Contingency Tables: Tests for independence and homogeneity (

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Contingency Tables Tests for independence and
homogeneity (10.5)
How to test hypotheses of independence
(association) and homogeneity (similarity) for
general two-way cross classifications of count
data.
Terms Contingency Table Cross-Classification
Table Measure of association
Independence in two-way tables Chi-Square Test
for Independence or Homogeneity
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Test of Independence or Association
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).
Data can be formatted into a cross-tabulation or
contingency table.
Each person has two categorical responses.
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What are we interested in from this two-way
classification table?
Is the level of teaching evaluation related to
rank?
Are Professors more likely to be judged above
average than other ranks?
Ho Teaching Evaluation and Rank are independent
variables.
Two variables that have been categorized in a
two-way table are independent if the probability
that a measurement is classified into a given
cell of the table is equal to the probability of
being classified into that row times the
probability of being classified into that column.
This must be true for all cells of the table.
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The independence assumption
Observed
Expected
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Observed Counts
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Expected Counts
Assumptions no Eij lt 1, and no more than 20 of
Eij lt 5.
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Individual Cell Chi Square Values
? Reject Ho
There is evidence of an association between rank
and evaluation. Note that we observed less
Assistant Professors getting below average
evaluations (13) than we would expect under
independence (26.2). Chi Square value is 6.67.
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Minitab
rank eval count 1 1 30 1 2 48 1 3 36 2 1 13 2 2 50
2 3 62 3 1 20 3 2 35 3 3 45 4 1 35 4 2 43 4 3 50
STAT gt TABLES gt Cross
Tabs Classification Variables rank eval Check
Chi-square Analysis, and Above and Std.
residual Frequencies are in count
Input data in this way
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Tabulated Statistics eval, rank Rows eval
Columns rank 1 2
3 4 All 1 30
13 20 35 98
23.92 26.23 20.99 26.86 98.00
1.24 -2.58 -0.22 1.57 --
2 48 50 35 43
176 42.96 47.11 37.69 48.24
176.00 0.77 0.42 -0.44
-0.75 -- 3 36 62
45 50 193 47.11
51.66 41.33 52.90 193.00 -1.62
1.44 0.57 -0.40 -- All
114 125 100 128
467 114.00 125.00 100.00 128.00
467.00 -- --
-- -- -- Chi-Square
17.435, DF 6, P-Value 0.008
Cell Contents -- Count Exp
Freq Std. Resid
Square roots of Individual Chi-square values
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SAS
options ls79 ps40 nocenter 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
dataeval weight number table jobrating /
chisq run
Table of job by rating job
rating Frequency Percent Row Pct Col Pct
Above Average Below Total ƒƒƒƒƒƒƒƒƒˆƒƒƒƒ
ƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ Assistan 62
50 13 125 13.28 10.71
2.78 26.77 49.60 40.00
10.40 32.12 28.41 13.27
ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ Associat
45 35 20 100
9.64 7.49 4.28 21.41 45.00
35.00 20.00 23.32 19.89
20.41 ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ Inst
ruct 36 48 30 114
7.71 10.28 6.42 24.41
31.58 42.11 26.32 18.65
27.27 30.61 ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒ
ƒƒƒˆ Professo 50 43 35
128 10.71 9.21 7.49 27.41
39.06 33.59 27.34
25.91 24.43 35.71 ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒ
ƒƒƒˆƒƒƒƒƒƒƒƒˆ Total 193 176 98
467 41.33 37.69 20.99
100.00
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The FREQ Procedure Statistics for Table of job
by rating Statistic DF
Value Prob ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ Chi-Square
6 17.4354 0.0078 Likelihood Ratio
Chi-Square 6 18.7430 0.0046 Mantel-Haens
zel Chi-Square 1 10.8814 0.0010 Phi
Coefficient
0.1932 Contingency Coefficient
0.1897 Cramer's V
0.1366 Sample Size 467
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SPSS
First you need to tell SPSS that each observation
must be weighted by the cell count.
DATA gt WEIGHT CASES
Then you choose the analysis. ANALYZE gt
DESCRIPTIVE STATISTICS gt CROSS TABS
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R

• gt score lt- c(36,48,30,62,50,13,45,35,20,50,43,35)
• gt mscore lt- matrix(score,3,4)
• gt mscore
• ,1 ,2 ,3 ,4
• 1, 36 62 45 50
• 2, 48 50 35 43
• 3, 30 13 20 35
• gt chisq.test(mscore)
• Pearson's Chi-squared test
• data mscore
• X-squared 17.4354, df 6, p-value 0.00781
• gt out lt- chisq.test(mscore)
• gt out1length(out)
• statistic
• X-squared
• 17.43537

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method 1 "Pearson's Chi-squared
test" data.name 1 "mscore" observed
,1 ,2 ,3 ,4 1, 36 62 45 50 2,
48 50 35 43 3, 30 13 20
35 expected ,1 ,2 ,3
,4 1, 47.11349 51.65953 41.32762
52.89936 2, 42.96360 47.10921 37.68737
48.23983 3, 23.92291 26.23126 20.98501
26.86081 residuals ,1 ,2
,3 ,4 1, -1.6191155 1.4386830
0.5712511 -0.3986361 2, 0.7683695 0.4211764
-0.4377528 -0.7544218 3, 1.2424774 -2.5834003
-0.2150237 1.5704402
Square roots of Individual Chi-square values
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Test of Homogeneity
Suppose we wish to determine if there is an
association between a rare disease and another
more common categorical variable (e.g. smoking).
We cant just take a random sample of subjects
and hope to get enough cases (subjects with the
disease). One solution is to choose a fixed
number of cases, and a fixed number of controls,
and classify each according to whether they are
smokers or not. The same chi square test of
independence applies here, but since we are
sampling within subpopulations (have fixed margin
totals), this is now called a chi square test of
homogeneity (of distributions).
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Homogeneity Null Hypothesis
In general, if the column categories represent c
distinct subpopulations, random samples of size
n1, n2, , nc are selected from each and
classified into the r values of a categorical
variable represented by the rows of the
contingency table. The hypothesis of interest
here is if there a difference in the distribution
of subpopulation units among the r levels of the
categorical variable, i.e. are the subpopulations
homogenous or not.
Subpop 1 Subpop 2 Subpop c ?11
?12 ...
?1c ?21 ?22 ...
?2c ?r1 ?r2
... ?rc
?ij proportion of subpop j subjects (j1,,c)
that fall in category i (i1,,r).
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Null hypothesis of homogeneity
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Example Myocardial Infarction (MI)
Data was collected to determine if there is an
association between myocardial infarction and
smoking in women. 262 women suffering from MI
were classified according to whether they had
ever smoked or not. Two controls (patients with
other acute disorders) were matched to every case.
Is the incidence of smoking the same for MI and
non-MI sufferers? Ho the incidence of MI is
homogenous with respect to smoking Ho ?11?12
and ?21?22
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Example MI results in MTB
Stat -gt Tables -gt Chi-Square Test ----------------
--------------------------------------------------
-------------------------- Chi-Square Test MI
Yes, MI No Expected counts are printed below
observed counts MI Yes MI No
Total 1 172 173 345
115.74 229.26 2 90 346
436 146.26 289.74 Total 262
519 781 Chi-Sq 27.352 13.808 21.643
10.926 73.729 DF 1, P-Value 0.000
Conclude there is evidence of lack of
homogeneity of incidence of MI with respect to
smoking.
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Odds and Odds Ratios

• Sometimes probabilities are expressed as odds,
  e.g.
• Gambling circles. (Why?)
• Biomedical studies. (Easy interpretation in
  logistic regression, etc.)

Odds of Event A P(A) ? (1-P(A)) P(A) Odds of
A / (1 Odds of A)
Ex A horse has odds of 3 to 2 of winning. This
means that in every 325 races the horse wins 3
and loses 2. So P(Wins) 3/5. To use the above
formula express the odds as d to 1, so 1.5 to 1
in this case. Thus P(Wins) 1.5 / (11.5) 1.5
/ 2.5 3/5.
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Example MI and Odds Ratios
For women sufferers of MI, the proportion who
ever smoked is 172/262 0.656. In other words,
the odds that a woman MI sufferer is a smoker are
0.656/(1-0.656) 1.9. For women non-sufferers
of MI, the proportion who ever smoked is 173/519
0.333. In other words, the odds that a woman
non-MI sufferer is a smoker are 0.333/(1-0.333)
0.5. We can now calculate the odds ratio of
being a smoker among MI sufferers OR 1.9/0.5
3.82 Among MI suffers, the odds of being a smoker
are about 4 times the odds of not being a smoker.
Put another way a randomly selected MI sufferer
is about twice as likely (.656/.333) of being a
smoker than of not being one.