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Chi Square Agreement

Posted by Josh On September - 14 - 2021

Weldon thought that Mendel`s data “were so remarkably in agreement with Mendel`s summary that the probability that the agreement between observation and hypothesis would be worse than what actually exists is about 16 to 1.” Weldon was so flummoxed that the empirical distribution of Mendel`s peas was so similar to a hypothetical (or expected) Mendelschen distribution that he couldn`t help but wonder if the results were too good. These conclusions were so confusing that a month after he began analyzing Mendel`s pea distribution, Weldon began to think that Mendel is “either a black liar or a wonderful man.” While later Weldon said the results were so good that they were suspicious, he wondered why Mendel`s laws were absolutely true for his peas and absolutely false for Laxton, while Chermak`s were partly true. You may be wondering why we can`t just assess the association between the two categories in the usual way, for example. B with a square of chi. A pearson-chi-square test on the square above gives an X2 value of 71.2, very significant (PConsider of this second record. Take the example of the assessment of alcohol consumption for a cohort study on the effect of alcohol on heart disease. The problem here is that people tend to underestimate the amount they drink. One way to get around this is to ask the spouse to estimate the number of drinks the husband consumes each day. The data is arranged in a square contingency table as follows.

While there is a high degree of concordance, most observations are along the diagonal. In this case, we have 106 of the 120 observations along the diagonal. We can obtain a measure of the degree of compliance by calculating the proportion of identical assessments between wives and husbands. In this example, the proportion is 0.883. The chi square, as we have seen, is a measure of the divergence between the expected and observed frequencies, and if there is no difference between the expected and observed frequencies, the value of the chi square is 0. This problem can be circumvented with the kappa weighted measurement. A weight, wi, is assigned to cases where the two evaluators differ according to categories i. . . .

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