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Summary Experimental Research Methods: extensive summary with images & examples!

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A very extensive summary of the course Experimental Research Methods. All of the information & visualizations are representations of and based on all of the study materials of the course ERM of Tilburg University. All of the tables/text with a black striped border are examples of questions/calcu...

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  • June 5, 2023
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  • 2021/2022
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Lecture 1 – Summary material previous courses and its relation to ERM
Descriptive statistics – summarize data. We use descriptive statistics because raw data can be very
confusing. There are two ways to summarize data;

1. With a distribution
2. With sample statistics

Data – numerical information about a population or sample.

- Population – all members of a defined group. Parameters are measures of characteristics of
the scores in a population and are indicated by Greek letters (𝜇, 𝜎, etc)
- Sample – subset of members of the population, because most of the time we are unable to
study the entire target group. Sample quantities are measures of characteristics in a sample
and are indicated by Latin letters (s, X, etc)

Distribution – summarizes data by grouping data with the same Creating a histogram and frequency
score. This can be done by using a frequency distribution or distribution in SPSS:
histogram. A histogram is less precise than a table. FREQUENCIES
VARIABLES=x
/HISTOGRAM
/ORDER=ANALYSIS
Sample statistics/quantities – data summarized using
characteristic features of the distribution of the data. Characteristic properties are;

1. Central tendency – the most characteristic score of a distribution. These measures are the
mean, mode and median.
- Mean – sum of all scores divided by the number of scores;
∑𝑁𝑖=1 𝑋𝑖
𝑋̅ =
𝑁
- Median – the middle number when the data is written down in a sequence from low
to high
- Mode – the number which occurs the most often in the data
2. Dispersion/spread – deviation of the scores from the characteristic score. Measures of
dispersion are;
- Range
- Variance – the sum of all squared deviation scores divided by the number of scores
minus one
2
𝑆𝑆 ∑𝑁 ̅ 2
𝑖=1(𝑋𝑖 − 𝑋 )
𝑠 = =
𝑁−1 𝑁−1
- Standard deviation – the square root of the variance

𝑠 = √𝑠 2
Inferential statistics – by using inferential statistics we can draw conclusions about a population
based on a sample. When we have data of the entire population, we do not need inferential
statistics. There are 3 different procedures in inferential statistics;

1. Hypothesis testing
2. Point estimation
3. Interval estimation




1

,There are several reasons why we mostly use data of a sample and not the population because it
either is too expensive, it takes way too much time to collect the data or because it simply is
impossible.

Hypothesis testing - examines whether the mean of the population is equal to a certain value or not.
Hypothesis are exclusive and exhaustive. Rules of thumb for creating a hypothesis are;

- 𝐻0 (=null hypothesis) contains “=”
- 𝐻1 (=alternative hypothesis) contains (mostly) expectations of the 𝐻0 : 𝜇 = 2.5
𝐻1 : 𝜇 ≠ 2.5
researcher

In the example on the right we discuss a two-sided test, because 𝐻1 contains ≠. In a one-sided test,
𝐻1 contains < or>.

The steps we take in hypothesis testing are;

1. Formulating the hypothesis 𝐻0 and 𝐻1
In this course we use an
2. Determining the decision rule to decide when a result is statistically
alpha level of 𝛼 = 0.05
significant; 𝑝 ≤ 𝛼.
3. Determine the p-value based on the SPSS output
4. Decision on significance and the conclusion T-test for the previous example:
T-TEST
/TESTVAL=2.5
/MISSING=ANALYSIS
Logic hypothesis testing /VARIABLES=x
/CRITERIA=CIN (.95)
- You make an assumption about the value of a parameter,
for example of the mean (=𝜇).
- Given that this value is true, you determine the possible values in the sample statistic (e.g. 𝑋̅)
can take in a simple random sample of 𝑁 cases
𝜎2
- The mean of the sample distribution is 𝜇, the variance is 𝑁
- Using that sampling distribution, you determine the probability (= the 𝑝-value), that the
value of 𝑋̅ or a more extreme value occurs
- In step 3 you determine the position of 𝑋̅ in the sampling distribution, so you also implicitly
determine the 𝑝 -value.
- If the 𝑝-value is lower than 𝛼, then you can conclude:

“If my 𝐻0 is true, then the probability that I observe this value for 𝑋̅ or an even more extreme value is
smaller than 𝛼. This probability is so small, that I do not trust my null hypothesis anymore. I reject
𝐻0 .”

- If the 𝑝-value is larger than 𝛼, then you can conclude:

“If my 𝐻0 is true, then the probability that I observe this value for 𝑋̅ or an even more extreme value is
quite large. I do not have enough reasons to doubt the correctness of 𝐻0 . Thus, I do not reject 𝐻0 .”

Note: one of the assumptions for hypothesis testing is that the sample is a simple random sample,
meaning that;

- All cases have an equal chance to be sampled
- Cases are selected independently of one another



2

,One-sided vs. two-sided testing

The logic for a one-sided and two-sided test is the same, but
SPSS output is always two-sided. Thus we need to convert the
two-sided Sig. in SPSS output to the correct one-sided p-
value. We use the schema in figure 1 to determine how we
should interprate the Sig.

Point estimation – used to answer the following question:
“What is the best guess of this parameter?”, thus in other
words: which values lie closest to the population value? In
case of the mean (𝜇), the best guess is 𝑋̅. In the case of the

variance (𝜎 2 ), the best guess is 𝑠 2 . Figure 1 Decisions tree hypothesis testing

Interval estimation – with confidence intervals (CI), we can
answer the following question: “What is the interval in which the value of the parameter lies with …%
confidence?” With a 95% confidence interval for 𝜇 we can state: “In 95% of the times I draw a sample
of N=50, the confidence interval will contain 𝜇”. The formula for the confidence interval is:
𝑠
𝑋̅ ± 𝑡𝑐𝑣 ×
√𝑁
We can use confidence intervals to test two-sided hypotheses;

- If 𝜇𝐻0 falls in the 𝐶𝐼(1−𝛼)×100% , you cannot reject 𝐻0 in favour of a two-sided alternative
- 𝜇𝐻0 does not fall in the 𝐶𝐼(1−𝛼)×100%, you can reject 𝐻0 in favour of a two-sided alternative

Assuming that 𝐻0 is true…

… 95% of all possible samples will produce a 𝐶𝐼95 in which 𝜇𝐻0 falls, thus we can correctly
reject 𝐻0

… 5% of all possible samples will produce a 𝐶𝐼95 in which 𝜇𝐻0 does not fall, thus in which we
incorrectly reject 𝐻0 (=Type I error)

An alternative interpretation of the CI in relation with hypothesis testing:

“The 𝐶𝐼95 gives all hypothetical possible values for 𝜇 that are not rejected by the sample statistics
(given 𝛼)”

Testing means

Five different tests of means have been discussed in the previous statistics courses:

One population:

1. 𝐻0 : 𝜇 = 𝜇0 , 𝜎 known (z-test)
2. 𝐻0 : 𝜇 = 𝜇0 , 𝜎 unknown (t-test)

Two populations:

3. 𝐻0 : 𝜇1 = 𝜇2 , 𝜎1 = 𝜎2 and unknown, independent samples (t-test)
4. 𝐻0 : 𝜇1 = 𝜇2 , 𝜎1 ≠ 𝜎2 and unknown, independent samples (t-test)
5. 𝐻0 : 𝛿 = 𝜇1 − 𝜇2 = 0, 𝜎𝐷 unknown, dependent samples (t-test)


3

, All of these tests above are similar, as the following formula always holds:
𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 − 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟
𝑡𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 =
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟


RQ: on average, do male and female students differ in their self-esteem?
1. Formulating hypotheses;
𝐻0 : 𝜇𝑀 = 𝜇𝐹
𝐻1 : 𝜇𝑀 ≠ 𝜇𝐹
2. Determining significance level;
𝑝 ≤ 𝛼 = 0.05
3. Determining p-value based on SPSS output;




Also always look at the Levene’s test! P-value is .062, which is larger than our alpha level of 0.05.
Thus we assume equal variances
4. Decision on significance and conclusion:
𝑝 = 0.105, which is larger than 𝛼 = .05, thus 𝐻0 cannot be rejected. Our conclusion is:
“Average self-esteem does not differ between male and female students”

In the course Experimental Research Methods tests 3 till 5 are of main importance.

Levene’s test – used to test equality of variances. Answers the question:

“Are the population variances of dominance equal for population A and population B?”

We test the following hypotheses for the Levene’s test;

𝐻0 : 𝜎𝐴2 = 𝜎𝐵2

𝐻1 : 𝜎𝐴2 ≠ 𝜎𝐵2
When we interpret the p-value, we can reject the null hypothesis, thus that there are no equal
variances, when the p-value is smaller than 𝛼 = .05.




In the SPSS output we can see a p-value of .468, which is larger than our alpha level; 𝑝 > 𝛼. Thus we
cannot reject 𝐻0 , which means that we assume equal variances. If the p-value would be smaller than
𝛼, we can reject 𝐻0 , which means that we cannot assume equal variances.



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