Two-Way Contingency Tables:
Tests for Independence and Homogeneity
1. Test for Homogeneity (product multinomial sampling):
Example 1 (2X2 table – test for equality of two population proportions,
independent samples, both large). Based on the following case-control data:
Smoker
15
8
Stroke
No Stroke
Non-Smoker
35
12
(a) Please test at the significance level of 0.05 whether smoking is a risk factor
for stroke.
(b) Please write up the entire SAS program necessary to answer the question
raised in (a), including the data step.
Solution:
(a)
Smoker
Non-Smoker
Total
Stroke (case)
15
35
50
No Stroke (control)
8
12
20
𝐻0 : 𝑃
𝑠𝑚𝑜𝑘𝑒𝑟𝑠
𝑎𝑚𝑜𝑛𝑔 𝑐𝑎𝑠𝑒
(𝑠𝑡𝑟𝑜𝑘𝑒)
𝐻𝛼 : 𝑃
𝑠𝑚𝑜𝑘𝑒𝑟𝑠
𝑎𝑚𝑜𝑛𝑔 𝑐𝑎𝑠𝑒
(𝑠𝑡𝑟𝑜𝑘𝑒)
𝑃̂ =
Z0 =
p̂1 − p̂2 − 0
1
1
√p̂(1 − p̂) ( + )
n1 n2
=
=𝑃
𝑠𝑚𝑜𝑘𝑒𝑟𝑠
𝑎𝑚𝑜𝑛𝑔 𝑐𝑜𝑛𝑡𝑟𝑜𝑙𝑠
(𝑛𝑜 𝑠𝑡𝑟𝑜𝑘𝑒)
≠𝑃
𝑠𝑚𝑜𝑘𝑒𝑟𝑠
𝑎𝑚𝑜𝑛𝑔 𝑐𝑜𝑛𝑡𝑟𝑜𝑙𝑠
(𝑛𝑜 𝑠𝑡𝑟𝑜𝑘𝑒)
15 + 8
≈ 0.33
50 + 20
0.30 − 0.40
√0.33(1 − 0.33) ( 1 + 1 )
50 20
≈ −0.80
|Z0 | ≈ 0.08 < Z0.025 = 1.96
We can not reject H 0 at
α = 0.05 and conclude that based on the given data,
smoking is not confirmed as a risk factor for stroke.
(a) SAS code:
1
Data smoking;
Input stroke $ smoker $ count;
Datalines;
stroke yes 15
stroke no 35
nostrk yes 8
nostrk no 12
;
Run;
Proc freq data=smoking;
Tables stroke*smoker/chisq;
Weight count;
Run;
Discussion: Now we examine the relations between the two
possible Z-tests, and the Chi-square test for homogeneity as follows.
The general format of the data is:
Smoker
Non-Smoker
Stroke
a (15)
b (35)
a+b (50)
No stroke
c (8)
d (12)
c+d (20)
a+c (23)
b+d (47)
a+b+c+d (70)
The two possible Z-tests are (*** although the one we have shown
above in part (a) is the reasonable choice – but they indeed both lead to
the same test statistic – which in turn is equivalent to the Chi-square test
for homogeneity in this 2X2 table):
𝐻0 : 𝑃
𝑠𝑚𝑜𝑘𝑒𝑟𝑠
𝑎𝑚𝑜𝑛𝑔 𝑐𝑎𝑠𝑒
(𝑠𝑡𝑟𝑜𝑘𝑒)
𝐻𝛼 : 𝑃
𝑠𝑚𝑜𝑘𝑒𝑟𝑠
𝑎𝑚𝑜𝑛𝑔 𝑐𝑎𝑠𝑒
(𝑠𝑡𝑟𝑜𝑘𝑒)
=𝑃
Since
𝑎𝑚𝑜𝑛𝑔
𝑠𝑚𝑜𝑘𝑒𝑟
≠𝑃
̂2 − 0
𝑃̂1 − 𝑃
1
1
√𝑃̂(1 − 𝑃̂)( + )
𝑛1 𝑛2
𝑠𝑡𝑟𝑜𝑘𝑒𝑠
𝑎𝑚𝑜𝑛𝑔
𝑛𝑜𝑛− 𝑠𝑚𝑜𝑘𝑒𝑟𝑠
𝐻𝛼 : 𝑃𝑠𝑡𝑟𝑜𝑘𝑒𝑠 ≠ 𝑃
𝑠𝑚𝑜𝑘𝑒𝑟𝑠
𝑎𝑚𝑜𝑛𝑔 𝑐𝑜𝑛𝑡𝑟𝑜𝑙𝑠
(𝑛𝑜 𝑠𝑡𝑟𝑜𝑘𝑒)
𝑎𝑚𝑜𝑛𝑔
𝑠𝑚𝑜𝑘𝑒𝑟
𝑠𝑡𝑟𝑜𝑘𝑒𝑠
𝑎𝑚𝑜𝑛𝑔
𝑛𝑜𝑛− 𝑠𝑚𝑜𝑘𝑒𝑟𝑠
𝑍0
(Large sample sizes)
𝑍0 =
𝐻0 : 𝑃𝑠𝑡𝑟𝑜𝑘𝑒𝑠 = 𝑃
𝑠𝑚𝑜𝑘𝑒𝑟𝑠
𝑎𝑚𝑜𝑛𝑔 𝑐𝑜𝑛𝑡𝑟𝑜𝑙𝑠
(𝑛𝑜 𝑠𝑡𝑟𝑜𝑘𝑒)
𝐻0
~ 𝑁(0,1)
=
∗
̂2 ∗ − 0
𝑃̂1 − 𝑃
1
1
√𝑃̂∗ (1 − 𝑃̂∗ )( ∗ + ∗ )
𝑛1
𝑛2
𝐻0
~ 𝑁(0,1)
Since
2
𝑎
𝑐
̂2 =
, 𝑃
,
𝑎+𝑏
𝑐+𝑑
𝑎+𝑐
𝑃̂ =
𝑎+𝑏+𝑐+𝑑
𝑛1 = 𝑎 + 𝑏, 𝑛2 = 𝑐 + 𝑑
𝑎
𝑏
̂2 ∗ =
, 𝑃
,
𝑎+𝑐
𝑏+𝑑
𝑎+𝑏
𝑃̂ ∗ =
𝑎+𝑏+𝑐+𝑑
𝑛1 ∗ = 𝑎 + 𝑐, 𝑛2 ∗ = 𝑏 + 𝑑
∗
𝑃̂1 =
𝑃̂1 =
then
then
𝑎
𝑐
−
−0
𝑎
+
𝑏
𝑐
+
𝑑
𝑍0 =
𝑎+𝑐
𝑏+𝑑
1
1
√
(
+
)
𝑎+𝑏+𝑐+𝑑𝑎+𝑏+𝑐+𝑑 𝑎+𝑏 𝑐+𝑑
𝑍0 =
𝑎
𝑏
−
−0
𝑎+𝑐 𝑏+𝑑
𝑎+𝑏
𝑐+𝑑
1
1
√
(
+
)
𝑎+𝑏+𝑐+𝑑𝑎+𝑏+𝑐+𝑑 𝑎+𝑐 𝑏+𝑑
𝑎+𝑏+𝑐+𝑑
= ⋯ = (𝑎𝑑 − 𝑏𝑐)√
(𝑎 + 𝑐)(𝑏 + 𝑑)(𝑎 + 𝑏)(𝑐 + 𝑑)
Now we denote the probabilities of the four table cells as follows:
Smoker
Non-Smoker
Stroke
𝑝11 (= 𝑝1 )
𝑝12 (= 1 − 𝑝1 )
No stroke
𝑝21 (= 𝑝2 )
𝑝22 (= 1 − 𝑝2 )
The original null hypothesis of equal population proportions:
𝐻0 : 𝑝1 = 𝑝2
is equivalent to the null hypothesis for the homogeneity test:
𝐻0 : 𝑝11 = 𝑝21 , 𝑝12 = 𝑝22
The test statistic is:
𝜒02 =
[𝑎−𝑃̂(𝑎+𝑏)]2
𝑃̂(𝑎+𝑏)
+
[𝑏−(1−𝑃̂)(𝑎+𝑏)]2
(1−𝑃̂)(𝑎+𝑏)
+
[𝑐−𝑃̂(𝑐+𝑑)]2
𝑃̂(𝑐+𝑑)
+
[𝑑−(1−𝑃̂)(𝑐+𝑑)]2
(1−𝑃̂)(𝑐+𝑑)
(𝑎 + 𝑐)(𝑎 + 𝑏) 2
(𝑏 + 𝑑)(𝑎 + 𝑏) 2
]
[𝑏 −
]
𝑎+𝑏+𝑐+𝑑
𝑎+𝑏+𝑐+𝑑
+
(𝑎 + 𝑐)(𝑎 + 𝑏)
(𝑏 + 𝑑)(𝑎 + 𝑏)
𝑎+𝑏+𝑐+𝑑
𝑎+𝑏+𝑐+𝑑
[𝑎 −
=
2
2
𝑎+𝑏+𝑐+𝑑
2
and 𝜒02 ~𝜒(2−1)(2−1)
(𝑎 + 𝑐)(𝑐 + 𝑑)
(𝑏 + 𝑑)(𝑐 + 𝑑)
[𝑐 −
]
[𝑑 −
]
𝑎+𝑏+𝑐+𝑑
𝑎+𝑏+𝑐+𝑑
2
+
+
= 𝑍0
(𝑎 + 𝑐)(𝑐 + 𝑑)
(𝑏 + 𝑑)(𝑐 + 𝑑)
𝑎+𝑏+𝑐+𝑑
𝑎+𝑏+𝑐+𝑑
Where 𝑃̂ =
𝑎+𝑐
[For 𝑚1 × 𝑚2 table, 𝑑𝑓 = (𝑚1 − 1)(𝑚2 − 1)]
3
The above example can be easily extended to other two-way
contingency tables for testing homogeneity as follows:
# of
0
1
2
≥3
Stroke
𝑝11
𝑝12
𝑝13
𝑝14
No stroke
𝑝21
𝑝22
𝑝23
𝑝24
Children
𝐻0 : 𝑝11 = 𝑝21 , 𝑝12 = 𝑝22 , 𝑝13 = 𝑝23 , 𝑝14 = 𝑝24
Note: The test for homogeneity is indeed a test of independence as
well, that is why the two tests have the same Chi-square test statistic.
2. Test for Independence (multinomial sampling):
Example 2. A university conducted a study concerning faculty teaching
evaluation classification by students. A sample of 467 faculty is randomly
selected, and each person is classified according to rank (Instructor, Assistant
Professor, etc. ) and teaching evaluation (Above, Average, Below).
Rank
Teaching
Evaluation
Above
Average
Average
Below
Average
Sum
Relative
Frequency
Instructor
Assistant Associate
Relative
Professor Professor Professor Sum Frequency
36
62
45
50
193
0.413
48
50
35
43
176
0.377
30
13
20
35
98
0.210
114
125
100
128
467
1.000
0.244
0.268
0.214
0.274
1.000
Ho: Teaching Evaluation and Rank are independent variables.
Teaching
Evaluation Instructor
Above
p11
Average
Average
p21
Below
p31
Average
Sum
n j 114
Relative
p.1
Frequency
Rank
Assistant Associate
Relative
ni
Professor Professor Professor Sum Frequency
p12
p13
p14
193
p1.
p22
p23
p24
176
p2.
p32
p33
p34
98
p3.
125
100
128
467
p.2
p.3
p.4
1.000
n
1.000
4
The independence assumption:
p ij  p ip  j for all ij
Obserrved :
nij
Expected :
Eij  n  pˆij  n  pˆi pˆ j 
𝑟
ni   n j
n
𝑐
(𝑛𝑖𝑗 −𝐸𝑖𝑗 )
𝜒02 = ∑ ∑
𝐸𝑖𝑗
2
𝑖=1 𝑗=1
df = (r-1)(c-1)
For this example: r=#rows=3,
c=#cols=4,
3X4 table
Expected counts:
Rank
Teaching
Evaluation
Above
Average
Average
Below
Average
Sum
Instructor
Assistant Associate
Professor Professor Professor Sum
47.113
51.660
41.328
52.899
193
42.964
47.109
37.687
48.240
176
23.923
26.231
20.985
26.861
98
114
125
100
128
467
Individual cell chi-square values:
Teaching
Evaluation
Above
Average
Instructor
Assistant Associate
Professor Professor Professor
2.6215
2.0698
0.3263
0.1589
Average
0.5904
0.1774
0.1916
0.5692
Below
Average
1.5438
6.6740
0.0462
2.4663
 2  2.62    2.47  17.44   62,0.95  12.59
5
 Reject Ho
There is evidence of an association between rank and evaluation.
SAS Program:
data eval;
input job $ rating $ number;
datalines;
Instructor Above 36
Instructor Average 48
Instructor Below
30
Assistant Above 62
Assistant Average 50
Assistant Below 13
Associate Above 45
Associate Average 35
Associate Below 20
Professor Above 50
Professor Average 43
Professor Below 35
;
run;
proc freq data=eval;
weight number;
tables job*rating / chisq ;
run;
The FREQ Procedure
Statistics for Table of job by rating
Statistic
DF
Value
Prob
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Chi-Square
6
17.4354
0.0078
Likelihood Ratio Chi-Square
6
18.7430
0.0046
Mantel-Haenszel Chi-Square
1
10.8814
0.0010
Phi Coefficient
0.1932
Contingency Coefficient
0.1897
Cramer's V
0.1366
Sample Size = 467
6