1. CJ 526 Statistical Analysis in Criminal Justice
Chi-Square
CJ 526 Statistical Analysis in Criminal Justice

2. Parametric vs Nonparametric
DV: Interval/Ratio

3. Nonparametric
Nonparametric
DV: Nominal/ordinal

4. 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

5. Null Hypothesis Population is evenly distributed Or
Some other distribution, such as the normal distribution

6. Observed Frequency
Number of individuals from the sample who are classified in a particular category

7. Expected Frequency
The frequency value for a particular category that is predicted from the null hypothesis and the sample size

8. Chi-Square Statistic
Sum of
(Observed - Expected)2
divided by
Expected

9. 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

10. Interpretation
If H0 is rejected, distribution is different from what is expected

11. 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.

12. Report Writing: Discussion Section
It appears as if the <sample> is <not> distributed as expected.

13. 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

14. Example O E
O-E
(O-E)^2 /E 60 40 20
400 10

15. Chi square Chi square = 20 D.f. = 2 See p. 726
Chi square = 20, p < .01
The distribution is significantly different from the expected distribution

16. Example
Dr. Zelda, a correctional psychologist, is interested in determining whether the intelligence of delinquents enrolled in a state training school is normally distributed

17. 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) +1 +2
131+ +3

18. Distribution of Intelligence in Dr. Zelda’s School
Below 60
119
60-85
150
86-100
687 32 12
131+

19. 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

20. 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

21. 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

22. 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
309
95481
280 12
124
15376
113
529

23. X2 (5) = 1169, p < .001
Reject H0

24. SPSS: Chi-Square Goodness of Fit Test
Weight Cases
Data, Weight Cases
Check Weight Cases by
Move weighted variable over to Frequency Variable
Analysis
Analyze, Nonparametric Statistics, Chi-Square
Move DV to Test Variable List
Enter Expected Values

25. 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) = , p <

26. 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.

27. Chi-Square Test for Independence
Used to assess the relationship between two or more variables

28. Null Hypothesis No relationship between the two variables Or
Alternative: the two variables are related to one another

29. 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

30. Example
Dr. Cyrus, a forensic psychologist, is interested in determining whether gender has an effect on the type of sentence that convicted burglars receive

31. Dr. Cyrus’ Results Probation Jail Prison Females 37 42 14 Males 23 1658

32. Nature of Samples: Independent N/A IV: Gender
Number of Samples: 2
Nature of Samples: Independent
N/A
IV: Gender
DV: Type of sentence received
Nominal
Target Population: all convicted burglars

33. 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

34. Create a bivariate table
probation
jail
total
male 14 80 94
female 46 20 66 60
100
160

35. Calculate expected values
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

36. O E
(O-E)
(O-E)^2 /E 14 35 21
441
12.6 80 59
7.5 46 25
17.6 20 41
10.6

37. If the p-value of the obtained test statistic is less than
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

38. Probation
Jail
Total
Male
14 (35)
80 (59) 94
Female
46 (25)
20 (41) 66 60
100
160

39. SPSS: Chi-Square Test of Independence
Analyze
Descriptive Statistics
Crosstabs
Move DV into Columns
Move IV into Rows
Statistics
Chi-Square
Cells
Percentage
Rows
Columns

40. 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) = , p < .001.

41. Discussion Section
It appears as if gender has an effect on the type of sentence received.

42. Assumptions
Independence of Observations