This paper presents how one can adjust the McNemar’s test for the analysis of clustered matched-pair data. A McNemar’s-like table for K clusters of matched-pair data is used

McCarthy, William F.

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Collection of Biostatistics Research ArchiveCOBRA Preprint SeriesYear  Paper Adjustment to the McNemar’s Test for theAnalysis of Clustered Matched-Pair DataWilliam F. McCarthy∗∗Maryland Medical Research Institute, dr.w.f.mccarthy@gmail.comThis working paper is hosted by The Berkeley Electronic Press (bepress) and may not be commer-cially reproduced without the permission of the copyright holder.http://biostats.bepress.com/cobra/art29Copyright c©2007 by the author.Adjustment to the McNemar’s Test for theAnalysis of Clustered Matched-Pair DataWilliam F. McCarthyAbstractThis paper presents how one can adjust the McNemar’s test for the analysis ofclustered matched-pair data. A McNemar’s-like table for K clusters of matched-pair data is used.                                                                       The following discussion is based on the statistical procedure developed by Nancy A. Obuchowski of the Cleveland Clinic [1] and Valerie Durkalski [2] of the Medical University of South Carolina.  Obuchowski’s approach avoids having to assume a constant within-cluster correlation (Durkalski’s approach [2] does assume a constant within-cluster correlation).  Obuchowski’s test statistic for clustered matched-pair data takes into account the possible variation in correlation between units within a cluster.  Obuchowski’s approach considers the homogeneity hypothesis  H0: p1 = p2 versus HA: the pi’s are unequal  This is the same homogeneity hypothesis as used in the unadjusted McNemar’s Test outlined by Conover [3].  Durkalski’s approach considers the null hypothesis  0 10 01: k kH p p− = 0  for all K clusters versus 10 01: 0A k kH p p− ≠   ______________________________________________________________________   Table 1. McNemar’s-Like Table for K Clusters of Matched-Pair Data.  Procedure 2  Procedure 1    Success  Failure  Success 1Kkka=∑   1Kkkb=∑  1( )Kk kka b=+∑  Failure 1Kkkc=∑   1Kkkd=∑    1( )Kk kka c=+∑    1Kkkn=∑ = N         Hosted by The Berkeley Electronic Press                                                                     Obuchowski’s test statistic for clustered matched-pair data:          2OX  =  21 21 2ˆ ˆ( )ˆ ˆvar( ) pp pp p−−  2OX  is asymptotically distributed as a chi-square with one degree of freedom under the null hypothesis.  Where  1 2ˆ ˆvar( ) pp p−  = 1 2 1ˆ ˆ ˆvar( ) var( ) 2cov( , )p p 2ˆ pp p p+ − p ;   1 2ˆ ˆ( )p p p= + / 2 ;   ˆ ip   =  11KikkKkkxn==∑∑     with  (procedure i=1,2);    2121( )ˆvar( )1( )Kik kki p Kkkx n pKpKn==⎡ ⎤⎢ ⎥−⎢ ⎥= ⎢− ⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∑∑ ⎥  with (procedure i=1,2);   '1' 21( ) (ˆ ˆcov( , )1Kik k i k kki i p Kkk)x n p x n pKp pKn==⎡ ⎤⎢ ⎥− −⎢ ⎥= ⎢− ⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∑∑ ⎥  with (procedure i=1,i] =2).   Note: The adjustment for clustering is taken into account by multiplying the variance and covariance by an inflation factor 1KK − .  http://biostats.bepress.com/cobra/art29                                                                     Where we have a random sample of K clusters from a population, with nk units in the k th cluster, k=1,...,K.  Each unit is given I treatments or tests.  Let xik denote the number of units in the k th cluster that respond to treatment i, i=1,...,I.  Let pi denote the probability that a randomly selected unit from the population will respond to treatment i.   ^ip  is an estimator of pi which is the overall proportion of responses to treatment.  Example:  Table 2. Specificity of PET versus SPECT (21 clusters; 51 matched-pairs).  Patient No. (cluster k) Number of glands (nk) x(PET)k x(SPECT)k1 3 0 [0,0,0] 2 [0,1,1] 2 3 2 [1,1,0] 3 [1,1,1] 3 3 3 [1,1,1] 3 [1,1,1] 4 1 1 [1] 1 [1] 5 3 2 [1,1,0] 3 [1,1,1] 6 4 4 [1,1,1,1] 4 [1,1,1,1] 7 3 3 [1,1,1] 3 [1,1,1] 8 2 2 [1,1] 2 [1,1] 9 2 2 [1,1] 1 [1,0] 10 1 1 [1] 1 [1] 11 3 2 [1,1,0] 2 [1,1,0] 12 2 2 [1,1] 2 [1,1] 13 3 3 [1,1,1] 3 [1,1,1] 14 2 2 [1,1] 2 [1,1] 15 2 0 [0,0] 2 [1,1] 16 3 2 [1,1,0] 2 [1,1,0] 17 3 2 [1,1,0] 2 [1,1,0] 18 3 2 [1,1,0] 3 [1,1,1] 19 2 2 [1,1] 2 [1,1] 20 1 1 [1] 1 [1] 21 2 2 [1,1] 2 [1,1]  xik is the number of true negative test results for patient (cluster) k, test i. The numbers in brackets correspond with the test results for each gland, where 1 indicates that the gland was called “true negative” for condition (success) and 0 indicates that the gland was called “false positive” (failure). The PET test result listed first in the brackets corresponds with the SPECT test result listed first, etc. For example, for patient number 1 we have 3 matched-pairs: (0,0), (0,1), and (0,1).       Hosted by The Berkeley Electronic Press                                                                     151Kkkn==∑                    ( )140KPET kkx==∑ ( )146KSPECT kkx==∑   ˆ 40 / 51 0.784PETp = =          ˆ 46 / 51 0.902SPECTp = =   ˆ ˆvar( ) 0.00484PET SPECTp p p− =             2 2.88OX =   p = 0.090     No rejection of the Ho If the McNemar’s Test is used without consideration of adjustment for clustered data, we get different results:  2 4.5McX =   p = 0.034     Rejection of the Ho   Durkalski’s Method:   This is a simple method for the analysis of clustered matched-pair data that adjusts for multiple units within a cluster, yet avoids correlation assumptions among and within clusters and avoids distributional assumptions.  The null hypothesis is  for all K clusters, where 0 10 01: k kH p p− = 0    10 kkkbp En⎡ ⎤= ⎢ ⎥⎣ ⎦   and 01 kkkcp En⎡ ⎤= ⎢ ⎥⎣ ⎦  A general form of the test statistic for H0: 10 01 0k p kp p− =  for all K clusters, asymptotically distributed as a chi-square with one degree of freedom, is    ( )( )210 01210 01ˆ ˆˆ ˆvarGp pXp p−= −   where   ( ) ( )10 0111 1ˆ ˆKk kk kp p bK n=− = −∑ c   http://biostats.bepress.com/cobra/art29                                                                     and  ( ) ( ) (210 01 10 01211 1ˆ ˆ ˆ ˆvarKk kk kp p b c p pK n=)⎡ ⎤− = − − −⎢ ⎥⎣ ⎦∑ .  Under the null hypothesis is H0: 10 01 0k p kp p− =  for all K clusters,    ( ) ( )210 01 211 1ˆ ˆvarKk kk kp p b cK n=⎡ ⎤− = −⎢ ⎥⎣ ⎦∑ .  Durkalski’s test statistic is:    ( )( )2122111Kk kk kVKk kk kb cnXb cn==⎛ ⎞−⎜ ⎟⎝ ⎠= ⎡ ⎤−⎢ ⎥⎣ ⎦∑∑.    Table 3. 2 x 2# Table Representation of the Specificity of PET versus SPECT (for use with the Durkalski Method). # Based on Table 1. McNemar’s-like table for K clusters of matched-pair data.  Patient No. (cluster k) Number of glands (nk) Matched- pair data ak bk ck dk 1/ nk bk - ck1 3 (0,0) (0,1) (0,1) 0 0 2 1 1/3 -2 2 3 (1,1) (1,1) (0,1) 2 0 1 0 1/3 -1 3 3 (1,1) (1,1) (1,1) 3 0 0 0 1/3 0 4 1 (1,1) 1 0 0 0 1/1 0 5 3 (1,1) (1,1) (0,1) 2 0 1 0 1/3 -1 6 4 (1,1) (1,1) (1,1) (1,1) 4 0 0 0 1/4 0 7 3 (1,1) (1,1) (1,1) 3 0 0 0 1/3 0 8 2 (1,1) (1,1) 2 0 0 0 1/2 0 9 2 (1,1) (1,0) 1 1 0 0 1/2 1 10 1 (1,1) 1 0 0 0 1/1 0 11 3 (1,1) (1,1) (0,0) 2 0 0 1 1/3 0 12 2 (1,1) (1,1) 2 0 0 0 1/2 0 13 3 (1,1) (1,1) (1,1) 3 0 0 0 1/3 0 14 2 (1,1) (1,1) 2 0 0 0 1/2 0 15 2 (0,1) (0,1) 0 0 2 0 1/2 -2 16 3 (1,1) (1,1) (0,0) 2 0 0 1 1/3 0 17 3 (1,1) (1,1) (0,0) 2 0 0 1 1/3 0 18 3 (1,1) (1,1) (0,1) 2 0 1 0 1/3 -1 19 2 (1,1) (1,1) 2 0 0 0 1/2 0 20 1 (1,1) 1 0 0 0 1/1 0 21 2 (1,1) (1,1) 2 0 0 0 1/2 0      Hosted by The Berkeley Electronic Press                                                                     Table 4. Key to Interpretation of Matched-Pair Data.  PET  SPECT   Success  Failure Success ak = (1,1)  bk = (1,0)  Failure ck = (0,1)  dk = (0,0)     Thus,  2 2.32VX = and  p = 0.128     No rejection of the Ho.   Table 5. Comparison of Methods.  Method Test Statistic p-value Rejection of H0Unadjusted 4.5 0.034 Yes Obuchowski 2.88 0.090 No Durkalski 2.32 0.128 No   Confidence Interval for a Common Difference based on the Durkalski Method:  It is useful to look at a non-null variance estimate if one is interested in calculating a confidence interval for the common difference 10 01ˆ ˆp p−  (this assumes that the difference is constant across the K clusters).  If one assumes a common difference, then the variance estimate under the non-null condition, , is: 10 01: 0A k kH p p− ≠  ( ) ( ) (210 01 10 01211 1ˆ ˆ ˆ ˆvarKk kk kp p b c p pK n=⎡ ⎤− = − − −⎢ ⎥⎣ ⎦∑ ) .  Then, a 95 per cent confidence interval for the true difference 10 01p p−  is:  ( ) ({ )}10 01 10 01ˆ ˆ ˆ ˆ1.96 varp p p p− ± − .      http://biostats.bepress.com/cobra/art29                                                                     References:  [1] Obuchowski NA (1998). On the comparison of correlated proportions for clustered data. Statistics in Medicine 17, 1495-1507.  [2] Durkalski VL, Palesch YY, Lipsitz SR and Rust PF (2003). Analysis of clustered matched-pair data. Statistics in Medicine 22: 2417-2428.  [3] Conover WJ (1980). Practical nonparametric statistics, 2nd Edition, Wiley, NY.  Hosted by The Berkeley Electronic Press

Adjustment to the McNemar’s Test for the Analysis of Clustered Matched-Pair Data

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Adjustment to the McNemar’s Test for the Analysis of Clustered Matched-Pair Data

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