1 Contingency Tables: Tests for independence and homogeneity (§10.5) How to test hypotheses of...

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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 TableCross-Classification TableMeasure of association

Independence in two-way tablesChi-Square Test for Independence

or Homogeneity

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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).

Person Rank Evaluation1 Professor Above2 Instructor Average3 Professor Below4 Assistant Professor Average5 Associate Professor Average

. . .

. . .

. . .

Each person has two categorical responses.

Rank

Teaching Evaluation Instructor

Assistant Professor

Associate Professor Professor

Above Average

36 62 45 50

Average 48 50 35 43

Below Average

30 13 20 35

Data can be formatted into a cross-tabulation or contingency table.

Test of Independence or Association

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Is the level of teaching evaluation related to rank?

Are Professors more likely to be judged above average than other ranks?

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.

Rank


Assistant Professor

Associate Professor Professor Sum

Relative Frequency

Above Average

36 62 45 50 193 0.413

Average 48 50 35 43 176 0.377

Below Average

30 13 20 35 98 0.210

Sum 114 125 100 128 467 1.000Relative Frequency

0.244 0.268 0.214 0.274 1.000

What are we interested in from this two-way classification table?

Ho: Teaching Evaluation and Rank are independent variables.

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Rank


Assistant Professor


Relative Frequency

Above Average

p11 p12 p13 p14 193 p1.

Average p21 p22 p23 p24 176 p2.Below

Averagep31 p32 p33 p34 98 p3.


p.1 p.2 p.3 p.4 1.000

The independence assumption: ijallforjiij ppp

jiij

ijij

nE

nn

pp

p

Expected

n

nnE ji

ij

n

in

jn

Observed

r

i

c

j ij

ijij

E

En

1 1

2

2

r=#rows=3, c=#cols=4, 3 4 table.

df = (r-1)(c-1)

Test Statistic:

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Rank


Assistant Professor


Relative Frequency

Above Average

36 62 45 50 193 0.413

Average 48 50 35 43 176 0.377

Below Average

30 13 20 35 98 0.210


0.244 0.268 0.214 0.274 1.000

Observed Counts

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Rank


Assistant Professor


Above Average

47.113 51.660 41.328 52.899 193

Average 42.964 47.109 37.687 48.240 176

Below Average

23.923 26.231 20.985 26.861 98

Sum 114 125 100 128 467

n

nnE ji

ij

Expected Counts

Assumptions: no Eij < 1, and no more than 20% of Eij < 5.

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Assistant Professor

Associate Professor Professor

Above Average

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

,44.1747.262.22 ,59.12295.0,6 Reject Ho

Individual Cell Chi Square Values

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

Input data in this way

STAT > TABLES > Cross TabsClassification Variables:rank evalCheck Chi-square Analysis, and Above and Std. residualFrequencies are in: count

rank eval count1 1 301 2 481 3 362 1 132 2 502 3 623 1 203 2 353 3 454 1 354 2 434 3 50

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Tabulated Statistics: eval, rankRows: 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:

ij

ijij

E

En

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SASoptions ls=79 ps=40 nocenter;data eval;input job $ rating $ number;datalines;Instructor Above 36Instructor Average 48Instructor Below 30Assistant Above 62Assistant Average 50Assistant Below 13Associate Above 45Associate Average 35Associate Below 20Professor Above 50Professor Average 43Professor Below 35;run;proc freq data=eval; weight number; table job*rating / chisq ; run;

Table of job by rating

job rating

Frequency‚Percent ‚Row Pct ‚Col Pct ‚Above ‚Average ‚Below ‚ TotalƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒÂssistan ‚ 62 ‚ 50 ‚ 13 ‚ 125 ‚ 13.28 ‚ 10.71 ‚ 2.78 ‚ 26.77 ‚ 49.60 ‚ 40.00 ‚ 10.40 ‚ ‚ 32.12 ‚ 28.41 ‚ 13.27 ‚ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒÂssociat ‚ 45 ‚ 35 ‚ 20 ‚ 100 ‚ 9.64 ‚ 7.49 ‚ 4.28 ‚ 21.41 ‚ 45.00 ‚ 35.00 ‚ 20.00 ‚ ‚ 23.32 ‚ 19.89 ‚ 20.41 ‚ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒÎnstruct ‚ 36 ‚ 48 ‚ 30 ‚ 114 ‚ 7.71 ‚ 10.28 ‚ 6.42 ‚ 24.41 ‚ 31.58 ‚ 42.11 ‚ 26.32 ‚ ‚ 18.65 ‚ 27.27 ‚ 30.61 ‚ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆProfesso ‚ 50 ‚ 43 ‚ 35 ‚ 128 ‚ 10.71 ‚ 9.21 ‚ 7.49 ‚ 27.41 ‚ 39.06 ‚ 33.59 ‚ 27.34 ‚ ‚ 25.91 ‚ 24.43 ‚ 35.71 ‚ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆTotal 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

ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ

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

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SPSSFirst you need to tell SPSS that each observation must be weighted by the cell count.

DATA > WEIGHT CASES

Then you choose the analysis. ANALYZE > DESCRIPTIVE STATISTICS > CROSS TABS

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> score <- c(36,48,30,62,50,13,45,35,20,50,43,35)> mscore <- matrix(score,3,4)> mscore [,1] [,2] [,3] [,4][1,] 36 62 45 50[2,] 48 50 35 43[3,] 30 13 20 35> chisq.test(mscore)

Pearson's Chi-squared test

data: mscore X-squared = 17.4354, df = 6, p-value = 0.00781

> out <- chisq.test(mscore)> out[1:length(out)]$statisticX-squared 17.43537

$parameterdf 6

$p.value[1] 0.00780959

R

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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:

ij

ijij

E

En

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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 can’t 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

p11 p12 ... p1c

p21 p22 ... p2c

: : : :

pr1 pr2 ... prc

pij = proportion of subpop j subjects (j=1,…,c) that fall in category i (i=1,…,r).

r

iij cj

1

,,1each for ,1 p

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Null hypothesis of homogeneity

rc

c

c

rr p

pp

p

pp

p

pp

2

1

2

22

12

1

21

11

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Myocardial Infarction Smoked Yes No Totals

Yes 172 173 355No 90 346 436

Totals 262 519 791

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: p11=p12 and p21=p22

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Example: MI results in MTB

Stat -> Tables -> Chi-Square Test--------------------------------------------------------------------------------------------Chi-Square Test: MI Yes, MI NoExpected 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 = 73.729DF = 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 3+2=5 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 / (1+1.5) = 1.5 / 2.5 = 3/5.

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Example: MI and Odds RatiosFor 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.

656.0ˆ11 p

333.0ˆ12 p

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