Institute for Medical Biometry Informatics and Epidemiology University of Bonn D-53105 Bonn, Germany
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INTRODUCTION |
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In a recent article in the Journal, Schaid and Jacobsen (3) quantified the effect on the type I error rate of Pearson's chi-square test induced by deviations from Hardy-Weinberg equilibrium. The authors provided a correction method based on determining the correct variance for the observed allele frequency difference
d -
c between cases and controls. They noted that this variance can be estimated by either pooling cases and controls (
pool) or summing the estimates obtained from cases and controls separately (
sep). In the paragraphs that follow, the resulting test statistics are denoted zpool and zsep, respectively.
The first purpose of this letter is to point out that the square of the test statistic based on pool (i.e., z2pool) is identical to the test statistic of Armitage's trend test (2
). This point can be observed almost directly when the formula given by Sasieni ((1
), p. 1258) is rewritten in the notation used by Schaid and Jacobsen (3
).
The second purpose is to comment on the conjecture expressed by Schaid and Jacobsen (3) that the zsep test may be more powerful than the zpool test. It can be shown that
pool -
sep = (
d -
c)2/(Nd + Nc) for samples with an equal number of cases and controls (i.e., Nd = Nc). Therefore, zsep
zpool, and, when inference about the null hypothesis is based on comparing zsep or zpools with the quantile of a standard normal distribution, the power (but also the true type I error rate) of the zsep test will be greater than the power of the zpool test. A numerical example is provided by considering a model with a population allele frequency of p = 0.02 and relative risks of
2 = 500 and
1 = 5 for AA homozygotes and AB heterozygotes, respectively. For an assumed error rate of
= 0.001 and Nd = Nc = 56, the zsep test rejects H0 with a probability of 0.808, whereas the power of the zpool test is only 0.738. These values have been obtained by summing the probabilities of the samples, leading to rejection of H0. If Nd
Nc, however, zpool can be greater than zsep. In addition, the zpool test can be more powerful than the zsep test. With Nd = 49 and Nc = 65, and against the same alternative considered before, the power of the zpool test is 0.811, whereas the power of the zsep test is only 0.716.
In summary, it may be supposed that the zpool statistic (3) is more attractive to the genetics community than the less intuitive, but equivalent statistic of Armitage's trend test. The example described in this letter indicates that the zsep test proposed by Schaid and Jacobsen (3
) may deserve further evaluation.
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REFERENCES |
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Department of Health Sciences Research Mayo Clinic Rochester, MN 55905
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INTRODUCTION |
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Dr. Knapp also points out (1) that the "separate" and "pooled" methods of calculating the variance of the difference in allele frequencies can result in statistical tests with different power and even type I error rates. To speculate further, we anticipate that the sample size will have a large effect on the relative power, and type I error rates, of these two approaches. For small samples, when the variance of allele frequencies is most erratic, it will likely be best to use the "pooled" variance to avoid erratic behavior of the statistic and hence an increased type I error rate. On the other hand, large samples may be robust to this erratic behavior, so that the "separate" variance may provide the correct type I error rate and yet furnish greater power for a range of alternatives.
Finally, we agree with the statement that the efficiency and accuracy of tests of association in the case of departure from Hardy-Weinberg equilibrium deserve further attention (1). In particular, the potential influence of sample size, ratio of cases to controls, and magnitude of genetic effects requires additional study.
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References |
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