15.3: Chi square (?2) analysis, hypothesis testing, and small numbers

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    15.3: Chi square (?2) analysis, hypothesis testing, and small numbers

    Mendel’s conclusions depended on the number of plants he examined. The various ratios he predicted are expected to be true and reproducibly observed only when the number of individuals examined is large. With smaller numbers, there can be serious divergences between what is observed and what the model predicts, a situation characteristic of stochastic processes. Which gametes contain which alleles and who they fused with are both stochastic events.[509] Consider the general question, how many rolls of a die would you need to convince yourself, with high confidence, that a particular die is fair? or perhaps better put, not unfair. While the stochastic nature of meiosis and fertilization does not effect the F1 offspring of a cross between homozygous dominant and recessive plants, in which all offspring are expected to have the same (heterozygous) phenotype, it will influence the 3:1 ratio (in the F2generation) of phenotypically dominant to recessive plants predicted to occur when F1 individuals are crossed. How do we decide whether what we observe is consistent with our model or contradicts it? A model that does not produce the observed results will need to be abandoned or revised.

    \[X^2=\sum^n \frac{(\text { observed }- \text { expected })^2}{\text { expected }}\nonumber\]

    The answer is a statistical test known as a 𝛘2 (chi square) analysis.[510] It uses the equation (↓) together with two other concepts: degrees of freedom and the null hypothesis.[511] If we are testing a model that makes a mathematically precise prediction, such as the frequency of the phenotypic classes observed, our null hypothesis is that the data are unlikely to be generated by chance. Remember, we are not trying to prove that our specific hypothesis is correct; we are trying to estimate the probability that the values observed could have occurred by chance rather than by the specific mechanisms proposed.

    To define the degrees of freedom of a study, we need to know how many independent variables there are. In our two phenotype system (wrinkled or round, purple or white, etc.), we assumed that all individuals have one or the other unambiguously distinguishable phenotype. If we know the number of individuals involved and the number of either phenotype, we automatically know the number of the other. In the case of two phenotypic classes, the degree of freedom is 1 (if there are four classes, the degree of freedom is 3, and so on). What is the degree of freedom for a six-sided die? By convention, currently under some discussion[512], we take an observation to be consistent with the null hypothesis if it can be expected to occur by chance at less that 1 time out of 20 (0.05) or one time out of one hundred (0.01); otherwise we have a good case to reject our hypothesis - that is, the data observed could well be due to chance alone.

    For any particular experiment, we make observations to test our null hypothesis, are our predictions supported or rejected? Just for fun, let us consider Mendel’s monohybrid crosses. The prediction of his model is that the ratio of round to wrinkled seeds in the F2 will be 3:1. Mendel reported that he examined 7324 plants. Given his model, he would Table displaying critical percentiles of the Y distribution, including values for various percentiles.have predicted that 5492 of these plants would have round seeds, while 1849 plants would have wrinkled seeds. We can now do our 𝛘2 calculation. We have (5474 (observed) – 5492 (expected))2 = (-18)2 = 324/5492 (expected) equals 0.059 and (1850 (observed) – 1849 (expected))2 = 12 = 1/1849 (expected) = 0.00054. The sum (∑) of these two numbers is 0.0595. To determine whether these observations are consistent with our null hypothesis, we consult a 𝛘2 probability table (↓). The higher the 𝛘2 value the more likely the difference between observed and expected data is due to chance, rather than because our assumption, our null hypothesis, is correct. Our value of 0.059 lies well below the 0.05 probability value of 3.841, suggesting that the observed numbers are consistent with our model and unlikely to be generated by chance. But keep in mind, consistency does not imply “truth.” In fact, there have been suggestions that Mendel’s observed numbers are too good, too close to what would be predicted from his model.[513] Be that as it may, Mendel’s conclusions for the behavior of the types of traits he chose to study have been repeatedly verified - we can trust his general conclusions given his assumptions.

    References

    [509] It is similar to the question of which unstable isotope atom will decay next.

    [510] Here is an alternative presentation from GENETICS AND GENE PROBLEMS

    [511] chi square tutorial: http://www.radford.edu/rsheehy/Gen_flash/Tutorials/Chi-Square_tutorial/x2-tut.htm

    [512] Statistical errors and Colquhoun. 2014. An investigation of the false discovery rate and the misinterpretation of p-values

    [513] see On Fisher's Criticism of Mendel's Results With the Garden Pea


    This page titled 15.3: Chi square (?2) analysis, hypothesis testing, and small numbers was last modified on Thu, 08 Jan 2026 23:40:12 GMT and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Michael W. Klymkowsky and Melanie M. Cooper.