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Summary Grade 9.6!! 2.5 Psychometrics: HOMEWORK solutions, notes and explanations FSWP2-052-A

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Course
Psychometrics: An Introduction (FSWP2052A)

Institution
Erasmus Universiteit Rotterdam (EUR)

Extensive document with answers to all homework exercises, including calculations and further explanations behind answers linked to reading material. Conclusions (what knowledge/fact to take from the exercise) written down after every exercise. Received grade 9.6 (average was 5.6)

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Uploaded on October 5, 2023
Number of pages 34
Written in 2022/2023
Type Summary

psychometrics
homework
fswp2 052 a
summary
notes
25

Institution
Erasmus Universiteit Rotterdam (EUR)
Education
Psychologie
Course
Psychometrics: An Introduction (FSWP2052A)

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homeworksh 3

3.
1
Central tendency and Variability
a) mean :

1 .1 :
*
=

7 , 21 . 3

:
k =

6
,75
E
ily
-

b) Variance
si
:
=

2)
10 18 ,
1 , 69 46 , 303 1 Standard deviation s
=

.
1 : =

1
=
1

G

1 10 , 20
, 699 , 99
=

, 304 deviation
+ Standard s
3

7
: =

.
=
1 1

G
d) why are the Variance + SD per definition non-negative ?

I both based on the sum of squared deviations (X-)" +
squares can't be negative ,
so sum of

Squares will always be a positive number +
SD is the square root of the Variance +
always a positive

number

Exercise 3
2
. distribution shapes and normal distributions

a) When a variable is
normally distributed the ,
mean is equal to the median

b) When a variable is distributed positively skewed the,
mean is larger than the median .

d When a variable is distributed negatively skewed the ,
mean is smaller than the median

Mean ,

Median Median +

Mode Mode Mode
Median

A
Mean

(+) 1-
Positive skew Normal distribution Negative Skew

Exercise 3.
3 covariance and correlation

a) covariance :

T)
)(y
+3
4
,

Ei(x - ,738
-

=

0

(xy
=

=

N

b) Correlation :

rxy
2xy =0 ,735 09
=

=

c) Sign it ort of the covariance tells us whether the relationship btw the Variables is positive or negative .

4 When positive scores on 1
1
. are accompanied by negative scores on exam 1
. 3, those devictions

will be negative as well .
When many deviation scores are
negative ,
the covariance is likely to be nega
tive .
Size of covariance is determined by the SD of the Variables , which are influenced by the scale on

which the variables are measured .
Size of the covariance doesn't say anything about the size of

the relationship bth the variables .

, a) Why are both the sigh and the size of the correlation informative ?

· sign tells us whether the relationship btw the variables is positive or negative .
The correlation is a

Standardized measure + bounded btw-1 and Can interpret sign + size
.

1

!

I . large
5

Correlation of small . medium and
3

1

: :
:
0
, 0

Exercise 3
.
4 Variance and covariance of composite scores

as calculate composite score Yi Xi +X; for each subject c) Yk Xk+X
1

.
=

=
:

,

Id :

.
1

) Yij =
G 4 .) Xij =
4 Id :

1 .1 Yki =
7

4 .)
Yk1
=

4

2 ) Vij 5) Yij 2) 9 5) 4
=

Yk1 Yk1
=
7 7

= =
.

3) 6. G 3) 3
5

Yij Vij Yk1 6) Yk1
=
=
7
=
=

b) Calculate the meantvariance of new Variable
Vi Using regular' formulas :

37
* *,
5

=
=

6
, 17
C) ,3
LT G
. 8334 25 , 33
gi
Z
.1395 , 867 4 , 225 . 054
6

5+
=

= =
= =
=
1 1 2

j
G G

e) covariance between Vij and Yki :

2 ,
67 =
,
444
0

G

#I
. f) Variance of composite scores
Vi
j

and Yik Using composite scores formula :

Scomp
=

Sp +
Sp+21 Si
1 , 67 , 14 + (2 , 33) 0 , 89 + , 33 + (2 50)
, 4 . 22
=

+
+ , 15
-
=

2 1 1 2 0

9) covariance :

Compacompz CikCijk Ci

=
0
, 17 + 0 ,33 + 0
,
44 + 1 -

0
, 17) =

0 43
,

4) Both the Variance and the covariance of a composite score can be calculated in 2 ways +traditional

formulas or those for composite scores , which you use depends on available information

· When scores on diff .
Variables given traditional
:

·

when COVGriance Variance matrix given :

composite formula

Exercise Binary
5
3
. :

items
EiX
=1
a) ·*
5

* ,
=
5

=

= ,
Px
=

0
N
0

2

5(1 5)
5

b) , , . 25
-
= =

0 0 0

5

s =

45 =

0
,

189
C k 726
,
=

=> 0

199 + 71

4) 50 ,726+ (1 -

0
, 726)
5 =

0
,
446

,Exercise 3
6
. 12-scores and converted standard scores)
a) Exam course 1
1
.

# ,5-7,
2
, 2302
7

1.
)
5 =

7 , + z-Score :
=
0 et ...

b) Correlation using -scores :

2 zxzy 2
. 6072 = .
434
Txy
=
0

=

N G

c) Convert exam course to a 1-100 scale with SD =

10 and mean I = 50

T =

/Snew) + Knew

0 , 2302 + 10 + 50 52 , 302
=

a) The Maximum possible score on exam 1
3
. is 10 . The z-score associated with a score of

10 is 2 49
. + T-score becomes 2
49 *10 +50 =74 .
. 9 (rounded to 75) .
Therefore ,
given

the Standard deviation and mean calculated on .
3

1,
it's not possible to get a T-score

Of above 75 .

The minimum possible score on exam 1 3
. is 1 +z-score of -4 .41 +T-Score becomes

-

4 4110
. + 50 =
5.
9 (rounded to 6)
.
Thus ,
given the SD + mean from 3
. 1,
it's not possible

to get a T-score below .
6

Percentile F(x)
Exercise 3
.
7

ranks and normalited scores
↑
a) percentile rank for score
3309
2

:

w
5 fx) 54) +100
px(Fx-0 600
. (6 -

,
=
0
=
1 , 294 flx)
N 305

55)
,

(11 -

0 +100 =

,75
2

Px
=

,
309

524)100
,

,
135 -

0 =

7 44
Px4
=

309

6) E =
100 and Sa =
15

Ilook up z-scores corresponding to the percentile ranks . Then , calculate T-scures with

formula T =

z(snew= 180) +
(Enew =

15)

, no m w or DC H4 :

Exercise 4 .1 :

NUMBER OF DIMENSIONS

a) 3 Ways to use the eigenvalues to assess #of dimensions :

1) Examine relative size of eigenvalues and find point at which the difference between values becomes

relatively small

2) Eigenvalue Greater than 1
0
. rule :
all dimensions that have eigenvalue greater than 1
.
8

3) Examine scree plot +
trying to find levelling-off point point of inflection) :

the number of dimension

We commonly consider is 1 less than the point where graph levels off

look for advantages and disadvantages in other notes

Exercise 4 . EIGENVALUES AND EXPLAINED VARIANCE
2

:

9) Calculate eigenvalue for the 6th factor :

·
the sum of all eigenvalues is always equal to the total number of variables (3)

Sum of the Other 7

factors is 7
.
4 1
consequently eigenvaire ,
of 6th factor is 1-7 .4 =

0
,
6

6) Calculate the percentage of Variance explained + cumulative percentage
:

Variance explained
,

& I factors in total :
Total Value for 6 = 2
2
. =
0
, 275 127 5%
of
7
Amount of factors S
# 108

1 X

I

6) which factor explains the most variance :

The first factor explains most of the Variance with 5
27 , % of the total variance of all s items

a) scatterplot :

·

Based on the screeplot we'd choose 2 factors ,
as the point of inflection is

Point of at the 3rd factor .

X inflection
There are 3 factors with eigenvalve greater than so this criterion would
·

1,

lead to 3 factors

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Summary Grade 9.6!! 2.5 Psychometrics: HOMEWORK solutions, notes and explanations FSWP2-052-A

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