Mindano
abroad
30
5
0
10
62.5
0
0
12.5
3=Less good
36
36
18
0
4=Bad
45
22.5
22.5
0
Sum of ranks R *)
173.5
63.5
40.5
22.5
Mean rank (=R/n)
11.57
15.88
20.25
7.5
1=Very good
2=Good
*) Sum of ranks and mean ranks are computed accordingly to the Mann-Whitney U-test.
The test value H is an approximation of the chisquare distribution with k-1
degrees of freedom, were k is the number of groups (here: 4 groups of
origin).
DF = 4 – 1 = 3
The formula below refers to a data set, where every value appears only one
time within the ordinal variable:
H=
12
*
N * ( N  1)
Ti2
 3 * ( N  1)
i 1 n
i

k
In our example it appears that the same rating was given more often that
only one time; in this case a correction term is used:
H´ =
H

1
m
(t 3
i 1 i
3
 ti )
N N
H´err.= 5.519
42
Due to the chisquare distribution approximation refer to the chisquare table
in order to find the critical value:
H´1-;DF = H´0,95;3 = 7.815
As our calculated H-value is smaller than the critical value from the
chisquare table, we cannot reject H0 .
According to the chisquare-test one can calculate the z-value and then the
exact level of significance.
SPSS-Output - Kruskal-Wallis-Test:
Ra nks
rat ing s port pos sibilities
in your area
origin
Luzon
Vis ayas
Mindanao
abroad
Total
N
15
4
2
3
24
Mean Rank
11,57
15,88
20,25
7,50
Test Statisticsa,b
Chi-Square
df
As ymp. Sig.
rating
sportpossi
bilities in
your area
5,519
3
,138
a. Kruskal Wallis Test
b. Grouping Variable: origin
To determine if H0 or H1 , we look at the „Asymptotic Significance”
(0.138), which is corrected for ties and we decide not to reject H0 .
Once again: If we would rejct H0 , we interpret the mean ranks or take
a look at the contingency-table.
43
 Analysis of Variance:
Analysis of Variance (ANOVA) is used to compare the means of two or
more groups (the t-test is used to compare only two groups) regarding a
metric variable. As we can see the F-test of the ANOVA is a more widely to
use test, so we will go deeper in his test-statistic.
Two assumptions must be met before you can run a oneway ANOVA.
Normaldistribution (Kolmogorov-Smirnoff Test)
Equal variances of the groups (Levene-statistic)
If these assumption are not given, we have to go back to the tests for the
ordinal scale and compute a Mann Whitney U-test or a Kruskal-Wallis test.
The Analysis of Variance partitions the total variance of a set of
observations into parts due to particular factors, for example, sex , origin,
age-groups, etc and comparing variances (mean squares) by way of F-tests,
differences between means can be assessed.
The following hypotheses will be verified or falsified:
H0 : 1 = 2 = ...... = n
H1 : 1 ≠ 2 ≠ ...... ≠ n
44
One approach to calculate the F-test statistic (Fcalculated by data) is:
W
T * 2
W
1
T
1
Fcalculated =
,
where T = sum of total-variation and W is the variation within the groups.
N
T =  (x
i
 x) 2
i 1
W……. to be computed accordingly for each level (group).
Degrees of freedom:
 1 = I – 1, where I is number of groups to be compared
 2 = N – I, where N is the sample-size
45
The following example demonstrates the application of an
ANOVA:
The question is: Is there a difference betwenn male and female persons
regarding how often they perform sport per week, if at all?
Formulate the hypotheses !!!
How often sport a
week
(if at all)
group-mean
T
W
1= male
1
3.7273
(1-3.0667)2
(1-3.7273)2
1
2
(2-3.0667)2
(2-3.7273)2
1
2
(2-3.0667)2
(2-3.7273)2
1
3
(3-3.0667)2
(3-3.7273)2
1
4
(4-3.0667)2
(4-3.7273)2
1
4
(4-3.0667)2
(4-3.7273)2
1
4
(4-3.0667)2
(4-3.7273)2
1
5
(5-3.0667)2
(5-3.7273)2
1
5
(5-3.0667)2
(5-3.7273)2
1
5
(5-3.0667)2
(5-3.7273)2
1
6
(6-3.0667)2
(6-3.7273)2
(1-3.0667)2
(1-1.25)2
(1-3.0667)2
(1-1.25)2
(1-3.0667)2
(1-1.25)2
(2-3.0667)2
(2-1.25)2
Sum = 42.933
Sum= 24.93
1
2=female
1.25
1
2
1
2
2
2
Total mean=3.0667
T .........42.933
W......... 24.93
46
1
2
=I–1=2–1=1
= N – I = 15 – 2 = 13
Fcomputed by data = (1-(W/T))/(W/T) * (  2 /  1 )= 9.387
As within the other statistical tests, the critical F-value has to be
taken from the table of the F-distribution:
F tab 0.05; 1; 13 = 4.75
As Fcomp. ≥ Ftab. we can reject H0 and we can state, that there is a
statistically significant difference in the means of the two groups
(male/female) regarding how often they perform sports per week.
47
SPSS – Output of ANOVA including K-S Test for normal distribution
and test of homogeneity of variances:
1) K-S Test for normal-distribution:
One-Sample Kolmogorov-Smirnov Test
N
Normal Parameters a,b
Most Extrem e
Di fferences
Mean
Std. Deviati on
Absolute
Positive
Negati ve
Kolmogorov-Sm irnov Z
As ymp. Sig. (2-tailed)
how often
sport per
week
15
3,0667
1,75119
,195
,195
-,170
,757
,615
a. Test di stribution is Norm al.
b. Calculated from data.
H0: normal distribution is given
H1: the data are not normally distributed
2) Test of Homogeneity of Variances
Test of Homogeneity of Variances
how often s port per week
Levene
Statistic
4,224
df1
df2
1
13
Sig.
,061
H0: Homogeneity of Variances is given
H1: Homogeneity of Variances is not given
48
3) ANOVA - results
Descriptives
how often sport per week
N
male
female
Total
11
4
15
Mean
3,7273
1,2500
3,0667
Std. Deviation
1,55505
,50000
1,75119
95% Confidence Interval for
Mean
Lower Bound Upper Bound
2,6826
4,7720
,4544
2,0456
2,0969
4,0364
Std. Error
,46887
,25000
,45216
Minimum
1,00
1,00
1,00
Maximum
6,00
2,00
6,00
ANOVA
how often s port per week
Between Groups
W ithin Groups
Total
Sum of
Squares
18,002
24,932
42,933
df
1
13
14
Mean Square
18,002
1,918
F
9,386
Sig.
,009
H0: The means of the groups male and female are equal
H1: The means of the groups male and female are not equal
As the level of sognificance is less/equal 0.05 (Sig=0.009) we can
reject H0 and we can say, that there are statistically significant
differences between the means of the groups (male and female
persons). The next step is the interpretation of the means from
above.
49
Measures of association / relationships:
 Contingency coeffizient and Cramer´s V: (0 bis 1)
Regarding nomnal variables an appropriate measure of relationship is the
contingency coeffizient or Cramer´s V. Its values always fall between 0 and
1 and, thus, can be interpreted as reflecting relationships of different
magnitudes
Both measures are chisquare-based as the following formulas show:

C=
2
n2
Cramer´s V =
2
n  ( k  1)
;
where n is the sample-size and k is the smaller of the number of rows or
columns in the contingency table.
We analyse the question: Is there a relationship between the own sportbehaviour and the smoking-behaviour
The hypotheses are:
H0 : There is no relationship between sport-behaviour and
the smoking-behaviour.
H1 : There is a significant relationship between sport-behaviour
and the smoking-behaviour.
50
The results of the contingency-coefficient and the Cramer´s V are:
chisqu
C=
chisqu  n
chisqu
 0.488
Cramer´s V =
n  ( k  1)
 0.559
SPSS-Output contingency coeffizient and Cramer´s V::
Symmetric Measures
Nominal by
Nominal
Cramer's V
Contingency Coefficient
N of Valid Cases
Value
,559
,488
24
Approx. Sig.
,024
,024
a. Not as suming the null hypothesis .
b. Us ing the asymptotic standard error assuming the null
hypothesis .
The level of significance (sig=0.024) indicates a significant relationship
between the two variables; the magnitude is 0.559 (Cramer´s V) and 0.488
(contingency-coefficient).
51
 Spearman´s Rank-Order Correlation: (-1 to +1)
If dealing with a situation in which both variables concerned are ordinal,
you can investigate not only the strength of the association but also its
direction (positive or negative relationship). Spearman´s rank-order
correlation coefficient ranges from –1 to +1, with values close to zero
indicating little or no association between the variables.
Having two variables x and y:
x1 ≤ ........ ≤xn, being ranked in an order, the rank (xi)=i, and
y1 ≤ ........ ≤yn, being ranked in an order, the rank (yi)=i;
Each observation gives a pair of ratings (xi , yi).As well within the x-values
as within the y-values identical values can appear and mean rank-values
will be calculated.
The formula for the computation of Spearman´s rank-order coefficient is:
d
rSP= 1 -  ;
2
i
6
(n 2  1)n
if we have to correct our data for ties, which is quite often the case, the
following formula has to be applied:
rSP corr. = 1 -
d
6
2
i
(n 2  1)n  (T x´  T y´ )
;
where di are the rank-differences and n is the sample-size.
Tx´ = 12  (t x3
´
i
 t x´ ) ;
i
considers the frequency of each single rating-value
of the ordinal variable
Ty´ to be computed likewise.
52
Example: Two variables are given….
x....rating of the possibilities of doing sport, where you alive
(1 = very good
2 = somehow good
3 = not so good)
y....rating of your own sporting-behaviour
(1 = very sporty 2 = somehow sporty 3 = not so sporty)
The hypotheses are:
H0: Between the rating of the possibilities of doing sport, where you live and
the rating of your own sporting-behaviour there is no relationship
H1: Between the rating of the possibilities of doing sport, where you live and the
rating of your own sporting-behaviour there is a significant relationship.
x
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
3
3
3
3
3
3
3
3
3
3
Rank-values rk(xi)
1+2+3+4+5+6+7+8+9+10=55/10=5.5
11+12+13+14+15=65/5=13
16+17+18+19+20+21+22+23+24+25=
= 205/10=20.5
y
1
1
1
1
1
1
2
2
2
2
2
2
2
3
3
1
2
2
2
3
3
3
3
3
3
rk (yi)
Rg (1) = 4
Rg(2) = 12.5
Rg(3) = 21.5
Computation of rank differences (di)
Rangdifferenzen
5.5 – 4= 1.5 (d)
5.5 – 4= 1.5
5.5 – 4= 1.5
5.5 - 4 = 1.5
5.5 – 4= 1.5
5.5. - 4 = 1.5
5.5 - 12.5 = 7
5.5 - 12.5 = 7
5.5 - 12.5 = 7
5.5 - 12.5 = 7
13 – 12.5= 0.5
13 – 12.5= 0.5
13 – 12.5 = 0.5
13 – 21.5= 8.5
13 – 21.5 = 8.5
20.5 – 4 = 16.5
20.5 – 12.5= 8
20.5 – 12.5= 8
20.5 – 12.5= 8
20.5 – 21.5 = 1
20.5 – 21.5 = 1
20.5 – 21.5 = 1
20.5 – 21.5 = 1
20.5 – 21.5 = 1
20.5 – 21.5 = 1
Sum:
d2
2.25
2.25
2.25
2.25
2.25
2.25
49
49
49
49
0.25
0.25
0.25
72.25
72.25
272.25
64
64
64
1
1
1
1
1
1
825
53
Tx´ = 12  (t x3
´
i
 t x´ ) =½*[(10
i
3
-10)+(53-5)+(103-10)] = 1050
Ty´ = ½*[(73-7)+(103-10)+(83-8)] = 915
6 * 825
rSP corr. = 1 -
(25  1)25  (1050  915 )
2
=1-
4950
113635
= + 0.6369
The relationship between the two variables is positive; we yet don´t know if
this relationship is significant or not. Via a z-transformation we can
calculate the exact level of significance:
z=ż*
n3 ,
where ż is the correlation-figure and can be computed via r as:
1 r
ż = ½*ln( 1  r ) = ½ * ln
1  0,6369
1  0,6369
= 0.753
following:
z = ż * n  3 = 0.753* 25  3 = 3.532
The level of significance is:
z
2
x
1
1
2

e
2
d x  0.001229270561
z
As the level of significance is ≤0.05, (sig=0.00123), we can reject H0 and
one can say, the relationship between the rating of the possibilities of doing
sport, where you live and the rating of your own sporting-behaviour is
statistically significant. The magnitude is positiv +0.637. This means: the
better somebody rates the sporting posibilities, where he/she lives, the better
he/she also rates his/her own sporting behaviour.
54
SPSS - Output: Spearman´s rank-order coefficient
Correl ations
Spearman's rho
rat ing the
sportpossibilities,
where one lives
rat ing the own
sporting-behaviour
Correlation Coefficient
Sig. (2-tailed)
N
Correlation Coefficient
Sig. (2-tailed)
N
rat ing the
sportpossibilities,
rat ing the own
where one lives
sporting-behaviour
1,000
,637**
.
,001
25
25
,637**
1,000
,001
.
25
25
**. Correlation is s ignificant at the 0.01 level (2-tailed).
How to read the output:
1) read the level of significance:
decide whether H0 or H1
2) if H1: how strong is the relationship ?
3) which direction is the relationship ?
4) interpretation of the result !
55
 Pearson´s Product Moment Correlation (-1 to +1)
This is the most widely used measure of association for examining
relationships between metric (interval/ratio) variables Also known as
Pearson´s r the product moment correlation coeffizient focuses specifically
on linear relationships. It ranges from –1 (a perfect negativ linear
relationship through 0 (no relationship) to +1 (a perfect positive linear
relationship). The emphasis on linear is important because if two variables
are linked to one another by means of a nonlinear relationship the Pearson
correlation coeffizient cannot detect it. It is always important and a good
idea to plot the relationship between the variables in a scattergram or
scatterplot before apllying the Pearson correlation coefficient.
Having two metric variables x and y from one observation (respondent)
means to have from:
x1, ........ ,xn, an appropriate y1, ........ ,yn ;
So each observation gives a pair of values (xi , yi).;
Within the following example we want to determine the possible
realtionship between the variable income (x) and the variable expenses for
x-mas presents (y).
The hypotheses are:
H0: Between income and expenses for x-mas presents is no relationship
H1: Between income and expenses for x-mas presents there is a significant
relationship
56
Formula for computing the Pearson´s product moment correlation
coeffizient:
n
rP =
 [( x
i
 x ) * ( y i  y )]
i 1
n
 (x
n
i
 x) 2 *
i 1
(y
 y) 2
i 1
income (x)
1000,00
1050,00
1100,00
1200,00
1250,00
1250,00
1250,00
1500,00
1400,00
1600,00
mean(x)=1260
rP =
i
37100
334000 * 4290
expenses (y)
75,00
80,00
85,00
90,00
100,00
110,00
110,00
130,00
120,00
140,00
mean(y)=104
[( xi  x ) * ( y i  y )]
7540,00
5040,00
3040,00
840,00
40,00
-60,00
-60,00
6240,00
2240,00
12240,00
Sum=37100
( xi  x ) 2
67600,00
44100,00
25600,00
3600,00
100,00
100,00
100,00
57600,00
19600,00
115600,0
Sum=334000
( yi  y) 2
841,00
576,00
361,00
196,00
16,00
36,00
36,00
676,00
256,00
1296,00
Sum=4290
= +0.98
The relationship between income and expenses for x-mas presents is a
positive with a magnitude of +0.98.
In other words:
The more somebody earns, the more he/she spends for x-mas presents.
57
SPSS-Output Pearson´s r:
Correlations
monthly income in USD
expens es for x-mas
presents in USD
Pearson Correlation
Sig. (2-tailed)
N
Pearson Correlation
Sig. (2-tailed)
N
monthly income
in USD
1
.
10
,980**
,000
10
expens es for x-mas
presents in USD
,980**
,000
10
1
.
10
**. Correlation is s ignificant at the 0.01 level (2-tailed).
This output shows, that there is a significant (sig=0.000 ≤ 0.05) relationship
between the two variables. The level of significance can be computed like
the significance within the Spearman rank order coefficient via z. H 0 can be
rejected.
The correlation coefficient +0.98 indicates a very strong positive
relationship between income and expenses for x-mas presents.
Again:
The more somebody earns, the more he/she spends for x-mas presents.
58
 Regression-Analysis
Sometimes it might be interesting in statistics to investigate the question:
How much does one quantity affect another quantity ? How can one
variable be explained through her relationship to another variable ? Within
the regression analysis we try to find a mathematical approach or
mathematical function (=equation) to predict the value of an dependent
variable if we know the value of the independent variable.
Usually a first step into regression analysis is to find out if there is any
relationship between two variables at all, using the Pearson correlation
coefficient. If so, a diagram (e.g. a scatterplot) would also show this kind of
relationship. Knowing more about that relationship, we would be able to
predict – refering to our data (income and expenses for x-mas presents) –
how much one will spend for x-mas presents if we´d know his/her income.
This form of relationship will be represented by a straight line in the
diagram. Such a line is called a regression line, and this type of analysis is
called regression analysis. We need to know two numbers to summarize the
line: the slope and the vertical intercept. We also would like some way to
measure whether the line fits the data very well or rather poorly. This could
be expressed by measure called R2 (ranges from 0 to 1). The closer to 1 the
better the fit, or the more we are able to predict the value of the dependent
variable with our applied model.
59
At this point we introduce the simple regression, which shows the
relationship between one dependent and one independent variable.
The relation between the two variables is expressed by a linear function:
ŷi
= α + βxi ,
where β is the slope and α is the vertical intercept. β is also known as the
regression-coefficient. The sign expresses the direction of the relationship:
Positive:
the more, the more / the less, the less
Negativ:
the more, the less / the less, the more
~ 0:
no pattern, no relationship.
The formula to compute β is:
n
β=
n
n
 (x * y )  n *  x *  y
1
i
i
i 1
i 1
n
n
x
i
2
i
i 1
i
i 1

1
 *(
xi ) 2
n i 1
The vertical intercept, after knowing β, is:
α = y  *x
60
Within the following example, the regression analysis will be demonstrated:
The variables of our last example will be used (income, expenses for x-mas
presents):
ni
income ( x)
expenses (y)
y2
x2
x iy i
1
2
3
4
5
6
7
8
9
10
1000.00
1050.00
1100.00
1200.00
1250.00
1250.00
1250.00
1500.00
1400.00
1600.00
75.00
80.00
85.00
90.00
100.00
110.00
110.00
130.00
120.00
140.00
5625.00
6400.00
7225.00
8100.00
10000.00
12100.00
12100.00
16900.00
14400.00
19600.00
1000000
1102500
1210000
1440000
1562500
1562500
1562500
2250000
1960000
2560000
75000.00
84000.00
93500.00
108000.0
125000.0
137500.0
137500.0
195000.0
168000.0
224000.0
Mean
1260
104
11245
1621000
134750
Sum
12600
1040
112450
16210000
1347500
1
*1040 *12600
10
1
16210000 
*12600 2
10
1347500 
β=
= 0.1110778
α = 104 – 0.1110778 * 1260 = -35.958
The regression-model for our data is:
ŷi = -35.958 + 0.1110778*xi
Remember: The relationship between our two variables has to be linear,
otherwise it makes no sense to compute the above model; so you better
check the data via a simple correlation diagram, which shows you roughly
the relationship.
61
Correlation-diagram of the variables income / expenses for x-mas presents:
correlation scatterplot: income vs. expenses
150
140
130
120
110
100
90
80
70
900
1000
1100
1200
1300
1400
1500
1600
1700
m onthl y i ncom e i n USD
As we can see from the diagram (scatterplot), there is a linear relationship
between the two variables.
The regression line, that should be applied to our data, fits best, when the
distance between the line and the single pair of observations is min!. The
method for finding that line is called least square method (to be calculated
by the euklidian distance).
The differences between the observed and the fitted or estimated values are
called residuals. The values of the residuals should be as minimal as
possible.
ˆi  yi  yˆi
62
If applying the above regression-model
ŷi = -35.958 + 0.1110778*xi
to the question:
How much will somebody spend for x-mas presents, if his monthly income
is 1250,-USD ? , the answer will be:
ŷi = 0.1110778 *1250 +(-35.958) = 102.8892 USD
We can expect a person that earns 1250,- USD to spend 102.89 USD for xmas presents. The difference for an actual observed income of 1250,between the actual expenses of 100,- USD in the data set and estimated
expenses of 102.89, is –2,8892; this is called the residual for the expenses
value of 100.- USD.
For our data set, the estimated values for y and the residuals are:
income (x)
1000.00
1050.00
1100.00
1200.00
1250.00
1250.00
1250.00
1500.00
1400.00
1600.00
expenses (y)
75.00
80.00
85.00
90.00
100.00
110.00
110.00
130.00
120.00
140.00
ŷ
75.11976
80.67365
86.22754
97.33533
102.8892
102.8892
102.8892
130.6587
119.5509
141.7665
residuals
-.11976
-.67365
-1.22754
-7.33533
-2.88922
7.11078
7.11078
-.65868
.44910
-1.76647
63
Partial SPSS Output: regression analysis
Model Summary
Model
1
R
,980a
R Square
,961
Adjusted
R Square
,956
Std. Error of
the Estimate
4,59636
a. Predictors: (Constant), monthly income in USD
Coeffi cientsa
Model
1
Unstandardized
Coeffic ients
B
St d. Error
-35,958
10,126
,111
,008
(Const ant)
monthly inc ome in USD
St andardiz ed
Coeffic ients
Beta
,980
t
-3, 551
13,966
Sig.
,007
,000
expenses for x-mas presents in USD
a. Dependent Variable: ex pens es for x -mas presents in USD
Linear Regression

expenses for x-m as presents in USD = -35,96 + 0,11 * incom e
R-Square = 0,96
140,00


120,00


100,00



80,00

1000,00
1200,00
1400,00
1600,00
monthly income in USD
64
 Multiple Regression:
The multiple regression is used if you analyse one dependent and several (a
minimum of two independent variables) independent variables, where the
independent variables should not correlate and the variance should not be
unequal. The relation between three variables will be demonstrated through
the following equation:
ŷi
= α + β1x1 + β2x2 ,
where β is the slope and α is the vertical intercept. β is also known as the
regression-coefficient.
The sign expresses the direction of the relationship:
Positive:
the more, the more / the less, the less
Negativ:
the more, the less / the less, the more
~ 0:
no pattern, no relationship.
The x-variables should not be correlated and the variances should be equal
or at least similar.
65
 Logistic Regression:
See example during the lecture !!!!
66