A test against an umbrella ordered alternative

Computational Statistics & Data Analysis 51 (2006) 1957–1964www.elsevier.com/locate/csda

A test against an umbrella ordered alternative

Parminder Singha,∗, Wei Liub

aDepartment of Mathematics, Guru Nanak Dev University, Amritsar, Punjab 143005, IndiabSchool of Mathematics, University of Southampton, SO17 IBJ UK

Received 27 April 2004; received in revised form 13 December 2005; accepted 14 December 2005Available online 6 January 2006

Abstract

Consider k(k�3) independent populations �1, . . . , �k such that cumulative distribution function (cdf) of an observation frompopulation �i is Fi(x) = F

[(x − �i

)/�i

], a member of general location scale family, where F(.) is any absolutely continuous

cdf, �i

(−∞ < �i < ∞)is the location parameter and �i (�i > 0) is the scale parameter, i = 1, . . . , k. It is assumed that scale

parameters �is satisfy the umbrella ordering �1 � · · · ��h � · · · ��k with at least one strict inequality, where 1�h�k is given.When h is equal to 1 or k the umbrella ordering is just the simple ordering. Gill and Dhawan [1999. A one-sided test for testinghomogeneity of scale parameters against ordered alternative. Comm. Statist. Theory Methods 28(10), 2417–2439] considered theproblem of testing the null hypothesis H0 : �1 = · · · = �k against the simple ordering alternative, and provided some critical pointsby using simulation technique. In this paper a one-sided test for testing null hypothesis H0 against the umbrella ordered alternativeHa : �1 � · · · ��h � · · · ��k with at least one strict inequality is proposed. A recursive method to compute the exact criticalpoints for the test procedure is proposed and selected tables of critical points for exponential probability model are provided. Theproposed test is inverted to obtain a set of one-sided simultaneous confidence intervals for all the ordered pairwise ratios �j /�i for1� i < j �h and for h�j < i �k, which enable the experimenter to infer which �is are different and by how much when H0 isrejected. Applications of the proposed test procedure to normal and inverse Guassian distributions are also discussed.© 2006 Elsevier B.V. All rights reserved.

Keywords: Pairwise comparisons; Contrasts; Critical points; Numerical integration; Simultaneous confidence intervals

1. Introduction

Let �1, . . . , �k be k (k�3) independent populations such that an observation from population �i follows a distributionwith cumulative distribution function (cdf) Fi(x) = F

[(x − �i

)/�i

], where F(.) is any absolutely continuous cdf,

�i

(−∞ < �i < ∞)is the location parameter and �i (�i > 0) is the scale parameter, i =1, . . . , k. The problem of testing

the null hypothesis of homogeneity against the simple ordered alternative hypothesis with at least one strict inequalityhas received considerable attention in the literature; for detailed references, the readers are referred to Barlow et al.(1972), Hochberg and Tamhane (1987) and Robertson et al. (1988). Hayter (1990) and Hayter and Liu (1996) proposeda one-sided studentized range test for normal population means. Dhawan and Gill (1997) discussed simultaneous

∗ Corresponding author. Tel.: +91 1832426967; fax: +91 1832258820.E-mail addresses: [email protected], [email protected] (P. Singh).

0167-9473/$ - see front matter © 2006 Elsevier B.V. All rights reserved.doi:10.1016/j.csda.2005.12.010

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mailto:[email protected]

mailto:[email protected]

1958 P. Singh, W. Liu / Computational Statistics & Data Analysis 51 (2006) 1957–1964

one sided confidence intervals for ordered pairwise differences of exponential location parameters. Gill and Dhawan(1999) and Singh and Gill (2004) proposed a test procedures for scale parameters for testing the null hypothesisH0 : �1 = · · · = �k against the simple ordered alternative HA : �1 � · · · ��k with at least one strict inequality. Theseprocedures have important applications for problems where the treatments can be assumed to satisfy a simple ordering,such as for a sequence of increasing dose-level of a drug. For testing the null hypothesis against the umbrella orderingwith at least one strict inequality, Bartholomew (1959, 1961) proposed the likelihood ratio test and Hayter and Liu(1996) proposed another test for the ANOVA model. More detailed references are available in Robertson et al. (1988),Marcus and Genizi (1994) and Marcus and Talpaz (1992). Umbrella ordering is important in dose–response experiment;e.g., see Simpson and Margolin (1986). In case where mode of action of a drug is related to its toxic effects, e.g., incase of life saving digitalization therapy of heart failure, umbrella behavior is anticipated and careful dosage planningis required.

The layout of the paper is as follows. A test procedure for testing null hypothesis H0 : �1 = · · · = �k against thealternative Ha : �1 � · · · ��h � · · · ��k with at least one strict inequality is proposed in Section 2. Section 3 containsa set of one-sided simultaneous confidence intervals for all the ordered pairwise ratios �j /�i for 1� i < j �h and forh�j < i�k which can be obtain by inverting the test.A recursive method for the exact computation of the critical pointsand a simulation study to assess the accuracy of the recursive procedure and Gill and Dhawan’s (1999) computationprocedure are provided for the exponential probability model in Section 4. Section 5 provides two situations where thenew procedure can readily be applied.

2. A new test

Let Xi1, . . . , Xin be a random sample of a common size n from the ith population �i , i = 1, . . . , k. Let Si be anysuitable estimator of parameter �i based on a random sample of size n. Our proposed test for testing the null hypothesisH0 : �1 =· · ·=�k against the alternative �1 � · · · ��h � · · · ��k with at least one strict inequality and a given 1�h�k

is based on the test statistic

Wh = max

(max

1� i<j �h

(Sj/Si

), maxh� j<i �k

(Sj/Si

)). (2.1)

H0 is rejected at level � iff

Wh �ck,h,�,�,

where c = ck,h,�,� is the critical point satisfying

P0 [Wh �c] = �. (2.2)

HereP0(A) means that the probability is computed under H0.The test of Gill and Dhawan (1999) is corresponding to the special case of h = k, which is for testing the null

hypothesis H0 against the simple ordered alternative.

3. Simultaneous confidence intervals

If the null hypothesis of homogeneity H0 is not rejected then further statistical analysis may not be necessary.However, if H0 is rejected then one may wish to determine which �is differ and by how much. In other words,the experimenter may be interested to explore the reasons underlying the rejection of the null hypothesis of equality.Simultaneous confidence intervals enable the experimenter to determine which �is are different. While the simultaneousconfidence intervals given in Gill and Dhawan (1999) are useful when there is prior information that �is have a simpleordering, the following simultaneous confidence intervals are more pertinent if there is prior information that the �isfollow the umbrella ordering.

P. Singh, W. Liu / Computational Statistics & Data Analysis 51 (2006) 1957–1964 1959

The above test can be inverted to obtain 1 − � the simultaneous one-sided bounds for the ordered pair-wise ratios�j /�i for 1� i < j �h and for h�j < i�k as follows:

1 − � = P0 (Wh �c)

= P

(Sj/�j

Si/�i

�c for 1� i < j �h andSj/�j

Si/�i

�c for h�j < i�k

)

= P

(Sj

Si

1

c� �j

�i

for 1� i < j �k andSj

Si

1

c� �j

�i

for h�j < i�k

)

= P

[�j /�i �

Sj

Si

1

c, 1� i < j �h, h�j < i�k

].

Hence

P

[�j /�i �

Sj

Si

1

c, 1� i < j �h, h�j < i�k

]= 1 − �. (3.1)

The set of simultaneous confidence intervals in Eq. (3.1) is not based on the assumption that the �is follow theumbrella ordering. However, if there is prior information about the ordering of �is: �1 � · · · ��h � · · · ��k , then thisinformation may be used to improve the confidence interval given by (3.1), e.g. a lower limit less than one will beuseless and may be truncated at one.

Below, we discuss the computation of the critical points by taking two parameter exponential probability model.

4. Computation of the critical points

The two parameter exponential probability density function (pdf)

f (x|�, �) = 1

�e−(x−�/�)I[�,∞)(x), � > 0, −∞ < � < ∞,

where IA(.) is the indicator function of event A and �(�) is the location (scale) parameter, is routinely used to modelthe life time in many scientific areas. The scale parameter � is the mean life in addition to the guaranteed life � in lifetesting and mean effective duration of dose in dose–response experiment. In biological and epidemiological studies thelocation parameter � is called the latency period, the time between the infection and the onset of the disease, and thescale parameter � is termed the mean duration of a disease in addition to the latency period.

Let Xi1, . . . , Xin be a random sample from ith exponential population with location parameter �i and scale parameter�i , i = 1, . . . , k. The minimum variance-unbiased estimator of �i is given by Si = ∑n

j=1

(Xij − Yi

)/�, where Yi =

min1� j �nXij and � = n − 1. The random variable �Si/�i follows the gamma distribution with shape parameter � andscale parameter unity.

While Gill and Dhawan (1999) obtained critical points using Monte Carlo simulation, below we show that recursivemethod similar to Hayter and Liu (1996, 1999) can be used to accurately calculate the critical points for any givenk, h, � and �.

Let G�(.) and g�(.) be the cdf and pdf respectively of gamma variate with shape parameter � and scale parameterunity. Define

f0(s, c) = 1,

f1(s, c) = P0

[S1

s�c

]= G�(cs)

and

fr(s, c) = P0

(max

1� i<j � r+1

Si

Sj

�c|Sr+1 = s

), r �2. (4.1)


Table 1Values of one-sided critical points ck,h,�,� when h = 1, � = .01

� k

3 4 5 6 7 8 9 10

3 12.342 15.458 18.148 20.541 22.722 24.747 26.630 28.4144 8.222 9.878 11.245 12.434 13.496 14.452 15.337 16.1585 6.345 7.432 8.310 9.061 9.719 10.310 10.848 11.3406 5.283 6.080 6.714 7.252 7.715 8.130 8.505 8.8487 4.600 5.227 5.721 6.133 6.491 6.806 7.089 7.3478 4.124 4.639 5.043 5.377 5.666 5.917 6.145 6.3519 3.774 4.212 4.552 4.832 5.073 5.283 5.473 5.643

10 3.505 3.886 4.181 4.422 4.629 4.809 4.970 5.11511 3.290 3.629 3.889 4.101 4.283 4.441 4.581 4.70712 3.117 3.421 3.654 3.844 4.005 4.145 4.269 4.38213 2.973 3.250 3.460 3.632 3.777 3.904 4.015 4.11614 2.851 3.105 3.298 3.454 3.587 3.702 3.804 3.89515 2.746 2.982 3.159 3.304 3.425 3.530 3.624 3.70720 2.384 2.558 2.688 2.792 2.880 2.955 3.021 3.08125 2.169 2.308 2.412 2.495 2.564 2.623 2.675 2.72230 2.023 2.141 2.228 2.298 2.355 2.405 2.448 2.48735 1.918 2.021 2.097 2.156 2.206 2.249 2.286 2.32040 1.837 1.929 1.996 2.050 2.094 2.131 2.164 2.193

Then, for r �2,

fr(s, c) = P0

(max

1� i<j � r

Si

Sj

�c, max1� i � r

Si

s�c

)

=∫ cs

0P0

(max

1� i<j � r

Si

Sj

�c, max1� i � r−1

Si

s�c|Sr = w

)g�(w) dw,

=∫ cs

0P0

(max

1� i<j � r−1

Si

Sj

�c, max1� i � r−1

Si � min(sc, wc)

)g�(w) dw,

=∫ s

0P0

(max

1� i<j � r−1

Si

Sj

�c, max1� i � r−1

Si � min(sc, wc)

)g�(w) dw

+∫ cs

s

P0

(max

1� i<j � r−1

Si

Sj

�c, max1� i � r−1

Si � min(sc, wc)

)g�(w) dw

=∫ s

0fr−1(w, c)g�(w) dw + fr−1(s, c) [G�(sc) − G�(s)] . (4.2)

Therefore, c = ck,h,�,� is the solution of the equation

Fk,h(c) = P0 [Wh �c] =∫ ∞

0g�(s)fk−h(c, s)fh−1(c, s) ds = 1 − �. (4.3)

From (4.2) the function fr(., c) can be computed recursively. To compute Fk,h(c) from (4.3), we start by calculatingf1(s, c) over a one-dimensional grid of values of s on the interval [0, y] and store these values as a one-dimensionalarray in the computer, where y is chosen such that 1 − G�(y)�10−7. Then (4.2) is used to find the values of f2(s, c)

over a similar grid of values of s. Continuing in this manner, we can find the values of fr(c, s), r > 2, over a finegrid of values of s. Then the integral in (4.3) is evaluated using Guass quardrature, and the critical value c canbe solved numerically. For � = .01, � = .05, h = 1(1), 3 and � = 3(1)15, 15(5)40, Tables 1–6 provide the criticalpoints c. Additional tables are available from first author. A one-dimensional grid of values s on the interval [0, y] of



� k

3 4 5 6 7 8 9 10

3 6.472 8.232 9.752 11.110 12.348 13.493 14.562 15.5674 4.854 5.927 6.819 7.594 8.285 8.913 9.489 10.0245 4.028 4.794 5.416 5.948 6.414 6.834 7.215 7.5676 3.525 4.121 4.598 4.999 5.349 5.661 5.942 6.2007 3.185 3.674 4.061 4.384 4.663 4.910 5.132 5.3358 2.939 3.355 3.681 3.951 4.183 4.388 4.571 4.7389 2.752 3.115 3.397 3.630 3.829 4.004 4.160 4.302

10 2.604 2.928 3.177 3.382 3.556 3.709 3.845 3.96811 2.485 2.777 3.001 3.184 3.340 3.475 3.596 3.70512 2.386 2.653 2.856 3.022 3.163 3.285 3.394 3.49213 2.303 2.548 2.735 2.887 3.016 3.127 3.226 3.31514 2.231 2.459 2.632 2.773 2.891 2.994 3.084 3.16615 2.169 2.382 2.544 2.674 2.784 2.879 2.963 3.03920 1.949 2.113 2.235 2.333 2.415 2.485 2.547 2.60325 1.813 1.948 2.048 2.128 2.194 2.251 2.301 2.34530 1.720 1.836 1.921 1.989 2.045 2.094 2.136 2.17335 1.651 1.754 1.829 1.888 1.938 1.980 2.016 2.04940 1.598 1.690 1.758 1.811 1.855 1.893 1.925 1.954


� k

3 4 5 6 7 8 9 10

3 10.332 13.206 15.967 18.474 20.782 22.906 24.888 26.7434 7.146 8.717 10.155 11.429 12.562 13.588 14.530 15.4015 5.637 6.685 7.620 8.431 9.142 9.779 10.356 10.8846 4.763 5.538 6.222 6.806 7.312 7.761 8.165 8.5337 4.190 4.806 5.340 5.791 6.183 6.526 6.832 7.1108 3.786 4.298 4.735 5.103 5.418 5.694 5.940 6.1639 3.487 3.922 4.292 4.603 4.867 5.098 5.303 5.487

10 3.255 3.636 3.957 4.225 4.453 4.652 4.827 4.98311 3.070 3.408 3.692 3.928 4.129 4.302 4.455 4.59212 2.919 3.224 3.479 3.690 3.868 4.022 4.159 4.28013 2.792 3.071 3.302 3.493 3.654 3.793 3.915 4.02514 2.685 2.941 3.153 3.328 3.475 3.602 3.713 3.81315 2.593 2.830 3.027 3.187 3.322 3.439 3.541 3.63220 2.272 2.448 2.592 2.709 2.806 2.890 2.963 3.02825 2.078 2.220 2.335 2.429 2.506 2.572 2.629 2.68030 1.947 2.068 2.165 2.243 2.307 2.362 2.410 2.45235 1.851 1.957 2.041 2.109 2.165 2.212 2.254 2.29040 1.778 1.872 1.947 2.008 2.057 2.099 2.135 2.167

equal width .05 is used. Linear interpolations in s are used to approximate the value of function hr (s) at any point ifrequired. The computation by using one-dimensional grid of values s on the interval [0, y] of equal width .025 is alsotried and difference in the critical points noticed at third decimal places only.

Remark. When �i is known, one may use Si = 1�∑n

j=1

(Xij − �i

)and � = n in the above discussion.


Table 4Values of two-sided critical points ck,h,�,� when h = 2, � = .05

� k

3 4 5 6 7 8 9 10

3 5.319 6.957 8.517 9.942 11.247 12.452 13.574 14.624 4.141 5.169 6.107 6.937 7.677 8.347 8.961 9.5285 3.516 4.263 4.926 5.501 6.007 6.459 6.868 7.2436 3.125 3.713 4.226 4.664 5.046 5.383 5.687 5.9637 2.856 3.343 3.761 4.116 4.422 4.691 4.931 5.1498 2.659 3.075 3.430 3.728 3.984 4.207 4.406 4.5869 2.507 2.873 3.181 3.439 3.658 3.850 4.020 4.173

10 2.387 2.713 2.987 3.214 3.407 3.575 3.723 3.85611 2.289 2.584 2.831 3.034 3.207 3.356 3.488 3.60612 2.207 2.478 2.702 2.887 3.043 3.178 3.297 3.40313 2.137 2.388 2.594 2.764 2.907 3.030 3.138 3.23414 2.077 2.311 2.502 2.659 2.791 2.904 3.004 3.09215 2.025 2.244 2.423 2.569 2.691 2.796 2.888 2.97020 1.839 2.008 2.144 2.254 2.346 2.424 2.492 2.55325 1.723 1.863 1.974 2.064 2.139 2.202 2.257 2.30530 1.642 1.763 1.859 1.935 1.998 2.052 2.099 2.14035 1.582 1.689 1.774 1.841 1.897 1.943 1.984 2.02040 1.536 1.632 1.708 1.769 1.818 1.860 1.897 1.928


� k

4 5 6 7 8 9 10

3 13.199 15.160 17.369 19.564 21.646 23.628 25.4974 8.717 9.744 10.870 11.960 12.972 13.921 14.8065 6.685 7.354 8.073 8.760 9.397 9.981 10.5196 5.538 6.029 6.547 7.039 7.489 7.900 8.2787 4.806 5.190 5.593 5.970 6.317 6.629 6.9158 4.298 4.611 4.940 5.246 5.526 5.777 6.0079 3.922 4.190 4.466 4.724 4.958 5.167 5.358

10 3.636 3.867 4.106 4.329 4.529 4.710 4.87311 3.408 3.613 3.825 4.020 4.196 4.353 4.49612 3.224 3.408 3.598 3.771 3.928 4.067 4.19413 3.071 3.238 3.409 3.567 3.708 3.834 3.94814 2.941 3.095 3.251 3.395 3.524 3.638 3.74215 2.830 2.972 3.117 3.249 3.367 3.473 3.56720 2.448 2.552 2.658 2.753 2.839 2.914 2.98225 2.220 2.304 2.388 2.464 2.532 2.591 2.64430 2.068 2.138 2.208 2.272 2.328 2.378 2.42235 1.957 2.018 2.079 2.135 2.183 2.226 2.26440 1.872 1.927 1.981 2.030 2.073 2.111 2.145

4.1. Simulated study

Gill and Dhawan (1999) calculated the critical points for the special case h=1 by simulation. To assess the accuracyof the critical points computed by Gill and Dhawan (1999) and by the recursive method here for h= 1, we have carriedout a simulation study. A fresh random sample of size n was generated from the ith exponential distribution with pdff (x|�)= 1

� exp(−x/�) in each repetition, i=1, . . . , k. We have computed the proportion of times out of 105 repetitions,say �̂, the test statistic Wh, defined in (2.1) for h = 1, exceeds the corresponding critical points of our and of Gill andDhawan’s (1999). The results are given below for n = 8, 12, 15.



� k

4 5 6 7 8 9 10

3 6.956 8.084 9.339 10.576 11.756 12.875 13.9354 5.169 5.849 6.585 7.293 7.956 8.574 9.1495 4.263 4.744 5.257 5.744 6.193 6.607 6.9906 3.713 4.086 4.479 4.847 5.184 5.493 5.7767 3.343 3.648 3.966 4.262 4.532 4.777 5.0018 3.075 3.334 3.602 3.850 4.075 4.278 4.4639 2.873 3.098 3.330 3.543 3.736 3.910 4.068

10 2.713 2.913 3.118 3.306 3.475 3.627 3.76511 2.584 2.765 2.948 3.117 3.268 3.403 3.52512 2.478 2.642 2.809 2.961 3.098 3.220 3.33013 2.388 2.539 2.692 2.832 2.957 3.068 3.16814 2.311 2.451 2.593 2.722 2.837 2.939 3.03115 2.244 2.375 2.507 2.627 2.734 2.829 2.91420 2.008 2.108 2.208 2.298 2.378 2.448 2.51125 1.863 1.945 2.027 2.100 2.164 2.221 2.27230 1.763 1.834 1.903 1.966 2.020 2.069 2.11135 1.689 1.752 1.813 1.868 1.916 1.958 1.99540 1.632 1.689 1.744 1.793 1.836 1.873 1.907

Table 7Estimated value of �̂ when true value of � is .05

n k

3 4 5 6 7 8 9

8 .0493 .0496 .0499 .0497 .0492 .0489 .05018 .0490 .0507 .0488 .0498 .0489 .0474 .0498

12 .0507 .0502 .0503 .0483 .0488 .0493 .050112 .0487 .0491 .0493 .0461 .0484 .0496 .0498

15 .0501 .0510 .0502 .0505 .0501 .0494 .049515 .0465 .0482 .0458 .0483 .0501 .0495 .0477

In the Table 7 the underlined values are the estimated values of � corresponding to Gill and Dhawan’s (1999) criticalpoints. The other values are corresponding to our critical points based on recursive integration. From the table, it is clearthat both tests maintain the Type-I error closely at level � = .05 and, in general, our points seem to be more accuratethan Gill and Dhawan’s (1999).

5. Applications to normal and inverse Guassian probability models

In this section we point out two situations where the inferential method of this paper can readily be applied.

5.1. Application to normal variances

Let observations from the ith population follow a normal distribution with mean �i and variance �i =�2i , i=1, . . . , k.

Fujino (1979) studied various tests for mean homoscedasticity when the variances are known to be in non-decreasingorder. Mudholkar et al. (1993) proposed robust finite-intersection tests for homogeneity of ordered variances. The

minimum variances unbiased estimator of �2i is �̂2

i = ∑nj=1

(Xij − X̄i

)2/�, where X̄i = ∑n

j=1Xij /n, i = 1, . . . , k and


�=n− 1. Here ��̂2i /2�2 follows gamma distribution with shape parameter �/2 and scale parameter unity, i = 1, . . . , k.

For testing null hypothesis H0 : �21 = · · · = �2

k against the alternative Ha : �21 � · · · ��2

h � · · · ��2k with at least one

strict inequality, the test statistic given in (2.1) with Si = �̂2i can readily be used.

5.2. Application to inverse Gaussian probability model

Let observations Xi1, . . . , Xin from the ith population follow a two-parameter inverse Gaussian distribution withprobability density function

fi

(x|�i , �i

) ={

�i

2�x3

}1/2

exp

{− �i

2�2i x

(x − �i

)2

}, x > 0, �i , �i > 0,

where �i is the location parameter and �i is the scale-like parameter, i = 1, . . . , k. The two parameter inverseGaussian distribution is widely used for modelling right-skewed data. When testing for mean homogeneity of in-verse Gaussian distributions, homogeneity of scale-like parameters �i’s is an important assumption in the analysis ofreciprocal (Chhikara and Folks, 1989). Natarajan et al. (2005) proposed methods for testing homogeneityof the scale-like parameters �i from k independent inverse Guassian populations subject to order restrictions.

The rescaled MLE’s of �−1i given by Si = ∑n

j=1

(1/Xij − 1/Xi

)/�, where Xi = ∑n

j=1Xij /n, and � = n − 1. Notethat ��iSi/2 follows a gamma distribution with shape parameter �/2 and scale parameter unity. Hence the test statisticsgiven in (2.1) with Si = ∑n

j=1

(1/Xij − 1/Xi

)/� can readily be used for testing null hypothesis H0 : �−1

1 = · · · = �−1k

against the alternative Ha : �−11 � · · · �−1

h � · · · ��−1k with at least one strict inequality.

Acknowledgements

The authors gratefully acknowledges the helpful suggestions of the referees and the associate editor, which led to anappreciable improvement in the quality of the paper.

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A test against an umbrella ordered alternative