Uniformly more powerful tests for hypotheses concerning linear inequalities and normal means

This article considers some hypothesis-testing problems regarding normal means. In these problems, the hypotheses are defined by linear inequalities on the means. We show that in certain problems the likelihood ratio test (LRT) is not very powerful. We describe a test that has the same size, α, as the LRT and is uniformly more powerful. The test is easily implemented, since its critical values are standard normal percentiles. The increase in power with the new test can be substantial. For example, the new test’s power is 1/2α times bigger (10 times bigger for α = .05) than the LRT’s power for some parameter points in a simple example. Specifically, let X = (X₁, …, X_p)′ (p ≥ 2) be a multivariate normal random vector with unknown mean μ = (μ₁, …, μ_p)′ and known, nonsingular covariance matrix Σ. We consider testing the null hypothesis H₀: B′_i,μ ≤ 0 for some i = 1, …, k versus the alternative hypothesis H₁: b′_iμ > 0 for all i = 1, …, k. Here b₁, …, b_k (k ≥ 2) are specified p-dimensional vectors that define the hypotheses. Many types of relationships among the means may be described with the linear inequalities. Two interesting types are those that specify the signs of the means and those that describe an order relationship. Some examples of alternative hypotheses that can be specified in this way are these: (Equation presented) (sign testing), (Equation presented) (simple order), (Equation presented) (simple loop), and (Equation presented) (simple tree). If μ_i = v_2i – v_1i, where v_ji is the average response of the ith patient subset to the jth treatment, then (Equation presented) states that Treatment 2 is better than Treatment 1 for all subsets. If the μ_i are regression coefficients, then (Equation presented) states that the mean response increases with each independent variable. In any case, these relationships would be the alternative hypothesis. Rejection of H₀ by a test with small size would be taken as strong evidence confirming that the specified sign or order relationship is true. Sasabuchi (1980) showed that the size-α LRT of H₀ versus H₁ is the test that rejects H₀ if Z_i = b′_iX/(b′_iΣb_i)^1/2 ≥ z_α for all i = 1, …, k, where z_α is the upper 100α percentile of a standard normal distribution. This test is biased and has very low power if all of the values b′_iμ (i = 1, …, k) are only slightly bigger than 0. We define an integer J and constants c₀, …, c_J that are certain standard normal percentiles. We show that, in many cases, a size-α test that is uniformly more powerful than the LRT is the test that rejects H₀ if X ∈ R₁ ∪ ··· ∪ R_J, where R_j = {x: c_j ≤ z_i ≤ c_j–1, i = 1, …, k} and z_i = b′_ix/(b′_iΣb_i)^1/2 is the LRT statistic. The set R₁ is the rejection region of the LRT, so this test is obviously more powerful than the LRT. But we show that if, for each i = 1, …, k, there exists an m ≠ i such that b,′_iΣb_m ≤ 0, then this test is also a size-α test. It is easy to verify that this condition is satisfied, for example, for all of the aforementioned H₁ hypotheses, except the simple tree, if Σ is diagonal. Tests that are even more powerful than those just described might exist. We discuss an example of such a test. But despite this test’s superior power properties, it has some counterintuitive properties. Thus tests such as in this example may be primarily of theoretical interest. All of the previously mentioned results are derived in the Σ-known case. Sasabuchi (1980) showed that, if Σ is unknown, the LRT is very similar. The differences are that Σ is replaced by an estimate and z_α is replaced by t_α, a t-distribution percentile. We show, in an example, that making the same modifications to this test does not give a size-α test. But in the example the size of the test converges to α quickly as the degrees of freedom for the estimate of Σ becomes large. So even for moderate degrees of freedom (≥ 10), this test might be preferable to the LRT, since its size is approximately α and it is much more powerful than the LRT. A two-sided version of this problem is obtained if we test (Equation presented) versus (Equation presented), where H₁ is a one-sided alternative as defined above. Sasabuchi (1980) showed that the LRT rejects (Equation presented) if Z_i ≥ c for all i = 1, …, k or Z_i ≤ – c for all i = 1, …, k, Sasabuchi gave some conditions under which c = z_α gives a size-α test. We consider only the special case in which H₁ is the sign-testing alternative and (Equation presented), a diagonal matrix. For constants c₀, …, c_2J, similar to those above, we show that the test that rejects (Equation presented) if X ∈ R₁ ∪ ··· ∪ R_2J, where R_j = {x: c_j ≤ x_i/σ_i ≤ c_j–1, i = 1, …, p}, is a size-α test that is uniformly more powerful than the LRT. For the special case of p = 2, this provides a test that is uniformly more powerful than a test proposed by Gail and Simon (1985) for testing for a qualitative interaction.