Class Notes (838,396)
Canada (510,879)
Sociology (3,264)
SOC202H1 (76)
Lecture

SOC202 Chi Square

3 Pages
63 Views
Unlock Document

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

Log In


OR

Join OneClass

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

Sign up

Join to view


OR

By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.


Submit