Causal Analysis Techniques (424024-B-6) - Summary lectures and tutorials - pre-master and bachelor
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Course
Causal Analysis Techniques (424024B6)
Institution
Tilburg University (UVT)
Hi, this summary consists of all the 14 lectures and tutorials for the course 'Causal Analysis Techniques'. It will give you all the information you need to pass your Causal Analysis Techniques exam, with tips on topics that the lecturer says are important for the exam. This summary is useful for b...
Lecture 1 – 21-10-2024
One-Way Between-Subjects Analysis of Variance (ANOVA)
When to use ANOVA
Substantive hypothesis: A person’s degree of organizational commitment (Y) depends on the
team in which the person works (X)
Team in which someone works (X) -> Organizational commitment (Y)
- Question: if the hypothesis is correct, what would you expect to find with regard to
diGerences in average commitment between the teams?
- Imagine that we have collected data of measurements of organizational commitment for
3 teams
In scenario 1 there is more variance and more significant diGerence. In scenario 2 there is less
variance.
Key idea of ANOVA = when there are 2 or more groups, can we make a statement about possible
significant diGerences between the mean scores of the groups? ANOVA is like a t-test but for
many groups (so it saves you time to just first do the ANOVA and if you can’t find any diGerences
you can conduct a t-test)
Fundamental principle of ANOVA = ANOVA analyses the ratio of the two components of
variance in data: between-group variance and within-group variance
𝑖𝑛𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑛 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑜𝑓 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑐𝑜𝑟𝑒𝑠 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑔𝑟𝑜𝑢𝑝𝑠
𝑖𝑛𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑛 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑜𝑓 𝑠𝑐𝑜𝑟𝑒𝑠 𝑤𝑖𝑡ℎ𝑖𝑛 𝑔𝑟𝑜𝑢𝑝𝑠
ANOVA analyses ratio in which:
- Between-group variance measures systematic diGerences between groups and all other
variables that influence Y, either systematically or randomly (residual variance or error)
o DiGerences between diGerent groups, can depend on the group membership,
what is the eGect of being in the team on the average commitment, there is a
systematic diGerence between groups due to the independent variable
o Systematic group eGect + error
1
, - Within-group variance measures influences of all other variables that influence Y either
systematically or randomly (residual variance or error)
o Cannot depend on group membership because everybody is in the same group,
cannot be systematic due to the independent variable (cannot be because of the
group they are in, because they are in the same group)
o Error
ANOVA cannot say something about the causal relationships
Important to realize
1. Any diGerences within a group cannot be due to diGerences between the groups
because everyone in a particular group has the same group score; so, within-group
diGerences must be due to systematic unmeasured factors (e.g. individual diGerences)
or random measurement error
2. Any observed diGerences between groups are probably not only pure between-group
diGerences, but also diGerences due to systematic unmeasured factors or random
measurement error.
Statistical null hypothesis of One-Way Between-Subjects ANOVA
- Mean scores of k populations corresponding to the groups in the study are all equal to
each other:
H0: µ1 = µ2 = … = µk
- When do we reject H0? -> when at least one mean is significantly diGerent from the other
means.
Why prefer One-Way Between-Subjects ANOVA instead of separate t-tests for means?
In the example with 3 teams, we could also conduct 3 separate t-tests for means:
1. H0: µ1 = µ2 versus HA: µ1 ¹ µ2
2. H0: µ1 = µ3 versus HA: µ1 ¹ µ3
3. H0: µ2 = µ3 versus HA: µ2 ¹ µ3
- Problem of this approach: the larger the number of test that is applied to a dataset, the
larger the chance of rejecting the null hypothesis while it is correct (Type I error)
- Why -> follows from logic of hypothesis testing: we reject the null hypothesis if a result is
exceptional, but the more tests we conduct, the easier it is to find an exceptional result
- One will easier make the mistake of concluding that there is an eGect, while there is not
-> inflated risk of Type I error
o The more tests you do, the more chance of a Type I error
Formula for calculation of chance of 1 or more Type I errors in a series of C tests with
significance level a = 1 – (1 - a)c
Therefore, with 3 separate tests with a a = 0.05, the chance of unjustified rejection of the null
hypothesis is: 1 – (1 – 0.05)3 = 0.143
2
,Solution: One-Way ANOVA -> one single omnibus test for the null hypothesis that the means of K
populations are equal, with chance of Type I error = 0.05
Calculations: F-statistic
If we want to test the statistical null hypothesis: H0: µ1 = µ2 = … = µk
With an ANOVA, the F-distribution is used
- In order to determine if a specific sample result is exceptional (significant) under the
assumption that the statistical null hypothesis is correct, the test-statistic F has to be
calculated.
Calculations: deviations
- Strategy: partition of scores into components
o Component of score that is associated with ‘group’
o Component of score that is not associated with ‘group’
How can you do this? Calculate deviation scores
1. Deviation of individual score from grand mean
𝑌!" − 𝑀#
2. Deviation of individual score from group mean
(𝑌!" − 𝑀! ) = 𝜀!"
3. Deviation of group mean from grand mean
(𝑀! − 𝑀# ) = 𝛼!
𝛼! denotes the ‘eGect of group i’ (do not confuse with significance level!)
Full calculation: (𝑌!" − 𝑀# ) = (𝑌!" − 𝑀! ) + (𝑀! − 𝑀# )
Total deviation = within-group deviation + between-group deviation
Residual/error ‘eGect’ for group
𝜀!" 𝛼!
3
, Lecture 2 – 22-10-2024
Total Deviation
Within-Group Deviation
Between-Group Deviation
The problem is that the sum of the deviations is 0, which says nothing.
In the end you don’t get any further because everything you will calculate will be 0, therefore you
take the squares of the deviations to get test statistics.
Solution: take the Sum of Squares (SS)
𝑆𝑆$%$&' = 𝑆𝑆()$*))+ + 𝑆𝑆*!$,!+
.
𝑆𝑆()$*))+ = > 𝑛! (𝑀! − 𝑀# )-
!/0
. +!
𝑆𝑆*!$,!+ = > >(𝑌!" − 𝑀! )-
!/0 "/0
. +"
𝑆𝑆$%$&' = > >(𝑌!" − 𝑀1 )-
!/0 "/0
When taking the square, you are not working with the diGerence anymore but with the area
k = number of groups
i = individual score from members of group 1
4
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