This package includes notes and practice question(s) with full calculated answers regarding nonparametric statistical tests relevant to the psychological and biological sciences. In particular, Chi-Square (goodness-of-fit and independence test), Mann-Whitney-U, and Wilcoxon-Signed-Rank-Test is cove...
-
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'
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)
"
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.
↓
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
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