The Statistical Technique Used in Studying Relationship of Caries Incidence in Deciduous and Permanent Dentitions

Abstract

A paper by Bruszt (J. dent. Res., 38:416, 1959) proposed a statistical technique of analyzing the relationship of caries incidence in deciduous and permanent dentitions from the following summarized data in Table 1. TABLE 1 AGE 11 AGE 5 TOTAL Caries-free PermaCarious nent Dentitions Permanent Dentitions Carious deciduous dentitions 4 70 74 Caries-free deciduous dentitions 12 11 23 Total .. .16 6 81 97The data consisted of 97 children, 74 with carious deciduous dentitions and 23 with caries-free deciduous dentitions at age five. The results of examination at age eleven are shown in the table. The author concluded that the probability that a caries-free permanent dentition at eleven came from a caries-free deciduous dentition was twelve to sixteen or 0.75. From this, the author "deduced the fact that children with caries-free deciduous dentition will probably have caries-free permanent dentition." The deduction was logically in error. The observation that 12 out of the 16 caries-free permanent dentitions at age eleven were from caries-free deciduous dentitions at age five might indicate that caries-free permanent dentitions are more likely to have come from caries-free deciduous dentitions, but it gave no cause to infer that caries-free deciduous dentition tends to give rise to caries-free permanent dentition. This latter inference could be drawn, however, from the comparison of the two proportions, namely:Number of caries-free permanent dentitions at age 11 that had carious = deciduous dentitions at age 5 4 Number of various deciduous dentitions at age 5 74 and Number of caries-free permanent dentitions at age 11 that had P caries-free deciduous dentitions at age 5 _ 12 2 = Number of caries-free deciduous dentitions at age 5 2 (2) To test the hypothesis that PI = P2, a simple x2 test with one degree ol f reedom may be performed: H[ 4 X 1 1-1 2 X 7 0 1 ( 9 7 )]29 7 v2 = 2 23 -O 2* = 16 X 81 X 74 X 23It was highly significant; consequently, we conclude that P2 ± Pi, and then infer P2 > Pl.The validity of the above test lies in the fact that the chi-square test, in a 2 X 2 contingency table, is effectively testing the significance of the difference of the respective frequency corresponding to PI and P2, since the expected values for the cells a and c are calculated on the assumption that these proportions are the same under the null hypothesis.