Class Notes (838,396)
Sociology (3,264)
SOC202H1 (76)
Lecture

# SOC202 Chi Square

3 Pages
63 Views

Department
Sociology
Course
SOC202H1
Professor
Scott Schieman
Semester
Winter

Description
SOC202 – Chi-Square Tests Exhaustion - 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 exhaustion - 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 only 3.84 - Why
More Less

Related notes for SOC202H1
Me

OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.