Sociology 5811: Lecture 16: Crosstabs 2 Measures of Association Plus Differences in Proportions Copyright © 2005 by Evan Schofer Do not copy or distribute without permission

Announcements Final project proposals due Nov 15 Get started now!!! Find a dataset figure out what hypotheses you might test Today: Wrap up Crosstabs If time remains, we’ll discuss project ideas…

Review: Chi-square Test Chi-Square test is a test of independence Null hypothesis: the two categorical variables are statistically independent There is no relationship between them H0: Gender and political party are independent Alternate hypothesis: the variables are related, not independent of each other H1: Gender and political party are not independent Test is based on comparing the observed cell values with the values you’d expect if there were no relationship between variables.

Review: Expected Cell Values If two variables are independent, cell values will depend only on row & column marginals Marginals reflect frequencies… And, if frequency is high, all cells in that row (or column) should be high The formula for the expected value in a cell is: fi and fj are the row and column marginals N is the total sample size

Review: Chi-square Test The Chi-square formula: Where: R = total number of rows in the table C = total number of columns in the table Eij = the expected frequency in row i, column j Oij = the observed frequency in row i, column j Assumption for test: Large N (>100) Critical value DofF: (R-1)(C-1).

Chi-square Test of Independence Example: Gender and Political Views Let’s pretend that N of 68 is sufficient Women Men Democrat O11: 27 E11: 23.4 O12 : 10 E12 : 13.6 Republican O21 : 16 E21 : 19.6 O22 : 15 E22 : 11.4

Chi-square Test of Independence Compute (E – O)2 /E for each cell Women Men Democrat (23.4 – 27)2/23.4 = .55 (13.6 – 10)2/13.6 = .95 Republican (19.6 – 16)2/19.6 = .66 (11.4 – 15)2/15 = .86

Chi-Square Test of Independence Finally, sum up to compute the Chi-square c2 = .55 + .95 + .66 + .86 = 3.02 What is the critical value for a=.05? Degrees of freedom: (R-1)(C-1) = (2-1)(2-1) = 1 According to Knoke, p. 509: Critical value is 3.84 Question: Can we reject H0? No. c2 of 3.02 is less than the critical value We cannot conclude that there is a relationship between gender and political party affiliation.

Chi-square Test of Independence Weaknesses of chi-square tests: 1. If the sample is very large, we almost always reject H0. Even tiny covariations are statistically significant But, they may not be socially meaningful differences 2. It doesn’t tell us how strong the relationship is It doesn’t tell us if it is a large, meaningful difference or a very small one It is only a test of “independence” vs. “dependence” Measures of Association address this shortcoming.

Measures of Association Separate from the issue of independence, statisticians have created measures of association They are measures that tell us how strong the relationship is between two variables Weak Association Strong Association Women Men Dem. 51 49 Rep. Women Men Dem. 100 Rep.

Crosstab Association:Yule’s Q Appropriate only for 2x2 tables (2 rows, 2 columns) Label cell frequencies a through d: a b c d Recall that extreme values along the “diagonal” (cells a & d) or the “off-diagonal” (b & c) indicate a strong relationship. Yule’s Q captures that in a measure 0 = no association. -1, +1 = strong association

Crosstab Association:Yule’s Q Rule of Thumb for interpreting Yule’s Q: Bohrnstedt & Knoke, p. 150 Absolute value of Q Strength of Association 0 to .24 “virtually no relationship” .25 to .49 “weak relationship” .50 to .74 “moderate relationship” .75 to 1.0 “strong relationship”

Crosstab Association:Yule’s Q Example: Gender and Political Party Affiliation Women Men Dem 27 10 Rep 16 15 Calculate “bc” bc = (10)(16) = 160 a b c d Calculate “ad” ad = (27)(15) = 405 -.48 = “weak association”, almost “moderate”

Association: Other Measures Phi () Very similar to Yule’s Q Only for 2x2 tables, ranges from –1 to 1, 0 = no assoc. Gamma (G) Based on a very different method of calculation Not limited to 2x2 tables Requires ordered variables Tau c (tc) and Somer’s d (dyx) Same basic principle as Gamma Several Others discussed in Knoke, Norusis.

Crosstab Association: Gamma Gamma, like Q, is based on comparing “diagonal” to “off-diagonal” cases. But, it does so differently Jargon: Concordant pairs: Pairs of cases where one case is higher on both variables than another case Discordant pairs: Pairs of cases for which the first case (when compared to a second) is higher on one variable but lower on another

Crosstab Association: Gamma Example: Approval of candidates Cases in “Love Trees/Love Guns” cell make concordant pairs with cases lower on both Hate Trees Trees OK Love Trees Love Guns 1205 603 71 Guns = OK 659 1498 452 Hate Guns 431 467 1120 All 71 individuals can be a pair with everyone in the lower cells. Just Multiply! (71)(659+1498+ 431+467) = 216,905 conc. pairs

Crosstab Association: Gamma More possible concordant pairs The “Love Guns/Trees are OK” cell and the “Trees = OK/Love Guns” cells also can have concordant pairs These 603 can pair with all those that score lower on approval for Guns & Trees (603)(659 + 431) = 657,270 conc. pairs Hate Trees Trees = OK Love Trees Love Guns 1205 603 71 Guns = OK 659 1498 452 Hate Guns 431 467 1120 These can pair lower too! (452)(431 + 467) = 405,896 conc. pairs

Crosstab Association: Gamma Discordant pairs: Pairs where a first person ranks higher on one dimension (e.g. approval of Trees) but lower on the other (e.g., app. of Guns) Hate Trees Trees = OK Love Trees Love Guns 1205 603 71 Guns = OK 659 1498 452 Hate Guns 431 467 1120 The top-left cell is higher on Guns but lower on Trees than those in the lower right. They make pairs: (1205)(1498 + 452 + 467 + 1120) = 4,262,085 discordant pairs

Crosstab Associaton: Gamma If all pairs are concordant or all pairs are discordant, the variables are strongly related If there are an equal number of discordant and concordant pairs, the variables are weakly associated. Formula for Gamma: ns = number of concordant pairs nd = number of discordant pairs

Crosstab Association: Gamma Calculation of Gamma is typically done by computer Zero indicates no association +1 = strong positive association -1 = strong negative association It is possible to do hypothesis tests on Gamma To determine if population gamma differs from zero Requirements: random sample, N > 50 See Knoke, p. 155-6.

Crosstab Association Final remarks: You have a variety of possible measures to assess association among variables. Which one should you use? Yule’s Q and Phi require a 2x2 table Larger ordered tables: use Gamma, Tau-c, Somer’s d Ideally, report more than one to show that your findings are robust.

Odds Ratios Odds ratios are a powerful way of analyzing relationships in crosstabs Many advanced categorical data analysis techniques are based on odds ratios Review: What is a probability? p(A) = # of outcomes that are “A” divided by total number of outcomes To convert a frequency distribution to a probability distribution, simply divide frequency by N The same can be done with crosstabs: Cell frequency over N is probability.

Odds Ratios If total N = 68, probability of drawing cases is: Women Dem 27 / 68 10 / 68 Rep 16 / 68 15 / 68 Women Men Dem .397 .147 Rep .235 .220

Odds Ratios Odds are similar to probability… but not quite Odds of A = Number of outcomes that are A, divided by number of outcomes that are not A Note: Denominator is different that probability Ex: Probability of rolling 1 on a 6-sided die = 1/6 Odds of rolling a 1 on a six-sided die = 1/5 Odds can also be calculated from probabilities:

Odds Ratios Conditional odds = odds of being in one category of a variable within a specific category of another variable Example: For women, what are the odds of being democrat? Instead of overall odds of being democrat, conditional odds are about a particular subgroup in a table Women Men Dem 27 10 Rep 16 15 Conditional odds of being democrat are: 27 / 16 = 1.69 Note: Odds for women are different than men

Odds Ratios If variables in a crosstab are independent, their conditional odds are equal Odds of falling into one category or another are same for all values of other variable If variables in a crosstab are associated, conditional odds differ Odds can be compared by making a ratio Ratio is equal to 1 if odds are the same for two groups Ratios much greater or less than 1 indicate very different odds.

Odds Ratios c d Formula for Odds Ratio in 2x2 table: Women Men Dem 27 10 Rep 16 15 a b c d Ex: OR = (10)(16)/(27)(15) = 160 / 405 = .395 Interpretation: men have .395 times the odds of being a democrat compared to women Inverted value (1/.395=2.5) indicates odds of women being democrat = 2.5 is times men’s odds

Odds Ratios: Final Remarks 1. Cells with zeros cause problems for odds ratios Ratios with zero in denominator are undefined. Thus, you need to have full cells 2. Odds ratios can be used to measure assocation Indeed, Yule’s Q is based on them 3. Odds ratios form the basis for most advanced categorical data analysis techniques For now it may be easier to use Yule’s Q, etc. But, if you need to do advanced techniques, you will use odds ratios.

Tests for Difference in Proportions Another approach to small (2x2) tables: Instead of making a crosstab, you can just think about the proportion of people in a given category More similar to T-test than a Chi-square test Ex: Do you approve of Pres. Bush? (Yes/No) Sample: N = 86 women, 80 men Proportion of women that approve: PW = .70 Proportion of men that approve: PM = .78 Issue: Do the populations of men/women differ? Or are the differences just due to sampling variability

Tests for Difference in Proportions Hypotheses: Again, the typical null hypothesis is that there are no differences between groups Which is equivalent to statistical independence H0: Proportion women = proportion men H1: Proportion women not = proportion men Note: One-tailed directional hypotheses can also be used.

Tests for Difference in Proportions Strategy: Figure out the sampling distribution for differences in proportions Statisticians have determined relevant info: 1. If samples are “large”, the sampling distribution of difference in proportions is normal The Z-distribution can be used for hypothesis tests 2. A Z-value can be calculated using the formula:

Tests for Difference in Proportions Standard error can be estimated as: Where:

Difference in Proportions: Example Q: Do you approve of Pres. Bush? (Yes/No) Sample: N = 86 women, 80 men Women: N = 86, PW = .70 Men: N = 80, PW = .78 Total N is “Large”: 166 people So, we can use a Z-test Use a = .05, two-tailed Z = 1.96

Difference in Proportions: Example Use formula to calculate Z-value And, estimate the Standard Error as:

Difference in Proportions: Example First: Calculate Pboth:

Difference in Proportions: Example Plug in Pboth=.739:

Difference in Proportions: Example Finally, plug in S.E. and calculate Z:

Difference in Proportions: Example Results: Critical Z = 1.96 Observed Z = .739 Conclusion: We can’t reject null hypothesis Women and Men do not clearly differ in approval of Bush