contingency tables

1. Contingency Table Analysis

2. • contingency tables show frequencies
produced by cross-classifying observations
• e.g., pottery described simultaneously
according to vessel form & surface
decoration
polished burnished matte
bowl 47 28 3
jar 30 42 8
olla 6 45 25

3. • most statistical tests for tables are designed
for analyzing 2-dimensions
– only examine the interaction of two variables at
one time…
• most efficient when used with nominal data
– using ratio data means recoding data to a lower
scale of measurement (ordinal)
– means ignoring some of the information
originally available…

4. • still, you might do this, particularly if you
are interested in association between metric
and non-metric variables
• e.g.: variation in pot size vs. surface
decoration…
• may decide to divide pot size into ordinal
classes…

5. large
medium
small
large
small

6. small large
specular 4 13
non-specular 15 18
rim diameter:
slip:
• other options may let you retain more of the
original information content

7. non-specular
slip
specular
slip
• could use a “t-test”
to test the equality
of the means
• makes full use of
ratio data…

8. • why do we work with contingency
tables??
polished burnished matte
bowl 47 28 3
jar 30 42 8
olla 6 45 25

9. • because we think there may be some kind
of interaction between the variables…
• basic question: can the state of one
variable be predicted from the state of
another variable?
• if not, they are independent
polished burnished matte
bowl 47 28 3
jar 30 42 8
olla 6 45 25

10. expected counts
• a baseline to which observed counts can be
compared
• counts that would occur by random chance
if the variables are independent, over the
long run
• for any cell
E = (col total * row total)/table total

11. M F
PP 4 1 5 45%  2.3 2.7 5
Pot 1 5 6 55% 2.7 3.3 6
Total 5 6 11 5 6 11
45% 55%

12. significance
• = probability of getting, by chance, a table
as or more deviant than the observed table,
if the variables are independent
– ‘deviant’ defined in terms of expected table
• no causality is necessarily implied by the
outcome
– but, causality may well be the reason for
observed association…
– e.g.: grave goods and sex

13. Fisher’s Exact Test
• just for 2 x 2 tables
• useful where chi-square test is inappropriate
• gives the exact probability of all tables with
• the same marginal totals
• as or more deviant
than the observed table…

14. P = (a+b)!(a+c)!(b+d)!(c+d)! / (N!a!b!c!d!)
P = 5!5!6!6! / 11!4!1!1!5! = 5*6!6! / 11!
P = 5*6!6! / 11! = 5*6! / 11*10*9*8*7
P = 5*6! / 11*10*9*8*7 = 3600 / 55440
P = .065
a b
c d
4 1
1 5

15. P = .065
use R (or Excel) if the counts aren’t too large…
> fisher.test(x)
a b
c d
4 1
1 5

16. 0 5 5
5 1 6
5 6 11
1 4 5
4 2 6
5 6 11
2 3 5
3 3 6
5 6 11
3 2 5
2 4 6
5 6 11
4 1 5
1 5 6
5 6 11
5 0 5
0 6 6
5 6 11

17. 0 5
5 1
1 4
4 2
2 3
3 3
3 2
2 4
4 1
1 5
5 0
0 6
2.3 2.7
2.7 3.3
0.013
0.162
0.433
0.325
0.065
0.002
• P = 0.065+0.002 = 0.067 or
• P = 0.067+0.013 = 0.080
(observed)
(expected)

18. • 2-tailed test = 0.067+0.013 = 0.080
• 1-tailed test = 0.065+0.002 = 0.067
M F
PP 4 1 5
Pot 1 5 6
5 6 11
> fisher.test(x, alt = "two.sided")
> fisher.test(x, alt = “greater”)
[i.e.: H1: odds ratio > 1]
in R:

19. Chi-square Statistic
• an aggregate measure (i.e., based on the
entire table)
• the greater the deviation from expected
values, the larger (exponentially!) the chi-
square statistic…
• one could devise others that would place
less emphasis on large deviations
 |o-e|/e
 




k
i i
i
i
E
E
O
1
2
2


20. • X2 is distributed approximately in accord
with the X2 probability distribution
• X2 probabilities are traditionally found in a
table showing threshold values from a CPD
– need degrees of freedom
– df = (r-1)*(c-1)
• just use R…

21. Status:
low intermed. high
Ritual arch.: altar 7 20 16 43
no altar 18 22 8 48
25 42 24 91
low intermed. high
altar 11.8 19.8 11.3 43
no altar 13.2 22.2 12.7 48
25 42 24 91
low intermed. high
altar 2.0 0.0 1.9 3.9
no altar 1.8 0.0 1.7 3.5
3.7 0.0 3.6 7.3
(43*24)
91
(7-11.8)2
11.8
=2
  .025

23. X2 assumptions & problems
• must be based on counts:
– not percentages, ratios or weighted data
• fails to give reliable results if expected
counts are too low:
2 3
3 3
2.27 2.72
2.72 3.27
obs. exp.
X2=0.74
P(Fishers)=1.0
5
6
6
5

24. rules of thumb
1. no expected counts less than 5
– almost certainly too stringent
2. no exp. counts less than 2, and 80% of
counts > 5
– more relaxed (but more realistic)

25. collapsing tables
• can often combine columns/rows to increase
expected counts that are too low
– may increase or reduce interpretability
– may create or destroy structure in the table
• no clear guidelines
– avoid simply trying to identify the combination
of cells that produces a “significant” result

26. 8 3 6 2 19
6 1 6 5 18
6 4 5 4 19
3 12 8 3 26
23 20 25 14 82
5.3 4.6 5.8 3.2 19
5.0 4.4 5.5 3.1 18
5.3 4.6 5.8 3.2 19
7.3 6.3 7.9 4.4 26
23 20 25 14 82
11 8 19
7 11 18
10 9 19
15 11 26
43 39 82
10.0 9.0 19
9.4 8.6 18
10.0 9.0 19
13.6 12.4 26
43 39 82
obs. counts
exp. counts
obs. counts
exp. counts

27. • chi-square is basically a measure of
significance
• it is not a good measure of strength of
association
• can help you decide if a relationship exists,
but not how strong it is

28. 17 13
13 17
60
34 26
26 34
120
X2=1.07
alpha=.30
X2=2.13
alpha=.14

29. • also, chi-square is a ‘global statistic’
• says nothing (directly) about which parts of
a table may be ‘driving’ a large chi-square
statistic
• ‘chi-square contributions’ from individual
cells can help:
low intermed. high
altar 2.0 0.0 1.9 3.9
no altar 1.8 0.0 1.7 3.5
3.7 0.0 3.6 7.3

30. Monte Carlo test of X2 significance
• based on simulated generation of cell-counts
under imposed conditions of independence
• randomly assign counts to cells:
23 14 8 45
15 6 13 34
38 20 21 79

31. • significance is simply the proportion of
outcomes that produced a X2 statistic >=
observed
• not based on any assumptions about the
distribution of the X2 statistic
• overcomes the problems associated with
small expected frequencies

32. G Test
• a measure of significance for any r x c table
• look up critical values of G2 in an ordinary
chi-square table; figure out degrees of
freedom the same way
• conforms to chi-square distribution better
than the chi-square statistic

















k
i i
i
e
i
E
O
O
G
1
2
log
*
2

33. an R function for G2
gsq.test  function(obs) {
df  (nrow(obs)-1) * (ncol(obs)-1)
exp  chisq.test(obs)$expected
G  2*sum(obs*log(obs/exp))
2*dchisq(G, df)
}

34. Measures of Association

35. Phi-Square (2)
• an attempt to remove the effects of sample
size that makes chi-square inappropriate for
measuring association
• divide chi-square by n
• 2=X2/n
• limits:
0: variables are independent
1: perfect association in a 2x2 table;
 no upper limit in larger tables

36. 17 13
13 17
60
34 26
26 34
120
2=0.18
2=0.18

37. Cramer’s V
• also a measure of strength of association
• an attempt to standardize phi-square
(i.e., control the lack of an upper boundary in tables
larger than 2x2 cells)
• V= 2/m
where m=min(r-1,c-1) ; i.e., the smaller of rows-1 or
columns-1)
• limits: 0-1 for any size table; 1=highest possible
association

38. Yule’s Q
• for 2x2 tables only
• Q = (ad-bc)/(ad+bc)
a b
c d

39. Yule’s Q
• often used to assess the strength of presence /
absence association
• range is –1 (perfect negative association) to 1
(perfect positive association); values near 0
indicate a lack of association
Bone needles
+ -
Male burial + 12 14
- 16 3
Q = -.72

40. Yule’s Q
• not sensitive to marginal changes (unlike Phi2)
• multiply a row or column by a constant;
cancels out…
jars ollas
Source A 19 10
Source B 6 15
jars ollas
Source A 19 20
Source B 6 30
(Q=.65 for both tables)

41. Yule’s Q
• can’t distinguish between different degrees of ‘complete’
association
• can’t distinguish between ‘complete’ and ‘absolute’
association
M F
RHS 60 20
LHS 0 20
100
M F
RHS 60 10
LHS 0 30
100
M F
RHS 60 0
LHS 0 40
100

42. “odds” ratio
• easiest with 2 x 2 tables
• what are the ‘odds’ of a man being buried
on his right side, compared to those of a
woman??
• if there is a strong level of association
between sex and burial position, the odds
should be quite different…

43. a b
c d
a
c
b
d
odds ratio =

44. 29/11=2.64
14/33=0.42
2.64/0.42=6.21
if there is no association, the odds ratio=1
departures from 1 range between 0 and infinity
>1 =‘positive association’
<1 =‘negative association’
M F
RHS 29 14 43
LHS 11 33 44
40 47 87

45. Goodman and Kruskal’s Tau ()
• “proportional reduction of error”
• how are the probabilities of correctly
assigning cases to one set of categories
improved by the knowledge of another set
of categories??

46. Goodman and Kruskal’s Tau ()
• limits are 0-1; 1=perfect association
• same results as Phi2 w/ 2x2 table
• sensitive to margin differences
• asymmetric
– get different results predicting row assignments
based on columns than from column
assignments based on rows

47. • =[P(error|rule 1)-P(error|rule 2)] / P(error|rule 1)
• rule 1: random assignments to one variable are
made with no knowledge of 2nd variable
• rule 2: random assignments to one variable are
made with knowledge of 2nd variable
B1 B2
A1
A2
6 14 20
B1 B2
A1 6 0
A2 0 14
6 14 20

48. Table Standardization
• even very large and highly significant X2 (or G2)
statistics don’t necessarily mean that all parts of the
table are equally “deviant” (and therefore interesting)
• usually need to do other things to highlight loci of
association or ‘interaction’
• which cells diverge the most from expected values?
• very difficult to decide when both row and column
totals are unequal…

49. Percent standardization
• highly intuitive approach, easy to interpret
• often used to control the effects of sample-
size variation
• have to decide if it makes better sense to
standardize based on rows, or on columns

50. • usually, you want to standardize whatever it
is you want to compare
– i.e., if you want to compare columns, base
percents on column totals
• you may decide to make two tables, one
standardized on rows, the other on
columns…

51. Site
Fauna A B C
bear 2 1 0 3
moose 15 5 10 30
coyote 2 0 0 2
rabbit 16 8 12 36
dog 2 3 0 5
deer 16 8 7 31
53 25 29 107
Site
Fauna A B C
bear 3.8 4.0 0.0
moose 28.3 20.0 34.5
coyote 3.8 0.0 0.0
rabbit 30.2 32.0 41.4
dog 3.8 12.0 0.0
deer 30.2 32.0 24.1
100 100 100
MNIs
Site
Fauna A B C
bear 66.7 33.3 0.0 100
moose 50.0 16.7 33.3 100
coyote 100.0 0.0 0.0 100
rabbit 44.4 22.2 33.3 100
dog 40.0 60.0 0.0 100
deer 51.6 25.8 22.6 100

52. Binomial Probabilities
• P(n,k,p):
“probability of k successes in n trials, with p
probability of success in any one trial”
5 3
1 4
13
3.7 4.3
2.3 2.7
13
n = 13
k = 5
p = 3.7/13

53. Binomial Probabilities
• in R:
> pbinom(k, n, p)
• easy to build into a function…

54. 10
20
30
40
50
60
70
80
90
100
percent
10
20
30
40
50
60
70
80
90
100
cumulative
percent
K-S test for cumulative percents

55. 10
20
30
40
50
60
70
80
90
100
cumulative
percent

56. 10
20
30
40
50
60
70
80
90
100
cumulative
percent
• some useful
statistical
measures
(ordinal or ratio scale)
• can be
misleading
when used with
nominal data
• good for
comparing data
sets
Cumulative Percent Graph

57. Percentages
Sites
A B C
Types 1 5 5 5
2 45 0 30
3 5 48 5
4 5 5 5
5 5 5 5
6 5 5 5
7 20 5 35
8 5 22 5
9 5 5 5
100 100 100
Cumulative Percents
Sites
A B C
Types 1 5 5 5
2 50 5 35
3 55 53 40
4 60 58 45
5 65 63 50
6 70 68 55
7 90 73 90
8 95 95 95
9 100 100 100
0
20
40
60
80
100
120
1 2 3 4 5 6 7 8 9
A
B
C

58. 0
20
40
60
80
100
120
1 2 3 4 5 6 7 8 9
A
B
C
0
20
40
60
80
100
120
1 5 3 4 2 6 7 8 9
A
B
C

59. K-S test
• find Dmax:
– maximum difference between 2 cumulative
proportion distributions
– compare to critical value for chosen sig. level
• C*((n1+n2)/(n1n2))^.5
– alpha =.05, C=1.36
– alpha =.01, C=1.63
– alpha =.001, C=1.95

60. example 2
• mortuary data (Shennan, p. 56+)
• burials characterized according to 2 wealth
(poor vs. wealthy) and 6 age categories
(infant to old age)
Rich Poor
Infans I 6 23
Infans II 8 21
Juvenilis 11 25
Adultus 29 36
Maturus 19 27
Senilis 3 4
Total 76 136

61. • burials for younger age-classes appear to be
more numerous among the poor
• can this be explained away as an example of
random chance?
or
• do poor burials constitute a different
population, with respect to age-classes, than
rich burials?

62. • we can get a visual sense of the problem
using a cumulative frequency plot:
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Infans
I
Infans
II
Juvenilis
Adultus
Maturus
Senilis
rich
poor

63. • K-S test (Kolmogorov-Smirnov test) assesses the
significance of the maximum divergence between two
cumulative frequency curves
H0:dist1=dist2
• an equation based on the theoretical distribution of
differences between cumulative frequency curves
provides a critical value for a specific alpha level
• observed differences beyond this value can be regarded
as significant at that alpha level

64. • if alpha = .05, the critical value =
1.36*(n1+n2)/n1n2
1.36*(76+136)/76*136 = 0.195
• the observed value = 0.178
• 0.178 < 0.195; don’t reject H0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Infans
I
Infans
II
Juvenilis
Adultus
Maturus
Senilis
rich
poor
Dmax=.178

65. age <= 30 age > 30
strongly disagree 8 7
mildly disagree 5 9
disagree 6 6
no opinion 0 1
agree 2 2
mildly agree 1 3
strongly agree 2 3
statement/question: “Oil exploration should be allowed in
coastal California…”
example 2

66. example 3
• survey data  100 sites
• broken down by location and time:
early late Total
piedmont 31 19 50
plain 19 31 50
Total 50 50 100

67. • we can do a chi-square test of independence
of the two variables time and location
• H0:time & location are independent
• alpha = .05
time
location
H0
location
time
H1

68. • 2 values reflect accumulated differences between
observed and expected cell-counts
• expected cell counts are based on the assumptions
inherent in the null hypothesis
• if the H0 is correct, cell values should reflect an
“even” distribution of marginal totals
early late Total
piedmont 50
plain 50
Total 50 50 100
25

69. • chi-square = ((o-e)2/e)
• observed chi-square = 4.84
• we need to compare it to the “critical value”
in a chi-square table:

71. • chi-square = ((o-e)2/e)
• observed chi-square = 4.84
• chi-square table:
 critical value (alpha = .05, 1 df) is 3.84
 observed chi-square (4.84) > 3.84
• we can reject H0
• H1: time & location are not independent

72. • what does this mean?
early late Total
piedmont 31 19 50
plain 19 31 50
Total 50 50 100