Transcript Chi-Square - Southeast Missouri State University

Chi-Square
CJ 526 Statistical Analysis in
Criminal Justice
Parametric vs Nonparametric

Parametric
1.
DV: Interval/Ratio
Nonparametric

Nonparametric
1.
DV: Nominal/ordinal
Chi-Square Test for Goodness
of Fit


One sample, DV is at Nominal/Ordinal
Level of Measurement
Determines whether the sample
distribution fits some theoretical
distribution
Null Hypothesis
1.
Population is evenly distributed


Or
Some other distribution, such as the normal
distribution
Observed Frequency

Number of individuals from the sample
who are classified in a particular
category
Expected Frequency

The frequency value for a particular
category that is predicted from the null
hypothesis and the sample size
Chi-Square Statistic


Sum of
(Observed - Expected)2


divided by
Expected
Degrees of Freedom



df = C - 1
where C is the number of categories
The degrees of freedom are the number
of categories that are free to vary
Interpretation

If H0 is rejected, distribution is different
from what is expected
Report Writing: Results
Section

The results of the Chi-Square Test for
Goodness of Fit involving <IV> were
(not) statistically significant, 2 (df) =
<value>, p < .05.
Report Writing: Discussion
Section

It appears as if the <sample> is <not>
distributed as expected.
Example


Concerned about health, neither
concerned or not concerned, not
concerned about health
Could assume that a sample would be
equally split among these three
categories i.e., 120 subjects, 40 would
say concerned, 40 neither, 40 not
concerned
Example
O
E
O-E
(O-E)^2 /E
60
40
20
400
10
40
40
0
0
0
20
40
20
400
10
Chi square
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Chi square = 20
D.f. = 2
See p. 726
Chi square = 20, p < .01
The distribution is significantly different
from the expected distribution
Example

Dr. Zelda, a correctional psychologist, is
interested in determining whether the
intelligence of delinquents enrolled in a
state training school is normally
distributed
Distribution of Intelligence in the
General Population
IQ Range
Z-score
Percentage of
General
Population
Below 60
-3
.0228 (23)
60-85
-2
.1359 (136)
86-100
-1
.3413 (341)
101-115
+1
.3413 (341)
116-130
+2
.1359 (136)
131+
+3
.0228 (23)
Distribution of Intelligence in
Dr. Zelda’s School
Below 60
119
60-85
150
86-100
687
101-115
32
116-130
12
131+
0
1.
2.
3.
4.
5.
6.
Number of Samples: 1
Nature of Samples: N/A
N/A
IV: School enrolled in
DV: IQ categories
Target Population: all delinquents
enrolled in the state training school
7.
8.
Inferential Test: Chi-Square Test for
Goodness of Fit
H0: The distribution of frequencies of
the IQ categories for the sample will
not be different from the population
distribution of frequencies of the IQ
categories
9.
10.
H1: The distribution of frequencies of
the IQ categories for the sample will
be different from the population
distribution of frequencies of the IQ
categories
If the p-value of the obtained test
statistic is less than .05, reject the null
hypothesis
Calculations
O
E
O-E
(O-E)^2 /E
119
23
96
9216
401
150
136
14
196
1
687
341
346
119716
351
32
341
309
95481
280
12
136
124
15376
113
0
23
23
529
23
11.
12.
X2 (5) = 1169, p < .001
Reject H0
SPSS: Chi-Square Goodness
of Fit Test

Weight Cases
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Data, Weight Cases
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
Check Weight Cases by
Move weighted variable over to Frequency Variable
Analysis

Analyze, Nonparametric Statistics, Chi-Square
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Move DV to Test Variable List
Enter Expected Values
Results Section

The results of the Chi-Square Test for
Goodness of Fit involving the
distribution of IQ categories for the
state training school were statistically
significant, X2 (6) = 682.646, p < . 001.
Discussion Section

It appears as if the distribution of
frequencies of the IQ categories for
students enrolled in the state training
school is different from the population
distribution of frequencies of the IQ
categories.
Chi-Square Test for
Independence

Used to assess the relationship between
two or more variables
Null Hypothesis

No relationship between the two
variables
Or
 Alternative: the two variables are related
to one another

Degrees of Freedom


df = (R - 1)(C - 1),
Where R is the number of rows and C is
the number of columns in a bivariate
table
Example

Dr. Cyrus, a forensic psychologist, is
interested in determining whether
gender has an effect on the type of
sentence that convicted burglars receive
Dr. Cyrus’ Results
Probation
Jail
Prison
Females
37
42
14
Males
23
16
58
1.
2.
3.
4.
5.
Number of Samples: 2
Nature of Samples: Independent
N/A
IV: Gender
DV: Type of sentence received
1.
6.
Nominal
Target Population: all convicted
burglars
7.
8.
9.
Inferential Test: Chi-Square Test for
Independence
H0: There is no relationship between
gender and type of sentence received
H1: There is a relationship between
gender and type of sentence received
Create a bivariate table
probation
jail
total
male
14
80
94
female
46
20
66
60
100
160
Calculate expected values


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For each cell, row total times column
total, divided by the total number of
subject
i.e., for the first cell, (94 x 60)/160 = 35
(66x60)/160 = 25, (94x100)/160 = 59,
(66x100)/160 = 41
O
E
(O-E)
(O-E)^2 /E
14
35
21
441
12.6
80
59
21
441
7.5
46
25
21
441
17.6
20
41
21
441
10.6
10.
11.
12.
If the p-value of the obtained test
statistic is less than .05, reject the null
hypothesis
X2 (2) = 48.3, p < .001
Reject H0
Probation
Jail
Total
Male
14 (35)
80 (59)
94
Female
46 (25)
20 (41)
66
60
100
160
SPSS: Chi-Square Test of
Independence

Analyze

Descriptive Statistics
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Crosstabs
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
Statistics
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
Move DV into Columns
Move IV into Rows
Chi-Square
Cells

Percentage
 Rows
 Columns
Results Section

The results of the Chi-Square Test for
Independence involving gender as the
independent variable and type of
sentence received as the dependent
variable were statistically significant, X2
(2) = 41.745, p < .001.
Discussion Section

It appears as if gender has an effect on
the type of sentence received.
Assumptions

Independence of Observations