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 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-scorePercentage 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. Number of Samples: 12. Nature of Samples: N/A3. N/A4. IV: School enrolled in5. DV: IQ categories6. Target Population: all delinquents

enrolled in the state training school

7. Inferential Test: Chi-Square Test for Goodness of Fit

8. 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. 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

10. 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. X2 (5) = 1169, p < .00112. Reject H0

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

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. Number of Samples: 22. Nature of Samples: Independent3. N/A4. IV: Gender5. DV: Type of sentence received

1. Nominal

6. Target Population: all convicted burglars

7. Inferential Test: Chi-Square Test for Independence

8. H0: There is no relationship between gender and type of sentence received

9. 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 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. If the p-value of the obtained test statistic is less than .05, reject the null hypothesis

11. X2 (2) = 48.3, p < .00112. 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 Crosstabs

Move DV into Columns Move IV into Rows

Statistics 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