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Samenvatting 2.5 Psychometrics Worked Examples & Uitwerkingen (FSWP2-052-A)
Dit document bevat de worked examples en de uitwerkingen hiervan voor de course 2.5 Psychometrics () van de bachelor Psychologie aan de EUR.
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12.5 PSYCHOMETRICS
Erasmus Universiteit Rotterdam, bachelor Psychologie
WORKED EXAMPLES
&
UITWERKINGEN
, 2
Paars: stappen te nemen in SPSS.
Groen: antwoorden.
Worked Example Chapter 3
Student’s Time Allocation
The exam of course 1_1 consisted of 60 multiple choice questions about social
psychology. Students needed 34 correct answers to get a grade 4.0 and they needed 42
correct answers for a grade 6.0, in which no compensation of other exams is required. In
the file exam1_1.sav the number correct scores and grades of 398 first year students are
given.
I. Standardized Scores
a. The total number of correct answers are transformed to T-scores which have
mean 50 and standard deviation 20. Between which T-scores does approximately
95% of the population scores lie?
ANTWOORD:
This is a theoretical question meant to make you think about the meaning of a 95%
interval. You should think of the answer without using SPSS, the only information
we have is the mean and the SD. If necessary, draw a normal distribution with M=50
and SD=20 and the upper and lower bounds for 95%, and think about what you
know about the distance from the mean to the upper and lower bounds in terms of
standard deviations.
We know that in a normal distribution (which is what we assume for now and will
come back later to at d.) the lower bound is at -1.96 SDs (or approximately 2 SDs)
below the mean and the upper bound is at 1.96 SDs (or approximately 2 SDs) above
the mean. The 95% interval lies therefore between a T-score of 50-(1.96*20)=10.8
and 50+(1.96*20)=89.2 (or approximately 10 and 90).
b. Describe the distribution of the number correct scores (Skewness, Kurtosis,
Kolmogorov-Smirnov, histogram, normal Q-Q plot). What is the mean, the
median and the standard deviation?
• Use Analyze->Descriptive Statistics->Explore and place
nr_cor under Dependent List.
• Click on Plots and select Histogram under Descriptive.
Also, select Normality plots with tests.
, 3
ANTWOORD:
You can start by looking at the histogram and describe what they see there.
The Skewness and Kurtosis values should be divided by their SEs, this value
should be compared to -2 and 2 (comparable to z-scores).
If the value for Skewness is larger than -2, this means that the distribution is
negatively skewed and if the value is greater than 2, this means that the
distribution is positively skewed.
If the Kurtosis value is greater than -2, this means that the peak of the distribution
is too flat and if this value is greater than 2, this means that the peak is too sharp.
In the output, we see that the Skewness is -0.119/0.122= -0.975, meaning that
the distribution is slightly negatively skewed (skewed to the left), but not
significantly.
The Kurtosis is 0.456/0.244= 1.869, meaning that the peak is sharp (but not
quite significantly).
The histogram looks approximately normal but the normal Q-Q plot shows some
deviations from normality for the low, middle and high scores (where the dots
deviate from the line).
The Kolmogorov-Smirnov test is significant, indicating a deviation from a
normal distribution.
As you can see, the different methods differ in the extent of detailed information
about the distribution of the scores (the K-S test is very conservative).
The mean is 38.68, the median is 40 and the standard deviation is 5.85.
c. Calculate the Z-scores and the T-scores using SPSS. Make sure the T-scores are
rounded.
F&B, p. 61: The Z-score is a standard score with a mean of 0 and a SD of 1, which is
calculated using the raw scores, the mean of the raw scores and the SD of the raw
scores. The T-score is a converted standard score, in order to obtain values that
people find easier to understand. The Z-score is converted into a new standard score
(T-score) by multiplying the Z-score with the SD of the new score (20) and adding
the mean of the new score (50). Therefore, in SPSS we take the following steps:
• Use Analyze->Descriptive Statistics->Descriptives and
select save standardized variables to calculate Z-scores.
• Use Transform -> Compute to calculate variable T. Use the
function RND to round the variable: T = RND(20 ×
Znr_cor + 50).
ANTWOORD:
, 4
You obtain the Z-scores (Znr_cor) and their converted standard scores (T) in the
dataset.
d. Which T-score corresponds to a grade of 4.0? Which T-score corresponds to a
grade of 6.0?
ANTWOORD:
If necessary, sort the data by ascending grade (right-click the column “grades” and
click “sort ascending”). In our data set there is no one with a grade of 4.0 (and no
one with a grade of 4.1 either), so what should we do now?
We look at the grades of 3.9 and 4.2. These grades correspond to a T-score of 31 and
34. We can see that a difference in 0.3 grade points corresponds to a difference of 3
T-score points. From this we can infer that a T-score of 32 corresponds to a grade of
4.0.
We can see in the data set that a T-score of 61 corresponds to a grade of 6.0.
e. Calculate the 95% interval of the scores using percentile ranks in SPSS. Explain
why this interval differs from your answer at a?
• Use Analyze>-Descriptive Statistics->Explore, Put T in the dependent
list. Then: Statistics->Percentiles->Paste
• In the syntax, change: /PERCENTILES(5,10,25,50,75,90,95)
HAVERAGE into /PERCENTILES(2.5, 97.5) HAVERAGE to get the
lowest and highest 2.5% percentiles.
ANTWOORD:
The output shows that the lower bound is 7 and the upper bound is 92.
This answer differs from the one at a., because at a. we assumed a normal
distribution of the scores, while we do not assume a normal distribution when we
use percentiles.
Extra question: Does the distribution of the T-scores approximate a normal
distribution? When we compare the interval obtained with the assumption of a
normal distribution with the interval obtained with percentiles; 10.8 – 89.92 vs. 7-
92, we can see that they differ, but this difference is not that large (taking into
consideration that the SD is 20). From this we can infer that the distribution of the
T-scores is fairly normal.
II. Percentile Ranks
In order to make norm scores, a test constructor can use percentile ranks and p-values
that stem from the standard normal distribution.
a. What is the difference between percentile ranks and p-values stemming from the
standard normal distribution?
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