Summary Grade 9.6!! 2.5 Psychometrics: Worked Example DETAILED Notes: answers + explanations FSWP2-052-A

Wo r k ed ex a m p l e 1 :( H 3




I. Standardized scores:

The total number of correctanswers is transformed to a T-score which has mean 50 andSD 20.

Between which I-scores will approx. 95% the
of population scores lie?

In normal distribution, the lower bound is at -1.96 SDs /orappr.2 SDS) below the mean and the

upper bound is at1.96 SDS above the mean. Thus, the 95%CI lies between a T-score of

50 11.96 28) 50 11.96 * IM*SDnew
-
=
10.8
=

and
+




20) 89.2 (10
+ =
and 90) new




Describe the distribution ofscores:Descriptives -
Explore +
"correct scores" as dependent

skeweness and kurtosis values should be divided by their
·




SES, this value should be compared to -

2 and 2.


Ifthe value for skeweness is larger than 2:distribution

is
negatively skewed
↳ If
t he value for skeweness is larger than 2:distribution


is positively skewed
-



8,975 If kurtosis value greater than -


2:peak
o fdistribution is
1,569
+00 flate


greater than 2:Peakis too sharp
skeweness 1-8,975):slightly negatively skewed butnotsign.!

->
Kurtosis(1,69):peak is sharp butn ot significantly!
Kolmogorox-smirnox test:is significantK.001), indicating a deviation from a normal distr.

ibution! unsignificant:normal distribution
1K.S.- testis very conservative!


calculate
the z-scores and the T-scores:

The Z-score is a standard score with a mean of0 and SD of 1, which is calculated using Raw
scores, the mean of the raw scores and sp of the raw scores. The T-score is a converted

Standardized score, intended to have values thatppl find easier to understand.

The Escore is convertedinto a new standard score IT-score) by multiplying the E-score with

the SD ofthe new score (28) and adding the mean ofthe new score (50)
1) Analyze, Descriptive Statistics' Descriptives X Save standardized variables (E- scores)
2.) Transform, compute to variable T
calculate

T RND
=

120x2-20r 50) +




T- scores based on raw scores

2-scores wereonly standardized, notnormalized

, percentile
2.5,97.5

calculate
9 5% interval of the scores using percentile ranks

1.) Analyze' Descriptive Statistics Explore: T-variable in dependentl ist

statistics X percentiles, Paste

2)I n syntax, change: / Percentiles (5,10,25,50,75,90,95) HAVERAGE
intO

↑Percentile (2.5,97.5) HAVERAGE to gett h e lowest and highest2.5% percentiles
The answer differs from a) bc lower bound is 7 andupper is 92 (VS.10190)

because ata) we assumed a normal distribution, while we DON'T assume a normal distrib-

ution when we use percentiles.



percentile rankS
#


To make norm scores, one can use percentile ranks andp-values thatstem from the standard normal

distribution:

both percentile ranks and p-values indicate the yof ppl with an equal or lower score.

percentiles:calculated using all the raw scores wo
making assumption aboutdistribution ofthe scores


p-values:calculated using only mean and sp ofraw scores and
assuming a standard normal

IP-values:normality assumption
distribution

whether you use percentiles or p-values depends on whether you can assume a normal distribution in the


population or not


p-value preferred by it's less influenced by sample fluctuations

When no information known aboutpopulation and whether it's normally distributed, percentile ranks best to use

calculate p-value stemming from the standard normal distribution (z-scores):

1) Transform, compute to calculate percentile ranks using the standard normal distribution

CDF. NORMAL (Enr cor, 0,1
mean SD

CDF:cumulative distribution function;needed to calculate
the p-value for a certain z- score.

we know thata standardnormal distribution is a perfectly norm. dist. With mean 0 andSD1



calculate
percentives for the number corrects cores (raw scores
1) Transform, Rank cases Variables:h r cor ranktypes:X
fractional rankas


INTERPRETATION:

The percentile rank for a grade of 3.9 is 18.34 and p-value is 0,17



3
15.34% ofthe students hada grade of3.9 or lower.
don't differ a lot:so we can say the distribution
1
17%of the students had a grade of3.9 or lower.
t he
of
grades is fairly normal


use p-values when we can safely assume the distribution ofscores to be normal!

Ifnot:use smith that's not
assuming normal distribution (e..:percentile ranks)

, .Normalized scores
#




normalized T-score has
A to be caculated: 50
Meannew= sDrew=20
O RMALIZATION:
M


1) compute directpercentile ranks Rank Cases


2) convert the percentile ranks into standard scores (Nar_20r)
where the actual normalization oft he scores takes place be the standard scores are now no


longer based on the raw scores (like z-scores), buton percentile ranks!

3) compute convertedstandard score with mean 50 and SD 20 +



ur -cor + 20 50
+




Difference btw F-Scores from I . andIII.:

·The T-scores and F-norm scores differ on same Grade (e.9.:Grace 6:61/63). They differ because

the T-scores were notn ormalized, they were only standardizedIZ-Scores). The T-norm scores were

both standardized and normalized. raw scores, z-scores transformation


·F scores:transformations oft he z-scores (Standardized, notnormalized) - use raw scores


Why Iscores have same distribution ofthe raw scores (thus notnecessarily norm all
·

F-norm scores:transformation t he
of normalized scores. perientile+ z-scures transformation
-




·Standard scores calculated based on assump. of a normal distribution

the course coordinator shouldprefer the normalized scores to be sure thatthe assumption of a


normal distribution ofthe scores is met.