a test for homogeneity of odds ratios in ordered 2 × 2 tables

A Test for Homogeneity of Odds Ratios in Ordered 2� 2 Tables

Arthur Cohen*; **; 1, John Kolassa***; 1, and H. B. Sackrowitz**; 1

1 Department of Statistics, Rutgers University, Piscataway NJ, USA

Received 3 February 2004, revised 10 July 2004, accepted 23 August 2004

Summary

Consider K ordered 2� 2 contingency tables. A new test of the null hypothesis that the odds ratios ofthese tables are equal vs the alternative hypothesis that the odds ratios are nondecreasing, is recom-mended. The test is exact (non-asymptotic), is easily carried out (software is available), and has otherfavorable properties.

Key words: Directed chi-square test; Conditional P-values; Linear test; Pool-adjacent-viola-tors algorithm.

1 Introduction

Consider K 2� 2 contingency tables. Let Xijk, i ¼ 1; 2; j ¼ 1; 2; k ¼ 1; . . . ;K be the cell frequency in theij-th cell of the k-th table. Let qk be the odds ratio for table k. Test the hypothesis H0 : q1 ¼ � � � ¼ qK

vs. H1 : q1 � � � � � qK , with at least one strict inequality. Let Xij: ¼PKk¼1

Xijk, Xi:k ¼P2j¼1

Xijk,

X:jk ¼P2i¼1

Xijk, X::k ¼Pi;j

Xijk. To test H0 vs. H1 we will condition on all marginal totals, i.e., we condi-

tion on X:jk, Xi:k, and X11:. Conditioning on these variables leaves ðX111;X112; . . . ;X11ðK�1ÞÞ0 ¼ Xð1Þ, thesubvector of X ¼ ðX111;X112; . . . ;X11KÞ0 as the random vector upon which to base a test statistic. This

is so, since X11K ¼ X11: �PK�1

k¼1X11k.

This testing problem has been studied by Hirotsu [1982]. Hirotsu considered a large sample test,based on a statistic which is a linear function of the efficient scores evaluated at qq, the maximumlikelihood estimator of the common log odds ratio parameter. Hirotsu’s test statistic is of the form u0Xwhere u1 � � � � � uk are constants. Agresti [1990, Section 7.5.3] discusses this problem. He notes thatif there is a concomitant variable and the alternative to H0 is that log qk changes linearly across thestrata (k tables), then Zelen [1971] gives an exact test and Breslow and Day [1980, p. 142] give anasymptotic chi-square test.

In this paper we offer an exact test (non-asymptotic) for testing H0 vs. H1. The test is a directedchi-square test based on the methodology of Cohen et al. [2003]. The test has desirable properties.Namely, it is exact (non-asymptotic), it is admissible, it is easily carried out (software is available),and it has a somewhat more robust power function than the linear test. The new test is described inthe next section. Section 3 contains some power comparisons with Hirotsu’s test and for contrast withthe two-sided test proposed by Breslow and Day [1980]. Section 3 contains two data sets which serveas examples for the use of the proposed test. A third artificial example is also offered to illustrate thesteps in carrying out the directed chi-square test as well as to illustrate its power advantages.

* Corresponding author: e-mail: [email protected]** Research supprted by NSF Grant DMS 0092659, R01 Grant CA 063050

*** Research supprted by NSA Grant MDA 904-02-1-0039

Biometrical Journal 46 (2004) 6, 633–641 DOI: 10.1002/bimj.200410065

# 2004 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

2 Directed Chi-Square Test

Note that the conditional distribution of X11k given X1:k;X:1k is the noncentral hypergeometric distribu-tion given by

PðX11k ¼ xÞ ¼ X1:k

x

� �X2:k

X:1k � x

� �qx

k

�PSu¼L

X1:k

u

� �X2:k

X:1k � u

� �qu

k ; ð1Þ

for Lk � x � Sk where Lk ¼ max ð0;X:1k � X::k þ X1:kÞ and Sk ¼ min ðX:1k;X1:kÞ. Now let q be thecommon odds ratio when H0 is true and let EkðqÞ ¼ EX11k when H0 is true. In order to determine thedirected chi-square test we need to first find estimators EEkðqÞ of EkðqÞ. Toward this end we seek themaximum likelihood estimator (MLE) of q. Consider the log likelihood ‘ðqÞ by taking the log of theproduct of terms in (1) (all qk ¼ q). Let n ¼ log q and let LðnÞ ¼ ‘ðqðnÞÞ. Next find the log likelihoodequation by differentiating LðnÞ with respect to n and setting the derivative equal to 0. The secantmethod is used to solve for n using the logit estimator of n as a starting point. (See Breslow and Day[1980, p. 139] for the logit estimator.) If nn is the MLE of n, then enn ¼ qq, and EEkðqÞ ¼ EkðqqÞ. Example3 in Section 3 illustrates the above method of obtaining the MLE of q.

Next let t ¼ ðt1; . . . ; tKÞ0 be the vector ðEE1; EE2; . . . ; EEKÞ0, Z ¼ ðZ1; . . . ; ZKÞ0, where Zk ¼ X11k=tk andPtðZ j SÞ be the weighted projection of Z onto S where S is the simple order coneS ¼ fq : q1 � � � � � qKg.

Robertson et al. [1988] state that Ptð� j SÞ can be obtained by using the pool-adjacent-violatorsalgorithm (PAVA). See Robertson et al. [1988, pp. 8–9]. We will illustrate PAVA in Example 3.

The directed chi-square statistic is

c2D ¼

PKk¼1

Z*k2tk ; ð2Þ

where Z*k is the k-th coordinate of PtðZ j SÞ. (See Example 3 again for an illustration.)One way to carry out the test based on (2) is to find the conditional P-value by Monte Carlo

methods. Tables with the fixed margins can be generated by making use of the algorithm of Patefield[1981]. Only those tables with the observed X11: are retained. The percentage of the retained tableswhose calculated value of (2) ties or exceeds the observed calculated value of (2) is the conditional P-value. A size a test, if desired, rejects if this P-value is less than or equal to a.

Another possible way to carry out the test based on (2) is by enumerating all sets of K contingencytables in which each table has the same row and column totals as observed and the same value of X11.This latter method will be applied in the next section. More details will be given when referring to thespecific examples.

Computer programs implementing the calculations for the directed chi-square test may be found atthe following website: http://stat.rutgers.edu/ kolassa/tables.tar.gz.

We close this section with a remark on the rationale for the directed chi-square test statistic.An appealing test statistic to use when testing H0 is the chi-square statistic. However the chi-square

statistic is basically a two-sided statistic and would be appropriate if the alternative was H2: not H0.Since our hypothesis is H1, some modification is needed that will, in effect, point the rejection regionmore in the direction of this one-sided hypothesis. Note that the usual chi-square statistic is

c2 ¼PKk¼1

ðX11k � EEkðqÞÞ2

EEkðqÞð3Þ

or the statistic which would lead to an equivalent test, namely

c2E ¼

PKk¼1

X211k

EEkðqÞ¼PKk¼1

Z2k tk : ð4Þ

634 A. Cohen, J. Kolassa and H. B. Sackrowitz: Test for Homogeneity of Odds Ratios


The theory dictates that any good test of H0 vs. H1 should be monotone in the partial sumsPrk¼1

X11k,

r ¼ 1; 2; . . . ;K � 1. Hence if the sample point ðx111; . . . ; x11ðK�1ÞÞ is a reject point, then

ðx*111; . . . ; x*11ðK�1ÞÞ should be a reject point ifPrk¼1

x*11k �Prk¼1

x11k, for all r ¼ 1; . . . ;K � 1. The directedchi-square test has this property.

Example 3.1 and Figure 3.1 can also help in understanding the distinction between the directed chi-square test based on (2) and the usual chi-square test based on (3).

3 Examples

Example 3.1: The data set to be analyzed appears in Le [1998]. The data were extracted from Kelseyet al. [1982]. The factors are estrogen replacement, cancer and weight.

Three tests were evaluated, the test of Breslow and Day [1980, equation 4.30] for testing H0 vs.H2, Hirotsu’s score test with u1 ¼ 3; u2 ¼ 2; u3 ¼ 1, and our directed chi-square test as given in (2).We examine the exact behavior of these tests. To do so we first enumerated all sets of three contin-gency tables in which each table has the same row and column totals as above (in this case, each ofour selected sets of tables has 81 and 195 as the row totals for the first table, 150 and 423 as the rowtotals for the second table, and 32 and 182 as the row totals for the third table, and similarly for thecolumn totals), and the same sum of upper left-hand entries (in our case, 66). One might considerperforming this enumeration by first forming all combinations of tables having the correct row andcolumn totals, and eliminating those whose sum of upper left-hand entries was incorrect; in fact,considerable effort is saved by first noting that potential values for the upper left-hand entry of thethird table are f0; . . . ; 32g. Hence after forming all combinations of the first two tables, one needonly retain those whose sum of upper left-hand entries is between 0 and 32. This approach is in thesame spirit as given by Hirji et al. [1987]. For each set of tables retained as part of the final condi-tional sample space, the product of the three individual hypergeometric probabilities was recorded aswell. Probabilities for each of the table sets are divided by the total probabilities to give the probabil-ities conditional on the sum of the upper left-hand corners.

This approach is similar to that of Agresti et al. [1990], who present a method for tabulating theconditional distribution of certain statistics linear in the table entries, except that we condition on thesum of the upper left-hand corners, and retain the values of all table entries and not just linear sum-mary, since some of the tests we evaluate are not linear combinations of the table entries. The methodof Agresti [1990] may be extended to a set of tables larger than 2� 2; the method employed in thismanuscript and described in the previous paragraph also generalizes to sets of larger tables. In thiscase, rather than generating a single upper left-hand table entry from a hypergeometric distribution ateach step, one might use the method of Pagano and Halvorsen [1981] to generate the sample space ofcontingency tables with given marginals, and then one might apply the feasibility test as above to allcells except those in the last row and column.

Biometrical Journal 46 (2004) 6 635

Table 1 Data collected to study the relationship between weight, estrogen ther-apy, and endometrial cancer.

Weight (kg)

< 57 Cancer 57–75 Cancer > 75 Cancer

Estrogen Replacement Yes No Total Yes No Total Yes No Total

Yes 20 61 81 37 113 150 9 23 32No 12 183 195 45 378 423 42 140 182

Total 32 244 276 82 491 573 51 163 214

Log odds ratio 1.6094 1.0118 0.2657


To calculate (2) we first calculate the maximum likelihood estimate of the common odds ratio. Aweighted pool-adjacent-violators algorithm was applied to the ratio of the upper left corners of thetables to their expectations under the null hypothesis using the common odds ratio estimate, and theresults were multiplied by these same expectations. All of the resulting upper left table entries wererescaled to make their sum satisfy the conditioning event. In general the pooled and rescaled upperleft table entries might correspond to tables with negative entries; this was observed for some mem-bers of the conditional sample space.

Table 2 contains critical values for a nominal .05 test, and the resulting actual test size.

All of these tests have actual sizes below the nominal level, because of the discreteness of theunderlying distribution. Because test sizes are not at their nominal level, and since actual test sizesvary among different tests applied to the same data sets, powers will be compared for randomizedversions of these tests calibrated to have exactly the nominal size. Alternatives chosen represent arange of values These powers are noted in Table 3. The first two columns of the table are the ratios ofthe odds ratios for the first and last table, and the ratios of the odds ratios for the second and lasttable. The next two columns are the logs of these values. The next three columns are powers of thevarious randomized tests:

The first line represents power under the null hypothesis; these powers are the nominal level, byconstruction. The next lines represent alternatives of interest to us; our test either does almost as wellas or clearly better than the score test. We expect better power than the Breslow-Day test which is notdesigned to detect a trend.

Example 3.2: Consider data presented by Breslow and Day [1980, p. 137]. Table 4 offers data onthe presence or absence of esophogeal cancer to level of alcohol consumption, dichotomized as highor low. Six tables are presented corresponding to six age strata. Below are the data with the first two,middle two, and last two age strata collapsed:


Table 2 Statistic values, P-values, critical values, and truesizes for Breslow-Day, score, and directed chi-square tests fordata in Table 1.

Test Names Breslow Score Directedchi-square

Statistic values 2.68812 0.44854 1.95854P-values 0.25498 0.38901 0.14614Critical values 6.07876 1.90463 3.81096True size 0.04972 0.03887 0.04658

Table 3 Power of Breslow-Day, score, and directed chi-square tests given mar-ginal totals of Table 1.

Odds Ratios Log odds ratios Breslow Score Directedchi-square

n1 n2

1.0000 1.0000 0.0000 0.0000 0.0500 0.0500 0.05004.0000 1.0000 1.3863 0.0000 0.6715 0.6789 0.80504.0000 4.0000 1.3863 1.3863 0.6984 0.7164 0.83165.0000 3.0000 1.6094 1.0986 0.6687 0.8350 0.8223

10.0000 3.0000 2.3026 1.0986 0.9130 0.9781 0.966110.0000 5.0000 2.3026 1.6094 0.9341 0.9778 0.978020.0000 5.0000 2.9957 1.6094 0.9907 0.9988 0.9977


The exact behavior of the Breslow test, the score test, and directed chi-square test, was investigatedas before. Table 5 contains critical values for a nominal .05 test, and the resulting actual test size.

Powers for randomized versions of these tests calibrated to have exactly the nominal size werecompared. This time several different alternatives are chosen so that the approximate power of theBreslow test is .75 at each alternative. These powers are given in Table 6. The first two columns ofthe table are the ratios of the odds ratios for the first and last table, and the ratios of the odds ratiosfor the second and last table. The next two columns are the logs of these values. The next threecolumns are powers of the various randomized tests:

The first line represents power under the null hypothesis; these powers are the nominal level, byconstruction. The next lines represent alternatives of interest to us; our test either does almost as wellor clearly better than the score test.


Table 4 Data collected to study the relationship among alcohol consumption,esophageal cancer, and age.

Age 25-44 Age 45-65 Age over 65

Alcohol level Alcohol level Alcohol level

High Low High Low High Low

Case 5 5 10 67 55 122 24 44 68Control 35 270 305 56 277 333 18 119 137

40 275 315 123 332 455 42 163 205

Log odds ratio 2.043 1.796 1.283

Table 5 Statistic values, P-values, critical values, and truesizes for Breslow-Day, score, and directed chi-square tests fordata in Table 4.

Test names Breslow Score Directedchi-square

Statistic values 1.81654 1.31025 1.81654P-values 0.41206 0.12692 0.15831Critical values 6.11989 1.98925 3.96921True size 0.04742 0.03546 0.04855

Table 6 Power of Breslow-Day, score, and directed chi-square tests given mar-ginal totals of Table 4.

Odds Ratios Log Odds Ratios Breslow Score Directedchi-square

1.00 1.00 0.000 0.000 0.0500 0.0500 0.05009.71 1.00 2.273 0.000 0.8228 0.5732 0.85923.29 3.29 1.192 1.192 0.7472 0.7989 0.83545.00 3.00 1.609 1.099 0.7354 0.8670 0.85607.08 2.54 1.957 0.934 0.7319 0.8898 0.86784.90 3.04 1.589 1.111 0.7389 0.8659 0.85716.40 2.71 1.857 0.998 0.7339 0.8880 0.8656


Example 3.3: To illustrate the test methodology and to gain further insight as to when the directedchi-square test will have higher power than the score test we consider an artificial example. Let K ¼ 3and let the margins for each 2� 2 table be 5. Take X11: ¼ 9. Possible sample pointsXð1Þ ¼ ðX111;X112Þ are given in Table 7 along with their null probabilities.

The first step in implementing the directed chi-square test is to find the MLE of n. In this example,since all marginals are equal the MLE can be formed from a single 2� 2 table with marginal row andcolumn totals equal to 15 and 9 as the number in the first row-first column. Writing the density (1) in

exponential family form the log likelihood LðnÞ ¼ 9n� logP15

j¼0exp ðjnÞ pj

!, for pj ¼ 1=j!ð15� jÞ!.

The likelihood equation sets

L0ðnÞ ¼ 0 ; or 9�

P15

j¼0jpj exp ðjnÞ

P15

j¼0pj exp ðjnÞ

¼ 0 :


Table 7 Null probabilities for the case ofthree 2� 2 tables with marginal row and col-umn totals of 5 and X11 ¼ 9:

Sample point Point Probability

0.000 4.000 0.0000096450.000 5.000 0.0000096451.000 3.000 0.0009644971.000 4.000 0.0060281051.000 5.000 0.0009644972.000 2.000 0.0038579872.000 3.000 0.0964496872.000 4.000 0.0964496872.000 5.000 0.0038579873.000 1.000 0.0009644973.000 2.000 0.0964496873.000 3.000 0.3857987483.000 4.000 0.0964496873.000 5.000 0.0009644974.000 0.000 0.0000096454.000 1.000 0.0060281054.000 2.000 0.0964496874.000 3.000 0.0964496874.000 4.000 0.0060281054.000 5.000 0.0000096455.000 0.000 0.0000096455.000 1.000 0.0009644975.000 2.000 0.0038579875.000 3.000 0.0009644975.000 4.000 0.000009645


Note that

‘00ðnÞ ¼�P15

j¼0j2pj exp ðjnÞ

P15

j¼0pj exp ðjnÞ

þ

P15

j¼0jpj exp ðjnÞ

P15

j¼0pj exp ðjnÞ

26664

37775

2

;

the negative of the variance of the distribution on the integers from 0 to 15 with weights proportionalto pj exp ðjnÞ. Hence ‘00ðnÞ < 0 for all n, and so the likelihood equation has a unique solution.

To solve this equation, one might try n ¼ 0. Note that ‘0ð0Þ ¼ 3=2; our next try ought to be higher.Note that ‘ð1Þ ¼ �0:40; the root lies between 0 and 1. Interpolation gives a next trial n of 0.789;‘0ð0:789Þ ¼ �0:01, and the root lies between 0 and 0.789. One might continue this process of inter-polation; however, because of the curvature of ‘0ðnÞ, the procedure is speeded up by interspersingsteps where the next trial n is the midpoint of the bracketing interval; hence as our next approximateroot we try ð0:789þ 0Þ=2 ¼ 0:395. Since ‘0ð0:395Þ ¼ 0:736, the root lies between 0.395 and 0.789.

One might continue the process of interchanging interpolation and averaging steps to obtain thesolution n ¼ 0:78315. There are also 15 complex roots.

Once n is obtained, estimates of the cell expectations may be determined from the separate hyper-geometric distributions for each table. In our case, since the table marginals are all the same, the

estimated expectations EEkðqÞ are allP5j¼0

jpj exp ð0:78315jÞ, P5

j¼0pj exp ð0:78315jÞ

" #¼ 3:04.

To continue illustrating how to derive the directed chi-square statistic we need to specify X11k,k ¼ 1; 2; 3. We’ll consider two cases. First, X111 ¼ 4, X112 ¼ 3, X113 ¼ 2. With t ¼ ð3:04; 3:04; 3:04Þ0,Z ¼ ð1=3:04Þð4; 3; 2Þ0. Since all components of t are equal PAVA yield Z* ¼ Z since 4 > 3 > 2. In thecase X111 ¼ 3, X112 ¼ 4, X113 ¼ 2, Z* ¼ ð1=3:04Þð3:5; 3:5; 2Þ0. Once Z* is found, use (2) to find c2

D.To gain insight as to when the directed chi-square test will have higher power than the score test

we consider Figure 3.1.The figure displays the sample points and critical regions of the score test and directed chi-square

test in terms of Y1 ¼ X111, Y2 ¼ X111 þ X112. For a test of size a ¼ :01280851 the score test rejectswhen Y1 þ Y2 ¼ 11 with probability .0197058853. We note that the directed chi-square test rejects forðY1; Y2Þ ¼ ð5; 5Þ (whereas the score test does not) and rejects with probability 1 for ðY1; Y2Þ ¼ ð5; 6Þand ð3; 8Þ. This suggests that the directed chi-square test should have higher power than the score testfor parameters in which (i) q1 is large relative to q3 and q2 is close to q3 and (ii) q1 and q2 are closeand large relative to q3.

In Table 8 the first two columns give the ratios of the odds ratios for the first and last table, and theratios of the second and last table. The next two columns are logs of these values. The next threecolumns are powers.


Table 8 Power of the Breslow-Day, score, and directed chi-square tests for thecase of three 2� 2 tables with marginal row and column totals of 5 and X11 ¼ 9:

Odds Ratios Log Odds Ratios Breslow Score Directedchi-square

1.0000 1.0000 0.0000 0.0000 0.0500 0.0500 0.05008.0000 1.0000 2.0794 0.0000 0.1969 0.2634 0.2745

77.71 77.71 4.3530 4.3530 0.7927 0.7762 0.8228156.59 31.49 5.0536 3.4496 0.7825 0.8402 0.8450234.42 13.51 5.4571 2.6037 0.7920 0.8581 0.8582151.35 33.40 5.0196 3.5086 0.7833 0.8387 0.8445206.93 17.55 5.3324 2.8648 0.7827 0.8509 0.8511


The power calculations confirm our expectations.The ellipse in Figure 1 represents an acceptance region of an undirected chi-square test of some

size that would be appropriate for testing H0 vs. H2. We note that the acceptance region of a directedchi-square test is obtained from the acceptance region of the undirected chi-square test by finding thehorizontal and vertical tangents to the ellipse.

References

Agresti, A. (1990). Categorical Data Analysis. Wiley, New York.Agresti, A., Mehta, C. R., and Patel, N. R. (1990). Exact inference for contingency tables with ordered categories.

Journal of the American Statistical Association 82, 619–623.Breslow, N. E. and Day, N. E. (1980). Statistical Methods in Cancer Research, Volume 1. International Agency

for Research on Cancer, Lyon.Cohen, A., Madigan, D., and Sackrowitz, H. B. (2003). Effective directed tests for models with ordered categori-

cal data. Australian and New Zealand Journal of Statistics 45, 285–300.Hirji, K. F., Mehta, C. R., and Patel, N. R. (1987). Computing distributions for exact logistic regression. Journal

of the American Statistical Association 82, 1110–1117.Hirotsu, C. (1982). Use of cumulative efficient scores for testing ordered alternatives in discrete models. Biometri-

ka 69, 567–577.Kelsey, J. L., LiVolsi, V. A., Holford, T. R., Fischer, D. B., Mostow, E. D., Schwartz, P. E., O’Connor, T., and

White, C. (1982). A case-control study of cancer of the endometrium. American Journal of Epidemiology116, 333–342.


0 1 2 3 4 5

Y 1

0

2

4

6

8

10

Y 2

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

. . .

. .

..........................

.......................

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.......................................................

.........................................

..................................

...............................

..........................

.........................

.......................

............................................................................

.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.......

........

........

........

.......

........

........

........

.......

........

........

........

........

........

.......

........

........

........

.......

........

........

........

.......

........

........

........

.......

........

........

........

.......

........

........

........

........

........

.......

........

........

........

.......

........

........

........

.......

........

........

........

.......

........

........

........

.......

........

........

........

........

........

.......

........

........

..

................................

................................

................................

................................

................................

................................

................................

................................

.................................

................................

................................

................................

................................

.................................

................................

................................

........

χ 2D boundary.............................................................................................................

........................................

score test boundary .......................................................................

............

Figure 1 Sample space and critical regions for score test and directed chi-square test.


Le, C. T. (1998). Applied Categorical Data Analysis. Wiley, New York.Pagano, M. and Halvorsen, K. T. (1981). An algorithm for finding the exact significance levels of r � c contin-

gency tables. Journal of the American Statistical Association 76, 931–934.Patefield, W. M. (1981). An efficient method of generating random r � c tables with given row and column totals.

Applied Statistics 30, 91–105.Robertson, T., Wright, F. T., and Dykstra, R. L. (1988). Order Restricted Inference. Wiley, New York.Zelen, M. (1971). The analysis of several 2� 2 contingency tables. Biometrika 58, 129–137.



a test for homogeneity of odds ratios in ordered 2 × 2 tables

Documents