Nonparametric Tests in Psychology

orparamaticstatisticatesa en
metic

ANOVA(-2)




time.A
-
Independent. V is categorical ->
qualitative

-

Dependent. V is continuous numerical - quantitative

-comparing sample means not a standardizeal population mean
N, N2
(parameters established) 1. V
2
Chi-square notation) 1. V
AWS CH

↑x
is




Chi-Square(x2-conditions for use (non-parametric Chi-square testing:Statistical
↓
significance
*

causation (only experimentally defined)
-Data measured nominally or
ordinally (qualitative measurements (



-
Chi-Square:Both variables are
categorical, not continous,
meaning no inherent
meaning
->
Qualitative


nonparametric fundamentally between observed (0) frequency (t) (expectedmatches observed),
2
tests for differences andexpected When E X




~
-
0
= 0
=




meaning to relationship exists


Raw score data the form frequencies, quantitative value depends

Do
in scores match w/statistical freedom
-
of not
thy of on degrees as
significance? ·Value of x depends on the size of discrepancy
between E/0
↳ (I) "Goodness of Fit" observed frequencies (0) frequencies
comparedto
predictedby a pre-establishedtheory/experimental prediction (E)

↳ (2) "Independence testing"testing the independence c al
of variables
to see if any relationship exists (= correlation w/ categorical variables)


-
Employ Chi-square analysis when each observation is independent
of other observation -> p aired
not (logically linked)

-
Samples are representative respective
of population (not comparisons) =
parametric)


Chi-Squares Theoretical Distribution -



measuring impact
on of ·
Shape of theoretical distribution depends on df'


df
* 19
=



· As of increases, a larger X value is neededto




*
successfully rejectthe
null hypothes is (Hol
Cast:how parametric?)
df 10



at
is this diferentfrom
B =




s
=




x2 x
=
x (E)2]dualsummation
↳ Beyond"x2"value is rejection region... x
2
av




Chi-Square Formulas, steps, and
computation -


Example Problem

A psychologist studying art appreciation selected an abstract that had no obvious top or bottom and tested to see if people had a preference
for how to hand the picture. Hangers were placed on each edge of the painting and the picture was shown to a sample of 50 participants.
Each was asked to hang the painting in whatever orientation looked best to them. The data is as follows:


leftside up Rightside I
It hen
e Data
-> is presentedin frequency counts


-> Sampla data is being compared to

an expectation - Goodness of fit



40. P1=P2 P3 P,... Pn H1:4, P2E43... n
=

=




Step 4 -



Statement
of Null Hypothesis H.:E O



NH(Ho):Observed data fits expecteddata (t 0) preference - Alternative Hypothesis (H.) States thatE F0,


meaning
the observed
no
meaning
=

-




inpopulation (equal
opportunity)
-
Wall Hypothesis proposes thatE 0, meaning that X
=
data (o)
not fitthe expectation, or, experimental predictions (E)
↳ Bleast misfit
one proportion is a




RejectHo
Fail
to (Retain Nul

Step 2 -




Sample size. ExpectedProportions, af, CV, alpha a (pre-established;given) 0.0 5



~
=




Be
-
N 50 (individual participants)
=
-
critical"value is found using theoretical
distribution table (given)


proportion expectedfor category 0.25 Notation: RejectHo
=




each
=



- -




"
adf
df # of categories (1 1 4 1 3 x (3)0.0s 7.815-c
= =
-
- =
=




i
-




chi-square

, Step 3 -




using formula
compute CALL E PN
=




Type of Central Tendency
=(0.25)(50); x =




(EE02]
defaceasshapeof date
-




Orientation Observed (o) Expected (E)


1. Top Up 18 12.5-
cate"arereasonina calculation
De




I sumgebutInee
2. Bottom Up 17 12.5

7.(8)2 2.h 3.
(57)2
2.42
-

=


=




3. LeftUp 7 12.5-



4. RightUp 8
12.5-2.(7)* 1.62 4.
(8)
=1.62
=




(obtained)

5 50
=

2 30
=

2 8.08
=




Step 4 Making Predictions, NHST
people weremorelikesto hamImage Feee
--
-




Recall:CV-> 7.815 < 8.08.. picture orientations were notall equally likely to be prefered, X 2(3) 8.08,4
=




I f testvalue (8.08) exceeds
=>

critical value (7.815), REJECT NULL claim statistical significance
=




↳ When you succesfully rejectmull hypothesis (Ho), meaning topt >
(V;p <0.05, thatm eans thato bserved frequencies

differ from expected frequencies;the endeavour to
researcher must accountfor discrepancies




Example-Khan Academy 100
-n
=




Over the years, MC options for a try question of a standardized test has had a equal distribution in four answers (A,B,C, or D). This is a
normal distribution so there is equal percent chance that each answer will be correct. A researcher of pedagogy wants to stistcically test
this using Chi Square testing. How would they proceed?

Step 1 -


Null Hypothesis Statement :8
Ho: There is a equal distribution ofH.:There is equal distribution, expectedfrequencies
thus the
correct choices, meaning the do not the observed.
match
expected matches observed.

↳ means:23%
of A, 23% of, 28% of C, 25% of D




I
4 25% (0.2st
Step 2 Sample Size, Expected (4), ·P 100% options 100 =

Proportions -
-




df, CV,
=
=
=




Given to
you df 2 =



1 4 1 3
(V 7.815
- =
- =
·


correct -
=




Observed
Choic
Exede (E)
(0) ·x 0.05
Frequencies
=




-




A 20 25 E P(N) 0.25(100)
=
=
23
=




Step 4 -

Conclusion statement
-




-
B 20 25
x2(3)0.0s 7.815
=



-> tobt > cr:RejectHo, tob+ <CV:Retain Ho

↓
C 25 25 Chi Greek Letter(!) ↳ tob+(6) < (r (7.815) =
Retain Ho


#$
25 conclusion:the (4) MCQ
rejectH)
(fail to
answers are equally

5 95
=
2 100
=




x=I [ EY likely be
to
chosen,

XF.os
~Recall: #
Adf 3
Step 3 Perform calculation for each
category
=




(43)2=1 (
I
7. Coption 1) -


3. Coption -
0
=




-
beyond
2 6 107
=




(Samething) 1 4.(option() Mee
-(32
2. Coption B) - =
=

4 x


↓!csisd
↳ Probability
of getting result - 6 IS

10%(from table)
2
x