1. Nominal Measures of Association2. 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 NOMINAL-LEVEL 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 (Φ) = 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 Φ= 1)

PHI– Example:

• 2 x 2 table– 2=5.28

• LIMITATION OF Φ:– Lack of clear upper limit

makes Φ an undesirable measure of association

FAVOR OR OPPOSE DEATH PENALTY FOR MURDER * RESPONDENTS SEXCrosstabulation

Count

52 43 95

10 22 32

62 65 127

1 FAVOR

2 OPPOSE

FAVOR OR OPPOSEDEATH PENALTYFOR MURDERTotal

1 MALE 2 FEMALERESPONDENTS SEX

Total

CRAMER’S V

• Cramer’s V = 2

√ (N)(Minimum of r-1, c-1)

– Unlike Φ, Cramer’s V will always have an upper limit of 1, regardless of # of cells in table • For 2x2 table, Φ & Cramer’s V will have the same value

– Cramer’s 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 Cramer’s V.

• ANSWER: – (Minimum of r-1, c-1) = 3-1 = 2– Cramer’s 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 Cramer’s V.

• ANSWER:– (Minimum of r-1, c-1) = 4 -1 = 3– Cramer’s V = .12– Interpretation: There is a very weak association

between type of vehicle driven & political affiliation.

SUMMARY: 2 -BASED MEASURES OF ASSOCIATION

– Limitation of Φ & Cramer’s 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 (λ)• 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

Re-arrested

Low 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

Re-arrested

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)

– E2 = (each column’s total – largest N in column)

= (75-50) + (40-20) + (90-75) = 25+20+15 = 60

Risk Classification

Re-arrested

Low Medium High Total

Yes 25 20 75 120

No 50 20 15 85

Total 75 40 90 205

CALCULATING LAMBDA: EXAMPLE– Calculate Lambda

λ = E1 – E2 = 85-60 = 25 = 0.294 E1 85 85

– Interpretation – when multiplied by 100, λ 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])

– Calculate & interpret lambda to answer this question

Attitudetowards police

RaceTotals

Black White Other

Positive 40 150 35 225

Negative 80 95 55 230

Totals 120 245 90 455

LAMBDA: EXAMPLE 2E1 = N – (largest row total) 455 – 230 = 225

E2 = (each column’s total – largest N in column) (120 – 80) + (245 – 150) + (90 – 55) =

40 + 95 + 35 = 170λ = E1 – E2 = 225 - 170 = 55 = 0.244

E1 225 225INTERPRETATION:

– 0. 244 x 100 = 24.4% - “Knowledge of an individual’s race improves our ability to predict attitude towards police by 24%”

Attitudetowards police

RaceTotalsBlack White Other

Positive 40 150 35 225

Negative 80 95 55 230

Totals 120 245 90 455

SPSS EXAMPLE

1. IS THERE A SIGNIFICANT RELATIONSHIP B/T GENDER & VOTING BEHAVIOR?

2. If so, what is the strength of association between these variables?

• ANSWER TO Q1: “YES”

PRES00 VOTE FOR GORE, BUSH, NADER * SEX RESPONDENTS SEX Crosstabulation

143 252 395

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%

Count% within SEX RESPONDENTS SEXCount% within SEX RESPONDENTS SEXCount% within SEX RESPONDENTS SEXCount% within SEX RESPONDENTS SEX

1 GORE

2 BUSH

3 NADER

PRES00 VOTEFOR GORE, BUSH,NADER

Total

1 MALE 2 FEMALE

SEX RESPONDENTSSEX

Total

Chi-Square Tests

17.730a 2 .00017.832 2 .000

17.295 1 .000

908

Pearson Chi-SquareLikelihood RatioLinear-by-LinearAssociationN of Valid Cases

Value dfAsymp. Sig.

(2-sided)

0 cells (.0%) have expected count less than 5. Theminimum expected count is 17.14.

a.

SPSS EXAMPLE• ANSWER TO

QUESTION 2:– By either measure, the

association between these variables appears to be weak

Directional Measures

.020 .027 .738 .461

.028 .050 .541 .588

.013 .016 .801 .423

.015 .007 .000c

.020 .009 .000c

SymmetricPRES00 VOTE FORGORE, BUSH, NADERDependentSEX RESPONDENTSSEX DependentPRES00 VOTE FORGORE, BUSH, NADERDependentSEX RESPONDENTSSEX Dependent

Lambda

Goodman andKruskal tau

Nominal byNominal

ValueAsymp.

Std. Errora Approx. Tb Approx. Sig.

Not assuming the null hypothesis.a.

Using the asymptotic standard error assuming the null hypothesis.b.

Based on chi-square approximationc.

Symmetric Measures

.140 .000

908

Cramer's VNominal byNominalN of Valid Cases

Value Approx. Sig.

Not assuming the null hypothesis.a.

Using the asymptotic standard error assuming the nullhypothesis.

b.

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 Lower Class Middle Class

Upper Class

Low Kenny Tim Kim

Middle Joey Deb Ross

High 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 Lower Class Middle Class

Upper Class

Low Kenny Tim Kim

Middle Joey Deb Ross

High 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 Lower Class Middle Class

Upper Class

Low Kenny Tim Kim

Middle Joey Deb Ross

High Randy Eric Barb

GAMMA• Logic: Applying PRE to PAIRS of individuals

– Formula:same – differentsame + 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:

9 – 9 = 0 = 0.0 18 18

Prejudice Lower Class Middle Class

Upper Class

Low Kenny Tim Kim

Middle Joey Deb Ross

High Randy Eric Barb

GAMMA• 3-case example

– Applying the Gamma formula, we would get:3 – 0 = 3 = 1.00

3 3

Prejudice Lower Class Middle Class

Upper Class

Low Kenny

Middle Deb

High Barb

Gamma: Example 1• Examining the relationship between:

– FEHELP (“Wife should help husband’s 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

14 8 0 0 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%

Count% within FEFAM BETTERFOR MAN TO WORK,WOMAN TEND HOMECount% within FEFAM BETTERFOR MAN TO WORK,WOMAN TEND HOMECount% within FEFAM BETTERFOR MAN TO WORK,WOMAN TEND HOMECount% within FEFAM BETTERFOR MAN TO WORK,WOMAN TEND HOMECount% within FEFAM BETTERFOR MAN TO WORK,WOMAN TEND HOME

1 STRONGLY AGREE

2 AGREE

3 DISAGREE

4 STRONGLY DISAGREE

FEHELP WIFESHOULD HELPHUSBANDS CAREERFIRST

Total

1 STRONGLYAGREE 2 AGREE 3 DISAGREE

4 STRONGLYDISAGREE

FEFAM BETTER FOR MAN TO WORK, WOMAN TEND HOME

Total

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

14 8 0 0 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%

Count% within FEFAM BETTERFOR MAN TO WORK,WOMAN TEND HOMECount% within FEFAM BETTERFOR MAN TO WORK,WOMAN TEND HOMECount% within FEFAM BETTERFOR MAN TO WORK,WOMAN TEND HOMECount% within FEFAM BETTERFOR MAN TO WORK,WOMAN TEND HOMECount% within FEFAM BETTERFOR MAN TO WORK,WOMAN TEND HOME

1 STRONGLY AGREE

2 AGREE

3 DISAGREE

4 STRONGLY DISAGREE

FEHELP WIFESHOULD HELPHUSBANDS CAREERFIRST

Total

1 STRONGLYAGREE 2 AGREE 3 DISAGREE

4 STRONGLYDISAGREE

FEFAM BETTER FOR MAN TO WORK, WOMAN TEND HOME

Total

GAMMA– Do we reject the null

hypothesis of independence between these 2 variables?• Yes, the Pearson chi

square p value (.000) is < alpha (.05)

– It’s worthwhile to look at gamma.• Interpretation:

– There is a strong positive relationship between these factors.

– Knowing someone’s view on a wife’s “first priority” improves our ability to predict whether they agree that women should tend home by 75.5%.

Chi-Square Tests

457.679a 9 .000383.933 9 .000

285.926 1 .000

846

Pearson Chi-SquareLikelihood RatioLinear-by-LinearAssociationN of Valid Cases

Value dfAsymp. Sig.

(2-sided)

2 cells (12.5%) have expected count less than 5. Theminimum expected count is 1.66.

a.

Symmetric Measures

.755 .029 18.378 .000846

GammaOrdinal by OrdinalN of Valid Cases

ValueAsymp.

Std. Errora Approx. Tb Approx. Sig.

Not assuming the null hypothesis.a.

Using the asymptotic standard error assuming the null hypothesis.b.

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?