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## Chi-Square

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**Chi-Square**CJ 526 Statistical Analysis in Criminal Justice**Parametric vs Nonparametric**• Parametric DV: Interval/Ratio • Nonparametric DV: nominal/ordinal**Chi-Square Test for Goodness of Fit**• One sample, DV is at Nominal/Ordinal Level of Measurement • This test , the chi-square good of fit, determines whether the sample distribution fits some theoretical distribution**Null Hypothesis**• Population is evenly distributed the uniform distribution • Or • Some other distribution, such as the normal distribution • The sample distribution is not different from the theoretical distribution (such as the uniform distribution or the normal distribution)**Observed and expected frequency**• Observed: number of individuals from the sample who are classified in a particular category • Expected frequency: the frequency for a particular category that is predicted from the null hypothesis**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 the null hypothesis is retained, the sample distribution is like that of the theoretical distribution • If H0 is rejected, distribution is different from what is expected**Report Writing: Results Section**• Null hypothesis retained: The results of the chi-square goodness of fit test were not statistically significant • Null hypothesis rejected • The results of the Chi-Square Test for Goodness of Fit involving <DV> were statistically significant, 2 (df) = <value>, p < .05.**Report Writing: Discussion Section**• It appears as if the <sample> is (or is not) distributed as expected. • Depends on the result**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 (uniform distribution)**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**Number of Samples: 1**• DV: IQ categories • Target Population: delinquents enrolled in the state training school**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**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**X2 (5) = 1169, p < .001**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 (5) = 1169, 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 (independent of one another) • 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 (review 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**Background**• Number of samples: 1 IV: Gender DV: Type of sentence received • Nominal Target Population: convicted burglars**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**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**X2 (1) = 48.3, p < .001**Reject H0**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 (1) = 48.3, p < .001.**Discussion Section**• It appears as if gender has an effect on the type of sentence received.**Assumptions**• Independence of Observations