Chapter 10

Categorical Data Analysis

Inference for a Single Proportion ()• Goal: Estimate proportion of individuals in a population with a

certain characteristic (). This is equivalent to estimating a binomial probability

• Sample: Take a SRS of n individuals from the population and observe y that have the characteristic. The sample proportion is y/n and has the following sampling properties:

)5)1(, : thumbof (Rule samples largefor normalely approximat :Shape

1 :Error Standard Estimated

)1( :ondistributi sampling of Dev. Std. andMean

:proportion Sample

^^

^

^

^^

nnn

SE

n

n

y

Large-Sample Confidence Interval for

• Take SRS of size n from population where is true (unknown) proportion of successes. – Observe y successes

– Set confidence level (1-) and obtain z/2 from z-table

mC

zmn

n

y

^

2/

^^

^

:for interval confidence %

SE :error ofMargin

1SE :Error Standard Estimated

:EstimatePoint

^

^

Example - Ginkgo and Azet for AMS• Study Goal: Measure effect of Ginkgo and

Acetazolamide on occurrence of Acute Mountain Sickness (AMS) in Himalayan Trackers

• Parameter: = True proportion of all trekkers receiving Ginkgo&Acetaz who would suffer from AMS.

• Sample Data: n=126 trekkers received G&A, y=18 suffered from AMS

)204,.082(.061.143.:for CI %95

061.)031(.96.1:%)95%100)1((error ofMargin

031.126

)86)(.14(.SE143.

126

18^

^

m

Wilson’s “Plus 4” Method• For moderate to small sample sizes, large-sample

methods may not work well wrt coverage probabilities• Simple approach that works well in practice (n10):

– Pretend you have 4 extra individuals, 2 successes, 2 failures

– Compute the estimated sample proportion in light of new “data” as well as standard error:

m

zmn

n

y

~

2/

~~

~

:for interval confidence %100)1(

SE :error ofMargin 4

1SE :Error Standard Estimated

4

2 :EstimatePoint

~

~

Example: Lister’s Tests with Antiseptic

• Experiments with antiseptic in patients with upper limb amputations (John Lister, circa 1870)

• n=12 patients received antiseptic y=1 died

)40,.0()3988,.0038.(1913.1875.:for CI %95

1913.)0976(.96.1:%)95)100%-1(error( ofMargin

0976.16

)8125(.1875.SE1875.

16

3

412

21~

~

Significance Test for a Proportion

• Goal test whether a proportion () equals some null value 0 H0:

)(2value-::

)(value-::

)(value-::

)1( :StatisticTest

2/0

0

0

0

0

^

obsobsa

obsobsa

obsobsa

o

obs

zZPPzzRRH

zZPPzzRRH

zZPPzzRRHn

z

Large-sample test works well when n0 and n(1-0) 5

Ginkgo and Acetaz for AMS

• Can we claim that the incidence rate of AMS is less than 25% for trekkers receiving G&A?

• H0: =0.25 Ha: < 0.25

0030.)75.2( value-

645.1:)05.(

75.2039.

107.

118)75(.25.

25.143. :StatisticTest

25.0143.0126

1818126

05.

0

^

ZPP

zzRR

z

yn

obs

obs

Strong evidence that incidence rate is below 25% (< 0.25)

Comparing Two Population Proportions

• Goal: Compare two populations/treatments wrt a nominal (binary) outcome

• Sampling Design: Independent vs Dependent Samples

• Methods based on large vs small samples

• Contingency tables used to summarize data

• Measures of Association: Absolute Risk, Relative Risk, Odds Ratio

Contingency Tables

• Tables representing all combinations of levels of explanatory and response variables

• Numbers in table represent Counts of the number of cases in each cell

• Row and column totals are called Marginal counts

2x2 Tables - Notation

n1+n2(n1+n2)-(y1+y2)

y1+y2Outcome

Total

n2n2-y2y2Group 2

n1n1-y1y1Group 1

Group

Total

Outcome

Absent

Outcome

Present

Example - Firm Type/Product Quality

17213438Outcome

Total

84795Vertically

Integrated

885533Not

Integrated

Group

Total

Low

Quality

High

Quality

• Groups: Not Integrated (Weave only) vs Vertically integrated (Spin and Weave) Cotton Textile Producers

• Outcomes: High Quality (High Count) vs Low Quality (Count)

Source: Temin (1988)

Notation• Proportion in Population 1 with the characteristic

of interest: 1

• Sample size from Population 1: n1

• Number of individuals in Sample 1 with the characteristic of interest: y1

• Sample proportion from Sample 1 with the characteristic of interest:

• Similar notation for Population/Sample 2

1

11

^

n

y

Example - Cotton Textile Producers

1 - True proportion of all Non-integretated firms that would produce High quality

2 - True proportion of all vertically integretated firms that would produce High quality

060.084

5584

375.088

333388

2

22

^

22

1

11

^

11

n

yyn

n

yyn

Notation (Continued)

• Parameter of Primary Interest: 1-2, the difference in the 2 population proportions with the characteristic (2 other measures given below)

• Estimator:

• Standard Error (and its estimate):

• Pooled Estimated Standard Error when :

2

^

1

^

D

2

2

^

2

^

1

1

^

1

^

2

22

1

11

11)1()1(

nnSE

nn DD

21

21^

21

^^ 111

nn

yy

nnSE

PD

Cotton Textile Producers (Continued)

• Parameter of Primary Interest: , the difference in the 2 population proportions that produce High quality output

• Estimator: • Standard Error (and its estimate):

• Pooled Estimated Standard Error when :

315.0060.0375.02

^

1

^

D

0577.003335.84

)94.0(060.0

88

)625.0(375.011

2

1

^

2

^

1

1

^

1

^

nn

SED

221.08488

5330633.

84

1

88

1779.0221.0

^

PDSE

Significance Tests for

• Deciding whether can be done by interpreting “plausible values” of from the confidence interval:

– If entire interval is positive, conclude ( > 0)

– If entire interval is negative, conclude ( < 0)

– If interval contains 0, do not conclude that

• Alternatively, we can conduct a significance test:– H0: Ha: (2-sided) Ha: (1-sided)

– Test Statistic:

– RR: |zobs| z/2 (2-sided) zobs z (1-sided)

– P-value: 2P(Z|zobs|) (2-sided) P(Z zobs) (1-sided)

21

^^

2

^

1

^

111

nn

zobs

Example - Cotton Textile Production

0)98.4(2 value-

96.1:

98.40633.0

315.0

841

881

)779.0(221.0

060.0375.0

111

:

)0(:

)0(:

025.

21

^^

2

^

1

^

2121

21210

ZPP

zzRR

nn

zTS

H

H

obs

obs

A

Again, there is strong evidence that non-integrated performs are more likely to produce high quality output than integrated firms

Associations Between Categorical Variables

• Case where both explanatory (independent) variable and response (dependent) variable are qualitative

• Association: The distributions of responses differ among the levels of the explanatory variable (e.g. Party affiliation by gender)

Contingency Tables• Cross-tabulations of frequency counts where the

rows (typically) represent the levels of the explanatory variable and the columns represent the levels of the response variable.

• Numbers within the table represent the numbers of individuals falling in the corresponding combination of levels of the two variables

• Row and column totals are called the marginal distributions for the two variables

Example - Cyclones Near Antarctica• Period of Study: September,1973-May,1975

• Explanatory Variable: Region (40-49,50-59,60-79) (Degrees South Latitude)

• Response: Season (Aut(4),Wtr(5),Spr(4),Sum(8)) (Number of months in parentheses)

• Units: Cyclones in the study area

• Treating the observed cyclones as a “random sample” of all cyclones that could have occurred

Source: Howarth(1983), “An Analysis of the Variability of Cyclones around Antarctica and Their Relation to Sea-Ice Extent”, Annals of the Association of American Geographers, Vol.73,pp519-537

Example - Cyclones Near Antarctica

Region\Season Autumn Winter Spring Summer Total40 -49S 370 452 273 422 151750 -59S 526 624 513 1059 272260 -79S 980 1200 995 1751 4926Total 1876 2276 1781 3232 9165

For each region (row) we can compute the percentage of storms occuring during each season, the conditional distribution. Of the 1517 cyclones in the 40-49 band, 370 occurred in Autumn, a proportion of 370/1517=.244, or 24.4% as a percentage.

Region\Season Autumn Winter Spring Summer Total% (n)

40 -49S 24.4 29.8 18.0 27.8 100.0 (1517)50 -59S 19.3 22.9 18.9 38.9 100.0 (2722)60 -79S 19.9 24.4 20.2 35.5 100.0 (4926)

Example - Cyclones Near Antarctica

40-49S

50-59S

60-79S

region

Bars show Means

Autumn Winter Spring Summer

season

10.00

20.00

30.00

40.00re

gp

ct

Graphical Conditional Distributions for Regions

Guidelines for Contingency Tables• Compute percentages for the response (column)

variable within the categories of the explanatory (row) variable. Note that in journal articles, rows and columns may be interchanged.

• Divide the cell totals by the row (explanatory category) total and multiply by 100 to obtain a percent, the row percents will add to 100

• Give title and clearly define variables and categories.

• Include row (explanatory) total sample sizes

Independence & Dependence

• Statistically Independent: Population conditional distributions of one variable are the same across all levels of the other variable

• Statistically Dependent: Conditional Distributions are not all equal

• When testing, researchers typically wish to demonstrate dependence (alternative hypothesis), and wish to refute independence (null hypothesis)

Pearson’s Chi-Square Test

• Can be used for nominal or ordinal explanatory and response variables

• Variables can have any number of distinct levels• Tests whether the distribution of the response

variable is the same for each level of the explanatory variable (H0: No association between the variables

• r = # of levels of explanatory variable• c = # of levels of response variable

Pearson’s Chi-Square Test

• Intuition behind test statistic– Obtain marginal distribution of outcomes for

the response variable– Apply this common distribution to all levels of

the explanatory variable, by multiplying each proportion by the corresponding sample size

– Measure the difference between actual cell counts and the expected cell counts in the previous step

Pearson’s Chi-Square Test

• Notation to obtain test statistic– Rows represent explanatory variable (r levels)

– Cols represent response variable (c levels)

n..n.c…n.2n.1Total

nr.nrc…nr2 nr1 r

………………

n2. n2c …n22 n212

n1.n1c …n12 n111

Totalc…21

Pearson’s Chi-Square Test

• Observed frequency (nij): The number of individuals falling in a particular cell

• Expected frequency (Eij): The number we would expect in that cell, given the sample sizes observed in study and the assumption of independence. – Computed by multiplying the row total and the

column total, and dividing by the overall sample size.

– Applies the overall marginal probability of the response category to the sample size of explanatory category

Pearson’s Chi-Square Test

• Large-sample test (all Eij > 5)

• H0: Variables are statistically independent (No association between variables)

• Ha: Variables are statistically dependent (Association exists between variables)

• Test Statistic:

• P-value: Area above in the chi-squared distribution with (r-1)(c-1) degrees of freedom. (Critical values in Table 8)

ij

ijijobs E

En 22 )(

2obs

Example - Cyclones Near Antarctica

Region\Season Autumn Winter Spring Summer Total40 -49S 370 452 273 422 151750 -59S 526 624 513 1059 272260 -79S 980 1200 995 1751 4926Total 1876 2276 1781 3232 9165

Note that overall: (1876/9165)100%=20.5% of all cyclones occurred in Autumn. If we apply that percentage to the 1517 that occurred in the 40-49S band, we would expect (0.205)(1517)=310.5 to have occurred in the first cell of the table. The full table of Eij:

Region\Season Autumn Winter Spring Summer Total40 -49S 310.5 376.7 294.8 535.0 151750 -59S 557.2 676.0 529.0 959.9 272260 -79S 1008.3 1223.3 957.3 1737.1 4926Total 1876 2276 1781 3232 9165

Observed Cell Counts (nij):

Example - Cyclones Near Antarctica

Region Season n_ij E_ij (n-E)^2 ((n-E)^2)/E40-49S Autumn 370 310.5 3540.25 11.401771340-49S Winter 452 376.7 5670.09 15.052004240-49S Spring 273 294.8 475.24 1.6120759840-49S Summer 422 535.0 12769 23.867289750-59S Autumn 526 557.2 973.44 1.7470208250-59S Winter 624 676.0 2704 450-59S Spring 513 529.0 256 0.4839319550-59S Summer 1059 959.9 9820.81 10.231076260-79S Autumn 980 1008.3 800.89 0.7942973360-79S Winter 1200 1223.3 542.89 0.4437913860-79S Spring 995 957.3 1421.29 1.484686160-79S Summer 1751 1737.1 193.21 0.11122561

71.2291706

Computation of 2obs

Example - Cyclones Near Antarctica

• H0: Seasonal distribution of cyclone occurences is independent of latitude band

• Ha: Seasonal occurences of cyclone occurences differ among latitude bands

• Test Statistic:

• RR: obs2 .05,6

2 = 12.59

• P-value: Area in chi-squared distribution with (3-1)(4-1)=6 degrees of freedom above 71.2

From Table 8, P(222.46)=.001 P< .001

2.712 obs

SPSS Output - Cyclone ExampleREGION * SEASON Crosstabulation

370 452 273 422 1517

310.5 376.7 294.8 535.0 1517.0

24.4% 29.8% 18.0% 27.8% 100.0%

526 624 513 1059 2722

557.2 676.0 529.0 959.9 2722.0

19.3% 22.9% 18.8% 38.9% 100.0%

980 1200 995 1751 4926

1008.3 1223.3 957.3 1737.1 4926.0

19.9% 24.4% 20.2% 35.5% 100.0%

1876 2276 1781 3232 9165

1876.0 2276.0 1781.0 3232.0 9165.0

20.5% 24.8% 19.4% 35.3% 100.0%

Count

Expected Count

% within REGION

Count

Expected Count

% within REGION

Count

Expected Count

% within REGION

Count

Expected Count

% within REGION

40-49S

50-59S

60-79S

REGION

Total

Autumn Winter Spring Summer

SEASON

Total

Chi-Square Tests

71.189a 6 .000

71.337 6 .000

23.418 1 .000

9165

Pearson Chi-Square

Likelihood Ratio

Linear-by-LinearAssociation

N of Valid Cases

Value dfAsymp. Sig.

(2-sided)

0 cells (.0%) have expected count less than 5. Theminimum expected count is 294.79.

a.

P-value

Misuses of chi-squared Test

• Expected frequencies too small (all expected counts should be above 5, not necessary for the observed counts)

• Dependent samples (the same individuals are in each row, see McNemar’s test)

• Can be used for nominal or ordinal variables, but more powerful methods exist for when both variables are ordinal and a directional association is hypothesized

Measures of Association

• Absolute Risk (AR):

• Relative Risk (RR):

• Odds Ratio (OR): o1 / o2 (o = /(1-))

• Note that if (No association between outcome and grouping variables):– AR=0– RR=1– OR=1

Relative Risk

• Ratio of the probability that the outcome characteristic is present for one group, relative to the other

• Sample proportions with characteristic from groups 1 and 2:

2

22

^

1

11

^

n

y

n

y

Relative Risk• Estimated Relative Risk:

2

^

1

^

RR

95% Confidence Interval for Population Relative Risk:

2

2

^

1

1

^

96.196.1

)1()1(71828.2

))(,)((

yyve

eRReRR vv

Relative Risk

• Interpretation– Conclude that the probability that the outcome

is present is higher (in the population) for group 1 if the entire interval is above 1

– Conclude that the probability that the outcome is present is lower (in the population) for group 1 if the entire interval is below 1

– Do not conclude that the probability of the outcome differs for the two groups if the interval contains 1

Example - Concussions in NCAA Athletes

• Units: Game exposures among college socer players 1997-1999

• Outcome: Presence/Absence of a Concussion• Group Variable: Gender (Female vs Male)• Contingency Table of case outcomes:

Outcome

GenderConcussion

NoConcussion Total

Female 158 74924 75082

Male 101 75633 75734

Total 259 150557 150816Source: Covassin, et al (2003)

Example - Concussions in NCAA Athletes

)13.2,27.1(1.62e,1.62e

:Risk Relative Populationfor 95%CI

1273.0162.101

0013.1

158

0021.1

62.10013.

0021.)/(

es)player/gam male 1000per sConcussion (1.3

0013.075734

101 :Males Among

es)player/gam female 1000per sConcussion (2.1

0021.075082

158 :Females Among

)1.96(.1273)1.96(.1273-

^

^

^

^

vv

MFRRM

F

M

F

There is strong evidence that females have a higher risk of concussion

Odds Ratio

• Odds of an event is the probability it occurs divided by the probability it does not occur

• Odds ratio is the odds of the event for group 1 divided by the odds of the event for group 2

• Sample odds of the outcome for each group:

22

22

11

1

111

111 /)(

/

yn

yodds

yn

y

nyn

nyodds

Odds Ratio

• Estimated Odds Ratio:

)(

)(

)/(

)/(

112

221

222

111

2

1

yny

yny

yny

yny

odds

oddsOR

95% Confidence Interval for Population Odds Ratio

222111

96.196.1

111171828.2

))(,)((

ynyynyve

eOReOR vv

Odds Ratio

• Interpretation– Conclude that the probability that the outcome

is present is higher (in the population) for group 1 if the entire interval is above 1

– Conclude that the probability that the outcome is present is lower (in the population) for group 1 if the entire interval is below 1

– Do not conclude that the probability of the outcome differs for the two groups if the interval contains 1

Osteoarthritis in Former Soccer Players

• Units: 68 Former British professional football players and 136 age/sex matched controls

• Outcome: Presence/Absence of Osteoathritis (OA)• Data:• Of n1= 68 former professionals, y1 =9 had OA, n1-y1=59 did not

• Of n2= 136 controls, y2 =2 had OA, n2-y2=134 did not

)80.48,14.2(23.10,23.10

:Ratio Odds Populationfor CI 95%

797.6355.134

1

2

1

59

1

9

1

23.100149.

1525.

0149.134

21525.

59

9

)797(.96.1)797(.96.1

2

1

211

11

ee

vv

odds

oddsOR

oddsXn

Xodds

Source: Shepard, et al (2003) Interval > 1