exam ii
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Exam II. 1. Nominal Measures of Association 2. Ordinal Measure s of Associaiton. 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 - PowerPoint PPT PresentationTRANSCRIPT
Exam II
1. Nominal Measures of Association2. Ordinal Measure s of Associaiton
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
Prejudice Lower Class Middle Class
Upper Class
Low Kenny Tim Kim
Middle Joey Deb Ross
High Randy Eric Barb
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?