SOC202 – Chi-Square Tests
- How could you turn this into a continuous variable?
Use some kind of scale
- How could you recode this into a categorical/nominal variable?
Take a spread and condense it into categories of: “low”, “medium”, and “high”. Drawback of
this: you lose information.
- Research Question: as levels of work stress increase, what happens to the likelihood of being in
the “low” versus “high” exhaustion groups?
- Hypothesis: high levels of work stress correspond with high levels of exhaustion
Null: variables are independent of the population; thus levels of exhaustion have no
correlation with work stress in the population
Work Stress and Exhaustion
- For each presentation, how could you recode this into a categorical/nominal variable?
Collapse 5 into 2 or 3 categories for the purpose of cross-classification
Chi-Square Test: Example #1
- The relationship between two nominal/categorical variables
- Cross-tabulation (cross-tab) table of:
1. The frequency of joint occurrences
2. Observed (joint) frequencies
3. Marginal totals
4. Appropriate percentages of the independent variable
Presented in columns
- 849: the joint frequency. 849 people reported low work stress and low exhaustion
- 417: 417 people reported high exhaustion and high work stress
- 74.73%: 74.73% of people who reported high work stress, also reported high levels of
- 990: row marginal total – 990 people reported low exhaustion
- 42.44%: row marginal percentage – 42.44% of people reported high exhaustion
- Build a case against the null that the X and Y are independent (chi-square statistic = 0)
- Obtain expected frequencies: these are the values of the joint cell frequencies if the null
hypothesis is true
- Obtain expected frequencies
- The difference between observed and expected frequencies builds the chi-square statistic
- 849: observed frequency
668.8: expected number of people who would report if null was true
Pearson chi2(1) = 352.5114 Pr = 0.000 352.5114 is a large chi-square statistic because the difference between observed and
expected is so large. This proves a strong case against the null. We are so confident that it
rejects it because at the level of .05, the chi-square statistic can reject the null at a number of