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Chapter 16: Chi Square PSY295-001—Spring 2003 Summerfelt.

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Presentation on theme: "Chapter 16: Chi Square PSY295-001—Spring 2003 Summerfelt."— Presentation transcript:

1 Chapter 16: Chi Square PSY295-001—Spring 2003 Summerfelt

2 Chapter 16 Chi-Square 2 Overview z, t, ANOVA, regression, & correlation have –Used at least one continuous variable –Relied on underlying population parameters –Been based on particular distributions Chi square (χ 2 ) is –Based on categorical variables –Non-parametric –Distribution-free

3 Chapter 16 Chi-Square 3 Categorical Variables Generally the count of objects falling in each of several categories. Examples: –number of fraternity, sorority, and nonaffiliated members of a class –number of students choosing answers: 1, 2, 3, 4, or 5 Emphasis on frequency in each category

4 Chapter 16 Chi-Square 4 Contingency Tables Two independent variables –Can be various levels similar to two-way ANOVA –Gender identity, level of happiness

5 Chapter 16 Chi-Square 5 Intimacy and Depression Everitt & Smith (1979) Asked depressed and non-depressed women about intimacy with boyfriend/husband Data on next slide

6 Chapter 16 Chi-Square 6 Data

7 7 What Do the Data Say? It looks as if depressed women are more likely to report lack of intimacy. What alternative explanations? Is the relationship reliably different from chance? –Chi-square test

8 Chapter 16 Chi-Square 8 Chi-Square on Contingency Table The formula Expected frequencies E = RT X CT GT RT = Row total, CT = Column total, GT = Grand total

9 Chapter 16 Chi-Square 9 Expected Frequencies E 11 = (37*138)/419 = 12.19 E 12 = (37*281)/419 = 24.81 E 21 = (382*138)/419 = 125.81 E 22 = (382*281)/419 = 256.19 Enter on following table

10 Chapter 16 Chi-Square 10 Observed and Expected Freq.

11 Chapter 16 Chi-Square 11 Degrees of Freedom For contingency table, df = (R - 1)(C - 1) For our example this is (2 - 1)(2 - 1) = 1 –Note that knowing any one cell and the marginal totals, you could reconstruct all other cells.

12 Chapter 16 Chi-Square 12 Chi-Square Calculation

13 Chapter 16 Chi-Square 13 Conclusions Since 25.61 > 3.84, reject H 0 Conclude that depression and intimacy are not independent. –How one responds to “satisfaction with intimacy” depends on whether they are depressed. –Could be depression-->dissatisfaction, lack of intimacy --> depression, depressed people see world as not meeting needs, etc.

14 Chapter 16 Chi-Square 14 Larger Contingency Tables Is addiction linked to childhood experimentation? Do adults who are, and are not, addicted to substances (alcohol or drug) differ in childhood categories of drug experimentation? One variable = adult addiction –yes or no Other variable = number of experimentation categories (out of 4) as children –Tobacco, alcohol, marijuana/hashish, or acid/cocaine/other

15 Chapter 16 Chi-Square 15

16 Chapter 16 Chi-Square 16 Chi-Square Calculation

17 Chapter 16 Chi-Square 17 Conclusions 29.62 > 7.82 –Reject H 0 –Conclude that adult addiction is related to childhood experimentation –Increasing levels of childhood experimentation are associated with greater levels of adult addiction. e.g. Approximately 10% of children not experimenting later become addicted as adults. Cont.

18 Chapter 16 Chi-Square 18 Conclusions--cont. Approximately 40% of highly experimenting children are later addicted as adults. These data suggest that childhood experimentation may lead to adult addiction.

19 Chapter 16 Chi-Square 19 Tests on Proportions Proportions can be converted to frequencies, and tested using  2. Use a z test directly on the proportions if you have two proportions From last example –10% of nonabused children abused as adults –40% of abused children abused as adults Cont.

20 Chapter 16 Chi-Square 20 Proportions--cont. There were 566 nonabused children and 30 heavily abused children. Cont.

21 Chapter 16 Chi-Square 21 Proportions--cont. z = 5.17 This is a standard z score. –Therefore.05 (2-tailed) cutoff = +1.96 –Reject null hypothesis that the population proportions of abuse in both groups are equal. This is just the square root of the  2 you would have with  2 on those 4 cells.

22 Chapter 16 Chi-Square 22 Independent Observations We require that observations be independent. –Only one score from each respondent –Sum of frequencies must equal number of respondents If we don’t have independence of observations, test is not valid.

23 Chapter 16 Chi-Square 23 Small Expected Frequencies Assume O would be normally distributed around E over many replications of experiment. This could not happen if E is small. Rule of thumb: E > 5 in each cell –Not firm rule –Violated in earlier example, but probably not a problem Cont.

24 Chapter 16 Chi-Square 24 Expected Frequencies--cont. More of a problem in tables with few cells. Never have expected frequency of 0. Collapse adjacent cells if necessary.

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