Contingency Table Analysis

• 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

• 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…

• 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…

largemediumsmall

largesmall

small large

specular 4 13

non-specular 15 18

rim diameter:

slip:

• other options may let you retain more of the original information content

non-specularslip

specularslip

• could use a “t-test” to test the equality of the means

• makes full use of ratio data…

• why do we work with contingency tables??

polished burnished matte

bowl 47 28 3

jar 30 42 8

olla 6 45 25

• 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

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

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%

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

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…

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

P = .065

use R (or Excel) if the counts aren’t too large…

> fisher.test(x)

a b

c d

4 1

1 5

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

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)

• 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:

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

ii

E

EO

1

22

• 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…

Status:low intermed. high

Ritual arch.: altar 7 20 16 43no altar 18 22 8 48

25 42 24 91

low intermed. highaltar 11.8 19.8 11.3 43no altar 13.2 22.2 12.7 48

25 42 24 91

low intermed. highaltar 2.0 0.0 1.9 3.9no 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

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

665

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)

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

8 3 6 2 196 1 6 5 186 4 5 4 193 12 8 3 26

23 20 25 14 82

5.3 4.6 5.8 3.2 195.0 4.4 5.5 3.1 185.3 4.6 5.8 3.2 197.3 6.3 7.9 4.4 2623 20 25 14 82

11 8 197 11 18

10 9 1915 11 2643 39 82

10.0 9.0 199.4 8.6 18

10.0 9.0 1913.6 12.4 26

43 39 82

obs. counts

exp. counts

obs. counts

exp. counts

• 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

17 1313 17

60

34 2626 34

120

X2=1.07alpha=.30

X2=2.13alpha=.14

• 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. highaltar 2.0 0.0 1.9 3.9no altar 1.8 0.0 1.7 3.5

3.7 0.0 3.6 7.3

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 4515 6 13 3438 20 21 79

• 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

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

iei E

OOG

1

2 log*2

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)

}

Measures of Association

Phi-Square (2)

• an attempt to remove the effects of sample size that makes chi-square inappropriate for measuring association

• divide chi-square by n2=X2/n

• limits:0: variables are independent

1: perfect association in a 2x2 table;

no upper limit in larger tables

17 1313 17

60

34 2626 34

120

2=0.18

2=0.18

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/mwhere 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

Yule’s Q

• for 2x2 tables only• Q = (ad-bc)/(ad+bc)

a b

c d

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

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)

Yule’s Q

• can’t distinguish between different degrees of ‘complete’ association

• can’t distinguish between ‘complete’ and ‘absolute’ association

M FRHS 60 20LHS 0 20

100

M FRHS 60 10LHS 0 30

100

M FRHS 60 0LHS 0 40

100

“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…

a b

c d

acbd

odds ratio =

29/11=2.64

14/33=0.42

2.64/0.42=6.21

if there is no association, the odds ratio=1departures 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

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??

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

=[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

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…

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

• 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…

SiteFauna A B Cbear 2 1 0 3moose 15 5 10 30coyote 2 0 0 2rabbit 16 8 12 36dog 2 3 0 5deer 16 8 7 31

53 25 29 107

SiteFauna A B Cbear 3.8 4.0 0.0moose 28.3 20.0 34.5coyote 3.8 0.0 0.0rabbit 30.2 32.0 41.4dog 3.8 12.0 0.0deer 30.2 32.0 24.1

100 100 100

MNIs

SiteFauna A B Cbear 66.7 33.3 0.0 100moose 50.0 16.7 33.3 100coyote 100.0 0.0 0.0 100rabbit 44.4 22.2 33.3 100dog 40.0 60.0 0.0 100deer 51.6 25.8 22.6 100

Binomial Probabilities

• P(n,k,p):“probability of k successes in n trials, with p probability of success in any one trial”

5 31 4

13

3.7 4.32.3 2.7

13

n = 13k = 5

p = 3.7/13

Binomial Probabilities

• in R:> pbinom(k, n, p)

• easy to build into a function…

10

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perc

ent

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cum

ulat

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perc

ent

K-S test for cumulative percents

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perc

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• some useful statistical measures

(ordinal or ratio scale)

• can be misleading when used with nominal data

• good for comparing data sets

Cumulative Percent Graph

PercentagesSitesA B C

Types 1 5 5 52 45 0 303 5 48 54 5 5 55 5 5 56 5 5 57 20 5 358 5 22 59 5 5 5

100 100 100

Cumulative PercentsSitesA B C

Types 1 5 5 52 50 5 353 55 53 404 60 58 455 65 63 506 70 68 557 90 73 908 95 95 959 100 100 100

0

20

40

60

80

100

120

1 2 3 4 5 6 7 8 9

A

B

C

0

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120

1 2 3 4 5 6 7 8 9

A

B

C

0

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1 5 3 4 2 6 7 8 9

A

B

C

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

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

• 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?

• 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

Infa

ns I

Infa

ns II

Juve

nilis

Adu

ltus

Mat

urus

Sen

ilis

rich

poor

• 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

• 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

Infa

ns I

Infa

ns II

Juve

nilis

Adu

ltus

Mat

urus

Sen

ilis

rich

poor

Dmax=.178

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

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

• we can do a chi-square test of independence of the two variables time and location

• H0:time & location are independent

• alpha = .05time

location

H0

location

time

H1

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 Totalpiedmont 50plain 50Total 50 50 100

25

• 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:

• 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

• what does this mean?

  early late Total

piedmont 31 19 50

plain 19 31 50

Total 50 50 100