A Studentized Range Test for the Equivalency of Normal Means under Heteroscedasticity

Transcription

1 A Studentized Range Test for the Equivalency of Normal Means under Heteroscedasticity Miin-Jye Wen and Hubert J. Chen Department of Statistics National Cheng-Kung University Tainan, Taiwan ABSTRACT A studentized range test using a two-stage and a one-stage sampling procedures, respectively, is proposed for testing the hypothesis that the average deviation of the normal means is falling into a practical indifference zone. Both the level and the power of the proposed test associated with the hypotheses are controllable and they are completely independent of the unknown variances. The two-stage procedure is a design-oriented procedure that satisfies certain probability requirements and simultaneously determines the required sample sizes for an experiment while the one-stage procedure is a data-analysis procedure after the data have been collected, which can supplement the two-stage procedure when the later has to end its experiment sooner than its required experimental process is completed. Tables needed for implementing these procedures are given. Key Words and Phrases : Indifference zone; Least favorable configuration; Power of a test; Studentized range test; t distribution; Heteroscedasticity. 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 (e.g., 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 (e.g., Berger (1985)). In many real world problems, the practical interest is often to examine whether the population means fall into an indifference zone, not just the equality of means. This idea leads to the consideration of equivalency hypothesis stated as H 0 : 1 ki=1 k µ i µ δ vs H a : 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 and the quantity stated in the null hypothesis is often regarded as the average deviation of µ i s from their grand mean. 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 difference among means within a small δ-value or there is practically equivalent among the means. 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 equivalency hypotheses are also referenced to as interval hypotheses or bioequivalence in pharmaceutical and medical studies (e.g., see Chow and Liu (1992)) when H 0 and H a are reversed. For the case of testing three or more means, the studentized range test for the equivalency hypotheses was studied by Chen and Lam (1991) and Chen, Xiong and Lam (1993) for the common unknown variances. For the case of testing the null hypothesis of the equality of means H0 : µ 1 =... = µ k against a specific alternative hypothesis, Ha : µ max µ min δ, where δ (> 0) is a prespecified constant, Chen and Chen (2000) proposed a range test when the variances are unknown and possibly unequal. In this paper, the case of H 0 vs. H a under unequal variances is studied. When the variances are known but unequal, one can divide the original random variables by their corresponding standard deviations so that the transformed variables satisfy the assumption of a common 2

3 (known) variance and then Chen, Xiong and Lam s (1993) method can be used to define a range test. Furthermore, in situations where the variances are unknown but possibly unequal, we use the two-stage sampling procedure as described by Bishop and Dudewicz (1978) and the one-stage sampling procedure proposed by Chen and Lam (1989), respectively, to formulate a modified studentized range test for testing the null hypotheses of equivalency H 0 against the alternative H a. In Section 2, we briefly discuss the case when variances are known but unequal. In Section 3, the two-stage sampling procedure is introduced and a modified studentized range test is defined, and the level and the power of the proposed range test are calculated. In Section 4, the numerical calculation of the level and the power is discussed. In Section 5, a numerical example is given to illustrate the use of the studentized range test. In Section 6, the single-stage sampling procedure is defined and the level and the power of the range test are determined. Finally, a summary and conclusion is given in Section Transformed Sampling Procedure and Range Test In a one-way layout model, let X ij (i = 1,..., k; j = 1,..., N) be a random sample of size N ( 2) drawn from the normal distribution π i having a unknown mean µ i and a known but unequal variance σi 2, i = 1,..., k. Since the heteroscedastic variances σ 2 i are known, divide the r.v. s X ij (i = 1,..., k; j = 1,..., N) by σ i, through i = 1,..., k, to obtain X ij = X ij/σ i, i = 1,..., k; j = 1,..., N. The purpose of transforming the original variables is to eliminate the influence of unequal variances. To see this, one can easily calculate the variance of X ij : V ar(x ij ) = V ar(x ij σ i ) = 1, i = 1,..., k, which is a constant and X ij is distributed as N(µ i, 1), where µ i = µ i/σ i. That is, the variance of the transformed variable X ij is now homoscedastic. Our goal is to test the following null hypothesis H 0 : 1 k µ i k µ δ i=1 3

4 against the alternative H a : 1 k µ i µ δ (δ > δ), k i=1 where µ = k i=1 µ i /k, and the quantity 1 ki=1 k µ i µ is interpreted as the average deviation of means from the grand mean µ. A range test is defined as X max X min, and then Chen, Xiong and Lam s (1993) result with respect to the transformed data can be used without difficulty. It can be shown that the probability of rejecting H 0 attains its maximum at the least favorable configuration (LFC) µ 0 = ( µ kδ N/2, µ,..., µ, µ + kδ N/2) or at its permutations when k 3 and at µ 0 = ( µ kδ N/2, µ + kδ N/2) or at its permutation when k = 2. When sample sizes are unequal, one can replace N by the harmonic mean to obtain an approximation. 3. Two-Stage Procedure and Modified Studentized Range Test Similarly, for the one-way layout model, let X ij (i = 1,..., k; j = 1,..., N i ) be a random sample of size N i ( 3) drawn from the normal distribution π i which has a unknown mean µ i and a possibly unknown variance σi 2, i = 1,..., k, where these variances may be highly unequal. Our goal is to test the null hypothesis against the alternative H 0 : 1 k µ i µ δ (1) k i=1 H a : 1 k µ i µ δ (δ > δ), k i=1 where µ = k i=1 µ i /k, and the quantity 1 k ki=1 µ i µ is interpreted as the average deviation of means from the grand mean µ. A modified studentized range test is to be proposed in such a way that both the level and the power of the test are controllable under the proposed two-stage sampling procedure and they are shown to be free of the unknown and unequal variances. The two-stage sampling procedure for this problem is stated as follows. 4

5 From each of the k populations takes an initial random sample of size n 0 (n 0 2, but Bishop and Dudewicz (1978) suggested that the initial sample of size n 0 be 10 or more giving better results. For economic reason, it is usually taken to be 25 or less.), let S 2 i be the usual unbiased estimate of σ 2 i based on the first n 0 observations from the i th population, and define N i = max{n 0 + 1, [ S2 i z ] + 1} (2) where z > 0 is a design constant to be determined by the power of the test under a specified H a and [X] stands for the largest integer less than X. Then take N i n 0 additional observations from the i th population so that we have a total of N i observations denoted by X i1,..., X in0,..., X ini. For each i, set the coefficients a i1,..., a in0,..., a ini, such that a i1 =... = a in0 = 1 (N i n 0 )b i a in0 +1 =... = a ini = 1 N i [1 + and computed the weighted sample mean as n 0 X i = a i X ij + b i j=1 n 0 N i j=n 0 +1 = a i n 0 (N i z Si 2) ] (N i n 0 )Si 2 = b i (3) X ij. It should be noted that these coefficients a ij s are so determined to satisfy the following conditions, N i j=1 a ij = 1, a i1 =... = a in0, and Si 2 Ni j=1 a2 ij = z. It is easy to show that the r.v. s T i = ( X i µ i )/ z, i = 1,..., k, have independent Student s t-distributions, free of the unknown variances σ 2 i (see, e.g., Chen and Chen (1998)). Let X [1]... X [k] be the order statistics of X 1,..., Xk and let the modified studentized range statistic be defined by R = X [k] X [1] z. (4) The null hypothesis H 0 is rejected at level α iff R > γ α (5) where γ α = γ α (δ, z, k, n 0 ) is the level-α critical value such that P δ (γ α ) = sup P (R > γ α H 0 : 1 Ω k 5 k µ i µ δ) = α, (6) i=1

6 where α (0, 1) is a predetermined level, and Ω is the set of all possible configurations of the µ i s and σ i s, and the design constant z is determined by the power of the test such that P δ (γ α ) = inf Ω P (γ > γ α H a : 1 k where P is taken to be a large value in advance in (0, 1). k µ i µ δ > δ) = P, (7) i=1 Here, it is necessary to find a LFC of the means which maximizes the level of the test under H 0 in (6) and a least favorable configuration which minimizes the power of the test under a specified H a in (7) such that the level and the power are not only independent of all means differences but also free of the unknown variances. The LFC s are determined by Theorem 1 of Chen, Xiong and Lam (1993). It is stated as follows without proof. 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 maximun at some θ for all θ A = {θ : θ i θ δ}, where θ = (θ θ k )/k. Then θ is the LFC 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. We now find these LFC s of (6) and (7) described below. 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 Ω (R > γ α ) = P Ω ( 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] 6

7 k = 1 P [T j + δ ji T i T j + δ ji + γ α, i = 1,..., k, i j] j=1 k k = 1 [F (y + δ ji + γ α ) F (y + δ ji )]f(y)dy, (8) j=1 i=1,i j 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, are i.i.d. Student s t r.v. s having p.d.f. f( ) and c.d.f. F ( ) with n 0 1 d.f.. As the n 0 becomes large, T i converges to the standard normal, or equivalently X i / z converges to N(µ i / z, 1). Thus, by Theorem 1, the asymptotic LFC of means which maximizes the integral in (8) occurs at µ 0 = ( µ kδ/2, µ,..., µ, µ + kδ/2) for k 3 and at µ 0 = ( µ kδ/2, µ + kδ/2) for k = 2. For computational reason (for detail, see Chen and Chen (1999).), we rewrite µ 0 as µ 0 = (( µ kvw/2, µ,..., µ, µ + kvw/2), where v = δ/δ and w = δ / z. Let the maximum of (8) subject to µ H 0 at the LFC µ 0 be P δ (γ α, w). Then P δ (γ α, w) = 1 { [F (y kvw/2 + γ α ) F (y kvw/2)] k 2 [F (y kvw + γ α ) F (y kvw)]f(y)dy +(k 2) [F (y + kvw/2 + γ α ) F (y + kvw/2)][f (y + γ α ) F (y)] k 3 [F (y kvw/2 + γ α ) F (y kvw/2)]f(y)dy + [F (y + kvw/2 + γ α ) F (y + kvw/2)] k 2 [F (y + kvw + γ α ) F (y + kvw)]f(y)dy}. (9) Note that in the special case where δ = 0, the P δ (γ α, w) reduces to P 0 (γ α, w) = 1 k [F (y + γ α ) F (y)] k 1 f(y)dy = P ( X [k] X [1] > γ α z H0 : µ 1 = = µ k ), (10) which is the range of i.i.d. Student s t r.v. s. by Wilcox(1983). Similarly, as n 0 becomes large, the minimum power in (8) under H a : 1 ki=1 k µ i µ δ > δ is attained at the asymptotic LFC µ 1 = ( µ kw/(2l),..., µ kw/(2l), µ + kw/(2(k l)),..., µ + kw/(2(k l))) or its permutations with the l such ( µ kw/2l) s and k l such ( µ + kw/(2(k l))) s, l = 1,..., k 1. (See Chen, Xiong and Lam (1993).) 7

8 Let β δ (γ α, w) denote the minimum power in (8) at the LFC µ = µ 1. Then β δ (γ α, w) = 1 {l + (k l) where b = k 2 /(2l(k l)). [F (y + γ α ) F (y)] l 1 [F (y bw + γ α ) F (y bw)] k l f(y)dy [F (y + bw + γ α ) F (y + bw)] l [F (y + γ α ) F (y)] k l 1 f(y)dy}, (11) The minimum power of (11) occurs at l = k/2 when k is even, and at l = (k 1)/2 or l = (k + 1)/2 when k is odd. It is trivial when k = 2. When k = 3, the power in (11) is the same for l = 1, 2. However, when k is four or large, it is not trivial to find its minimum by analytical method. By numerical quadrature as illustrated in Table 2 we confirm the above claim, and it is true in general. Table 2. The Behavior of Power Function of Expression (11). k n α P ν γ ω l Power Power Power Power Computation of the level and the power Using a grid-searching method combined with Newton-Raphson s iterations as described by Chen and Chen (1999), the critical values of γ α and the values of w in (9)-(11) are calculated. Given the level α and the required power P, the equivalent constant δ and the dedective difference δ (or the delta ratio = δ/δ ) under H 0 and H a, respectively, for a specified number of treatments k and an initial sample size n 0, one inputs a small guess of 8

9 ω (= δ / z) into (9) and uses Newton-Raphson s iterations to find a critical value γ α such that the absolute error of P δ (γ α, ω) α is bounded by Then, substitute these values into (11) to see if the probability inequality β δ (γ α, ω) P (12) is satisfied. Increase or decrease ω such that the smallest ω (accurate up to two decimal figures) satisfies (12). Tables 3-9 present the critical values γ α (upper entry) and the values of w = δ / z (lower entry) for the levels at 5% and 1%, and the powers at 90% and 95%, when k = 2(1)6, 8, 10, n 0 = 5, 10, 15, 25 and various ratio δ/δ 0.6. For example, if α = 5% with required power being at least 90% and k = 3, n 0 = 10, δ/δ = 0.2, then the γ α =4.52 and w = δ / z=2.54 are found from Table 4. If the δ value in the null hypothesis H 0 : 1 ki=1 k µ i µ δ is set to be 0.2 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=2.54 for z to obtain the z value of A Numerical Example The data in Table 8 come from an experiment reported by Bishop and Dudewicz (1978). The experiment involved testing four types of solvents for their effects on the ability of a fungicide methyl-2-benzimidazole-carbamate to destroy the fungus penicillium expansum. The fungicide was diluted in exactly the same manner in the four different types of solvents and sprayed on the fungus, and the percentage of fungus destroyed was measured. The mean percentage of fungus destroyed by solvent i is denoted by µ i. Wen and Chen (1994) found that the data from solvents 1 and 3 are not normally distributed using SAS PROC Univariate testing procedure. However, Dudewicz and van der Meulen (1983) has shown the robustness results that apply to the two-stage sampling procedure for nonnormal distributions when population variances are not all equal. Chen, Chen and Ding (2000) had conducted a robust modified Levene s test for homogeneity of variances and found a significant difference (p-value < ) among variances. So, the two-stage sampling procedure can be applied. 9

10 Table 3. Percentage Point γ (upper entry) of a Range Test R and its Power- Related Ratio, δ / z (lower entry) when k = 2 and P =.90 and.95. n 0 = 5 n 0 = 10 n 0 = 15 n 0 = 25 level level level level δ/δ P = P =

11 Table 4. Percentage Point γ (upper entry) of a Range Test R and its Power- Related Ratio, δ / z (lower entry) when k = 3 and P =.90 and.95. n 0 = 5 n 0 = 10 n 0 = 15 n 0 = 25 level level level level δ/δ P = P =

12 Table 5. Percentage Point γ (upper entry) of a Range Test R and its Power- Related Ratio, δ / z (lower entry) when k = 4 and P =.90 and.95. n 0 = 5 n 0 = 10 n 0 = 15 n 0 = 25 level level level level δ/δ P = P =

13 Table 6. Percentage Point γ (upper entry) of a Range Test R and its Power- Related Ratio, δ / z (lower entry) when k = 5 and P =.90 and.95. n 0 = 5 n 0 = 10 n 0 = 15 n 0 = 25 level level level level δ/δ P = P =

14 Table 7. Percentage Point γ (upper entry) of a Range Test R and its Power- Related Ratio, δ / z (lower entry) when k = 6 and P =.90 and.95. n 0 = 5 n 0 = 10 n 0 = 15 n 0 = 25 level level level level δ/δ P = P =

15 Table 8. Percentage Point γ (upper entry) of a Range Test R and its Power- Related Ratio, δ / z (lower entry) when k = 8 and P =.90 and.95. n 0 = 5 n 0 = 10 n 0 = 15 n 0 = 25 level level level level δ/δ P = P =

16 Table 9. Percentage Point γ (upper entry) of a Range Test R and its Power- Related Ratio, δ / z (lower entry) when k = 10 and P =.90 and.95. n 0 = 5 n 0 = 10 n 0 = 15 n 0 = 25 level level level level δ/δ P = P =

17 Suppose that the experimenter regards an average deviation of 0.25% (δ value) among the µ i s to be irrelevant and wishes to detect the average deviation to be at least 1.0% (δ value). This can be stated by the null hypothesis H 0 : 1 4i=1 4 µ i µ 0.25 against the alternative hypothesis H a : 1 4i=1 4 µ i µ 1.0. If the experimenter choose the level of the test to be 5% and a power of at least 0.90, and one takes the initial sample of size n 0 = 15 observations from each solvent, then, the critical value of the modified studentized range test produces a γ α = 5.71 and δ / z = 3.04 at δ/δ = 0.25 from Table 5, so z = The remaining N i 15 observations taken at the second stage are given in Table 11. The summary statistics Si 2, a i, b i and N i defined in Section 3 are given in Table 12. The final weighted sample means X 1, X2, X3 and X 4 are given in the last row of Table 12, and finally, the calculated studentized range test statistic using (4) is R=6.68. Hence, H 0 is rejected in asserting H a with a power of being at least Pairwise multiple comparisons given by Chen and Chen (1999) can be employed to detect the differences among mean percentages. 17

18 Table 10. Bacterial Killing Ability Example (First 15 Observations) Solvent 1 Solvent 2 Solvent 3 Solvent Table 11. Bacterial Killing Ability (Second-Stage Observations) Solvent 1 Solvent 2 Solvent 3 Solvent Table 12. Summary Statistics Statistic Solvent 1 Solvent 2 Solvent 3 Solvent 4 Si a i b i N i X i

19 6. Single-Stage Sampling procedure and Modified Studentized Range Test The two-stage sampling procedure discussed in Section 3 is a design-oriented method which determines the necessary sample sizes of N i and to simultaneously meet a prespecified power requirement at a specified level of the test. In some situations where the two-stage experiment is terminated earlier due to some unexpected reasons such as budget shortage or patients drop-out, the required total sample size N i in (2) cannot be achieved, one only has to use the available m i observations ((n 0 + 1) m i N i ) on hand and recalculate the coefficients a ij s according to the so-called one-stage sampling procedure proposed by Chen and Lam (1989) such that the statistical inference theory can still work. The one-stage procedure is briefly described as below. From each of the k populations one uses an initial sample of size n 0 (2 < n 0 < m i ). Calculate the usual unbiased sample mean and unbiased sample variance, respectively, by X i = n0 j=1 X ij n 0 and S 2 i = n0 j=1 (X ij X i ) 2. n 0 1 Then, construct the new coefficients as U i = 1 m i + 1 m i mi n 0 n 0 (m i z /S 2 i 1), V i = 1 m i 1 m i where U i and V i satisfy the following conditions: n0 m i n 0 (m i z /S 2 i 1), n 0 U i + (m i n 0 )V i = 1, S 2 i [ n 0 U 2 i + (m i n 0 )V 2 i ] = z, where z is the maximum of {S 2 i /m i, i = 1,..., k }. Let the final weighted sample mean be defined by n 0 X i = U i X ij + V i j=1 m i n 0 +1 X ij. (13) 19

20 It is known (Chen and Chen (1998)) that ( X i µ i )/ z, i = 1,..., k, have independent Student s t-distributions each with n 0 1 df.. Similarly, the modified studentized range statistic is defined as R = X [k] X [1] z (14) which is used to test the H 0 against H a in (1). The probability statement P Ω (R > γ α ) can be similarly derived by replacing z by z and it is given by P Ω (R > γ α) = P Ω ( X [k] X [1] > γ α z ) = 1 P ( X [k] X [1] + γ α z ) k = 1 P ( X [k] X [1] + γα z, X (j) = X [1] ) j=1 k = 1 P ( X (j) X (i) X (j) + γα 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 +γα, i = 1,..., k, i j] k = 1 P [T j + δ ji T i T j + δ ji + γα, i = 1,..., k, i j] j=1 k k = 1 [F (y + δ ji + γα ) F (y + δ ji)]f(y)dy, (15) j=1 i=1,i j 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 are i.i.d. Student s t r.v. s having p.d.f. f( ) and c.d.f. F ( ) with n 0 1 d.f.. As the n 0 becomes large, T i converges to the standard normal, or equivalently X i / z converges to N(µ i / z, 1). Thus, by Theorem 1, we can obtain the asymptotic LFC of means which maximizes the integral in (15) at µ 0 = ( µ kδ/2, µ,..., µ, µ + kδ/2) or its permutations for k 3 and at µ 0 = ( µ kδ/2, µ + kδ/2) for k = 2. For computational reason, let d = δ/ z, the maximum level of (8) subject to µ H 0 occurs at the asymptotic LFC µ 0 and its value can be approximated by 20

21 P δ (γα, d ) = 1 { [F (y kd /2 + γα ) F (y kd /2)] k 2 [F (y kd + γ α ) F (y kd )]f(y)dy +(k 2) [F (y + kd /2 + γ α ) F (y + kd /2)][F (y + γ α ) F (y)]k 3 [F (y kd /2 + γ α ) F (y kd /2)]f(y)dy + [F (y + kd /2 + γ α) F (y + kd /2)] k 2 [F (y + kd + γ α) F (y + kd )]f(y)dy}. (16) Note that in the special case where δ = 0, the above P δ (γ α, d ) reduces to P 0 (γα, d ) = 1 k [F (y + γα) F (y)] k 1 f(y)dy = P (T i > γ α H 0 : µ 1 = = µ k ). (17) Similarly, as n 0 becomes large, the minimum power in (15) under H a : 1 ki=1 k µ i µ δ > δ is attained at the asymptotic LFC µ 1 = ( µ kd /(2l),..., µ kd /(2l), µ+kd /(2(k l)),..., µ+kd /(2(k l))) with the l such (µ kd /2l) s and k l such (µ + kd /(2(k l))) s, l = 1,..., k 1, where d = δ/ z. (See Chen, Xiong and Lam (1993).) Let β δ (γ α, d ) denote the minimum power in (15) at the LFC µ = µ 1. Then, the minimum power can be calculated by β δ (γα, d ) = 1 {l + (k l) where b = k 2 /(2l(k l)). [F (y + γ α ) F (y)]l 1 [F (y bd + γ α ) F (y bd )] k l f(y)dy [F (y + bd + γ α) F (y + bd )] l [F (y + γ α) F (y)] k l 1 f(y)dy} (18) By numerical calculation we confirm that the minimum power of (18) 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 δ at H 0 and d, k, and n 0, the critical value γ α can be obtained by solving the equation g 1 (γ α, d ) = P δ (γ α, d ) α = 0. (19) The solution γ α in (19) is data dependent on d. 21

22 Using Gaussian quadrature to evaluate the integrals in (16) and the Newton-Raphsons iteration one can find the solution γ α to (19) at the nth iteration by the formula γ n = γ n 1 P δ(γn 1, d ) α P δ (γn 1,, (20) d ) where P δ (γ, d ) is the derivative of P δ w.r.t. γ, given by P δ (γ, d ) = {(k 2)[F (y kd /2 + γ ) F (y kd /2)] k 3 f(y kd /2 + γ ) [F (y kd + γ ) F (y kd )] + [F (y kd /2 + γ ) F (y kd /2)] k 2 f(y kd + γ )}f(y)dy (k 2) [F (y + γ ) F (y)] k 4 {f(y + kd /2 + γ )[F (y + γ ) F (y)] [F (y kd /2 + γ ) F (y kd /2)] + (k 3)[F (y + kd /2 + γ ) F (y + kd /2)]f(y + γ ) [F (y kd /2 + γ ) F (y kd /2)] + [F (y + kd /2 + γ ) F (y + kd /2)][F (y + γ ) F (y)] f(y kd /2 + γ )}f(y)dy {(k 2)[F (y + kd /2 + γ ) F (y + kd /2)] k 3 f(y + kd /2 + γ ) [F (y + kd + γ ) F (y + kd )] + [F (y + kd /2 + γ ) F (y + kd /2)] k 2 f(y + kd + γ )}f(y)dy. The solution (20) is unique because P δ (γα, d ) is monotonically decreasing in γ. When the required sample sizes listed in Table 10 cannot be met due to early termination of the experiment, one can only have the total samples m 1 = 19, m 2 = 25, m 3 = 49 and m 4 = 16. At this time the one-stage procedure can be applied. By (13) the weighted sample means are calculated as X 1 = 96.93, X2 = 95.67, X3 = 95.32, X4 = and z = max{si 2/m i} = By (14) the test statistic is γ = Given δ = 0.25, δ = 1.0, k = 4, n 0 = 15 as previously specified, the p-value of γ (= 5.23) is At 10% level 22

23 of significance, the critical value is γ.10 = 4.97 and the estimated power is Therefore, at α = 0.10, we can reject the null hypothesis of all means being within an equivalent average deviation of.25 units in favor of the difference of being at least 1 unit. This calculation was done by a Fortran program named chenwen7.for available from the authors. 7. Summary and Conclusion 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 average deviation of the means, which extends the idea of equivalency among means. The test of equivalency receives more attention in health sciences, pharmaceutical industry, and other areas. When the variances are unknown and unequal, a studentized range test using a two-stage and a one-stage sampling procedures, respectively, is proposed for testing the hypothesis that the average deviation of the normal means is falling into a practical indifference zone. Both the level and the power of the proposed test by the two-stage procedure are controllable and they are completely independent of the unknown variances. Statistical tables to implement the procedure are provided for practitioners, and an example is given to demonstrate the use of the procedure. The two-stage procedure is a design-oriented procedure that satisfies certain probability requirements and simultaneously determines the required sample sizes (which can be large at the second stage) in an experiment while the one-stage procedure is a dataanalysis procedure after the data have been collected, which can supplement the two-stage procedure when the later has to end its experiment sooner than its required experimental process is completed. At that time the level and power can be approximated, and the onestage sampling procedure is shown to be quite feasible under heterocedasticity. 23

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