Communications in Statistics—Simulation and Computation® , 37: 14–24, 2008 Copyright © Taylor & Francis Group, LLC ISSN: 0361-0918 print/1532-4141 online DOI: 10.1080/03610910701420065
Inference
Which Exact Test is More Powerful in Testing the Hardy–Weinberg Law? SEUNG-HO KANG Department of Statistics, Ewha Womans University, Seoul, Korea The asymptotic chi-square test for testing the Hardy–Weinberg law is unreliable in either small or unbalanced samples. As an alternative, either the unconditional or conditional exact test might be used. It is known that the unconditional exact test has greater power than the conditional exact test in small samples. In this article, we show that the conditional exact test is more powerful than the unconditional exact test in large samples. This result is useful in extremely unbalanced cases with large sample sizes which are often obtained when a rare allele exists. Keywords Exact power; Rare allele; Unbalanced sample. Mathematics Subject Classification 62F03; 62P10.
1. Introduction The Hardy–Weinberg law is a fundamental concept of population genetics. According to the law, the genotype frequencies and gene frequencies of a large, randomly mating population remain constant if immigration, mutation, and selection do not take place. Maiste and Weir (1995) reviewed and compared several tests of independence of allelic frequencies within and between loci. Testing deviation from the Hardy–Weinberg law is often performed by using the asymptotic chi-square test. The asymptotic chi-square test was developed based on the assumption that the expected frequencies in all cells are larger than a predetermined number (for example, five; Weir, 1996). The assumption may be violated in the following cases. The first is in a small sample case in which the expected frequencies are small in some cells. The second is in an unbalanced sample case which is often obtained when there is a rare allele at a locus. In such unbalanced sample cases, the expected frequencies can be small even if the total sample size is large. As an example of an unbalanced sample case, we consider genotype data of the Received January 01, 2007; Accepted April 25, 2007 Address correspondence to Seung-Ho Kang, Department of Statistics, 11-1 DaeHyundong, SeoDaeMun-Ku, Seoul 120-750, Korea; E-mail:
[email protected]
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Table 1 Genotypes data of the Scarlet tiger moth A1 A1 A1 A2 A2 A2 = 5 138 1469
Scarlet tiger moth in Table 1 (Ford, 1971). Since the data is heavily unbalanced, the asymptotic chi-square test may have inflated Type I error. Actually, since the expected frequency in the first cell is less than 5 as shown below, the asymptotic chi-square test is not recommended. E11 = npˆ 2 = 1612 × 004592 = 34 pˆ =
2 × 5 + 138 = 00459 2 × 1612
Furthermore, Kang and Shin (2004) proved that the supremum of the Type I error rate of the asymptotic chi-square test is always greater than the nominal level in large samples. For example, Table 2 in Kang and Shin (2004) shows that the sizes of the 5%-level asymptotic chi-square test at n = 1000 and n = 5000 are 0.085 and 0.103, respectively. Wigginton et al. (2005) also showed that the chi-square test can have inflated Type I error rates, even in relatively large samples. As alternatives, we consider the conditional and the unconditional exact test (Aoki, 2003; Chapco, 1976; Elston and Forthofer, 1977; Guo and Thompson, 1992; Haber, 1994; Haldane, 1954; Lazzeroni and Lange, 1997; Louis and Dempster, 1987; Vithayasai, 1973). Especially, Zaykin et al. (1995) examined the conditional exact tests with arbitrary numbers of loci. A main advantage of exact tests (both conditional and unconditional) is that it is guaranteed to bound the Type I error rate to any desired level. Since two exact tests can control the Type I error rate under a nominal level, the next natural question is: “Which exact test is more powerful?” A disadvantage of exact tests is that they are often conservative. Therefore, comparing the power of two exact tests is essentially the same as comparing the conservativeness of the tests. In general, the conservativeness of a conditional exact test might be due to the discreteness of the conditional distribution of a test statistic, while the conservativeness of an unconditional exact test might be due to the need for eliminating a nuisance parameter by considering the worst possible case (Mehta and Hilton, 1993). For testing the homogeneity of two independent binomial proportions, the powers of the two exact tests were compared (Kang and Ahn, 2007; Mehta and Hilton, 1993; Suissa and Shuster, 1985). Suissa and Shuster (1985) showed that the exact unconditional test has greater power than the exact conditional test when the sample size in each group is the same and less than 150. But, Mehta and Hilton (1993) also concluded that the discreteness of the conditional distribution makes the result of Suissa and Shuster (1985) and the powers of the two exact tests become close as the sample size increases. Recently, Kang and Ahn (2007) showed that the opposite conclusion holds in extremely unbalanced data; that is, the exact conditional test has greater power than the exact unconditional test for testing the homogeneity of two independent binomial proportions. The reason is that the peak of the Type I error rate increases as the imbalance in sample sizes does (Kang et al., 2006).
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For testing the Hardy–Weinberg law in biallele cases, Haber (1994) argued that the unconditional exact test is more powerful than the conditional exact test. He compared the powers of the two exact tests in sample size of n = 10–100 by increments of 10. The results are displayed in Table 2 in Haber (1994). However, careful investigation of Table 2 in Haber (1994) reveals that there are some cases where the unconditional exact test is less powerful than the conditional exact test. For example, when n = 80 p = 05, and f = 02, the powers of both the conditional and unconditional exact tests are 0.419 and 0.282, respectively. In this article, we will investigate why such phenomena occur. Specifically, limiting the discussion to biallele cases, in this article we will compare the powers of the two exact tests in testing the Hardy–Weinberg law as Kang and Ahn (2007) did for testing the homogeneity of two independent binomial proportions. Although any test statistic can be used in exact tests, in this article we employ the chi-square test statistic because it has been widely used. Using the chi-square test statistic, in small samples we confirm the results of Haber (1994). That is, the unconditional exact test is more powerful than the conditional exact test. The reason is that the conditional distribution of the test statistic has only a few distinct values, so that the conditional exact test of nominal size may have an actual size much less than and result in a loss of power. On the other hand, Kang and Shin (2004) showed that the supremum of the Type I error rate of the asymptotic chi-square test are greater than the nominal level and deviates more upwardly from the nominal level at the edges of the nuisance parameter space as the sample size gets larger, although the Type I error rates are close to the nominal level in most values of the nuisance parameter. Since the unconditional exact test removes the nuisance parameter by taking supremum over the nuisance parameter space, we expect that in large samples the conditional exact test will be more powerful than the unconditional exact test. In Sec. 2, we review the asymptotic chi-square test in testing the Hardy– Weinberg law. In Secs. 3 and 4, we provide the formulae of the powers of the conditional and the unconditional exact test, respectively. In Sec. 5, we present the numerical comparison of the powers of the two exact tests. In Sec. 6, we analyze genotype data of the Scarlet tiger moth in Table 1.
2. The Asymptotic Chi-Square Test Let X11 X12 X22 denote the observed frequencies of the genotypes A11 A12 A22 in a sample of size n. Then, X11 X12 X22 follow the multinomial distribution Mn p11 p12 p22 . The null hypothesis of the Hardy–Weinberg law is H0 p11 = p2 , p12 = 2p1 − p, p22 = 1 − p2 , where 0 < p < 1. Let pˆ =
2X11 + X12 2n
The asymptotic chi-square test statistic for the Hardy–Weinberg law is given by
X2 =
2 2 Xij − Eij 2 Eij i=1 j≥i
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where E11 = npˆ 2 , E12 = 2np1 ˆ − p, ˆ and E22 = n1 − p ˆ 2 . The chi-square test rejects 2 2 2 H0 if X > 1 where 1 is the upper -quantile of the chi-square distribution with one degree of freedom. For a nonnegative number h, let Eh = minij Eij > h . Since the value of X 2 is not defined if min Eij = 0, we consider the only cases in which the event Eh occurs. The size of a test is defined by the supremum of the Type I error rate over the nuisance parameter space under the null hypothesis. Hence, the size represents the Type I error rate in the worst possible case. Kang and Shin (2004) proved that the size of the asymptotic chi-square test is greater than the nominal level in infinite samples. The size of the level-5% asymptotic chi-square test is greater than 16.1% and 7.1%, respectively, when h = 0 and h = 4 (see Table 1 in Kang and Shin, 2004). However, the Type I error rate of the asymptotic chi-square test is close to the nominal level in most values of p and the peak of the Type I error rate takes place when the value of p is very close to either 0 or 1. In other words, Kang and Shin (2004) showed theoretically that the asymptotic chi-square test has an inflated Type I error rate in extremely unbalanced samples. Wigginton et al. (2005) also showed the anticonservativeness of the asymptotic chi-square test from a different standpoint. They found that the chi-square test can sometimes be very anticonservative even in a sample of size of 1,000 and also found some periodicity of the Type I error rate.
3. The Conditional Exact Test The conditional exact test originated with work by Haldane (1954). Some statistical properties were examined in small samples (Chapco, 1976; Elston and Forthofer, 1977). The conditional exact test was extended into multiple alleles cases. And efficient algorithms to compute the exact p-value were developed by Aoki (2003), Guo and Thompson (1992), Lazzeroni and Lange (1997), and Louis and Dempster (1987). In this article, we limit the discussion to biallele cases for power comparison with the unconditional exact test. Here we present the formula for the power of the conditional exact test. First, the null likelihood is given by PX11 = x11 X12 = x12 X22 = x22 H0 = =
n! p2 x11 2p1 − px12 1 − p2 x22 x11 !x12 !x22 ! n! 2x12 p2x11 +x12 1 − px12 +2x22 x11 !x12 !x22 !
Then, M = 2X11 + X12 is a sufficient statistics for p under the null hypothesis. In the conditional exact tests, we use the conditional distributions of the observed sample given a sufficient statistic for the nuisance parameter because the conditional distribution is free from the unknown nuisance parameter p. For 0 ≤ x11 ≤ m/2 PX11 = x11 X12 = x12 X22 = x22 H0 M = m =
n!m!2n − m!2m−2x11 x11 !m − 2x11 !n − m + x11 !2n!
When the sample space S is given by S = x11 x12 x22 x11 + x12 + x22 = n 0 ≤ x11 x12 x22 ≤ n
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let T be a test statistic on the sample space S whose large value is significant. When m (an observed value of M) is given, the conditional exact tests consider the following set as sample spaces. Sm = x11 x12 x22 ∈ S 2x11 + x12 = m 0 ≤ m ≤ 2n Since the large value of T is significant, for the given m, the rejection region is Sm t = x11 x12 x22 ∈ Sm Tx11 x12 x22 ≥ t We evaluate the exact null distribution of T
PT ≥ t m H0 =
x11 x12 x22 ∈Sm
n!m!2n − m!2m−2x11 x !m − 2x11 !n − m + x11 !2n! t 11
for each possible value of t. Let t m be the smallest possible value such that PT ≥ t m m H0 ≤ Then the power of the conditional exact test is given by
c p11 p12 p22 =
2n
PT ≥ t m m p11 p12 p22 PM = m p11 p12 p22
m=0
=
x11 x12 x22 ∈S
n! x x x p 11 p 12 p 22 x11 !x12 !x22 ! 11 12 22
(1)
where S =
2n
Sm t m
m=0
Although any test statistic T can be employed, in this article we use the chisquare test statistic X 2 to construct the conditional exact test for the comparison of the powers.
4. The Unconditional Exact Test For a given 0 < p < 1, we evaluate the exact null distribution of T under the null hypothesis. PT ≥ t p H0 =
x11 x12 x22 ∈S
n! 2x12 p2x11 +x12 1 − px12 +2x22 I Tx11 x12 x22 ≥t x11 !x12 !x22 !
for each possible value of t. Let t p be the smallest possible value such that PT ≥ t p p H0 ≤
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In unconditional exact tests, the unknown nuisance parameter p is removed by taking supremum over the nuisance parameter space. So, let t = sup0
