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Chi-Square

Chi-Square. CJ 526 Statistical Analysis in Criminal Justice. Parametric vs Nonparametric. Parametric DV: Interval/Ratio. Nonparametric. Nonparametric DV: Nominal/ordinal. Chi-Square Test for Goodness of Fit. One sample, DV is at Nominal/Ordinal Level of Measurement

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Chi-Square

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

  2. Parametric vs Nonparametric • Parametric • 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

  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

  18. Distribution of Intelligence in Dr. Zelda’s School

  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

  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) = 682.646, p < . 001.

  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

  32. 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

  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. 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

  37. 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

  38. 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.

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

  40. Assumptions • Independence of Observations

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