A Practical Approach, Second Edition=Ronald D. Ho.pdf

HUMAN STUDIES 827Table 20.4CaseSmoking among women who hadan infant with Down syndromeand age-matched controlsControlNumberof PairsSmoking Smoking n 1 = 38Not smoking Not smoking n 2 = 411Smoking Not smoking n 3 = 145Not smoking Smoking n 4 = 158Source: Data from the Swedish Medical BirthRegistry, 1983–1990.is “broken,” and the two sets of data (cases and controls) are analyzed as described above. Thisusually reduces the efficiency (if the matching has been meaningful). If the matching is donecorrectly, the variation between individuals within the strata used for matching is less than the totalvariation in the population. Differences between cases and controls can be revealed against thebackground of the low variation within the stratum but may not be observable when compared withthe large variation in the total population. Effectively matched materials should therefore beanalyzed without breaking the matching.There are many methods to analyze such data sets. Table 20.4 shows a matched sample ofDown syndrome infants and maternal age-matched controls, with information on maternal smoking.The material can be divided into four groups: (1) both case and control mothers smoked (++), (2)neither smoked (--), (3) the case mother but not the control mother smoked (+-), and (4) the controlbut not the case mother smoked (-+). If smoking was unrelated to the risk of having a Down infant,n 3 = n 4 , and the probability can be estimated that the observed values of n 3 and n 4 are taken fromthe binomial distribution of 50% of the total, n 3 + n 4 (which can be expressed as Bin(n 3 + n 4 , 0.5)].This probability can easily be evaluated (e.g., from a binomial table).The exact formula is: p = n!*p x *(1 – p) n–x /(x!*(n – x)!), where n is the total number of pairs,p = 1 – p = 0.50, and x is the observed number of pairs with characteristics +-. Note that the pvalue also has to be calculated for still more skewed distributions (x + 1/n – x – 1; x + 2/n – x –2, etc., if x > n/2). It should be noted that the two concordant groups (++ and – –) do not contributeto the evaluation of the effect studied.A similar exact analysis can be made with triplets (or a mixture of pairs and triplets). Whennumbers are large, approximate methods can be used instead. One much used technique is theMcNemar test, which can be used both for pairs and triplets. The formula for pairs is: If there aren 3 +– pairs and n 4 –+ pairs, then χ 2 = (n 3 – n 4 ) 2 /(n 3 + n 4 ). The odds ratio is then OR = n 3 /n 4 . Fromthe example in Table 20.4, we see that χ 2 = (145 158) 2 /(145 + 158) = 0.50, which is far fromsignificant, and the odds ratio is 0.92. Its 95% CI, calculated as described above (see p. 827), is0.72 to 1.16.With triplets, the following will be informative (triplets with all exposed or all nonexposed arenot informative): +–– (n 1 ), ++- (n 2 ), –++ (n 3 ), –+– (n 4 ). Then χ 2 = (2*n 1 – n 4 + n 2 – 2*n 3 ) 2 /(2*n 1+ n 2 + n 3 + n 4 ). The estimate of the OR is complicated but can be made.Another possibility is to use the Mantel-Haenszel test described above. There is no minimumsize limit for a stratum in that technique, and in principle a stratum can be a pair (or a triplet or aquadruplet). If pairs are analyzed, a 2×2 table for a +- pair (see above) will be:Exposed Unexposed TotalCase 1 0 1Control 0 1 1Total 1 1 2© 2006 by Taylor & Francis Group, LLC