# EXAM - University of Minnesota Duluth

1. Nominal Measures of Association 2. Ordinal Measure s of Association ASSOCIATION Association The strength of relationship between 2 variables Knowing how much variables are related may enable you to predict the value of 1 variable when you know the value of another As with test statistics, the proper measure of association depends on how variables are measured Significance vs. Association Association = strength of relationship Test statistics = how different findings are from null They do capture the strength of a relationship t = number of standard errors that separate means Chi-Square = how different our findings are from what is expected under null If null is no relationship, then higher Chi-square values indicate stronger relationships. HOWEVER --- test statistics are also influenced by other stuff (e.g., sample size) MEASURES OF ASSOCIATION FOR NOMINALLEVEL VARIABLES Chi-Square Based Measures 2 indicates how different our findings are from what is expected under null 2 also gets larger with higher sample size (more confidence in larger samples) To get a pure measure of strength, you have to remove influence of N Phi Cramer's V

PHI Phi () = ) = 2 N Formula standardizes 2 value by sample size Measure ranges from 0 (no relationship) to values considerably >1 (Exception: for a 2x2 bivariate table, upper limit of ) = = 1) PHI Example: 2 x 2 table 2=5.28 FAVOR OR OPPOSE DEATH PENALTY FOR MURDER * RESPONDENTS SEX Crosstabulation Count RESPONDENTS SEX 1 MALE 2 FEMALE FAVOR OR OPPOSE DEATH PENALTY FOR MURDER Total LIMITATION OF ) = : Lack of clear upper limit makes ) = an undesirable measure of association Total 1 FAVOR 52 43

95 2 OPPOSE 10 22 32 62 65 127 CRAMERS V Cramers V = 2 (N)(Minimum of r-1, c-1) Unlike ) = , Cramers V will always have an upper limit of 1, regardless of # of cells in table For 2x2 table, ) = & Cramers V will have the same value Cramers V ranges from 0 (no relationship) to +1 (perfect relationship) 2-BASED MEASURES OF ASSOCIATION Sample problem 1: The chi square for a 5 x 3 bivariate table examining the relationship between area of Duluth one lives in & type of movie preference is 8.42, significant at .05 (N=100). Calculate & interpret Cramers V. ANSWER: (Minimum of r-1, c-1) = 3-1 = 2 Cramers V = .21 Interpretation: There is a relatively weak association between area of the city lived in and movie preference.

2-BASED MEASURES OF ASSOCIATION Sample problem 2: The chi square for a 4 x 4 bivariate table examining the relationship between type of vehicle driven & political affiliation is 12.32, sig. at .05 (N=300). Calculate & interpret Cramers V. ANSWER: (Minimum of r-1, c-1) = 4 -1 = 3 Cramers V = .12 Interpretation: There is a very weak association between type of vehicle driven & political affiliation. SUMMARY: 2 -BASED MEASURES OF ASSOCIATION Limitation of ) = & Cramers V: No direct or meaningful interpretation for values between 0-1 Both measure relative strength (e.g., .80 is stronger association than .40), but have no substantive meaning; hard to interpret Rules of Thumb for what is a weak, moderate, or strong relationship vary across disciplines LAMBDA ()) PRE (Proportional Reduction in Error) is the logic that underlies the definition & computation of lambda Tells us the reduction in error we gain by using the IV to predict the DV Range 0-1 (i.e., proportional reduction) E1 Attempt to predict the category into which each case will fall on DV or Y while ignoring IV or X E2 Predict the category of each case on Y while taking X into account The stronger the association between the variables the greater the reduction in errors LAMBDA: EXAMPLE 1 Does risk classification in prison affect the likelihood of being rearrested after release? (2=43.7)

Risk Classification ReLow arrested Medium High Total Yes 25 20 75 120 No 50 20 15 85 Total 75 40 90 205

LAMBDA: EXAMPLE Find E1 (# of errors made when ignoring X) E1 = N (largest row total) = 205 -120 = 85 Risk Classification Rearrested Low Medium High Total Yes 25 20 75 120 No 50 20 15 85 Total 75

40 90 205 LAMBDA: EXAMPLE Find E2 (# of errors made when accounting for X) E = (each columns total largest N in column) 2 = (75-50) + (40-20) + (90-75) = 25+20+15 = 60 Risk Classification Rearrested Low Medium High Total Yes 25 20 75 120 No 50 20

15 85 Total 75 40 90 205 CALCULATING LAMBDA: EXAMPLE Calculate Lambda ) = E1 E2 = 85-60 = 25 = 0.294 E1 85 85 Interpretation when multiplied by 100, ) indicates the % reduction in error achieved by using X to predict Y, rather than predicting Y blind (without X) 0.294 x 100 = 29.4% - Knowledge of risk classification in prison improves our ability to predict rearrest by 29%. LAMBDA: EXAMPLE 2 What is the strength of the relationship between citizens race and attitude toward police? (obtained chi square is > 5.991 (2[critical]) critical]) ) Calculate & interpret lambda to answer this question Attitude towards police

Positive Negative Totals Black Race White Other Totals 40 80 120 150 95 245 35 55 90 225 230 455 LAMBDA: EXAMPLE 2 E1 = N (largest row total) 455 230 = 225 E2 = (each columns total largest N in column) (120 80) + (245 150) + (90 55) = 40 + 95 + 35 = 170 ) = E1 E2 = 225 - 170 = 55 = 0.244 E1 225 225 INTERPRETATION:

0. 244 x 100 = 24.4% - Knowledge of an individuals race improves our ability to predict attitude towards police by 24% Attitude towards police Race Black White Other Totals Positive 40 150 35 225 Negative 80 95 55 230 Totals 120 245

90 455 SPSS EXAMPLE PRES00 VOTE FOR GORE, BUSH, NADER * SEX RESPONDENTS SEX Crosstabulation 1. IS THERE A SIGNIFICANT PRES00 VOTE RELATIONSHIP B/FOR GORE, BUSH, NADER T GENDER & VOTING BEHAVIOR? 2. If so, what is the strength of association between these variables? ANSWER TO Q1: YES Total 1 GORE 2 BUSH 3 NADER Count % within SEX RESPONDENTS SEX Count % within SEX RESPONDENTS SEX

Count % within SEX RESPONDENTS SEX Count % within SEX RESPONDENTS SEX SEX RESPONDENTS SEX 1 MALE 2 FEMALE 143 252 35.8% 49.5% 43.5% 234 240 474 58.6% 47.2% 52.2% 22 17 39 5.5% 3.3%

4.3% 399 509 908 100.0% 100.0% 100.0% Chi-Square Tests Pearson Chi-Square Likelihood Ratio Linear-by-Linear Association N of Valid Cases Value 17.730a 17.832 17.295 Total 395 2 2 Asymp. Sig. (2-sided) .000 .000 1

.000 df 908 a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 17.14. SPSS EXAMPLE Symmetric Measures ANSWER TO QUESTION 2: By either measure, the association between these variables appears to be weak Value Nominal by Nominal Cramer's V Approx. Sig. .140 N of Valid Cases .000 908 a. Not assuming the null hypothesis. b. Using the asymptotic standard error assuming the null hypothesis.

Directional Measures Nominal by Nominal Lambda Goodman and Kruskal tau Symmetric PRES00 VOTE FOR GORE, BUSH, NADER Dependent SEX RESPONDENTS SEX Dependent PRES00 VOTE FOR GORE, BUSH, NADER Dependent SEX RESPONDENTS SEX Dependent Value .020 Asymp. a Std. Error .027 Approx. T .738 Approx. Sig. .461 .028 .050

.541 .588 .013 .016 .801 .423 .015 .007 .000 .020 .009 .000 a. Not assuming the null hypothesis. b. Using the asymptotic standard error assuming the null hypothesis. b c c 2 LIMITATIONS OF LAMBDA 1. Asymmetric Value of the statistic will vary depending on which variable is taken as independent

2. Misleading when one of the row totals is much larger than the other(s) For this reason, when row totals are extremely uneven, use a chi square-based measure instead ORDINAL MEASURE OF ASSOCIATION GAMMA For examining STRENGTH & DIRECTION of collapsed ordinal variables (<6 categories) Like Lambda, a PRE-based measure Range is -1.0 to +1.0 GAMMA Logic: Applying PRE to PAIRS of individuals Prejudice Low Middle High Lower Class Middle Class Upper Class Kenny Tim Kim Joey

Deb Ross Randy Eric Barb GAMMA CONSIDER KENNY-DEB PAIR In the language of Gamma, this is a same pair direction of difference on 1 variable is the same as direction on the other If you focused on the Kenny-Eric pair, you would come to the same conclusion Prejudice Low Middle High Lower Class Middle Class Upper Class Kenny Tim

Kim Joey Deb Ross Randy Eric Barb GAMMA NOW LOOK AT THE TIM-JOEY PAIR In the language of Gamma, this is a different pair direction of difference on one variable is opposite of the difference on the other Prejudice Low Middle High Lower Class Middle Class Upper Class Kenny Tim

Kim Joey Deb Ross Randy Eric Barb GAMMA Logic: Applying PRE to PAIRS of individuals Formula: same different same + different GAMMA If you were to account for all the pairs in this table, you would find that there were 9 same & 9 different pairs Applying the Gamma formula, we would get: 99= 0 = 0.0 18 18 Prejudice Low Middle High Lower Class Middle

Class Upper Class Kenny Tim Kim Joey Deb Ross Randy Eric Barb GAMMA 3-case example Applying the Gamma formula, we would get: 30= 3 = 1.00 3 3 Prejudice Low Middle High Lower Class

Middle Class Upper Class Kenny Deb Barb Gamma: Example 1 Examining the relationship between: FEHELP (Wife should help husbands career first) & FEFAM (Better for man to work, women to tend home) Both variables are ordinal, coded 1 (strongly agree) to 4 (strongly disagree) FEHELP WIFE SHOULD HELP HUSBANDS CAREER FIRST * FEFAM BETTER FOR MAN TO WORK, WOMAN TEND HOME Crosstabulation FEHELP WIFE SHOULD HELP HUSBANDS CAREER FIRST Total 1 STRONGLY AGREE Count % within FEFAM BETTER FOR MAN TO WORK, WOMAN TEND HOME 2 AGREE Count % within FEFAM BETTER FOR MAN TO WORK, WOMAN TEND HOME 3 DISAGREE Count % within FEFAM BETTER FOR MAN TO WORK, WOMAN TEND HOME

4 STRONGLY DISAGREE Count % within FEFAM BETTER FOR MAN TO WORK, WOMAN TEND HOME Count % within FEFAM BETTER FOR MAN TO WORK, WOMAN TEND HOME FEFAM BETTER FOR MAN TO WORK, WOMAN TEND HOME 1 STRONGLY 4 STRONGLY AGREE 2 AGREE 3 DISAGREE DISAGREE 14 8 0 0 Total 22 21.9% 3.8% .0% .0% 2.6% 26 72 26 3

127 40.6% 34.3% 6.4% 1.8% 15.0% 21 111 307 45 484 32.8% 52.9% 75.2% 27.4% 57.2% 3 19 75 116

213 4.7% 9.0% 18.4% 70.7% 25.2% 64 210 408 164 846 100.0% 100.0% 100.0% 100.0% 100.0% Gamma: Example 1 Based on the info in this table, does there seem to be a relationship between these factors? Does there seem to be a positive or negative relationship between them? Does this appear to be a strong or weak relationship? FEHELP WIFE SHOULD HELP HUSBANDS CAREER FIRST * FEFAM BETTER FOR MAN TO WORK, WOMAN TEND HOME Crosstabulation

FEHELP WIFE SHOULD HELP HUSBANDS CAREER FIRST Total 1 STRONGLY AGREE Count % within FEFAM BETTER FOR MAN TO WORK, WOMAN TEND HOME 2 AGREE Count % within FEFAM BETTER FOR MAN TO WORK, WOMAN TEND HOME 3 DISAGREE Count % within FEFAM BETTER FOR MAN TO WORK, WOMAN TEND HOME 4 STRONGLY DISAGREE Count % within FEFAM BETTER FOR MAN TO WORK, WOMAN TEND HOME Count % within FEFAM BETTER FOR MAN TO WORK, WOMAN TEND HOME FEFAM BETTER FOR MAN TO WORK, WOMAN TEND HOME 1 STRONGLY 4 STRONGLY AGREE 2 AGREE 3 DISAGREE DISAGREE 14

8 0 0 Total 22 21.9% 3.8% .0% .0% 2.6% 26 72 26 3 127 40.6% 34.3% 6.4% 1.8% 15.0% 21 111

307 45 484 32.8% 52.9% 75.2% 27.4% 57.2% 3 19 75 116 213 4.7% 9.0% 18.4% 70.7% 25.2% 64 210

408 164 846 100.0% 100.0% 100.0% 100.0% 100.0% GAMMA Do we reject the null hypothesis of independence between these 2 variables? Yes, the Pearson chi square p value (.000) is < alpha (.05) Its worthwhile to look at gamma. Interpretation: There is a strong positive relationship between these factors. Knowing someones view on a wifes first priority improves our ability to predict whether they agree that women should tend home by 75.5%. Chi-Square Tests

Pearson Chi-Square Likelihood Ratio Linear-by-Linear Association N of Valid Cases Value 457.679 a 383.933 9 9 Asymp. Sig. (2-sided) .000 .000 1 .000 df 285.926 846 a. 2 cells (12.5%) have expected count less than 5. The minimum expected count is 1.66. Symmetric Measures Ordinal by Ordinal Gamma N of Valid Cases Value .755 846

Asymp. a b Std. Error Approx. T Approx. Sig. .029 18.378 .000 a. Not assuming the null hypothesis. b. Using the asymptotic standard error assuming the null hypothesis. USING GSS DATA Construct a contingency table using two ordinal level variables Are the two variables significantly related? How strong is the relationship? What direction is the relationship?

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