Chapter 10
Categorical Data Analysis
Inference for a Single Proportion (p)
• Goal: Estimate proportion of individuals in a population with a
certain characteristic (p). 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:
^
Sample proportion : p 
y
n
Mean and Std. Dev. of sampling distributi on :  ^  p
p
 
^
p
p (1  p )
n
 ^
p 1  p 


Estimated Standard Error : SE ^ 
p
n
Shape : approximat ely normal for large samples (Rule of thumb : np , n(1  p )  5)
^
Large-Sample Confidence Interval for p
• Take SRS of size n from population where p is true
(unknown) proportion of successes.
– Observe y successes
– Set confidence level (1-a) and obtain za/2 from z-table
y
Point Estimate : p 
n
^
^

p 1  p 


Estimated Standard Error : SE ^ 
p
n
Margin of error : m  za / 2SE ^
^
p
^
C % confidence interval for p : p  m
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: p = 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
18
(.14)(.86)
p
 .143
SE ^ 
 .031
p
126
126
Margin of error ((1  a )100%  95%) : m  1.96(.031)  .061
^
95% CI for p : .143  .061  (.082,.204)
Wilson-Agresti-Coull Method
• For moderate to small sample sizes, large-sample
methods may not work well wrt coverage probabilities
• Simple approach that works well in practice:
– Adjust observed number of Successes (y) and sample size (n)
~
y  y  0.5 za / 2
2
~
2
Note that for a  0.05, z.025
 1.962  4
n  n  za2 / 2
~
~
Point Estimate: p 
y
~
n
~

p 1  p 


~
Estimated Standard Error: SE ~ 
p
~
n
Margin of error: m  za / 2SE ~
p
~
(1  a )100% confidence interval for p : p  m
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
~
y  1  0.5 1.96   1  0.5  3.84   2.92
~
2
n  12  1.96   12  3.84  15.84
2
2.92
.1843(.8157)
p
 .1843
SE ~ 
 .0974
p
15.84
15.84
Margin of error((1-a )100%  95%) : 1.96(.0974)  .1910
95% CI for p : .1875  .1910  ( .0035,.3985)  (0,.40)
~
Sample Size for Margin of Error = E
• Goal: Estimate p within E with 100(1-a% Confidence
• Confidence Interval will have width of 2E
p 1  p 
za2 /2p 1  p 
m  E  za /2
 n
n
E2
Since p is unknown, an educated guess can be used or set p  0.5
This is most conservative as p 1  p  is largest for p  0.5
a  0.05, p  0.5  za2 /2p 1  p   22  0.5 1  0.5   1
1
 n 2
E
Significance Test for a Proportion
• Goal test whether a proportion (p) equals some null
value p0 H0: pp0
^
p p0
Test Statistic : zobs 
p o (1  p 0 )
Ha :p  p 0
n
RR : zobs  za
Ha :p  p 0
RR : zobs   za
Ha :p  p 0
RR : zobs  za / 2
P - value  P( Z  zobs )
P - value  P( Z  zobs )
P - value  2 P( Z  zobs )
Large-sample test works well when np0 and n(1-p0)  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: p=0.25 Ha: p < 0.25
18
 0.143 p 0  0.25
126
.143  .25  .107
Test Statistic : zobs 

 2.75
.039
.25(.75)
126
RR (a  .05) : zobs   z.05  1.645
n  126
^
y  18 p 
P - value  P( Z  2.75)  .0030
Strong evidence that incidence rate is below 25% (p < 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
Group 1
Outcome
Present
y1
Outcome
Absent
n1-y1
Group
Total
n1
Group 2
y2
n2-y2
n2
Outcome
Total
y1+y2
(n1+n2)(y1+y2)
n1+n2
Example - Firm Type/Product Quality
Not
Integrated
Vertically
Integrated
Outcome
Total
High
Quality
Low
Quality
Group
Total
33
55
88
5
79
84
38
134
172
• 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: p1
• 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:
y1
p1 
n1
• Similar notation for Population/Sample 2
Example - Cotton Textile Producers
 p1 - True proportion of all Non-integretated
firms that would produce High quality
 p2 - True proportion of all vertically integretated
firms that would produce High quality
n1  88
n2  84
y1  33
y1 33
p 1    0.375
n1 88
y2  5
y2 5
p 2    0.060
n2 84
^
^
Notation (Continued)
• Parameter of Primary Interest: p1-p2, the difference
in the 2 population proportions with the
characteristic (2 other measures given below)
^
^
• Estimator:
D  p 1p 2
• Standard Error (and its estimate):
 ^  ^  ^ 
p 1 1  p 1  p 2  1  p 2 

 

SED 
n1
n2
^
D 
p 1 (1  p 1 ) p 2 (1  p 2 )
n1

n2
• Pooled Estimated Standard Error when p1p2p:
 ^  1 1 
 p 1  p   

 n1 n2 
^
SEDP
y1  y2
p
n1  n2
^
Cotton Textile Producers (Continued)
• Parameter of Primary Interest: p1p2, the difference
in the 2 population proportions that produce High
quality output
^
^
• Estimator:
D  p 1  p 2  0.375  0.060  0.315
• Standard Error (and its estimate):
 ^  ^  ^ 
p 1 1  p 1  p 2 1  p 2 

 
  0.375(0.625)  0.060(0.94)  .003335  .0577
SED 
n1
n2
88
84
^
• Pooled Estimated Standard Error when p1p2p:
SEDP
1 1
 0.2210.779    .0633
 88 84 
^
p
33  5
 0.221
88  84
Significance Tests for p1p2
• Deciding whether p1p2 can be done by interpreting
“plausible values” of p1p2 from the confidence interval:
– If entire interval is positive, conclude p1  p2 (p1p2 > 0)
– If entire interval is negative, conclude p1  p2 (p1p2 < 0)
– If interval contains 0, do not conclude that p1  p2
• Alternatively, we can conduct a significance test:
– H0: p1  p2 Ha: p1  p2 (2-sided)
^
^
– Test Statistic:
p 1p 2
zobs 
Ha: p1  p2 (1-sided)
 ^  1 1 
p 1  p   

 n1 n2 
^
– RR: |zobs|  za/2 (2-sided)
zobs  za (1-sided)
– P-value: 2P(Z|zobs|) (2-sided)
P(Z zobs) (1-sided)
Example - Cotton Textile Production
H 0 : p1  p 2
(p 1  p 2  0)
H A : p1  p 2
(p 1  p 2  0)
^
TS : zobs
^
p 1p 2
0.375  0.060
0.315



 4.98
^
 1 1  0.0633
 ^  1 1 
0.221(0.779)  
p 1  p   
 88 84 

 n1 n2 
RR : zobs  z.025  1.96
P - value  2 P( Z  4.98)  0
Again, there is strong evidence that non-integrated performs are
more likely to produce high quality output than integrated firms
Fisher’s Exact Test
• Method of testing for testing whether p2=p1 when
one or both of the group sample sizes is small
• Measures (conditional on the group sizes and
number of cases with and without the
characteristic) the chances we would see
differences of this magnitude or larger in the
sample proportions, if there were no differences in
the populations
Example – Echinacea Purpurea for Colds
• Healthy adults randomized to receive EP (n1=24)
or placebo (n2=22, two were dropped)
• Among EP subjects, 14 of 24 developed cold after
exposure to RV-39 (58%)
• Among Placebo subjects, 18 of 22 developed cold
after exposure to RV-39 (82%)
• Out of a total of 46 subjects, 32 developed cold
• Out of a total of 46 subjects, 24 received EP
Source: Sperber, et al (2004)
Example – Echinacea Purpurea for Colds
• Conditional on 32 people
developing colds and 24
receiving EP and 22
receiving placebo, the
following table gives the
outcomes that would have
been as strong or stronger
evidence that EP reduced
risk of developing cold (1sided test). P-value from
SPSS is .079 (next slide).
EP/Cold
14
13
12
11
10
Sum
Placebo/Cold Probability
18
0.059808
19
0.016025
20
0.002604
21
0.000229
22
0.000008
0.078674
 nEP  nPL 



yEP  yPL 

Probabilities:p  yEP , yPL  
 nEP  nPL 


 yEP  yPL 
 24  22 
   1961256 7315
14 18


  .059808
p 14,18      
23987744005
 46 
 
 32 
...
 24  22 
  
10 22
1961256 1  .0000082
p 10, 22      
23987744005
 46 
 
 32 
Example - SPSS Output
r
C
O
L
N
o
e
o
T
E
4
P
2
T
6
a
r
c
c
p
t
t
s
a
s
d
i
i
i
l
d
d
d
f
u
b
P
0
1
4
a
C
4
1
9
L
1
1
0
F
4
9
N
6
a
C
b
0
6
McNemar’s Test for Paired Samples
• Common subjects (or matched pairs) being observed
under 2 conditions (2 treatments, before/after, 2
diagnostic tests) in a crossover setting
• Two possible outcomes (Presence/Absence of
Characteristic) on each measurement
• Four possibilities for each subject/pair wrt outcome:
–
–
–
–
Present in both conditions
Absent in both conditions
Present in Condition 1, Absent in Condition 2
Absent in Condition 1, Present in Condition 2
McNemar’s Test for Paired Samples
Condition 1\2
Present
Absent
Present
n11
n12
Absent
n21
n22
McNemar’s Test for Paired Samples
• Data: n12 = # of pairs where the characteristic is present
in condition 1 and not 2 and n21 # where present in 2 and
not 1
• H0: Probability the outcome is Present is same for the 2
conditions (p1 = p2)
• HA: Probabilities differ for the 2 conditions (p1 ≠ p2)
Large-Sample Test (Normal Approximation to Binomial)
n12  n21
T .S .: zobs 
n12  n21
P  val  2 P ( Z | zobs |)
Example - Reporting of Silicone Breast
Implant Leakage in Revision Surgery
• Subjects - 165 women having revision surgery involving
silicone gel breast implants
• Conditions (Each being observed on all women)
– Self Report of Presence/Absence of Rupture/Leak
– Surgical Record of Presence/Absence of Rupture/Leak
L
C
G
p
o
u
t
t
S
R
9
8
7
N
5
3
8
T
4
1
5
Source: Brown and Pennello (2002), “Replacement Surgery and Silicone Gel Breast Implant Rupture”,
Journal of Women’s Health & Gender-Based Medicine, Vol. 11, pp 255-264
Example - Reporting of Silicone Breast
Implant Leakage in Revision Surgery
• H0: Tendency to report ruptures/leaks is the same
for self reports and surgical records
• HA: Tendencies differ
T .S .: zobs
n12  n21
28  5


 4.00
n12  n21
28  5
P  val  2 P ( Z | zobs |)  2 P ( Z  4)  2(.0000317)  0
Exact P-value: 2P Y  28 | Y ~ B  n  33, p  0.5  
Multinomial Experiment / Distribution
• Extension of Binomial Distribution to
experiments where each trial can end in
exactly one of k categories
• n independent trials
• Probability a trial results in category i is pi
• ni is the number of trials resulting in
category I
• p1+…+pk = 1
• n1+…+nk = n
Multinomial Distribution / Test for Cell Probabilities
p  n1 ,..., nk  
k
n
i 1
i
 n,
n!
p 1n1 ...p knk
n1 !...nk !
k
p
i 1
i
 1, ni  0, p i  0
Testing whether the category probabilities are specific values:
H 0 : p 1  p 10 ,..., p k  p k 0
k
p
i 1
i0
1
H A : At least one cell probability is not as specified
Expected cell counts under H 0 : Ei  np i 0
k
2
Test Statistic:  obs

i 1
 ni  Ei 
Ei
i  1,..., n
2
2
2
Rejection Region:  obs
  a2,k 1 P-value: P   k21   obs

Goodness of Fit Test for a Probability Distribution
• Data are collected and wish to be determined whether it
comes from a particular probability distribution (e.g.
Poisson, Normal, Gamma)
• Estimate any unknown model parameters (p estimates)
• Break down the range of data values into k > p intervals
(typically where ≥ 80% have expected counts ≥ 5) obtain
observed (n) and expected (E) values for each interval
k
2
Test Statistic:  obs

i 1
2
P-value: P   k2 p   obs

 ni  Ei 
2
Ei
Assessing quality of fit to hypothesized distribution:
P-value
Quality of Fit
> .25
Excellent
.15-.25
Good
.05-.15 Moderately Good
.01-.05
Poor
<.01
Unacceptable
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
40-49S
50-59S
60-79S
Total
Autumn
370
526
980
1876
Winter
452
624
1200
2276
Spring
273
513
995
1781
Summer
422
1059
1751
3232
Total
1517
2722
4926
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
40-49S
50-59S
60-79S
Autumn
24.4
19.3
19.9
Winter
29.8
22.9
24.4
Spring
18.0
18.9
20.2
Summer
27.8
38.9
35.5
Total% (n)
100.0 (1517)
100.0 (2722)
100.0 (4926)
Example - Cyclones Near Antarctica
40.00
region
40-49S
50-59S
60-79S
30.00
regp ct
Bars show Means
20.00
10.00
Autumn
Winter
Spring
Summer
season
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)
1
2
…
c
Total
1
n11
n12
…
n1c
n1.
2
n21
n22
…
n2c
n2.
…
…
…
…
…
…
r
nr1
nr2
…
nrc
nr.
Total
n.1
n.2
…
n.c
n..
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 (at least 80% of Eij > 5)
• H0: Variables are statistically independent
(No association between variables)
• Ha: Variables are statistically dependent
(Association exists between variables)
• Test Statistic:  2  (nij  Eij )2
obs

Eij
2
• P-value: Area above  obs
in the chi-squared
distribution with (r-1)(c-1) degrees of
freedom. (Critical values in Table 8)
Example - Cyclones Near Antarctica
Observed Cell Counts (nij):
Region\Season
40-49S
50-59S
60-79S
Total
Autumn
370
526
980
1876
Winter
452
624
1200
2276
Spring
273
513
995
1781
Summer
422
1059
1751
3232
Total
1517
2722
4926
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
40-49S
50-59S
60-79S
Total
Autumn
310.5
557.2
1008.3
1876
Winter
376.7
676.0
1223.3
2276
Spring
294.8
529.0
957.3
1781
Summer
535.0
959.9
1737.1
3232
Total
1517
2722
4926
9165
Example - Cyclones Near Antarctica
Computation of
Region
40-49S
40-49S
40-49S
40-49S
50-59S
50-59S
50-59S
50-59S
60-79S
60-79S
60-79S
60-79S
2
 obs
Season
Autumn
Winter
Spring
Summer
Autumn
Winter
Spring
Summer
Autumn
Winter
Spring
Summer
n_ij
E_ij
370
452
273
422
526
624
513
1059
980
1200
995
1751
310.5
376.7
294.8
535.0
557.2
676.0
529.0
959.9
1008.3
1223.3
957.3
1737.1
(n-E)^2
3540.25
5670.09
475.24
12769
973.44
2704
256
9820.81
800.89
542.89
1421.29
193.21
((n-E)^2)/E
11.4017713
15.0520042
1.61207598
23.8672897
1.74702082
4
0.48393195
10.2310762
0.79429733
0.44379138
1.4846861
0.11122561
71.2291706
2
 obs
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
2
• Test Statistic:
 obs  71.2
• RR: obs2  .05,62 = 12.59
• P-value: Area in chi-squared distribution with (31)(4-1)=6 degrees of freedom above 71.2
From Table 8, P(2 22.46)=.001  P< .001
Likelihood Ratio Statistic
Alternative statistic provided by many computer packages:
Test Statistic: 
 nij
 2 nij ln 
E
i 1 j 1
 ij
r
2
LR
c
r
c
 n  nij  


  2 nij ln 
 ni n j 
i 1 j 1



2
Rejection Region:  LR
 a2 , r 1 c 1

P-value: P  2r 1 c 1   L2R
Note: The formula on
page 512 of textbook is
incorrect

Row(i)
1
1
1
1
2
2
2
2
3
3
3
3
Sum
Column(j) n_ij
n_i●
n_●j
X2(LR)
1
370
1517
1876 129.6947
2
452
1517
2276 164.6768
3
273
1517
1781 -41.9335
4
422
1517
3232 -200.192
1
526
2722
1876 -60.5646
2
624
2722
2276 -99.8393
3
513
2722
1781 -31.4258
4
1059
2722
3232 208.0912
1
980
4926
1876 -55.8208
2
1200
4926
2276 -46.1601
3
995
4926
1781 76.9673
4
1751
4926
3232 27.84263
71.33672
SPSS Output - Cyclone Example
REGION * SEASON Crosstabulation
REGION
40-49S
50-59S
60-79S
Total
Count
Expected Count
% within REGION
Count
Expected Count
% within REGION
Count
Expected Count
% within REGION
Count
Expected Count
% within REGION
Autumn
370
310.5
24.4%
526
557.2
19.3%
980
1008.3
19.9%
1876
1876.0
20.5%
SEASON
Winter
Spring
452
273
376.7
294.8
29.8%
18.0%
624
513
676.0
529.0
22.9%
18.8%
1200
995
1223.3
957.3
24.4%
20.2%
2276
1781
2276.0
1781.0
24.8%
19.4%
Summer
422
535.0
27.8%
1059
959.9
38.9%
1751
1737.1
35.5%
3232
3232.0
35.3%
Total
1517
1517.0
100.0%
2722
2722.0
100.0%
4926
4926.0
100.0%
9165
9165.0
100.0%
Chi-Square Tests
Pears on Chi-Square
Likelihood Ratio
Linear-by-Linear
Ass ociation
N of Valid Cas es
Value
71.189a
71.337
23.418
6
6
Asymp. Sig.
(2-s ided)
.000
.000
1
.000
df
9165
a. 0 cells (.0%) have expected count less than 5. The
minimum expected count is 294.79.
P-value
Misuses of chi-squared Test
• Expected frequencies too small (at least
80% of 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
Ordinal Explanatory and Response Variables
• Pearson’s Chi-square test can be used to test
associations among ordinal variables, but more
powerful methods exist
• When theories exist that the association is
directional (positive or negative), measures exist
to describe and test for these specific alternatives
from independence:
– Gamma
– Kendall’s tb
Concordant and Discordant Pairs
• Concordant Pairs - Pairs of individuals where one
individual scores “higher” on both ordered
variables than the other individual
• Discordant Pairs - Pairs of individuals where one
individual scores “higher” on one ordered variable
and the other individual scores “higher” on the
other
• C = # Concordant Pairs D = # Discordant Pairs
– Under Positive association, expect C > D
– Under Negative association, expect C < D
– Under No association, expect C  D
Example - Alcohol Use and Sick Days
• Alcohol Risk (Without Risk, Hardly any
Risk, Some to Considerable Risk)
• Sick Days (0, 1-6, 7)
• Concordant Pairs - Pairs of respondents
where one scores higher on both alcohol
risk and sick days than the other
• Discordant Pairs - Pairs of respondents
where one scores higher on alcohol risk and
the other scores higher on sick days
Source: Hermansson, et al (2003)
Example - Alcohol Use and Sick Days
A
C
D
d
o
d
d
a
t
A
W
7
3
5
5
H
4
3
6
3
S
2
5
4
1
T
3
1
5
9
• Concordant Pairs: Each individual in a
given cell is concordant with each individual
in cells “Southeast” of theirs
•Discordant Pairs: Each individual in a given
cell is discordant with each individual in
cells “Southwest” of theirs
Example - Alcohol Use and Sick Days
A
C
D
d
o
d
d
a
t
A
W
7
3
5
5
H
4
3
6
3
S
2
5
4
1
T
3
1
5
9
C  347(63  56  25  34)  113(56  34)  154(25  34)  63(34)  83164
D  145(154  63  52  25)  113(154  52)  56(52  25)  63(52)  73496
Measures of Association
• Goodman and Kruskal’s Gamma:
CD
 
CD
^
^
1     1
• Kendall’s tb:
^
tb 
0.5

CD
n 2   ni.
2

n 2   n. j
2

When there’s no association between the ordinal variables,
the population based values of these measures are 0.
Statistical software packages provide these tests.
Example - Alcohol Use and Sick Days
C  D 83164  73496


 0.0617
C  D 83164  73496
^
c
y
m
a
b
o
r
l
E
x
o
u
O
K
5
0
7
5
O
G
2
2
7
5
N
9
a
N
b
U
Measures of Association
•
•
•
•
Absolute Risk (AR): p1p2
Relative Risk (RR): p1 / p2
Odds Ratio (OR): o1 / o2 (o = p/(1-p))
Note that if p1  p2 (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:
y1
p1 
n1
^
y2
p2 
n2
^
Relative Risk
• Estimated Relative Risk:
^
RR 
p1
^
p2
95% Confidence Interval for Population Relative Risk:
( RR (e
1.96 v
1.96 v
) , RR (e
^
))
^
(1  p 1 ) (1  p 2 )
e  2.71828 v 

y1
y2
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
Source: Covassin, et al (2003)
No
Concussion Concussion
Total
Gender
Female
158
74924
75082
Male
101
75633
75734
Total
259
150557
150816
Example - Concussions in NCAA Athletes
158
 0.0021
75082
(2.1 Concussion s per 1000 female player/gam es)
^
101
 0.0013
Among Males : p M 
75734
(1.3 Concussion s per 1000 male player/gam es)
^
Among Females : p F 
^
RR ( F / M ) 
pF
^
pM
.0021
 1.62

.0013
1  .0021 1  .0013
v  .1273
 .0162

101
158
95%CI for Population Relative Risk :
v
1.62e
-1.96(.1273)
,1.62e1.96(.1273)

 (1.27,2.13)
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:
y1 / n1
y1
odds1 

( n1  y1 ) / n1 n1  y1
y2
odds2 
n2  y2
Odds Ratio
• Estimated Odds Ratio:
odds1
y1 /( n1  y1 ) y1 (n2  y2 )
OR 


odds2 y2 /( n2  y2 ) y2 (n1  y1 )
95% Confidence Interval for Population Odds Ratio
( OR (e 1.96 v ) , OR (e1.96 v ) )
e  2.71828
1
1
1
1
v



y1 n1  y1 y2 n2  y2
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
odds1 
OR 
X1
9
2

 .1525 odds2 
 .0149
n1  X 1 59
134
odds1 .1525

 10.23
odds2 .0149
1 1 1 1
  
 .6355
v  .797
9 59 2 134
95% CI for Population Odds Ratio :
v
Source: Shepard, et al (2003)
10.23e
1.96(.797)
,10.23e1.96(.797)

 (2.14,48.80)
Interval > 1
Mantel-Haenszel Test / CI for Multiple Tables
• Data collected from q studies or strata in 2x2
contingency tables with common groupings/outcomes
• Each table has 4 cells: nh11, nh12, nh21, nh21 h=1,…,q
• They can be combined for an overall Chi-square statistic
or odds ratio and confidence Interval
Table 1
Trt\Response
1
1
n_111
2
n_121
Total
n_1●1
2
n_112
n_122
n_1●2
Total
n_11●
n_12●
n_1●●
Table q
Trt\Response
1
1
n_q11
2
n_q21
Total
n_q●1
2
n_q12
n_q22
n_q●2
Total
n_q1●
n_q2●
n_q●●
Mantel-Haenszel Computations
 
nh1 nh1  
   nh11 

n
h 1 
h 

 q
nh1 nh 2 nh1nh2

2
h 1 nh   nh  1
q
2
Test Statisic:  MH
2
Rejection Region:  MH
 a2 ,1
^
OR MH
R

S
q
n n
R   h11 h 22
nh
h 1
2
2
P-value: P  12   MH

q
nh12 nh 21
nh
h 1
S 
 1
1
1
1 
S 





n
n
n
n
h 1
h12
h 21
h 22 
 h11
^
 ^

1.96 v
95% CI for Overall Odds Ratio:  OR MH e
, OR MH e1.96 v 


 ^
 1
v  V  OR MH   2

 S
^
q
2
h