The Level and Power of a Studentized Range Test for Testing the Equivalence of Means

Transcription

1 The Level and Power of a Studentized Range Test for Testing the Equivalence of Means Miin-Jye Wen and Hubert J. hen Department of Statistics National heng-kung University Tainan, Taiwan ABSTRAT A studentized range test is proposed to test the hypothesis of equivalence of normal means in terms of the standardized average deviation among means. Both the level and the power of the test are each controlled at a given probability. The least favorable configuration (LF) of means to guarantee the maximum level at a null hypothesis and the LF to guarantee the minimum power at an alternative hypothesis are obtained. These level and power are fully independent of the unknown means and variances. Therefore, for a given level and a given power, the critical value and the required sample size for an experiment can be simultaneously determined. When the common population variance is unknown and the equivalence measure is the average deviation of means, a one-stage sampling procedure and a two-stage sampling procedure are employed, and their relative merits are discussed. Key Words and Phrases: Studentized range test; Least favorable configuration; Level and power. 1

2 1. Introduction The problem of statistical hypothesis testing concerning several normal means has long been a major concern for statisticians. In classical hypothesis testing the interest is often to test the null hypothesis that the population means are all equal ( Lehmann (1986)). It is well known that, for a large enough sample size, the classical test will almost always reject the null hypothesis as pointed out by many researchers (Berger (1985)). In many real world problems, the practical interest is frequently to examine whether the population means fall into an indifference zone, not just the point of equality of means. This idea leads to the consideration of equivalence hypothesis stated as H0 : 1 ki=1 k µ i µ δ vs the inequivalence alternative Ha : 1 ki=1 k µ i µ δ > δ, where µ is the grand average of the means µ 1,..., µ k ; δ ( 0) is a predetermined indifference zone and δ ( 0) is a detective amount specified in advance. The constant δ can be interpreted as the average deviation about which we are indifferent and the null hypothesis H 0 can be interpreted as saying that there is little or not much difference among means within a small δ-value or there is practically equivalent among the means. The quantity δ stated in the alternative is regarded as a difference of interest. This type of null hypothesis appears to be more useful and meaningful in the analysis of means among several treatment populations under fixed-effect analysis of variance models. When there are only two populations, the equivalence hypothesis is also referred to as interval hypothesis or bioequivalence in pharmaceutical and medical studies where the null and alternative are reversed (how and Liu (1992)). For the case of testing three or more means, the studentized range test for the equivalence hypothesis H 0 was studied by hen and Lam (1991), hen, Xiong and Lam (1993) for the common known variances, and hen and hen (1999) for unknown variances. None of them considered the level and the power simultaneously in a multiple-equivalence testing problem when variances are common but unknown. In this paper we study the case where the level and the power of a test are specified simultaneously to determine the critical value of the test and the necessary sample size for testing a null hypothesis against a given alternative. In Section 2, we develop a usual one-sample procedure to determine the level and the power for the test. In Section 3, a two- 2

3 stage sampling procedure is employed to handle the unknown variance problem. Statistical tables of critical values and sample sizes required are provided. Finally, a summary is given in Section The Statistical Procedure Let X ij (j = 1,..., n) be a random sample of size n ( 2) drawn from the ith normal distribution π i (i = 1,..., k) having a unknown mean µ i and a common variance σ 2. The goal is to test the null hypothesis of equivalence H 0 : 1 k k i=1 µ i µ σ against the alternative hypothesis of inequivalence δ (1) H a : 1 k k i=1 µ i µ σ δ, where δ is referred to as an amount of equivalence while δ as a detective difference with δ > δ, and µ is the average of µ 1,..., µ k. The null hypothesis claims that the k means are practically equal or equivalent while the alternative specifies some important difference of interest. The equivalence constant δ must be determined or chosen in advance by an expert in his/her field of expertise. This type of hypothesis appears to be more meaningful and realistic than the traditional ones of equal means in dealing with testing multiple populations problem. For example, when k = 2, the equivalence measure of a test product within a 20% of a reference product can be expressed as 0.80 < µ T /µ R < 1.20, where µ T is the mean of a test product and µ R is the mean of a reference product. This is a commonly used bioequivalence measure acceptable to FDA in USA. If, furthermore, the standard deviation σ is required to be not more than 25% of the reference mean (or equivalently, the coefficient of variation is at most 25%), then the equivalence µ T µ R 0.2µ R, can be translated into the hypothesis in (1) as H 0 : 1 2i=1 2 µ i µ /σ 0.4, or δ = 0.4. Assuming σ = 1 or known, and µ = 0, hen, Xiong and Lam (1993) proposed a simple range test associated with its critical region and showed that the probability of critical 3

4 region under H 0 attains its supremum at the least favorable configuration, (µ 1,..., µ k ) = ( kδ/2, 0,..., 0, kδ/2). When σ 2 is unknown, however, hen and Lam (1991) proposed a studentized range test for testing H 0 where the critical values were obtained at an arbitrarily given sample size; no power was reported in their work. In this paper we attempt to control both the level and the power of the test for testing a null hypothesis of equivalence against an alternative and simultaneously to determine the critical value and sample size for the equivalence testing problem. Denote the sample means and pooled sample variance, respectively, by X i = and n X ij /n, i = 1,..., k, j=1 k n S 2 = (X ij X i ) 2 /(k(n 1)). i=1 j=1 Let X [1] X [k] be the ordered means, X1,..., Xk and let the studentized range be defined as R = n( X [k] X [1] )/S. (2) The null hypothesis H 0 is rejected if R > γ where the critical value γ is determined by the equation of the maximum level being equal to a preassigned value α as given by sup P (R > γ H 0 ) = α, (3) Ω where α (0, 1), and Ω is the set of all possible means µ i and variance σ 2. Furthermore, the sample size can be determined by the following equation of the minimum power associated with the test (2) inf Ω P (R > γ H a) = P, (4) where P (0, 1). It should be noted that when k = 2 the range test is equivalent to a likelihood ratio test and the null hypothesis (1) becomes a standardized mean difference less than 2δ. 4

5 It is necessary to find a least favorable configuration (LF) of the means which maximizes the level of the test {R > γ} under H 0 such that equation (3) is satisfied and a LF which minimizes the power of the test under a specified H a such that equation (4) is satisfied. Both the level and the power are not only independent of all means differences but also free of the unknown variances. The LF for the maximum level is determined by the following Theorem without proof. (See Theorem 1 of hen, Xiong and Lam (1993).) Theorem 1: Let φ(x i θ i ) be the standard normal density of the independent normal r.v. X i with mean θ i, i = 1,..., k and let g c (θ) = P θ (R c) be the tail probability where R is the range of X 1,..., X k and θ = (θ 1,..., θ k ). Assume that the g c (θ) attains its minimum at some θ for all θ A = {θ : θ i θ δ}, where θ = (θ θ k )/k. Then θ is the LF and θ must be in B 1 where B 1 = {( θ δ/2, θ + δ/2), ( θ + δ/2, θ δ/2)} for k = 2, and B 1 = {(θ 1,..., θ k ) : one of the θ i s is θ + δ/2, one of the θ i s is θ δ/2, and the other θ i s are θ} for k 3. Let X (j) be associated with µ [j] where µ [1] µ [k] are the ordered values of the µ s. Then we can obtain P (R > γ) = P ( X [k] X [1] > γs/ n) = 1 P ( X [k] X [1] + γs/ n) k = 1 P ( X [k] X [1] + γs/ n, X (j) = X [1] ) j=1 k = 1 P [ X (j) X (i) X (j) + γs/ n, i = 1,..., k, i j] j=1 k = 1 P [Z j + δ ji Z i Z j + γu + δ ji, i = 1,..., k, i j] j=1 k k = 1 [Φ(y + δ ji + γu) Φ(y + δ ji )]φ(y)q m (u)dydu, (5) j=1 0 i=1 i j where δ ji = n(µ [j] µ [i] )/σ; Z i = n( X (i) µ [i] )/σ, i = 1,..., k are i.i.d. N(0, 1) r.v. s with p.d.f. φ(y) and c.d.f. Φ(y) and U = S/σ is distributed as χ m / m with m = k(n 1) d.f. and q m ( ) is the p.d.f. of U. Let X i = n X i /σ and θ i = nµ i /σ. onditioning on 5

6 S = s the inner integral in (5) is log concave in θ = (θ 1,, θ k ) over the convex set {R γ} (Prekopa (1973)). Then, by Theorem 1 the LF of θ which minimizes the inner integral in (5) is θ 0 = ( θ kδ n/2, θ,..., θ, θ +kδ n/2) for k 3 and is θ 0 = ( θ kδ n/2, θ +kδ n/2) for k = 2. Since θ 0 is independent of the value u assumed by U, θ 0 is the LF for P (R γ) being minimum under H 0 for all values of σ 2. Or equivalently, θ 0 is the LF which maximizes the tail probability P (R > γ). Let the maximum of P (R > γ) over µ H 0 be P δ (γ, n). Then { P δ (γ, n) = 1 0 [Φ(y kδ n 2 + γu) Φ(y kδ n )] k 2 2 [Φ(y kδ n + γu) Φ(y kδ n)]φ(y)q m (u)dydu +(k 2) 0 [Φ(y kδ n 2 [Φ(y + kδ n 2 + γu) Φ(y + kδ n )][Φ(y + γu) Φ(y)] k γu) Φ(y kδ n )]φ(y)q m (u)dydu 2 + [Φ(y + kδ n + γu) Φ(y + kδ n )] k [Φ(y + kδ n + γu) Φ(y + kδ } n)]φ(y)q m (u)dydu. (6) We note that in the special case where δ = 0 the P δ (γ, n) reduces to P 0 (γ, n) = 1 k 0 [Φ(y + γu) Φ(y)] k 1 φ(y)q m (u)dydu = P ( X [k] X [1] > γs/ n H 0 : µ 1 = = µ k ) (7) which is the tail probability of the studentized range statistic with m d.f. arising from k i.i.d. N(0, σ 2 ) r.v. s. Note that P δ (γ, n) in (6) is the tail probability of the studentized range statistic with m d.f. arising from independent sample means of size n from k normal populations all having variance σ 2, and k-2 of the normal populations have zero mean, one has kδσ/2 and the other has mean kδσ/2. It is also interesting to note that the p.d.f. of R is equal to (-1) times the derivative of P δ (γ, n) with respect to γ. The power of the test under H a can be obtained by applying the following lemma. (See hen, Xiong and Lam (1993, p.23).) Lemma 1. Let φ(x i θ i ), g c (θ), R and θ be defined as of Theorem 1. Let A = {θ : ki=1 θ i θ δ } where θ = (θ θ k )/k. Then the minimum of P (R > γ) occurs at 6

7 some θ for all θ A, and θ is the LF, where θ = ( θ δ /(2l),, θ δ /(2l), θ + δ /(2(k l)),, θ + δ /(2(k l))) for some l = 1, 2,..., k 1, with l such ( θ δ /(2l)) s and k l such ( θ + δ /(2(k l))) s. Let X i = n X i /σ and θ i = nµ i /σ. By Lemma 1, the minimum power in (5) under H a : 1 k ki=1 µ i µ /σ δ > δ is attained at the LF θ 1 = ( θ kδ n/(2l),, θ kδ n/(2l), θ + kδ n/(2(k l)),, θ + kδ n/(2(k l))) with l such ( θ kδ n/(2l)) s and k l such ( θ + kδ n/(2(k l))) s for some l = 1, 2,..., k 1. Let β δ (γ, n) denote the minimum power in (5) over H a at the LF θ 1. Then, the minimum power can be calculated by { β δ (γ, n) = 1 l 0 [Φ(y + γu) Φ(y)] l 1 [Φ(y bδ n + γu) Φ(y bδ n)] k l φ(y)q m (u)dydu +(k l) [Φ(y + bδ n + γu) Φ(y + bδ n)] l 0 } [Φ(y + γu) Φ(y)] k l 1 φ(y)q m (u)dydu, (8) where b = k 2 /(2l(k l)) for some l = 1, 2,..., k 1. Gaussian quadrature is used to calculate the double integrals in (6) and (8) for the level and power evaluation. Because the double integrals in (6) and (8) involve infinite integration limits, it was necessary to truncate the limits such that the truncation error in each double integral over the tails is controllable by a negligible but small number. It is for chi/ m distribution and for normal distribution. So, the maximum total truncation error for the double integrals in (6) or (8) is less than 4k Furthermore, due to the wide range of the integration limits, (a, b) for normal and (c, d) for chi/ m, it was needed, for the purpose of accuracy, to partition the normal integral over (a, b) into six subintervals (-12, -2.33), (-2.33, -.66), (-.66, 0), (0,.66), (.66, 2.33), and (2.33, 9) so that the area in each subinterval (except for the first and the last ones) is approximately.245, and to split the chi/ m integral into two subintervals (c, c 1 ), (c 1, d) where the integral limits c, c 1, and d were chosen at the 10 9 th percentile, median and ( )th percentile depending on the degrees of freedom m. These integration limits used are given in Appendix A. Then a 32-by 32-point 7

8 Gauss-Legendre quadrature over each of the twelve subrectangles of each double integral in (6) or (8) was performed and the results were summed up. A 64-by 64-point quadrature was used to check the solution and they agree to the tabled values. By numerical calculation we confirm that the minimum power of (8) occurs at l = k/2 when k is even, and at l = (k 1)/2 or l = (k + 1)/2 when k is odd. Given the level α and equivalence constant δ under H 0, for specified k and sample size n, one finds the critical value γ α by Newton-Raphson s iteration by solving the integral equation P δ (γ α, n) α = 0, (9) so that the stopping rule for (9) is less than The critical values γ α are given in Tables 1-4 for k = 2, 3, 4, 5, and various sample sizes n = 2(1)10(2)30(5)40(20)120, 200, 300, 400 at α = ritical values for other combinations of k, α and n can be obtained from TES- TRANGE7.FOR available from the authors, or from TESTRANGE6.FOR given in Appendix B. Further, if the power is also required, then substitute the obtained γ α into the power function (8) under H a at a given δ and sample size n to see if the calculated power exceeds a pre-specified P, i.e., to check if the probability inequality β δ (γ α, n) P (10) is satisfied. Increase or decrease n until the critical value γ α in (9) and the smallest n in (10) are satisfied simultaneously. Tables of critical values and sample sizes are given in Tables 5-6 for k = 2(1)5, P = 0.80, 0.90, 0.95 and various δ = 0(0.1)0.6 and selected values of δ. For example, at α = 0.05, P = 0.90, k = 2, δ = 0.4 and δ = 1.0, the required sample size is n = 16 and the critical value of the test is γ = 5.88 from Table 5. The sample sizes and critical values for other combinations of k, α, P, δ and δ can be evaluated by a FORTRAN program named TESTRANGE6.FOR given in Appendix B. 8

9 Table 1. ritical Values of Studentized Range Test R for k=2 at α = δ n

10 Table 2. ritical Values of Studentized Range Test R for k=3 at α = δ n

11 Table 3. ritical Values of Studentized Range Test R for k=4 at α = δ n

12 Table 4. ritical Values of Studentized Range Test R for k=5 at α = δ n

13 Table 5. Sample Size (left) and ritical Values (right) of Studentized Range Test for k=2 and 3 at a Given Power P and α = k=2 k=3 δ δ P = P =

14 Table 5. (continued) α = 0.05 k=2 k=3 δ δ P = P =

15 Table 6. Sample Size (left) and ritical Values (right) of Studentized Range Test for k=4 and 5 at a Given Power P and α = k=4 k=5 δ δ P = P =

16 Table 6. (continued) α = 0.05 k=4 k=5 δ δ P = P =

17 3. Two-Stage Procedure When the equivalence hypothesis is measured only as the average deviation of means without standardization, we state null hypothesis as against the alternative H0 : 1 k µ i µ δ (11) k i=1 Ha : 1 k µ i µ δ, (12) k i=1 where δ and δ (δ > δ > 0) are specified in advance. In situations where the common population variance σ 2 is unknown, there does not exist a one-sample testing procedure for handling the hypothesis H 0 in (11) vs H a in (12) such that both the level and the power are independent of the unknown parameter σ 2 as argued by Stein (1945). In this section we employ a two-stage sampling procedure similar to that of Bechhofer, Dunnett and Sobel (1954) to perform the equivalence test in (11). The sampling procedure is stated as follows. 1. Take from population π i (i = 1,..., k) a first random sample of n 0 ( 2) observations X ij (j = 1,..., n 0 ), calculate the unbiased sample mean and pooled sample variance, respectively, by and n 0 X i (n 0 ) = X ij /n 0, i = 1,, k j=1 k S0 2 n 0 = (X ij X i (n 0 )) 2 /k(n 0 1). i=1 j=1 Note that an initial sample of size n 0 ( 2) will work in theory, but Bishop and Dudewicz (1978) suggested that n 0 be 10 or more giving better results. For economical reason it is suggested that one take n 0 to be 10 or more but less than Define the total sample size from each population N = max {n 0 + 1, [ S2 0 z ]}, 17

18 where [y] denotes the smallest integer greater than or equal to y; here z is a positive design constant (depending on the level of significance, power, k, and n 0 ), which will be determined later. Then, take N n 0 additional observations from the i th population so that we have a total of N observations denoted by X i1,..., X in0,..., X in. For each i, set the coefficients a 1,..., a n0,..., a N, such that a 1 =... = a n0 = 1 (N n 0)b n 0 = a a n0 +1 =... = a N = 1 N [ 1 + n 0 (Nz S0 2) ] (N n 0 )S0 2 = b and computed the weighted sample mean as n 0 X i = a X ij + b N X ij. (13) j=1 j=n 0 +1 It should be denoted that these coefficients a j s are so determined to satisfy the following conditions, N j=1 a j = 1, a 1 =... = a n0, and S 2 0 Nj=1 a 2 j = z. It is well known that the r.v. s T i = ( X i µ i )/ z, i = 1,..., k, have a k-variate t-distribution with m 0 = k(n 0 1) d.f. and zero correlation coefficient, free of the unknown variances σ Denote the ranked values of Xi by X [1] < X [2] < < X [k], and formulated the test statistic R = X [k] X [1] z. (14) 4. Find the critical region {R > γ α } of level at most α and the design constant z so that the level of the test (14) under H 0 in (11) is maximized and the power of the test (14) under a specified alternative H a in (12) is minimized as stated below: P δ (γ α, w) = sup P (R > γ α H0 ) (15) Ω and P δ (γ α, w) = inf Ω P (R > γ α H a), (16) 18

19 where w = δ / z and the supremum in (15) (infimum in (16)) is taken over the set Ω of all possible configurations of the means µ i and σ 2. The critical value γ α and the design constant z can be obtained by solving the simultaneous equations P δ (γ α, w) = α (17) and P δ (γ α, w) = P, (18) where α (0, 1) is usually taken to be a small value, say 0.05, P taken to be a large value, say We now find the LF of the means which maximizes the level of the test in (11) under H0 and the LF of the means which minimizes the power of the test in (12) under H a such that the level (15) and the power (16) are not only independent of all mean differences but also free of the common unknown variance. Let µ [1]... µ [k] be the ordered values of µ 1,..., µ k and let X (j) be associated with µ [j]. We have P (R > γ α ) = P ( X [k] X [1] > γ α z) = 1 P ( X [k] X [1] + γ α z) = k 1 P ( X [k] X [1] + γ α z, X(j) = X [1] ) j=1 = k 1 P ( X (j) X (i) X (j) + γ α z, i = 1,..., k, i j) j=1 = k X (j) µ [j] + µ [j] µ [i] X (i) µ [i] X (j) µ [j] + µ [j] µ [i] 1 P [ j=1 z z z +γ α, i = 1,..., k, i j] = k 1 P [T j + δ ji T i T j + δ ji + γ α, i = 1,..., k, i j], (19) j=1 where δ ji = (µ [j] µ [i] )/ z, i, j = 1,..., k, i j, and T i = ( X (i) µ [i] )/ z, i = 1,..., k. Given S 0, the vector (T i, i = 1,, k) has a conditional multivariate normal distribution 19

20 with a zero mean vector, a common variance σ 2 /S0 2 and a common zero correlation coefficient. Following Stein s (1945) argument, it has an unconditional multivariate t distribution with m 0 = k(n 0 1) d.f., a common zero correlation coefficient and components with T i = Y i /U, i = 1,..., k, where Y 1,, Y k are i.i.d. N(0, 1) r.v. s and U = S 0 /σ = χ/ m 0 is a χ/ m 0 r.v. with m 0 d.f., stochastically independent of Y i s. Thus, the expression (19) can be written as P (R > γ α ) = 1 k j=1 P [ Y j U + δ ji < Y i U < Y j U + δ ji + γ α, i = 1,..., k, i j], (20) As the m 0 becomes large, T i converges to the standard normal, or equivalently X i / z converges independent to N(θ i, 1) r.v. s, where θ i = µ i / z. Thus, by Theorem 1, we can obtain the asymptotic LF of means or δ ji which maximizes the integral in (20) at θ 0 = ( θ kδ/(2 z), θ,..., θ, θ+kδ/(2 z)) for k 3 and at θ 0 = ( θ kδ/(2 z), θ+kδ/(2 z)) for k = 2, where θ = µ/ z. For computational reason, we rewrite θ 0 as θ 0 = (( θ kvw/2, θ,..., θ, θ + kvw/2), where v = δ/δ and w = δ / z. Let the maximum of (20) subject to µ H 0 at the LF θ 0 be P δ (γ α, w). Then { P δ (γ α, w) = 1 0 [Φ(y kvwu/2 + γ α u) Φ(y kvwu/2)] k 2 [Φ(y kvwu + γ α u) Φ(y kvwu)]φ(y)q m0 (u)dydu +(k 2) 0 [Φ(y + γ α u) Φ(y)] k 3 [Φ(y + kvwu/2 + γ α u) Φ(y + kvwu/2)] [Φ(y kvwu/2 + γ α u) Φ(y kvwu/2)]φ(y)q m0 (u)dydu + [Φ(y + kvwu/2 + γ α u) Φ(y + kvwu/2)] k 2 0 } [Φ(y + kvwu + γ α u) Φ(y + kvwu)]φ(y)q m0 (u)dydu. (21) Note that in the special case where δ = 0, the P δ (γ α, w) reduces to P 0 (γ α, w) = 1 k 0 [Φ(y + γ α u) Φ(y)] k 1 φ(y)q m0 (u)dydu = P (( X [k] X [1] )/ z > γ α H 0 : µ 1 = = µ k ), which is the range of i.i.d. Student s t r.v. s. 20

21 Similarly, by Lemma 1, as m 0 becomes large, the minimum power in (20) under H a : 1 k ki=1 µ i µ δ > δ is attained at the asymptotic LF θ 1 = ( θ kw/(2l),..., θ kw/(2l), θ + kw/(2(k l)),..., θ + kw/(2(k l))) with the l such ( θ kw/2l) s and k l such ( θ + kw/(2(k l))) s, l = 1,..., k 1. Let P δ (γ α, w) denote the minimum power in (20) at the LF θ 1. Then { P δ (γ α, w) = 1 l 0 +(k l) [Φ(y + γ α u) Φ(y)] l 1 [Φ(y bwu + γ α u) Φ(y bwu)] k l φ(y)q m0 (u)dydu 0 [Φ(y + bwu + γ α u) Φ(y + bwu)] l } [Φ(y + γ α u) Φ(y)] k l 1 φ(y)q m0 (u)dydu, (22) where b = k 2 /(2l(k l)). By numerical calculation we find that the minimum power of (22) occurs at l = k/2 when k is even, and at l = (k 1)/2 or l = (k +1)/2 when k is odd. Given the level α at H 0 and the power P at Ha, for specified δ, δ, k, and n 0, the critical value γ α and the w (= δ / z) value are obtained by grid search such that (17) and (18) hold by numerical quadrature similar to that discussed in Section 2. These critical values γ α and the power-related values of w = δ / z can be found in Tables 7 to 8 for level at 5% and the powers at 80%, 90% and 95% when k = 2(1)5, n 0 = 5, 10, 15, 25 and various δ/δ ratio 0.5. The 80% power requirement is a commonly accepted criterion by health science and U.S. FDA. For example, if α = 5% with required power being 80% and k = 2, n 0 = 15, δ/δ = 0.1, then the γ α =3.01 and w = δ / z=2.11 are found from Table 7. If the δ value in the null hypothesis H 0 : 1 ki=1 k µ i µ δ is set to be 0.1 unit and the δ value in the alternative hypothesis H a : 1 ki=1 k µ i µ δ is set to be 1 units, then solve δ / z=2.11 for z to obtain the z value of The additional critical values γ α and power-related w values for other combinations of α, P, k, n 0 and δ/δ ratio can be calculated from a fortran program named NE- TRANGE2.FOR available from the authors. 21

22 Table 7. Percentage Point of γ α (upper entry) of a Range Test R and its Power-Related Design onstant w (lower entry) for k=2, 3 and α = 0.05 k=2 n P P P P δ/δ k=3 n P P P P δ/δ

23 Table 8. Percentage Point of γ α (upper entry) of a Range Test R and its Power-Related Design onstant w (lower entry) for k=4, 5 and α = 0.05 k=4 n P P P P δ/δ k=5 n P P P P δ/δ

24 4. Summary and onclusion Testing the null hypothesis of equal treatment means is sometimes impractical in real applications, as pointed out by Berger (1985). An alternative measure to detect the difference among means is the range of the means, which extends the idea of equivalence among means. The test of equivalence receives more attention in health science, pharmaceutical industry, and other applied areas. When the common variance σ 2 is unknown, a studentized range test using a traditional one-sample sampling procedure is proposed for testing the hypothesis that the standardized average deviation of the normal means is falling into a practical indifference zone. Both the level and the power of the proposed test are controllable and they are completely independent of the unknown variance. Statistical tables to implement the procedure are provided for practitioners. When the equivalence hypothesis is expressed purely as the average deviation of means (11) when the common variance is unknown, a two-stage sampling procedure is used and a modified range test (14) is proposed for testing the hypothesis (11). Both the level and power by the two-stage procedure are free of the unknown variance so that the critical value and sample size can be determined simultaneously. Thus, the two-stage procedure gives the equivalence hypothesis (11) a possible solution in practice. 24

25 REFERENES Bechhofer, R. E., Dunnett,. W. and Sobel, M. (1954). A Two-Sample Multiple Decision Procedure for Ranking Means of Normal Populations with a ommon Unknown Variance. Biometrika, 41, Berger, J. O. (1985). Statistical Decision Theory, 2nd edition, Springer-Verlag, N.Y. Bishop, T. A. and Dudewicz, E. J. (1978). Exact Analysis of Variance with Unequal Variances : Test Procedures and Tables. Technometrics, 20, hen, S. Y. and hen, H. J. (1999). A Range Test for the Equivalency of Means under Unequal Variances. Technometrics, Vol. 41, No. 3, hen, H. J. and Lam, K. (1991). Percentage Points of a Studentized Range Statistic Arising from Non-identical Normal Random Variables. ommunications in Statistics : Simulation and omputation, 20(4), hen, H. J., Xiong, M. and Lam, K. (1993). Range Tests for the Dispersion of Several Location Parameters. Journal of Statistical Planning and Inference, 36, how, S.. and Liu, J. P. (1992). Design and Analysis of Bioavailability and Bioequivalence Studies, New York: Marcel Dekker. Hayter, A. J. and Liu, W. (1990). The Power Function of the Studentized Range Test. The Annals of Statistics, 18, Lehmann, E. L. (1986). Testing Statistical Hypothesis, 2nd edition, Wiley, N. Y. Prekopa, A. (1973). On Logarithmic oncave Measures and Functions. Acta Scientiarum Mathematicarum (Szeged). 36, Stein,. (1945). A Two-Sample Test for a Linear Hypothesis whose Power is Independent of Variance. Annals of Mathematical Statistics, 16, Wen, M. J. and hen, H. J. (2004). A Studentized Range Test for the Equivalency of Normal Means under Heteroscedasticity. Technical Report No. 64, October 2004, Department of 25

26 Statistics, National heng Kung University, Tainan, Taiwan. 26

27 APPENDIX A Table of Integral Limits Used in chi/ m Distribution m c c 1 d m c c 1 d

28 APPENDIX B Program For Evaluating The ritical Value and the Sample Size "The Level and Power of a Studentized Range Test for Testing the Equivalence of Means" (Average Deviation ase) (NEWTON ITERATION TO FIND RITIAL VALUE (9). THEN ALULATE THE POWER (10) FOR GIVEN K,N,ALPHA,DELTA,DELTAS,GAMMA) (6 INTERVALS FOR INNER NORMAL AND 2 INTERVALS FOR HI/RT(DF)) ********************************************************** * * * MAIN PROGRAM : TESTRANGE6.FOR (11/01/2004 EDITION). * * AUTHORS : M J WEN AND H J HEN, Revised 03/03/2005 * * THIS PROGRAM FINDS THE LEVEL (6) AND POWER (8) OF A * * RANGE TEST UNDER A LEAST FAVORABLE ONFIGURATION * * ONERNING TESTING THE AVERAGE DEVIATION OF SEVERAL * * NORMAL MEANS WITH A OMMON UNKNOWN VARIANE. * * THIS PROGRAM USES A 32 BY 32-POINTS GAUSSIAN * * QUADRATURE IN EAH OF THE SIX BY TWO SUBRETANGLES * * IN THE DOUBLE INTEGRAL. * * THE TRUNATION ERROR IS < 4k x 1.D-9. * * K = THE NUMBER OF POPULATIONS (INPUT). * * N = THE OMMON SAMPLE SIZE (INPUT) * * DX = THE GAUSSIAN POINTS (INPUT). * * DW = THE GAUSSIAN WEIGHTS (INPUT). * * DBN = THE BOUNDARY OF NORMAL DISTRIBUTION * * FOR EAH SUBINTERVAL. ALSO DIN (INPUT). * * DB = THE BOUNDARY OF HI/ROOT(DF) DISTRIBUTION * * FOR EAH SUBINTERVAL (INPUT). * * GAMMA = RITIAL VALUE (OUTPUT) * * POW = POWER OF THE TEST (OUTPUT) * * SUBPROGRAMS REQUIRED : * * (1) STANDARD NORMAL DISTRIBUTION FUNTION (DNORMX) * * (2) STANDARD NORMAL DENSITY FUNTION (DPDF), (3) * * UNDERFLOW ONTROL FUNTION (DAK), (4) INTEGRAL * * SUBROUTINE FUNTIONS (LEVEL, POWER, PL1 AND PL2), * * (5) LOG GAMMA FUNTION (DLGGM), AND (6) DENSITY * * FUNTION OF HI/ROOT(V) (DPGF). * * * ********************************************************** MAIN PROGRAM -- THE OMMON VARIANE IS UNKNOWN. IMPLIIT REAL*8 (A-H,O-Z) 28

29 DIMENSION DX(32),DW(32),DDX(16),DDW(16),DB(3), + DB2(3),DB3(3),DB4(3),DB5(3),DB6(3),DB8(3), + DB10(3),DB12(3),DB14(3),DB16(3),DB18(3),DB20(3), + DB25(3),DB30(3),DB35(3),DB40(3),DB50(3),DB60(3), + DB70(3),DB80(3),DB90(3),DB100(3),DB120(3), + DB140(3),DB160(3),DB180(3),DB200(3),DB250(3), + DB300(3),DB350(3),DB400(3),DB450(3),DB500(3), + DB750(3),DB1K(3),DB2K(3),DB4K(3),DB6K(3),DB8K(3), + DIN(8),DBN(7) INPUT GAUSSIAN POINTS (32-POINT FORMULA). DATA DDX/ D-01, D+00, D+00, D+00, D+00, D+00, D+00, D+00, D+00, D+00, D+00, D+00, D+00, D+00, D+00, D+00/ INPUT GAUSSIAN WEIGHTS (32-POINT FORMULA). DATA DDW/ D-01, D-01, D-01, D-01, D-01, D-01, D-01, D-01, D-01, D-01, D-01, D-01, D-01, D-01, D-01, D-02/ SIX BY TWO RETANGLES USED IN DOUBLE INTEGRALS. BOUNDARY PTS. FOR NORMAL DISTRIBUTION. DATA DBN/-66.D0,-7.D0,-1.D0,0.D0,1.D0,7.D0,60.D0/ BOUNDARY PTS.(P=1.D-9,MEDIAN,1-1.D-9) FOR HI/RT(DF) DIST. DATA DB2/ D0,0.833D0,4.553D0/ DATA DB3/ D0,0.888D0,3.867D0/ DATA DB4/ D0,0.916D0,3.460D0/ DATA DB5/0.0127D0,0.933D0,3.185D0/ DATA DB6/0.0246D0,0.944D0,2.982D0/ DATA DB8/0.0558D0,0.958D0,2.670D0/ DATA DB10/0.0912D0,0.967D0,2.509D0/ DATA DB12/0.1264D0,0.972D0,2.370D0/ DATA DB14/0.1603D0,0.976D0,2.262D0/ DATA DB16/0.1909D0,0.979D0,2.175D0/ DATA DB18/0.2194D0,0.981D0,2.104D0/ DATA DB20/0.2454D0,0.983D0,2.044D0/ DATA DB25/0.3014D0,0.987D0,1.927D0/ 29

30 DATA DB30/0.3469D0,0.989D0,1.842D0/ DATA DB35/0.3848D0,0.991D0,1.776D0/ DATA DB40/0.4168D0,0.992D0,1.723D0/ DATA DB50/0.4682D0,0.993D0,1.643D0/ DATA DB60/0.5082D0,0.994D0,1.584D0/ DATA DB70/0.5401D0,0.995D0,1.539D0/ DATA DB80/0.5665D0,0.996D0,1.502D0/ DATA DB90/0.5888D0,0.996D0,1.473D0/ DATA DB100/0.6070D0,0.997D0,1.447D0/ DATA DB120/0.6390D0,0.997D0,1.407D0/ DATA DB140/0.6638D0,0.998D0,1.376D0/ DATA DB160/0.6841D0,0.998D0,1.350D0/ DATA DB180/0.7011D0,0.998D0,1.330D0/ DATA DB200/0.7156D0,0.998D0,1.312D0/ DATA DB250/0.7442D0,0.999D0,1.312D0/ DATA DB300/0.7656D0,0.999D0,1.312D0/ DATA DB350/0.7824D0,0.999D0,1.312D0/ DATA DB400/0.7957D0,0.999D0,1.312D0/ DATA DB450/0.8024D0,0.999D0,1.312D0/ DATA DB500/0.8167D0,0.999D0,1.312D0/ DATA DB750/0.8524D0,1.000D0,1.312D0/ DATA DB1K/0.8695D0,1.000D0,1.312D0/ DATA DB2K/0.9075D0,1.000D0,1.312D0/ DATA DB4K/0.9354D0,1.000D0,1.312D0/ DATA DB6K/0.9465D0,1.000D0,1.312D0/ DATA DB8K/0.9470D0,1.000D0,1.312D0/ DATA DIN/-70.D0,-7.D0,-2.D0,-.8D0,.8D0,2.D0,7.D0,40.D0/ OPEN (7,FILE= TESTRANGE6.DD,STATUS= NEW ) DO 10 I=1,16 DX(I)=DDX(I) DW(I)=DDW(I) J=32-I+1 DX(J)=-DDX(I) DW(J)=DDW(I) 10 ONTINUE 20 ONTINUE POWTEST=.7D0 WRITE (*,25) 25 FORMAT(/, RITIAL VALUES OF STUDENTIZED RANGE BEGIN:,/, + FOR TESTRANGE6.FOR ) WRITE (*,28) 28 FORMAT (/, ENTER NUMBER OF POPULATIONS, K. STOP IF -99. ) READ (*,*) K IF (K.LT. 2) STOP 30

31 WRITE (*,29) K 29 FORMAT (1X, K =,I6) 30 WRITE (*,31) 31 FORMAT (/, ENTER P* IN (0,1) ) READ(*,*) PSTAR IF (PSTAR.LT. 0.D0.OR. PSTAR.GT. 0.99D0) GOTO 30 WRITE (*,32) PSTAR 32 FORMAT ( PSTAR=,F5.3) 40 WRITE (*,45) 45 FORMAT (/, ENTER DELTA, DELTA_*; TRY AGAIN IF < 0 ) READ (*,*) DELTA,DELTAS WRITE (*,48) DELTA,DELTAS 48 FORMAT (1X, DELTA =,F7.3, DELTA_* =,F7.3) IF (DELTA.LT. 0.D0.OR. DELTA.GT. DELTAS) GOTO WRITE (*,52) 52 FORMAT (/, ENTER ALPHA. TRY AGAIN IF NOT IN (0,1). ) READ (*,*) ALPHA WRITE (*,53) ALPHA 53 FORMAT (1X, ALPHA =,F6.3) IF (ALPHA.LE. 0.D0.OR. ALPHA.GE. 1.D0) GOTO 50 ALPHA=0.05D0 54 WRITE (*,55) 55 FORMAT (/, ENTER OMMON SAMPLE SIZE, N.,/, + IF VARIANE IS KNOWN, ENTER 999 FOR N.,/) READ (*,*) N WRITE (*,56) N 56 FORMAT ( N =,I6) IF (N.LE. 1) GOTO WRITE (*,60) 60 FORMAT(/, ENTER INITIAL GUESS OF GAMMA, OR TRY 3.,/, + TRY AGAIN IF IT IS < 0. ) 65 READ (*,*) DGAMMA WRITE (*,66) DGAMMA 66 FORMAT ( GAMMA=,F8.3) IF (DGAMMA.LT. 0.D0) GOTO 58 DN=N DK=K search begins 70 DF=DK*(DN-1.D0) IF (DN.GT. 2.D3.OR. DF.GE. 1.D4) GOTO 2050 IF (DF.GE. 2.D0.AND. DF.LT. 3.D0) THEN DO 75 I=1,3 75 DB(I)=DB2(I) ELSE IF (DF.GE. 3.D0.AND. DF.LT. 4.D0) THEN 31

32 DO 80 I=1,3 80 DB(I)=DB3(I) ELSE IF (DF.GE. 4.D0.AND. DF.LT. 5.D0) THEN DO 85 I=1,3 85 DB(I)=DB4(I) ELSE IF (DF.GE. 5.D0.AND. DF.LT. 6.D0) THEN DO 90 I=1,3 90 DB(I)=DB5(I) ELSE IF (DF.GE. 6.D0.AND. DF.LT. 8.D0) THEN DO 95 I=1,3 95 DB(I)=DB6(I) ELSE IF (DF.GE. 8.D0.AND. DF.LT. 10.D0) THEN DO 97 I=1,3 97 DB(I)=DB8(I) ELSE IF (DF.GE. 10.D0.AND. DF.LT. 12.D0) THEN DO 100 I=1,3 100 DB(I)=DB10(I) ELSE IF (DF.GE. 12.D0.AND. DF.LT. 14.D0) THEN DO 102 I=1,3 102 DB(I)=DB12(I) ELSE IF (DF.GE. 14.D0.AND. DF.LT. 16.D0) THEN DO 104 I=1,3 104 DB(I)=DB14(I) ELSE IF (DF.GE. 16.D0.AND. DF.LT. 18.D0) THEN DO 106 I=1,3 106 DB(I)=DB16(I) ELSE IF (DF.GE. 18.D0.AND. DF.LT. 20.D0) THEN DO 108 I=1,3 108 DB(I)=DB18(I) ELSE IF (DF.GE. 20.D0.AND. DF.LT. 25.D0) THEN DO 115 I=1,3 115 DB(I)=DB20(I) ELSE IF (DF.GE. 25.D0.AND. DF.LT. 30.D0) THEN DO 120 I=1,3 120 DB(I)=DB25(I) ELSE IF (DF.GE. 30.D0.AND. DF.LT. 35.D0) THEN DO 125 I=1,3 125 DB(I)=DB30(I) ELSE IF (DF.GE. 35.D0.AND. DF.LT. 40.D0) THEN DO 130 I=1,3 130 DB(I)=DB35(I) ELSE IF (DF.GE. 40.D0.AND. DF.LT. 50.D0) THEN DO 140 I=1,3 140 DB(I)=DB40(I) 32