McNemar’s TestPrevious test only valid if the two samples are independent; for example, students in SPHTM vs. students in SOM. Sometimes, it isobvious they are not independent. Perhaps we have matched pairs or perhaps each subject is his/her own control. The outcome of interest is categorical, however. We might want to test whether two skin lotions are equally effective at relieving a poison ivy rash.We test the two lotions on 100 persons with a poison ivy rash on each arm, applying lotion one to one arm and lotion two to the second arm. Let’s construct a different kind of 2 by 2 table. Let the matched pair be the unit of analysis and classify by the occurrence of the event or variable of interest for each member of the pair. Pairs fall into two groups. A concordantpair is one in which the event is the same for each member of the pair. A discordant pair is one in which the event is different for the two members of the pair.Other member of pair TotalOne member of pair+ -+ na- nbTotalnD = na + nbNote: Concordant pairs provide no information about differences between the two groups in the occurrence of the event. We focus on the discordant pairs only in the analysis.Type A discordant pairs. First member of the pairhas the event and the second does not.Type B discordant pairs. Second member of the pair has the event and the first does not.Let p = probability that a discordant pair is of type A. If the event of interest occurs equally in the two groups, we expect p to be 0.5. H0: p = 0.5H1: p  0.5Two analytical approaches. Normal-Theory test vs. an exact test. We will focus first on the normal theory test.Suppose there are nD discordant pairs, nA of type A. Under H0, E(nA) = nD / 2 and Var(nA) = nD/4. These come from the mean and varianceof a binomial distribution. Assume the normal approximation to the binomial distribution is appropriate, and we will include a correction for continuity. This approach is valid if npq = nD / 4 ≥ 5 or nD > 20.The following test is called McNemar’s Test1. Create a 2 x 2 table of matched pairs2. Count the total number of discordant pairs (nD) and the number of type A discordant pairs (nA)3. Compute the test statistic BABADDAnnnn/nnn2221or4212X4. For a two-sided test at level of significance , reject H0 if X2 > 211,5. Exact p-value = Pr 2X21Example: We might want to test whether two skinlotions are equally effective at relieving a poison ivy rash. We test the two lotions on 100 persons with a poison ivy rash on both arms, applying one lotion to the first arm and the other lotion to the second arm.H0: The proportion of persons experiencing relief is the same with both lotions. p1 = p2This is equivalent to half of the discordant pairs being of type A and half of type BH1: The proportion of persons experiencing relief is not the same with both lotions. p1  p 2Let  = 0.05Reject H0 if X2 > 29501 ., = 3.84Of the 100 subjects, 30 experienced relief from both lotions, 34 experienced relief from lotion 2 only, 18 experienced relief from lotion 1 only, and 18 experienced relief from neither lotion.Lotion 1Lotion 2 Relief No relief TotalRelief 30 34 64No relief 18 18 36Total 48 52 100nD = 34 + 18 = 52nA = 34100641)lotionPr(Relief,ˆ100481)lotionPr(Relief,ˆ21pp 33413571350263445221252344212X22222...//nnnDDAAlternative Formula  33452151834118341X2222.nnnnBABA84333402529501297501....,.,2XTherefore, 0.025 < p < 0.05. There is evidence thatthe proportions of persons getting relief differs for the two Lotions.Exact TestIf nD / 4 < 5 (nD < 20), the previous test cannot be used. Instead, a test based on the exact binomial probabilities is needed. 1. Create a 2 x 2 table of matched pairs2. Count the total number of discordant pairs (nD) and the number of type A discordant pairs (nA)3. The exact p-value is given by a. 2/if21*20DAnnkDnnknpDAb. 2/if21*2DAnnnkDnnknpDDAc. p = 1 if nA = nD / 2Example: Twenty subjects volunteered to participate in a trial to test two methods for detecting hypercholesterolemia. Each subject gave a blood sample which was then split into two equal parts. One part was tested by Method A and the other part was tested by Method B. Note: An ordinary chi-square test for two independent samples cannot be used.Suppose 4 people are designated hypercholesterolemic by both methods, 8 people are designated as having normal cholesterol by both methods, 6 people hare designated as hypercholesterolemic by Method A but not B and 2 people are designated as hypercholeserolemic by MethodB but not A. Is there a significant difference in the proportion found hypercholesterolemic by the two methods?Method AMethod B High TC Normal TotalHigh TC 4 2 6Normal 6 8 14Total 10 10 20Note: nD = 2 + 6 = 8 < 20H0: The proportion found hypercholesterolemic by the two methods is the same. p1 = p2H1: The proportion found hypercholesterolemic by the two methods is not the same. p1 ≠ p2Note: nA = 2 < nD / 2. Therefore,  289.1445313.*2109375.03125.0039063.*2212821182108*2218*221*28888200knnkDkknpDANote: We can also evaluate this by using Table 1 for n = 8, p = 0.50, and k = 0, 1, 2.Conclusion: There is no evidence that the two methods differ in designating an individual as hypercholesterolemic.Example122 pairs of brothers, one member of each pair overweight and other of normal weight, were examined for presence of varicose veins. Test whether there is no relationship between being overweight and developing varicose veins. Note: thisis equivalent to testing that the same proportion of overweight men as normal weight men possesses varicose veins. Overweight brotherNormal weightbrotherWithvaricoseveinsWithoutvaricoseveinsTotalWith varicoseveins19 4 23Withoutvaricose veins13 86 99Total 32 90 122H0: The proportion of overweight brothers with varicose veins is the same as the proportion of normal weight brothers with varicose veins. p1 = p2H1: The