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Ch 13: Chi-square tests Note – only read p. 542-553 (stop at “Chi-Square test of independence” section) Dec. 3, 2013
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Chi-square tests Use with categorical, nominal data –T-tests, ANOVA, corr all require quantitative data Chi-square focuses on frequency data –Also called ‘non-parametric’ tests –Observed frequencies (from data) –Expected frequencies (based on theory or chance…) –We compare how well observed frequencies fit an expected frequency breakdown Called ‘goodness of fit’ test –Another use of chi-square is the ‘test of independence’ Determine if 2 categorical variables are correlated (this won’t be on the exam – but read p. 553 if interested)
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Chi-sq Hypotheses Null hyp states that the distribution of observed & expected are the same (no difference) Research hyp states that the 2 distributions are different. –What does it mean to reject the null here?
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Chi-square test for single nominal variable –3 attachment styles (secure, anxious, avoidant) –Get sample data w/observed frequencies for these 3 categories Might be interested in whether your sample has more anxious people than would be expected by chance Example: If equal distribution of sample into the 3 categories, you’d expect 1/3 rd secure, 1/3 rd anxious, 1/3 rd avoid –What numbers would be expected in each category?
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Comparing observed (O) & expected (E) frequencies: –Find the amount of ‘mismatch’ between the two …and determine whether it’s more than expected by chance –‘mismatch’ involves difference betw O & E –“O” is found based on your data (observed) –“E” is found for each category by using either: Theory (what # is expected in each category?) or.. Chance (if expect an equal distribution across categories, divide sample size by # categories)
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Χ 2 = (O – E) 2 E Σ Example? Is this ‘mismatch’ larger than expected by chance? Figure chi-sq: Find O-E for each category, square it, divide by E for that category, then add across all categories: O = observed frequency for a category E = expected frequency for a category
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Chi-sq distribution Need a comparison distribution to find critical value and compare our observed X 2. Chi-sq distribution depends on degrees of freedom (based on # categories we have) –df = N categories – 1 –Use Appendix A-4 to find critical value based on df and your chosen alpha level –For.05 alpha, 2 df, X 2 critical = ? –Can draw comparison distribution w/rejection region and if our sample X 2 is > critical X 2 reject null.
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APA format What do we conclude for this example? APA format sentence to report results:
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Getting “E” from theory… Instead of expecting equal frequencies in each category, you might refer to theory to find different E’s for each category: –Theory that a certain mineral affects mental health. In region w/mineral, 1000 people surveyed: 134 had anxiety disorder, 160 drug/alc abuse, 97 mood disorder, 12 schizoid, 597 no disorders. In general pop, 14.6% anxiety disorder, 16.4% drug/alc, 8.3% mood disorder, 1.5% schiz, 59.2% no disorder
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Null hyp: the distribution of people in the 5 mental health categories is the same in our sample as the general pop. Research hyp: the sample distribution frequencies differ from the general pop.
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So, based on % in general pop, in our sample of 1000, we’d expect: Anxiety disorder: ? people Drug/alc abuse: ? people Mood disorder: ? people Schiz: ? people No disorder: ? People Find X 2 Find critical value Reject null?
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