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Summary Grade 9.6!! 2.5 Psychometrics: HOMEWORK solutions, notes and explanations FSWP2-052-A €9,49   In winkelwagen

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

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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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  • 5 oktober 2023
  • 34
  • 2022/2023
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christinauhlenbruck
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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