COMPARISON OF RANDOMIZATION METHOD FOR RV COEFFICIENT AND PERMUTATION METHOD FOR HOTELLING T-SQUARE

SOP TRANSACTIONS ON STATISTICS AND ANALYSISISSN(Print): 2373-843X ISSN(Online): 2373-8448

Volume 1, Number 2, July 2014

SOP TRANSACTIONS ON STATISTICS AND ANALYSIS

Comparison of Randomization method forRV coefficient and Permutation Method forHotelling T-SquaredCharles Okechukwu Aronu1*, Godday Uwawunkonye Ebuh2, Cecilia NchedoOkoli3, Rose Chinwe Nwosu4

1 Department of Statistics, Anambra State University, Uli, Nigeria2 Monetary & policy department, Central bank of Nigeria, Abuja, Nigeria3 Department of Statistics, Anambra State University, Uli, Nigeria4 Department of Statistics, Nnamdi Azikiwe University, Awka, Nigeria

*Corresponding author: [email protected]

Abstract:This study compared the permutation method for Hotelling T-squared and the randomizationmethod for RV coefficient in terms of relative efficiency of the test statistic value. The sourceof data employed was simulation form distributions such as Cauchy distribution and Normaldistribution. Also, secondary data was used to evaluate the methods. The findings of this studyrevealed that the randomization method for RV coefficient performed better than the permutationmethod for Hotelling T-squared test in terms of relative efficiency of the test statistic value forboth Cauchy distribution and Normally distributed data since the standard deviation of the teststatistic measure for the randomization test for RV coefficient (Standard deviation=0.22 and 0.16respectively) is lesser than that of the permutation method for Hotelling T-squared test (standarddeviation= 1.09 and 0.39 respectively). This result implies that the variation on the test statisticvalue for the randomization method for RV coefficient is smaller than that of the permutationmethod for Hotelling T-squared test; hence the randomization method for RV coefficient wasconcluded to be the most relatively efficient method between the two methods under observation.The result of applying the permutation method for Hotelling T-squared analysis for measuringthe equality in weight of two groups of broiler birds showed that the reference value of thetest statistic measure T 2 was 6.04 and a corresponding P-value of 0.00 which falls on therejection region of the hypothesis assuming a 95% confidence level. Also, result of applying therandomization method for RV coefficient for measuring the equality in weight of two groups ofbroiler birds gave a test statistic value of 0.99 and a corresponding p-value of 0.00 which fall onthe rejection region of the hypothesis assuming a 95% confidence level. This result implies thatthere exists evidence of statistical significance in the weight of the two groups of broiler birdsusing the two methods.

Keywords:Randomization; Permutation; Cauchy distribution; Normal distribution; Relative efficiency

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1. INTRODUCTION

Biologists, Engineers, Economist, Accountants, and Ecologists are faced with increasingly complexcircumstances for the statistical analysis of data [1]. In experimental and observational studies, theassumptions that errors are independent and identically distributed as normal random variables withcommon variance and an expectation of zero, as required by traditional statistical methods, are no longergenerally considered realistic in many practical situations [2–4]. The traditional approach often relies onthe assumptions that the statistical distribution of a test statistic, such as t, χ2, or F, that is known under aspecified null hypothesis, for the calculation of a probability (a P- value), commonly relying on tabulatedvalues. An alternative to this traditional approach that does not rely on such strict assumptions is to use apermutation test. A permutation test calculates the probability of getting a value equal to or more extremethan an observed value of a test statistic under a specified null hypothesis by recalculating the test statisticafter random re-orderings (shuffling) of the data. Such tests are computationally intensive, however, andthe use of these tests as opposed to the traditional normal-theory tests did not receive much attention in thenatural and behavioral sciences until much later, with the emergence of widely accessible computer power[5–7]. The basic idea behind permutation methods is to generate a reference distribution by recalculatinga statistic for many permutations of the data. Ref [8], reported that the statistician does not carry out thisvery simple and very tedious process, but his conclusions have no justification beyond the fact that theyagree with those which could have been arrived at by this elementary method. Interestingly, the probabilitybasis for statistical inference in this area of statistics can largely be grouped into two probability models:where available subjects are randomly assigned to treatments and where subjects are randomly sampledfrom some population(s). Ref [9], called the former situation the randomization model and the latterthe population model. Ref [5], pointed out that the procedures used under the randomization model arecommonly called randomization tests or randomization intervals, while the same procedures used underthe population model are called permutation tests or permutation intervals. He added that the distinctionbetween the two models is often overlooked, and the terms “randomization” and “permutation” are oftenused interchangeably. The distinction in terminology may not be that important, rather, understandingthe underlying probability model is most important. Ref [10], in his study reported that the sole basis forinference in the randomization model is the random assignment of available subjects to treatment groups.It is not necessary to have random sampling from some population with a specified distribution. Strictlyspeaking, normal theory methods are not appropriate since their distribution theory depends on randomsampling. The consequence of this is that any inferences in the randomization model are limited to thesubjects in the study. However, [11], noted that in the permutation model, an expected distribution of agiven association statistic is generated by repeatedly randomizing one of the initial distance matrices andcalculating a null distribution of expected values from the randomly generated inter-matrix associations.Instead of assuming an asymptotically normal distribution, the observed statistic is contrasted with theexpected distribution and evaluated according to the probability of obtaining as extreme a measure ofassociation. The most convincing reason to choose the permutation test instead of the other test statistic isbecause the p-value will be exact instead of approximated, thus yielding a more accurate prediction of howrandom a given result is ([10–13]). Ref [14], in his study tested whether there are nodes with levels lowerthan would be expected from dendograms constructed on random permutations of the data. Ref [15] , intheir study used a permutation method known as Mantel test analysis to measure the extent of monotonicresemblance between symptoms of occupational stress experienced by traders and public civil workersin Nigeria. Also, they tested the equality of factors influencing traders and public civil workers stress.The result of their study revealed that there exist a positive resemblance on the symptoms of occupationalstress observed by traders and public civil Workers with an association of 59%. They equally found that

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Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared

there exist a strong positive association on the factors influencing stress level between traders and publiccivil workers with a resemblance of 70%. The result of the Mantel test enabled them to concluded that thesymptoms of occupational stress, factors influencing stress level and stress coping skills experienced bytraders and public civil workers has a positive resemblance. In their own contribution, [16] employed thepermutation method for Hotelling T-squared analysis in assessing the knowledge, attitude and factorsaffecting team building activities amongst health workers in Nigeria. The result of the permutation methodfor Hotelling T-squared analysis gave a test statistic value of 8073.7 and a p-value of 0.00 for 10,000permutations which implies rejection of the null hypothesis assuming a 95% confidence level. However,the present study seeks to determine between the Randomization method and the permutation method forHotelling T-Squared the method that is relatively efficient. The result from this study will be significantsince there exists little or no literatures on evaluating between the randomization and the permutation testfor Hotelling T- Square the test that is relatively efficient.

2. METHODOLOGY AND MATERIAL

2.1 Source of Data

The source of data for this study was primary and secondary source of data collection. The primarysource was from simulation from Cauchy distribution and Normal distribution for sample sizes 5, 10, 20,30, 40 and 50. The Secondary data was obtained from a journal publication.

2.2 Randomization Method for RV Coefficient

According to [17], a complete description of association preference of the randomization methodinclude measures of: (1) the observed level of association, (2) the level of association expected under thenull hypothesis of random association, (3) a measure of how much the observed association level deviatesfrom the expected, and (4) the probability under the null hypothesis of obtaining a value as extreme asthat observed. To obtain all but the first of these, he added that one must specify the frequency distributionof the association index under the null hypothesis. This can be generated by repeatedly randomizing thedata using Monte Carlo simulation.

The RV coefficient is a similarity coefficient between positive semi-definite matrices [18]. This is theresemblance coefficient between two cross-product matrices measure about the same object. Ref [19],noted that the RV coefficient had important mathematical properties because it can be shown that mostmultivariate analysis techniques amount to maximizing this coefficient with suitable constraints. The RVcoefficient considers that two sets of variables are correlated relative to the position of the samples inone set and similar to the relative position of the samples in the other set. The matrices representing therelative positions of the samples are the cross product matrices: WA =AAT and WB= BBT . They are ofsize n X n and so can be compared directly. To measure their proximity, the inner product between thematrices is computed:

<WA,WB >= tr(AAT BBT ) =p

∑l=1

q

∑m=1

Cov2(A.l ,B.m) (1)

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Since these two matrices may have different norms, a generalized correlation coefficient is computedby normalizing by the matrix norms. Hence, the RV coefficient can be expressed as given

RV (A,B) =<WA,WB >

‖WA‖×‖WB‖=

tr(AAT BBT )√tr(AAT )2× tr(BBT )2

(2)

This represents the cosine of the angle between the two vectors representing the cross-product matrices.For convenient sake RV can be written in a form that can help to better understand its properties, forinstance using the covariance matrices:

RV (A,B) =tr(SABSBA)√

tr(S2AA)× tr(S2

BB)(3)

Where SAB=AT B is the empirical covariance matrix between A and B.

However, the idea of using randomization methods to ascertain a link between two sets of variables isthe earliest instances of multi- tables resemblance testing

2.3 The testing procedure of the randomization method for RV coefficient

1. Considering two symmetric resemblance matrices (similarities) A andB, of size (n×n), whose rowsand columns correspond to the same set of objects. Compute the RV coefficient RV (AB) as thereference value in test using Equation (3).

2. Randomize the rows and corresponding columns of one of the matrices, sayA, obtaining a random-ized matrixA∗. This procedure is called ‘randomization of matrix’.

3. Compute the RV coefficient RV (A∗B) between matricesA∗andB, obtaining a value RV∗of the teststatistic under randomization.

4. Repeat steps 2 and 3 a large number of times to obtain the distribution of RV∗under randomization;then, add the reference value RV (AB) to the distribution.

5. For a one – tailed test involving the upper tail (i.e., H1+: distances in matrices A and B are positivelycorrelated), calculate the probability (p – value) as the proportion of values RV∗greater than orequal toRV (AB). For a test in the lower tail, the probability is the proportion of values RV∗smallerthan or equal toRV (AB).

Computing the exact randomization distribution is computationally costly when n > 15which makes itdifficult for users who will wish to obtain a stabilized distribution under randomization which is when thenumber of randomization increases to 10, 000 to 50, 000 randomizations. To overcome this challenge,RV coefficient can be implemented in r-programming package using the library ade4 o call upon the RV.randtest function [20].

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2.4 Permutation Method for Hotelling T 2

The permutation test is a conditional test since it generates the permutation distribution conditional onthe observed values of the random variables (unlike the randomization model where the observed valueswere not random, only their treatment assignments). The test is also conditionally distribution-free since,conditional on the observed data, the permutation distribution of test statistic does not depend on thepopulation distributions. Also, the permutation test is conditionally exact for the same reasons that therandomization test is exact, but it is also unconditionally exact since the probability of a Type I error iscontrolled for all possible samples [10, 16]. The permutation method for Hotelling T 2measures whetherthe mean vectors of the two groups differ significantly. One measure of the difference in mean vector isthe two-sample Hotelling T 2 statistic which is given as

T 2 =n1n2

n1 +n2(X− Y )S−1(X− Y ) (4)

The pooled estimate of the variance-covariance matrix S is given as

S =A+B

n1 +n2−2(5)

Where, X and Y are the mean vector of the two samples, A and B are the covariance matrix of the twosamples and n1 and n2 are the samples size of the two samples.

2.5 The Testing Procedure of the Permutation Method for Hotelling T 2

1. Consider two independent random samples {Xi, i = 1,2, · · · ,n1} and {Yi, i = 1,2, · · · ,n2}. To testthe hypothesis that both samples came from the sample distribution we shall obtain the mean vectorof the two samples X and Y .

2. We obtain the reference value of the test statistic, T 2 using Equation (4) and Equation (5)

3. Permute at random the rows and corresponding columns of the covariance matrix of one of thesamples; A.

4. Compute the test statisticT 2∗i , obtaining a value T 2∗of the test statistic under permutation.

5. Repeat steps 2 and 3 a large number of times to obtain the distribution of T 2∗under permutation;add the reference value T 2to the distribution.

6. For a one-tailed test involving the upper tail, calculating the probability (P-value) as the proportionof values T 2∗greater than or equal toT 2. Conversely, for a test in the lower tail, the probability isthe proportion of values T 2∗less than or equal toT 2.

According to [12], the distribution under permutation stabilizes when the number of permutationincreases to 10, 000 to 50, 000 permutations.

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3. DATA ANALYSIS AND RESULT

The two methods (Randomization method for RV coefficient and the Permutation method for HotellingT –Squared) discussed in the methodology shall be evaluated in this section using Cauchy distributed dataand Normally distributed data for samples 5, 10, 20, 30, 40 and 50.

3.1 Permutation method for Hotelling T-squared Analysis and Randomization methodfor RV coefficient Analysis using Cauchy Distributed data Set

In putting data generated using Cauchy distribution for sample size 5 on the R 2.13.0 command window,where matrix A and B are the objects or samples to be measured. The function hotelling.test was employedto call the Hotelling T-Squared test for 10, 000 permutations.

R>A=matrix(c(0.34, 1.83, -1.31, 0.83, -4.57, 0.58, 3.63, 2.78, 0.85, 1.06, 0.20, -3.40, -0.71, 0.51, 2.10,4.04, 2.88, 0.60, 0.63, -0.40, 0.14, 0.87, 0.06, 0.06, 1.34), nrow=5, byrow=TRUE)

R>B=matrix(c(0.25, -2.55, -4.23, 1.84, 0.10, 0.11, -0.61, -0.82, 1.03, 3.14, -1.94, -0.36, -1.61, 2.58,0.06, -0.25, -0.63, 0.55, 3.34, 2.36, 6.61, -0.64, 1.23, 1.30, 0.45), nrow=5, byrow=TRUE)

R >Test=hotelling.test(A, B, shrinkage = FALSE, perm = TRUE, B= 10000)

The result of the Hotelling T-squared test statistic for 10,000 permutations was obtained as given:

Test stat: 3.5025

Numerator df: 5

Denominator df: 4

Permuation P-value: 0.1497

Number of permutations : 10000

In putting data generated using Cauchy distribution for sample size 5 on the R 2.13.0 command window,where matrix A and B are the objects or samples to be measured. Also, it should be noted that objects A1,A2, A3, A4 and A5 are objects of matrix A while objects B1, B2, B3, B4,and B5 are objects of matrix B.The class distance of matrices A and B based on the canonical measure is labeled DA and DB respectively.The function RVdist.randtest was employed to call the randomization method for RV coefficient for 10,000 randomizations.

R>A1=c(0.34, 0.58, 0.20, 4.04, 0.14)

R>A2=c(1.83, 3.63, -3.40, 2.88, 0.87)

R>A3=c(-1.31, 2.78, -0.71, 0.60, 0.06)

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R>A4=c(0.83, 0.85, 0.51, 0.63, 0.06)

R>A5=c(-4.57, 1.06, 2.10, -0.40, 1.34)

R>B1=c(0.25, 0.11, -1.94, -0.25, 6.61)

R>B2=c(-2.55, -0.61, -0.36, -0.63, -0.64)

R>B3=c(-4.23, -0.82, -1.61, 0.55, 1.23)

R>B4=c(1.84, 1.03, 2.58, 3.34, 1.30)

R>B5=c(0.10, 3.14, 0.06, 2.36, 0.45)

R>A=matrix(c(A1, A2, A3, A4, A5), nrow=5, byrow=TRUE)

R>B=matrix(c(B1, B2, B3, B4, B5), nrow=5, byrow=TRUE)

R>DA<-dist.quant(A, method = 1)

R> DB<-dist.quant(B, method = 1)

R>RVdist.randtest(DA, DB, nrepet = 10000)

The result of the Randomization method for RV coefficient statistic for 10,000 randomizations wasobtained as given:

Monte-Carlo test

Call: as.randtest(sim = res[-1], obs = obs)

Observation: 0.4426036

Based on 10000 replicates

Simulated p-value: 0.7086291

Alternative hypothesis: greater

Std.Obs Expectation Variance

-0.52447773 0.52001051 0.02178237

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In putting data generated using Cauchy distribution for sample size 10 on the R 2.13.0 commandwindow, where matrix A and B are the objects or samples to be measured. The function hotelling.test wasemployed to call the Hotelling T-Squared test for 10, 000 permutations.

R>A=matrix(c(-11.93, 1.30, -0.51, 3.11, -0.81, -1.31, -0.18, 3.19, -0.82, -0.18, 0.17, 0.32, 2.74, 0.50,3.86, -0.56, -0.10, 2.42, -1.52, -0.98, -1.03, -0.94, 0.02, 0.76, -1.95, 2.74, 0.72, 1.65, -7.16, 0.76, -1.69,-0.33, -5.47, -1.94, -1.94, 0.31, 3.46, -1.25, -12.98, -12.98, -1.68, -1.22, 1.92, -0.01, -0.01, -1.97, -0.73,-0.41, -0.65, -0.65), nrow=10, byrow=TRUE)

R>B=matrix(c(-0.56, 0.13, -0.05, -0.08, 0.87, 1.91, -0.66, -7.86, 1.16, -2.30, -2.64, -1.75, -1.53, -2.79,0.03, -0.76, 0.52, -40.88, 6.1, -0.68, 0.82, 0.75, 0.21, -0.51, -0.62, -2.18, 0.78, -1.09, -0.72, 14.60, 0.72,0.80, -2.20, -0.22, 1.41, 0.93, 0.52, 0.27, 4.43, 1.22, -0.27, -0.21, 0.58, 1.32, 1.63, -1.55, -3.02, -1.22,17.95, 0.50), nrow=10, byrow=TRUE)

R>Test=hotelling.test(A, B, shrinkage = FALSE, perm = TRUE, B= 10000)


Test stat: 1.9366

Numerator df: 5

Denominator df: 14



In putting data generated using Cauchy distribution for sample size 10 on the R 2.13.0 commandwindow, where matrix A and B are the objects or samples to be measured. Also, it should be noted thatobjects A1, A2, A3, A4 and A5 are objects of matrix A while objects B1, B2, B3, B4,and B5 are objectsof matrix B. The class distance of matrices A and B based on the canonical measure is labeled DA andDB respectively. The function RVdist.randtest was employed to call the randomization method for RVcoefficient for 10, 000 randomizations.

R>A1=c(-11.93, -1.31, 0.17, -0.56, -1.03, 2.74, -1.69, 0.31, -1.68, -1.97)

R>A2=c(1.30, -0.18, 0.32, -0.10, -0.94, 0.72, -0.33, 3.46, -1.22, -0.73)

R>A3=c(-0.51, 3.19, 2.74, 2.42, 0.02, 1.65, -5.47, -1.25, 1.92, -0.41)

R>A4=c(3.11, -0.82, 0.50, -1.52, 0.76, -7.16, -1.94, -12.98, -0.01, -0.65)

R>A5=c(-0.81, -0.18, 3.86, -0.98, -1.95, 0.76, -1.94, -12.98, -0.01, -0.65)

R>B1=c(-0.56, 1.91, -2.64, -0.76, 0.82, -2.18, 0.72, 0.93, -0.27, -1.55)44


R>B2=c(0.13, -0.66, -1.75, 0.52, 0.75, 0.78, 0.80, 0.52, -0.21, -3.02)

R>B3=c(-0.05, -7.86, -1.53, -40.88, 0.21, -1.09, -2.20, 0.27, 0.58, -1.22)

R>B4=c(-0.08, 1.16, -2.79, 6.10, -0.51, -0.72, -0.22, 4.43, 1.32, 17.95)

R>B5=c(0.87, -2.30, 0.03, -0.68, -0.62, 14.60, 1.41, 1.22, 1.63, 0.50)




R>DB<-dist.quant(B, method = 1)


The result of the Randomization method statistic for 10,000 randomizations was obtained as given:

Monte-Carlo test







-0.80701939 0.46562060 0.01702808


R>A=matrix(c(-1.28, -0.77, -0.14, -0.27, -1.26, 0.90, -0.52, 3.39, -10.23, -1.54, 0.01, -2.00, 0.42, 0.15,-18.45, -5.98, 0.72, -0.72, 1.91, 3.40, 3.34, -1.45, -0.63, 7.83, 5.17, -0.20, -1.04, 0.63, 1.00, -0.38, -4.54,1.65, 0.91, -0.37, 6.46, 0.80, 1.34, -2.77, 1.96, 0.91, -4397.36, -2.86, -0.03, -0.19, -0.58, 2.81, 0.13, 1.30,

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0.18, 6.37, 0.30, -0.25, -0.73, 1.07, -1.16, -0.29, -1.06, 1.36, 0.63, -2.02, -0.01, 0.07, 2.14, -4.58, -0.3,1.22, -0.87, 0.26, 49.28, -1.14, 0.94, -0.90, 1.66, 1.32, -1.00, -3.20, 0.61, 0.63, 1.20, -4.05, 2.94, -0.50,-0.33, -1.35, -6.79, 1.34, -0.81, 0.57, -0.46, 0.46, -0.75, -1.67, 2.61, 2.36, 0.13, -1.35, -90.18, 0.32, 4.84,0.96), nrow=20, byrow=TRUE)

R>B=matrix(c(1.39, -0.11, -0.11, 0.86, -0.19, 11.49, -2.86, -2.88, 0.33, 0.16, -3.58, 1.29, 1.07, 2.48,4.24, -1.07, -1.39, -0.12, -1.65, -3.80, -0.96, 1.23, -0.30, 1.57, 0.53, -7.24, -24.26, 0.66, 0.64, -1.83, -2.83,-1.75, 1.81, 0.59, 0.11, -2.11, 0.55, -1.14, -0.41, 0.81, -7.29, -0.33, 3.63, 0.03, -0.19, 2.04, 0.39, 0.81,0.86, -0.15, -2.72, -1.00, 0.20, -0.16, 2.72, -4.81, -0.05, -0.01, -0.13, -2.10, -2.28, 0.54, -0.49, -1.62, 1.72,-6.45, 8.59, -0.46, 4.80, 0.78, 0.16, 0.96, -0.50, -0.11, 0.44, 0.64, 0.55, -0.04, 0.19, 2.47, -0.16, 0.11, 1.72,-0.73, 0.73, 1.73, -1.62, 17.81, 0.35, 129.86, 5.28, -0.23, -0.88, 0.59, -1.99, -2.85, 0.43, 1.32, 0.46, -0.73),nrow=20, byrow=TRUE)



Test stat: 0.94862

Numerator df: 5

Denominator df: 34




R>A1=c(-1.28, 0.90, 0.01, -5.98, 3.34, -0.20, -4.54, 0.80, -4397.36, 2.81, 0.30, -0.29, -0.01, 1.22, 0.94,-3.20, 2.94, 1.34, -0.75, -1.35)

R>A2=c(-0.77, -0.52, -2.00, 0.72, -1.45, -1.04, 1.65, 1.34, -2.86, 0.13, -0.25, -1.06, 0.07, -0.87, -0.90,0.61, -0.50, -0.81, -1.67, -90.18)

R>A3=c(-014, 3.39, 0.42, -0.72, -0.63, 0.63, 0.91, -2.77, -0.03, 1.30, -0.73, 1.36, 2.14, 0.26, 1.66, 0.63,-0.33, 0.57, 2.61, 0.32)

R>A4=c(-0.27, -10.23, 0.15, 1.91, 7.83, 1.00, -0.37, 1.96, -0.19, 0.18, 1.07, 0.63, -4.58, 49.28, 1.32,1.20, -1.35, -0.46, 2.36, 4.84)

46


R>A5=c(-1.26, -1.54, -18.45, 3.40, 5.17, -0.38, 6.46, 0.91, -0.58, 6.37, -1.16, -2.02, -0.80, -1.14, -1.00,-4.05, -6.79, 0.46, 0.13, 0.96)

R>B1=c(1.39, 11.49, -3.58, -1.07, -0.96, -7.24, -2.83, -2.11, -7.29, 2.04, -2.72, -4.81, -2.28, -6.45, 0.16,0.64, -0.16, 1.73, 5.28, -2.85)

R>B2=c(-0.11, -2.86, 1.29, -1.39, 1.23, -24.26, -1.75, 0.55, -0.33, 0.39, -1.00, -0.05, 0.54, 8.59, 0.96,0.55, 0.11, -1.62, -0.23, 0.43)

R>B3=c(-0.11, -2.88, 1.07, -0.12, -0.30, 0.66, 1.81, -1.14, 3.63, 0.81, 0.20, -0.01, -0.49, -0.46, -0.50,-0.04, 1.72, 17.81, -0.88, 1.32)

R>B4=c(0.86, 0.33, 2.48, -1.65, 1.57, 0.64, 0.59, -0.41, 0.03, 0.86, -0.16, -0.13, -1.62, 4.80, -0.11,0.19, -0.73, 0.35, 0.59, 0.46)

R>B5=c(-0.19, 0.16, 4.24, -3.80, 0.53, -1.83, 0.11, 0.81, -0.19, -0.15, 2.72, -2.10, 1.72, 0.78, 0.44,2.47, 0.73, 129.86, -1.99, -0.73)







Monte-Carlo test






47



-0.4505670 0.2651695 0.1278144


R>A=matrix(c(1.67, -25.15, -0.33, 3.39, 0.20, -0.23, -0.40, -2.18, -3.22, -1.00, -0.75, 0.24, -2.31,-14.76, 0.79, -0.28, -0.65, -0.31, 0.61, 3.69, -1.19, -1.17, -2.26, -94.3, 0.22, 1.87, 18.1, -0.99, -3.12, 1.58,1.23, 2.08, 1.69, -0.11, -2.36, -1.93, 0.74, 0.18, 0.86, -1.40, -0.53, 4.31, 0.43, -1.61, 2.54, 3.04, 0.07, -5.02,-1.24, 0.74, 0.35, 0.43, 2.48, 0.46, -0.22, -0.22, 1.00, 1.57, -4.63, -0.97, -0.61, -0.08, 0.10, 0.19, 1.61, 0.16,0.05, -0.28, -2.52, 1.74, -0.49, 0.18, -2.63, -0.09, -0.03, -2.05, 0.17, 0.24, -2.39, -1.23, 0.17, 1.00, 83.85,-2966.75, 1.26, -1.13, -0.75, 1.22, 0.38, 0.30, 1.14, 0.04, 0.42, 0.25, 0.68, 0.32, 0.25, -22.46, -656.69,2.15, -0.58, 0.21, 0.84, 0.65, -3.93, -1.39, 0.76, 1.26, -1.58, 5.40, -1.22, 0.14, 0.19, 1.56, -1.29, -0.47,-0.39, 1.71, 0.26, -0.36, -1.96, -0.82, -1.26, -0.46, 1.18, 4.01, 0.72, -0.91, 0.55, -1.81, -1.41, -1.57, 3.59,2.10, 0.68, 3.63, -3.62, 0.12, 18.13, -3.07, 1.92, -1.06, -2.52, -0.57, 1.87, 8.48, -0.10, 2.33, -0.67, 0.44 ),nrow=30, byrow=TRUE)

R>B=matrix(c(-0.46, -0.41,-1.23, 1.01, 0.50, -1.66, -0.83, 0.09, 0.34, 1.12, -0.17, -0.30, 0.03, -0.70,1.42, 0.55, 34.52, 8.92, -0.25, 0.58, -0.92, -5.86, -0.11, -2.14, 0.69, -0.92, -37.73, -0.49, 0.69, 0.93, 0.96,-0.94, -31.85, -2.16, -235.05, 1.32, 0.69, 0.61, 0.25, -0.21, 0.30, -3.61, 0, -0.80, 0.38, -1.84, 1.21, 13.33,0.01, 0.64, -12.85, -4.01, 22.62, 2.35, -1.46, 0.37, -1.86, 4.53, -2.86, 1.43, 1.88, 2.81, 2.16, 0.61, -0.47,-1.48, -0.48, -3.55, -0.43, -0.75, 11.65, 1.14, 1.54, 1.51, -1.17, -0.18, 1.02, -0.70, 0.20, -1.46, 1.20, 0.81,0.53, -0.06, -0.33, 0.4, 6.61, 3.03, -0.77, 3.66, 2.33, -2.19, 0.08, -10.07, 3.49, 21.11, 0.20, -0.40, 1.37,0.41, 1.79, 2.97, 2.12, 0.83, 0.11, 1.07, 6, 39.69, 0.79, 1.87, -2.45, -1.43, -3.52, 1.86, -0.13, -0.70, 12.74,0.29, 0.24, 3.63, -3.28, -1.63, 0.33, 0.28, 0.23, 7.49, 1.09, -0.36, -0.02, 0.72, 3.76, 1.51, 0.14, 113.33, 2.48,-1.47, 3.99, 6.6, -3.00, 0.76, -0.54, 3.28, -0.18, -0.39, -0.29, -0.85, 1.41, -9.21, -0.07, -0.20), nrow=30,byrow=TRUE)

R> Test=hotelling.test(A, B, shrinkage = FALSE, perm = TRUE, B= 10000)


Test stat: 1.3247

Numerator df: 5

Denominator df: 54




48


R>A1=c(1.67, -0.23, -0.75, -0.28, -1.19, 1.87, 1.23, -1.93, -0.53, 3.04, 0.35, -0.22, -0.61, 0.16, -0.49,-2.05, 0.17, -1.13, 1.14, 0.32, -0.58, -1.39, -1.22, -0.47, -1.96, 4.01, -1.41, 3.63, 1.92, 8.48)

R>A2=c(-25.15, -0.4, 0.24, -0.65, -1.17, 18.1, 2.08, 0.74, 4.31, 0.07, 0.43, 1.00, -0.08, 0.05, 0.18, 0.17,1.00, -0.75, 0.04, 0.25, 0.21, 0.76, 0.14, -0.39, -0.82, 0.72, -1.57, -3.62, -1.62, -0.1)

R>A3=c(-0.33, -2.18, -2.31, -0.31, -2.26, -0.99, 1.69, 0.18, 0.43, -5.02, 2.48, 1.57, 0.1, -0.28, -2.63,0.24, 83.85, 1.22, 0.42, -22.46, 0.84, 1.26, 0.19, 1.71, -1.26, -0.91, 3.59, 0.12, -2.52, 2.33)

R>A4=c(3.39, -3.22, -14.76, 0.61, -94.3, -3.12, -0.11, 0.86, -1.61, -1.24, 0.46, -4.63, 0.19, -2.52, -0.09,-2.39, -2966.75, 0.38, 0.25, -656.69, 0.65, -1.58, 1.56, 0.26, -0.46, 0.55, 2.1, 18.13, -0.57, -0.67)

R>A5=c(0.2, -1, 0.79, 0.39, 0.22, 1.58, -2.36, -1.4, 2.54, 0.74, -0.22, -0.97, 1.61, 1.74, -0.03, -1.23,-1.26, 0.3, 0.68, 2.15, -3.93, 5.4, -1.29, -0.36, 1.18, -1.81, 0.68, -3.07, 1.87, 0.44)

R>B1=c(-0.46, -1.66, -0.17, 0.55, -0.92, -0.92, 0.96, 1.32, 0.30, -1.84, -12.85, 0.37, 1.88, -1.48, 11.65,-0.18, 1.20, 0.40, 2.33, 21.11, 1.79, 1.07, -2.45, -0.7, -3.28, 7.49, 3.76, -1.47, -0.54, -0.85)

R>B2=c(-0.41, -0.83, -0.3, 34.52, -5.86, -37.73, -0.94, 0.69, -3.61, 1.21, -4.01, -1.86, 2.81, -0.48, 1.14,1.02, 0.81, 6.61, -2.19, 0.2, 2.97, 6, -1.43, 12.74, -1.63, 1.09, 1.51, 3.99, 3.28, 1.41)

R>B3=c(-1.23, 0.09, 0.03, 8.92, -0.11, -0.49, -31.85, 0.61, 0, 13.33, 22.62, 4.53, 2.16, -3.55, 1.54, -0.7,0.53, 3.03, 0.08, -0.4, 2.12, 39.69, -3.52, 0.29, 0.33, -0.36, 0.14, 6.6, -0.18, -9.21)

R>B4=c(1.01, 0.34, -0.7, -0.25, -2.14, 0.69, -2.16, 0.25, -0.8, 0.01, 2.35, -2.86, 0.61, -0.43, 1.51, 0.2,-0.06, -0.77, -10.07, 1.37, 0.83, 0.79, 1.86, 0.24, 0.28, -0.02, 113.33, -3, -0.39, -0.07)

R>B5=c(0.5, 1.12, 1.42, 0.58, 0.69, 0.93, -235.05, -0.21, 0.38, 0.64, -1.46, 1.43, -0.47, -0.75, -1.17,-1.46, -0.33, 3.66, 3.49, 0.41, 0.11, 1.87, -0.13, 3.63, 0.23, 0.72, 2.48, 0.76, -0.29, -0.2)




49





Monte-Carlo test







-0.04104809 0.32172092 0.10560875


R>A=matrix(c(0.67, 0.33, -0.12, 0.37, -1.67, 3.93, 0.8, -0.39, -0.09, 1.51, 0.66, 0.77, -0.06, -0.07,-2.83, -0.66, -1.08, -0.59, -0.05, -1.15, 1.09, 2.34, -1.59, 0.72, 0.17, 0.09, -0.18, 1.38, -13.55, 0.28, 0.18,-3.79, 0.07, -2.78, 1.75, -0.07, 1.23, -10.33, -11.09, 1.36, 0.64, -0.21, -1.30, -6.63, 1.25, -2.14, -1.54, 1.83,-0.97, 2.42, 0.63, -0.81, -20.20, -1.39, 1.55, -0.28, -1.69, 7.63, 0.32, 3.77, -0.37, 2.44, 0.04, 0.32, 1.01,-0.44, 3.12, 0.80, -1.34, -0.81, -8.05, 3.60, -0.14, 0.43, 0.25, 11.23, -14.12, 0.47, -0.73, -833.1, 0.18, -0.21,-10.24, 1.00, -1.74, -0.15, 0.32, -14.43, -0.63, -0.77, 0.59, -1.50, -0.57, -0.86, 5.11, -1.43, 0.76, 0.05, 0.72,0.26, -0.09, -0.01, -1.52, -1.16, -1.46, -3.88, 0.31, -0.48, -0.85, -2.4, 2.75, 2.38, -1.39, 0, -0.4, 0.48, -0.56,0, 1.3, 0.23, 0.98, 3.29, -3.03, 0.37, 0.53, 3.47, 0.7, 11.02, -20.04, -3.65, -2.34, -0.8, -2.04, 0.25, 0.79,-0.1, -1.37, 0.39, 1.01, 228.55, 1.29, 3.02, -0.22, -0.25, 0.94, 0.97, -1.81, -0.09, -11.32, -0.99, 8.33, 2.97,-0.3, 0.35, 0.91, 4.38, -0.67, -0.01, -1.27, 5.74, -0.72, -8.27, -3.43, -16.53, -0.12, 1.59, -7.52, -22.48, -0.74,-3.6, -0.27, 0.62, -3.02, -12.31, 2.65, 0.26, -0.55, 2.59, -0.91, 3.48, 0.58, 31.6, 1.21, -0.95, 1.44, -1.37,0.98, 7.78, 3.69, 2.1, -1.7, 1.01, 0.82, -0.55, 0.18, 2.05, 0.31, 4.64, 3.88, -0.67 ), nrow=40, byrow=TRUE)

R>B=matrix(c(2.01, 2.2, -1.69, -34.54, -0.65, -0.35, -2.49, -0.09, 25.63, 0.6, -2.5, -0.41, -889.36, 3.53,0.87, -0.58, -0.75, 0.68, -0.32, 0.87, 0.86, -1.6, -0.51, -13, 0.27, -3.31, -1.16, 0.22, -0.7, 1.38, -0.85, 1.56,-1.01, 0.77, -6.71, 2.25, -0.2, -0.21, -0.3, -0.53, 0.89, 0.87, 0.27, 2.52, -1.85, 0.79, -0.07, 0.33, -2.96, -0.42,1.37, 0.57, 0.19, 0.49, 2.88, 0.91, 0.8, -3.24, -0.76, -0.36, 0.06, -0.4, -2.76, 0.12, 0.52, 1.81, -6.93, 0.23,0.51, -9.06, -0.63, 2.16, 0.46, -0.46, -0.96, 4.75, -14.21, -0.55, -0.31, -0.52, 0.29, 908.52, -0.48, -0.83,-2.86, 0.24, -1.7, 1.43, -1.19, -0.76, -1.57, -0.71, 8.47, -0.18, -0.67, -2.97, -7.17, -2.32, -26.88, -2.4, -8.87,9.94, 0.71, -6.72, 9.49, -0.46, 1.64, -0.68, -17.5, 2.58, 17.77, 18.08, 3.3, 2.02, 0.28, 7.18, 0.84, -2.69,-0.02, 3.13, 1.61, 2.62, 12.67, -1.88, -0.55, -0.23, -1.57, -2.7, -0.83, 0.07, 5.9, -0.46, 0, 0.02, 0.6, 2.3, 0.27,0.18, 0.96, 2.7, -0.12, 0.39, -0.69, 1.33, -0.26, 1.83, 1.81, 0.28, 0.15, -120.58, 0.43, 0.21, -0.1, 0.01, -8.23,-0.43, 0.45, 0.1, 0.12, 0.91, -0.19, -6.88, -0.03, 0.17, -0.4, 0.31, -0.6, -1.38, 0.19, -3.84, -2.42, -0.07, 0.35,-1.9, -2.62, 0.85, -5.01, 0.33, -5.25, 1.4, -1.27, -0.79, 2.29, -1.5, 0.37, 1.12, 1.48, 0.58, 0.87, -1.72, -0.59,-0.53, -2.83, -0.3, -1.85, -0.5, -0.17, 0.71, -0.59, 24.03), nrow=5, byrow=TRUE)

50




Test stat: 0.44659

Numerator df: 5

Denominator df: 74



In putting data generated using Cauchy distribution for sample size 40 on the R 2.13.0 commandwindow, where matrix A and B are the objects or samples to be measured. Also, it should be noted thatobjects A1, A2, A3, A4 and A5 are objects of matrix A while objects B1, B2, B3, B4,and B5 are objectsof matrix B. The class distance of matrices A and B based on the canonical measure is labelled DA andDB respectively. The function RVdist.randtest was employed to call the randomization method for RVcoefficient for 10, 000 randomizations.

R>A1=c(0.67, 3.93, 0.66, -0.66, 1.09, 0.09, 0.18, -0.07, 0.64, -2.14, 0.63, -0.28, -0.37, -0.44, -8.05,11.23, 0.18, -0.15, 0.59, -1.43, -0.09, -3.88, 2.75, 0.48, 0.98, 3.47, -2.34, -0.1, 1.29, 0.97, 8.33, 4.38, -0.72,1.59, -0.27, 0.26, 0.58, -1.37, -1.7, 2.05)

R>A2=c(0.33, 0.8, 0.77, -1.08, 2.34, -0.18, -3.79, 1.23, -0.21, -1.54, -0.81, -1.69, 2.44, 3.12, 3.6,-14.12, -0.21, 0.32, -1.5, 0.76, -0.01, 0.31, 2.38, -0.56, 3.29, 0.7, -0.8, -1.37, 3.02, -1.81, 2.97, -0.67, -8.27,-7.52, 0.62, -0.55, 31.6, 0.98, 1.01, 0.31)

R>A3=c(-0.12, -0.39, -0.06, -0.59, -1.59, 1.38, 0.07, -10.33, -1.3, 1.83, -20.2, 7.63, 0.04, 0.8, -0.14,0.47, -10.24, -14.43, -0.57, 0.05, -1.52, -0.48, -1.39, 0, -3.03, 11.02, -2.04, 0.39, -0.22, -0.09, -0.3, -0.01,-3.43, -22.48, -3.02, 2.59, 1.21, 7.78, 0.82, 4.64)

R>A4=c(0.37, -0.09, -0.07, -0.05, 0.72, -13.55, -2.78, -11.09, -6.63, -0.97, -1.39, 0.32, 0.32, -1.34,0.43, -0.73, 1, -0.63, -0.86, 0.72, -1.16, -0.85, 0, 1.3, 0.37, -20.04, 0.25, 1.01, -0.25, -11.32, 0.35, -1.27,-16.53, -0.74, -12.31, -0.91, -0.95, 3.69, -0.55, 3.88)

R>A5=c(-1.67, 1.51, -2.83, -1.15, 0.17, 0.28, 1.75, 1.36, 1.25, 2.42, 1.55, 3.77, 1.01, -0.81, 0.25,-833.1, -1.74, -0.77, 5.11, 0.26, -1.46, -2.4, -0.4, 0.23, 0.53, -3.65, 0.79, 228.55, 0.94, -0.99, 0.91, 5.74,-0.12, -3.6, 2.65, 3.48, 1.44, 2.1, 0.18, -0.67)

R>B1=c(2.01, -0.35, -2.5, -0.58, 0.86, -3.31, -0.85, 2.25, 0.89, 0.79, 1.37, 0.91, 0.06, 1.81, -0.63, 4.75,0.29, 0.24, -1.57, -2.97, -8.87, -0.46, 17.77, 7.18, 1.61, -0.23, 5.9, 2.3, -0.12, 1.83, 0.43, -0.43, -0.19, 0.31,-2.42, 0.85, -1.27, 1.12, -0.59, -0.5)

R>B2=c(-2.2, -2.49, -0.41, -0.75, -1.6, -1.16, 1.56, -0.2, 0.87, -0.07, 0.57, 0.8, -0.4, -6.93, 2.16, -14.21,908.52, -1.7, -0.71, -7.17, 9.94, 1.64, 18.08, 0.84, 2.62, -1.57, -0.46, 0.27, 0.39, 1.81, 0.21, 0.45, -6.88,-0.6, -0.07, -5.01, -0.79, 1.48, -0.53, -0.17)

R>B3=c(1.69, -0.09, -889.36, 0.68, -0.51, 0.22, -1.01, -0.21, 0.27, 0.33, 0.19, -3.24, -2.76, 0.23, 0.46,-0.55, -0.48, 1.43, 8.47, -2.32, 0.71, -0.68, 3.3, -2.69, 12.67, -2.7, 0, 0.18, -0.69, 0.28, -0.1, 0.1, -0.03,-1.38, 0.35, 0.33, 2.29, 0.58, -2.83, 0.71)

R>B4=c(-34.54, 25.63, 3.53, -0.32, -13, -0.7, 0.77, -0.3, 2.52, -2.96, 0.49, -0.76, 0.12, 0.51, -0.46,-0.31, -0.83, -1.19, -0.18, -26.88, -6.72, -17.5, 2.02, -0.02, -1.88, -0.83, 0.02, 0.96, -1.33, 0.15, 0.01, 0.12,0.17, 0.19, -1.9, -5.25, -1.5, 0.87, -0.3, -0.59)

51


R>B5=c(-0.65, 0.6, 0.87, 0.87, 0.27, 1.38, -6.71, -0.53, -1.85, -0.42, 2.88, -0.36, 0.52, -9.06, -0.96,-0.52, -2.86, -0.76, -0.67, -2.4, 9.49, 2.58, 0.28, 3.13, -0.55, 0.07, 0.6, 2.7, -0.26, -120.58, -8.23, 0.91, -0.4,-3.84, -2.62, 1.4, 0.37, -1.72, -1.85, 24.03)







Monte-Carlo test







-0.80756583 0.34875727 0.09690798


R>A=matrix(c(0.53, -8.13, 0.99, -0.56, -8.75, -1, 3.27, 65.61, -1.84, -13.43, -9.33, -3.18, -0.15, 1.96,-0.55, -0.46, -2.65, 0.48, 0.27, -0.17, 0.38, 1.17, -7.32, -6.55, 0.58, -8.38, 0.26, -0.87, -3.5, 9.57, 0.08,-0.27, 0.27, 10.22, -1.29, -0.61, 0.11, -0.16, -0.51, -0.22, -0.07, -0.25, 0.03, 1.04, -13.14, 1.31, 6.8, 0.48,8.31, -0.21, -1.18, -0.08, -1.31, -19.71, -7.06, -0.07, 0.09, -1.95, -0.56, 2.86, -2.82, 5.16, 2.26, -10.37,-1.86, 1.51, 0.69, -1.64, -2.34, -1.95, 0.36, 3.1, 12.55, -2.14, 0.7, 6.75, 1.14, -0.47, 0.55, -0.12, -0.58, -3.76,0, -1.22, 2.26, -7.53, -0.35, -0.53, 0.38, 0.36, -0.19, -0.67, -1.21, -1.49, -0.04, -0.26, 5.82, -0.64, 0.99,-0.43, 2.69, 0.13, 0.09, -1.4, -0.85, 1.09, 0.69, -0.39, 0.65, 0.06, 103.66, -1.33, 0.76, 8.76, 0.18, -0.11,-0.69, -3.51, -1.24, 0.63, 0.37, -1.69, -0.98, -3.92, -0.63, 2.24, 0.37, -0.43, 0.21, 0.97, -0.66, -4.76, -1.4,-0.93, -0.54, 0.35, -0.4, -0.58, 1.03, 0.17, 1.02, 5.08, -1.08, 5.01, -20.97, 0.43, 2.25, 0.47, 0.54, 0.81, 40.81,-14.16, 4.09, 0.23, 1.7, 0.34, -0.85, 2.65, 2.87, -0.69, 0.55, 0.43, 6.83, -1.62, 1.74, 0.03, -0.62, -0.64, 1.59,1.19, 0.47, 3.17, 0.23, 1.05, 0.54, -0.61, 0.43, 1.82, -7.55, -0.26, 1.18, 7.14, -0.97, 0.39, -6.24, 2.94, -0.44,0.4, -6.48, -0.2, 0.35, -11.61, 2.28, 0.26, 0.49, 2.63, 0.18, -0.46, 1.76, -0.98, -1.09, -2.45, 0.27, 4.06, -0.41,-0.72, -0.45, 1.06, -1.25, 19.12, -6.87, 0.28, -0.51, -0.35, 2.67, -0.99, 0.64, 8.84, -1.24, 0.73, 4.33, -0.06,-1.21, -0.36, 1.04, 6.66, 7.52, 81.95, -2.63, -0.12, 3.69, 6.86, -0.23, 2.12, 5.88, -4.19, 0.42, -0.22, -1.12,-0.13, -0.54, -0.61, -0.23, 1.39, -0.46, -0.95, -0.34, 1.16, 0.05, -0.2), nrow=50, byrow=TRUE)

R>B=matrix(c(-0.7, -13.1, -1.46, 0.27, 0.71, -1.62, 0.21, 27.57, 2.55, 0.05, 0.28, 0.07, -5.76, -0.33,40.39, 0.19, 0.52, 0.27, -0.12, -0.65, 0.35, 0.2, 0.39, 75, 0.94, -1.9, -0.55, -3.38, 0.65, 14.1, 0.29, -0.83,

52


198.2, 0.49, -1.16, -0.48, 2.87, 1.44, -1.5, 6.31, 1.55, 72.48, 1.49, -0.13, 2.71, 2.21, -1, -0.6, -0.27, 5.73,1.94, -0.57, 1.19, -0.54, -0.87, -0.93, -2.81, -1.71, 31.23, 1.92, 3.27, -1.35, 0.74, -4.77, 6.4, 11.02, 27.2,1.16, -0.2, -1.44, -0.12, -0.28, 0.47, -44.55, 1.53, -0.66, -0.67, -0.43, -1.34, 0.05, -2.89, 3.26, 1.6, 4.73,-0.39, -3.88, 0.64, 0.71, 0.89, 2.81, 1.58, -1.08, 0.58, -1.01, -0.09, 0.05, -0.13, 1.47, 1.1, 0, -0.38, 0.17,-4.09, 0.95, -0.22, 6.36, -0.26, -2.12, 54.7, 1.17, -8.65, -8.31, 0.91, 0.11, -2.41, -0.62, -0.28, 8.31, -1.9,0.55, -1.48, -1.08, 0.48, 3.02, 1.34, -0.45, -0.04, -2.1, -2.51, -0.24, -0.81, -5.12, 51.63, -0.59, 2.71, -1.29,6.39, 0.37, 1.6, 0.01, -3.92, 5.68, -3.61, -1.88, -0.68, -0.31, 1.24, 1.38, -2.64, -0.42, -0.45, -1.49, 0.57,-1.78, 6.77, 1.49, 15.68, 1.84, 1.58, -57.49, -0.58, -6, -0.69, -7.93, 1.6, -0.71, -0.26, 5.05, 2.36, -1.01,-0.36, -1.06, 0.03, 0.92, -1.03, 0.26, 0.78, -0.93, 1.9, 2.47, -2.51, -0.48, -0.37, 3.67, 1.56, 0.58, 2.9, 0.11,1.34, 0.67, 1.85, -4.4, 1.9, 0.6, -1.71, -0.13, -75.68, -0.07, -0.22, 1.41, -0.54, -0.01, 5.96, -6.7, 2.81, -1.4,3.22, 0.74, 0.89, -1.05, 5.79, -6.93, -0.34, 11.61, -0.61, -0.03, 0.34, -1.12, -0.6, 0.19, 0.83, -2.23, -2.82,-0.02, 0.55, 0.78, -1.46, -0.46, -1.61, 8.02, 0.48, 22.46, 2.26, 1.42, 0.1, 3.16, -0.06, -1.45, 3.22, 0.04, -1.1,0.7, 0.47, -1.56, -1.22, 0.42, -0.24, -0.34, -1.06, 3.6), nrow=50, byrow=TRUE)



Test stat: 0.90172

Numerator df: 5

Denominator df: 94



In putting data generated using Cauchy distribution for sample size 50 on the R 2.13.0 commandwindow, where matrix A and B are the objects or samples to be measured. Also, it should be noted thatobjects A1, A2, A3, A4 and A5 are objects of matrix A while objects B1, B2, B3, B4,and B5 are objectsof matrix B. The class distance of matrices A and B based on the canonical measure is labelled DA andDB respectively. The function RVdist.randtest was employed to call the randomization method for RVcoefficient for 10, 000 randomizations.

R>A1=c(-0.53, -1, -9.33, -0.46, 0.38, -8.38, 0.08, -0.61, -0.07, 1.31, -1.18, -0.07, -2.82, 1.51, 0.36,6.75, -0.58, -7.53, -0.19, -0.26, 2.69, 1.09, 103.66, -0.11, 0.37, 2.24, -0.66, 0.35, 1.02, 0.43, 40.81, 0.34,0.55, 0.03, 0.47, -0.61, 1.18, 2.94, 0.35, 2.63, -1.09, -0.72, -6.87, -0.99, 4.33, 6.66, 3.69, -4.19, -0.54,-0.95)

R>A2=c(-8.13, 3.27, -3.18, -2.65, 1.17, 0.26, -0.27, 0.11, -0.25, 6.8, -0.08, 0.09, 5.16, 0.69, 3.1,1.14, -3.76, -0.35, -0.67, 5.82, 0.13, 0.69, -1.33, -0.69, -1.69, 0.37, -4.76, -0.4, 5.08, 2.25, -14.16, -0.85,0.43,-0.62, 3.17, 0.43, 7.14, -0.44, -11.61, 0.18, -2.45, -0.45, 0.28,0.64, -0.06, 7.52, 6.86, 0.42, -0.61,-0.34)

R>A3=c(0.99, 65.61, -0.15, 0.48, -7.32, -0.87, 0.27, -0.16, 0.03, 0.48, -1.31, -1.95, 2.26, -1.64, 12.55,-0.47, 0, -0.53, -1.21, -0.64, 0.09, -0.39, 0.76, -3.51, -0.98, -0.43, -1.4, -0.58, -1.08, 0.47, 4.09, 2.65, 6.83,-0.64, 0.23, 1.82, -0.97, 0.4, 2.28, -0.46, 0.27, 1.06, -0.51, 8.84, -1.21, 81.95, -0.23, -0.22, -0.23, 1.16)

R>A4=c(-0.56, -1.84, 1.96, 0.27, -6.55, -3.5, 10.22, -0.51, 1.04, 8.31, -19.71, -0.56, -10.37, -2.34,-2.14, 0.55, -1.22, 0.38, -1.49, 0.99, -1.4, 0.65, 8.76, -1.24, -3.92, 0.21, -0.93, 1.03, 5.01, 0.54, 0.23, 2.87,1.62, 1.59, 1.05, -7.55, 0.39, -6.48, 0.26, 1.76, 4.06, -1.25, -0.35, -1.24, -0.36, -2.63, 2.12, -1.12, 1.39,0.05)

53


R>A5=c(-8.75, -13.43, -0.55, -0.17, 0.58, 9.57, -1.29, -0.22, -13.14, -0.21, -7.06, 2.86, -1.86, -1.95,0.7, -0.12, 2.26, 0.36, -0.04, -0.43, -0.85, 0.06, 0.18, 0.63, -0.63, 0.97, -0.54, 0.17, -20.97, 0.81, 1.7, -0.69,1.74, 1.19, 0.54, -0.26, -6.24, -0.2, 0.49, -0.98, -0.41, 19.12, 2.67, 0.73, 1.04, -0.12, 5.88, -0.13, 0.46, -0.2)

R>B1=c(-0.7, -1.62, 0.28, 0.19, 0.35, -1.9, 0.29, -0.48, 1.55, 2.21, 1.94, -0.93, 3.27, 11.02, -0.12, -0.66,-2.89, -3.88, 1.58, 0.05, -0.38, 6.36, -8.65, -0.62, -1.48, -0.45, -0.81, -1.29, -3.92, -0.31, -0.45, 1.49, -0.58,-0.71, -0.36, 0.26, -2.51, 0.58, 1.85, -0.13, -0.54, -1.4, 5.79, -0.03, 0.83, 0.78, 0.48, 3.16, -1.1, 0.42)

R>B2=c(-13.1, 0.21, 0.07, 0.52, 0.2, -0.55, -0.83, 2.87, 72.48, -1, -0.57, -2.81, -1.35, 27.2, -0.28, -0.67,3.26, 0.64, -1.08, -0.13, 0.17, -0.26, -8.31, -0.28, -1.08, -0.04, -5.12, 6.12, 5.68, 1.24, -1.49, 15.68, -6,-0.26, -1.06, 0.78, -0.48, 2.9, -4.4, -75.68, -0.01, 3.22, -6.93, 0.34, -2.23, -1.46, 22.46, -0.06, 0.7, -0.24)

R>B3=c(-1.46, 27.57, -5.76, 0.27, 0.39, -3.38, 198.2, 1.44, 1.49, -0.6, 1.19, -1.71, 0.74, 1.16, 0.47,-0.43, 1.6, 0.71, 0.58, 1.47, -4.09, -2.12, 0.91, 8.31, 0.48, -2.1, 51.63, 0.37, -3.61, 1.38, 0.57, 1.84, -0.69,5.05, 0.03, -0.93, -0.37, 0.11, 1.9, -0.07, 5.96, 0.74, -0.34, -1.12, -2.82, -0.46, 2.26, -1.45, 0.47, -0.34)

R>B4=c(0.27, 2.55, -0.33, -0.12, 75, 0.65, 0.49, -1.5, -0.13, -0.27, -0.54, 31.23, -4.77, -0.2, -44.55,-1.34, 4.73, 0.89, -1.01, 1.1, 0.95, 54.7, 0.11, -1.9, 3.02, -2.51, -0.59, 1.6, -1.88, -2.64, -1.78, 1.58, -7.93,2.36, 0.92, 1.9, 3.67, 1.34, 0.6, -0.22, -6.7, 0.89, 11.61, -0.06, -0.02, -1.61, 1.42, 3.22, -1.56, -1.06)

R>B5=c(0.71, 0.05, 40.39, -0.65, 0.94, 14.1, -1.16, 6.31, 2.71, 5.73, -0.87, 1.92, 6.4, -1.44, 1.53, 0.05,-0.39, 2.81, -0.09, 0, -0.22, 1.17, -2.41, 0.55, 1.34, -0.24, 2.71, 0.01, -0.68, -0.42, 6.77, -57.49, 1.6, -1.01,-1.03, 2.47, 1.56, 0.67, -1.71, 1.41, 2.81, -1.05, -0.61, 0.19, 0.55, 8.02, 0.1, 0.04, -1.22, 3.6)







Monte-Carlo test





Alternative hypothesis: greater54



0.57775533 0.57476152 0.04278372

3.2 Permutation method for Hotelling T-squared Analysis and Randomization methodfor RV coefficient Analysis using Normally Distributed data Set

In putting data generated using Normal distribution for sample size 5 on the R 2.13.0 command window,where matrix A and B are the objects or samples to be measured. The function hotelling.test was employedto call the Hotelling T-Squared test for 10, 000 permutations.

R>A=matrix(c(-0.6, -1.24, -1.93,0, -0.34, 0.62, -0.38,-0.54, 0.74, 0, 0.05, 2.09, 1.07, 1.6, 1.93, 1.33,-1.55, -0.89, -0.3, 1.71, 1.03, -0.17, 0.69, 0.55, -0.34), nrow=5, byrow=TRUE)

R>B=matrix(c(-0.4, 0.17, -0.28, 0.67, 1.38, 0.15, 0.07, -0.59, 0.15, 1.15, -0.8, -0.59, -1.86, 0.20, -1.25,0.14, 0.96, -0.19, 0.24, -2.25, 0.79, -1.56, 0.26, 0.42, -0.74), nrow=5, byrow=TRUE)



Test stat: 0.89616

Numerator df: 5

Denominator df: 4



In putting data generated using Normal distribution for sample size 5 on the R 2.13.0 command window,where matrix A and B are the objects or samples to be measured. Also, it should be noted that objects A1,A2, A3, A4 and A5 are objects of matrix A while objects B1, B2, B3, B4,and B5 are objects of matrix B.The class distance of matrices A and B based on the canonical measure is labeled DA and DB respectively.The function RVdist.randtest was employed to call the randomization method for RV coefficient for 10,000 randomizations.

R>A1=c(-0.6, 0.62, 0.05, 1.33, 1.03)

R>A2=c(-1.24, -0.38, 2.09, -1.55, -0.17)

R>A3=c(-1.93, -0.54, 1.07, -0.89, 0.69)

R>A4=c(0, 0.74, 1.6, -0.3, 0.55)

55


R>A5=c(-0.34, 0, 1.93, 1.71, -0.34)

R>B1=c(-0.4, 0.15, -0.8, 0.14, 0.79)

R>B2=c(0.17, 0.07, -0.59, 0.96, -1.56)

R>B3=c(-0.28, -0.59, -1.86, -0.19, 0.26)

R>B4=c(0.67, 0.15, 0.2, 0.24, 0.42)

R>B5=c(1.38, 1.15, -1.25, -2.25, -0.74)







Monte-Carlo test







0.54055829 0.58165711 0.01004145

In putting data generated using Normal distribution for sample size 10 on the R 2.13.0 commandwindow, where matrix A and B are the objects or samples to be measured. The function hotelling.test wasemployed to call the Hotelling T-Squared test for 10, 000 permutations.

56


R > A=matrix(c(-0.78, 0.15, 0.30, -0.39, 0.9, -0.28, 0.15, -0.04, 0.87, 1.53, -0.6, 1.53, 0.50, 0.96, 0.23,-1.02, 0.48, -0.04, -2.28, -1.35, 0.48, 1.11, -0.65, 0.63, -0.66, 0.02, 0.88, -0.35, -0.2, 0.13, -0.59, -0.88,-1.38, -2.07, 0.29, -1.47, 2.30, 0.06, -1.28, -0.58, 0.41, 0.87, 1.94, -0.8, 0.53, -1.06, -0.53, -0.45, 0.45,0.07), nrow=10, byrow=TRUE)

R > B=matrix(c(0.48, 0.06, 0.03, 0.31, -0.03, 1.76, -1.59, 1, -0.13, 0.26, 0.39, 0.4, -2.22, -0.48, -0.35,-0.17, 1.81, -2.66, 1.06, -0.23, 0.77, 1.25, 1.15, 0.05, -0.97, -1.26, 0.28, 0.49, -0.32, -2.02, -1.37, 1.04,0.36, -0.64, 0.35, -0.5, 0.25, -0.64, 0.77, -0.10, -0.63, 0.26, 0.49, -0.04, -1.68, -0.35, 0.2, 0.07, -0.82,-1.14), nrow=10, byrow=TRUE)

R > Test=hotelling.test(A, B, shrinkage = FALSE, perm = TRUE, B= 10000)


Test stat: 1.7077

Numerator df: 5

Denominator df: 14



In putting data generated using Normal distribution for sample size 10 on the R 2.13.0 commandwindow, where matrix A and B are the objects or samples to be measured. Also, it should be noted thatobjects A1, A2, A3, A4 and A5 are objects of matrix A while objects B1, B2, B3, B4,and B5 are objectsof matrix B. The class distance of matrices A and B based on the canonical measure is labeled DA andDB respectively. The function RVdist.randtest was employed to call the randomization method for RVcoefficient for 10, 000 randomizations.

R>A1=c(-0.78, -0.28, -0.6, -1.02, 0.48, 0.02, -0.59, -1.47, 0.41, -1.06)

R>A2=c(0.15, 0.15, 1.53, 0.48, 1.11, 0.88, -0.88, 2.30, 0.87, -0.53)

R>A3=c(0.30, -0.04, 0.5, -0.04, -0.65, -0.35, -1.35, 0.06, 1.94, -0.45)

R>A4=c(-0.39, 0.87, 0.96, -2.28, 0.63, -0.2, -2.07, -1.28, -0.8, 0.45)

R>A5=c(0.9, 1.53, 0.23, -1.35, -0.66, 0.13, 0.29, -0.58, 0.53, 0.07)

R>B1=c(0.48, 1.76, 0.39, -0.17, 0.77, -1.26, -1.37, -0.5, -0.63, -0.35)

R>B2=c(0.06, -1.59, 0.4, 1.81, 1.25, 0.28, 1.04, 0.25, 0.26, 0.20)

R>B3=c(0.03, 1, -2.22, -2.66, 1.15, 0.49, 0.36, -0.64, 0.49, 0.07)

57


R>B4=c(0.31, -0.13, -0.48, 1.06, 0.05, -0.32, -0.64, 0.77, -0.04, -0.82)

R>B5=c(-0.03, 0.26, -0.35, -0.23, -0.97, -2.02, 0.35, -1.1, -1.68, -1.14)




R > DB<-dist.quant(B, method = 1)



Monte-Carlo test







-0.27193770 0.70666089 0.01009138


R>A=matrix(c(-0.38, 0.09, 0.68, 0.71, -0.152, -0.49, -0.62, 0.17, 1.44, 2.08, -0.18, 0.35, 0.71, -0.15,0.54, -1.72, -0.29, -0.47, -1.26, -0.26, 0.20, 0.71, 0.57, -1.15, 1.73, -0.05, -2.35, 0.16, 0.54, 0.56, -0.6,-1.51, -1.66, -0.63, -0.16, 1, -0.76, -0.31, -0.6, 0.50, 1.55, -1.54, -0.24, -1.94, -1.43, -0.17, -0.67, 0.01,-0.54, -0.10, -0.12, 1.66, 1.13, 2.08, -1.57, -2.07, -0.22, -1.07, 0.87, 0.29, -0.40, -0.2, -1.51, -0.04, -0.48,0.53, -1.16, 0.74, -0.5, -0.4, -1, -0.38, 0, -1.11, 0.04, 0.15, 0, -1.46, 0.44, -0.96, 0.54, -0.12, -0.83, -0.39,-0.15, 0.67, 1.92, -0.89, 0.71, -0.25, -1.48, -0.3, 0.43, -0.15, -0.17, 2.12, -0.86, -0.29, -0.49, 0.19), nrow=20,byrow=TRUE)

R>B=matrix(c(-0.44, 2.43, -0.66, 0.21, 0.15, 0.55, 0.68, -1.18, 0.17, -1.57, -0.67, 0.48, -0.05, 0.85,0.74, 0.1, 0.78, -0.42, 0.33, 1.17, 0.74, -0.53, -1.60, 0.32, -0.95, 1.11, -0.95, -0.48, 1.98, -0.3, 0.55, -0.68,

58


-1.12, 0.11, 0.21, 0.26, -0.68, 1.12, 1.51, 1.82, -0.91, -1.29, 3.07, 0.70, 0.16, -0.99, 0.27, 0.58, -0.28, 0.43,-0.31, 0.94, 0.27, 0.74, -3.13, -1.28, 1.12, 2.23, 0, -1.05, 0.84, 0.40, -0.35, -0.76, 0.76, 1.13, -0.42, -0.23,-0.54, -0.24, -0.40, -0.85, 1.01, 1.06, -0.19, 0.37, 1.35, -0.47, 0.15, -0.06, 0.88, -0.94, -0.26, -1, -1.19,-0.92, 0.42, -1.36, -1.99, -1.87, -0.39, 0.47, 0.77, 0.04, -0.72, 0.59, 0.14, 0.14, 1.03, -0.62), nrow=20,byrow=TRUE)



Test stat: 1.0641

Numerator df: 5

Denominator df: 34



In putting data generated using Normal distribution for sample size 20 on the R 2.13.0 commandwindow, where matrix A and B are the objects or samples to be measured. Also, it should be noted thatobjects A1, A2, A3, A4 and A5 are objects of matrix A while objects B1, B2, B3, B4,and B5 are objectsof matrix B. The class distance of matrices A and B based on the canonical measure is labeled DA andDB respectively. The function RVdist.randtest was employed to call the randomization methods for RVcoefficient for 10, 000 randomizations.

R>A1=c(-0.38, -0.49, -0.18, -1.72, 0.20, -0.05, -0.6, 1, 1.55, -0.17, -0.12, -2.07, -0.40, 0.53, -1, 0.15,0.54, 0.67, -1.48, 2.12)

R>A2=c(0.09, -0.62, 0.35, -0.29, 0.71, -2.35, -1.51, -0.76, -1.54, -0.67, 1.66, -0.22, -0.2, -1.16, -0.38,0, -0.12, 1.92, -0.3, -0.86)

R>A3=c(0.68, 1.7, 0.71, -0.47, 0.57, 0.16, -1.66, -0.31, -0.24, 0.01, 1.13, -1.07, -1.51, 0.74, 0, -1.46,0.83, -0.89, 0.43, -0.29)

R>A4=c(0.71, 1.44, -0.15, -1.26, -1.15, 0.54, -0.63, -0.6, -1.94, -0.54, 2.08, 0.87, -0.04, -0.5, -1.11,0.44, -0.39, 0.71, -0.15, -0.49)

R>A5=c(-1.52, 2.08, 0.54, -0.26, 173, 0.56, -0.16, 0.50, -1.43, -0.10, -1.57, 0.29, -0.48, -0.40, 0.04,-0.96, -0.15, -0.25, -0.17, 0.19)

R>B1=c(-0.44, 0.55, -0.67, 0.1, 0.74, 1.11, 0.55, 0.26, -0.91, -0.99, -0.31, -1.28, 0.84, 1.13, -0.40, 0.37,0.88, -0.92, -0.39, 0.59)

R>B2=c(2.43, 0.68, 0.48, 0.78, -0.53, -0.95, -0.68, -0.68, -1.29, 0.27, 0.94, 1.12, 0.40, -0.42, -0.85,1.35, -0.94, 0.42, 0.47, 0.14)

59


R>B3=c(-0.66, -1.18, -0.05, -0.42, -1.60, -0.48, -1.12, 1.12, 3.07, 0.58, 0.27, 2.23, -0.35, -0.23, 1.01,-0.47, -0.26, 1.36, 0.77, 0.14)

R>B4=c(0.21, 0.17, 0.85, 0.33, 0.32, 1.98, 0.11, 1.51, 0.70, -0.28, 0.74, 0, -0.76, -0.54, 1.06, 0.15, -1,-1.99, 0.04, 1.03)

R>B5=c(0.15, -1.57, 0.74, 1.17, -0.95, -0.3, 0.21, 1.82, 0.16, 0.43, -3.13, -1.05, 0.76, -0.24, -0.19, 0.06,-1.19, -1.87, -0.72, -0.62)







Monte-Carlo test







0.23490131 0.46477650 0.01440696


R >A=matrix(c(1.49, 0.32, 0.79, 0.77, -0.2, -0.42, 0.93, -0.57, -0.85, -1.36, 0.23, -0.3, -0.23, -1.11,-0.29, 1.48, -0.49, -0.03, 1.42, -0.8, -0.92, 1.93, 0.59, -0.19, -0.94, 1.15, -1.57, -0.45, -0.9, 0.18, -0.19, -2,-1.09, -0.3, -1.23, -1.88, -0.65, -0.79, -0.19, 1.72, 0.06, 1.1, -0.47, 0.88, -0.8, 0.55, 1.09, 0.79, -0.18, 0.02,

60


1.23, 0.05, 0.79, 0.27, -0.43, -0.7, -0.65, -0.06, -1.36, -0.37, 0.09, 3.29, -0.41, 1.02, -2.53, -0.49, -1.16,0.01, -0.52, -0.41, -0.55, -0.74, -1.33, -1.08, 0.55, -0.52, -0.42, 0.93, 0.95, 0.91, 0.52, 0.2, -0.81, -1.06,-1.5, 0.24, 0.77, -0.49, 0.48, 0.47, 0.08, 1.19, 0.5, 0.99, -0.45, -0.41, 0.41, -0.01, 1.39, 1.2, -0.69, 0.88,1.09, -0.8, 2.05, -0.29, 1.54, -0.13, -1.39, 0.66, -1.14, -0.14, -1.56, 0.88, 0.41, 0.68, -1.68, -0.73, -0.77,0.26, -1.69, -1.45, -0.33, 0.83, 0.04, 0.76, -1.72, -0.43, 0, -0.84, -1.55, 0.32, -0.63, -1.91, 0.4, 0.67, -1.12,1.37, -0.5, 2.02, -1.26, 0.12, -1.18, -0.13, -0.03, -0.67, -0.41, -0.4, -0.91, 0.02), nrow=30, byrow=TRUE)

R>B=matrix(c(0.28, -0.81, -1.83, -2.35, 0.37, -0.97, -1.22, 2.66, 0.96, 0.68, -0.61, 0.74, 0, 0.41, -0.19,-1.8, -1.16, -1.09, -0.24, 0.37, 1.02, -0.33, -0.07, -2.31, -1.46, 0.2, -1.17, -0.55, 0.89, 0.42, -0.89, -0.25,-2.16, 0.16, 0.81, -0.22, -1.24, -1.21, -0.36, 1.62, -0.03, 1.29, -0.03, 1.38, 0.4, 0.5, -1.43, -0.63, -0.38,-0.19, 1.29, 0.91, -0.72, -0.95, -0.37, 0.66, -1.75, -0.27, -0.76, -0.38, -0.62, 0.62, 0.2, -1.44, -0.11, 1.84,-0.89, 2.66, 0.53, -0.61, 1.01, -0.5, -0.66, 1, 0.56, 1, 0.66, -1.04, 0.85, 0.43, 0.85, -2.96, -0.3, -0.49, -1.43,-0.61, 0.6, 1.43, 0.32, 0.15, 1.24, 0.07, -1.11, -0.19, 0.21, 2.03, -0.07, -0.51, -1.65, 1.47, 1.21, 0.45, 0.23,0.75, -1.55, 2.69, -0.54, -0.62, -0.25, 0.71, -2.21, 1.47, -0.83, -0.52, -0.8, -1.83, 0.73, -0.57, 1.16, -0.45,-0.98, 0.25, -1.51, 0.75, 0.01, 0.1, -0.03, -1.03, -0.1, 0.87, -1.33, -0.48, 0.2, 1.29, 1.51, 0.38, -0.65, 0.3,-0.46, -1.43, 0.45, -0.87, -1.07, -0.76, 0.11, -0.15, 0.48, -1.41, -1.16, 1.16), nrow=30, byrow=TRUE)



Test stat: 0.63577

Numerator df: 5

Denominator df: 54




R > A1=c(1.49, -0.42, 0.23, 1.48, -0.92, 1.15, -1.19, -1.88, 0.06, 0.55, 1.23, -0.7, 0.09, -0.49, -0.55,-0.52, 0.52, 0.24, 0.08, -0.41, 0.69, -0.29, -1.14, 0.68, -1.69, 0.76, -1.55, 0.67, -1.26, -0.67)

R >A2=c(0.32, 0.93, -0.3, -0.48, 1.93, -1.57, -2, -0.65, 1.1, 1.09, 0.05, -0.65, 3.29, -1.16, -0.74, -0.42,0.2, 0.77, 1.19, 0.41, 0.88, 1.54, -0.14, -1.68, -1.45, -1.72, 0.32, -1.12, 0.12, -0.41)

R >A3=c(0.79, -0.57, -0.23, -0.03, 0.59, -0.45, -1.09, -0.79, -0.47, 0.79, 0.79, -0.06, -0.41, 0.01, -1.33,0.93, -0.81, -0.49, 0.5, -0.01, 1.09, -0.13, -1.56, -0.73, -0.33, -0.43, -0.63, 1.37, -1.18, -0.4)

61


R >A4=c(0.77, -0.85, -1.11, 1.42, -0.19, -0.9, -0.3, -1.19, 0.88, -0.18, 0.27, -1.36, 1.02, -0.52, -1.08,0.95, -1.06, 0.48, 0.99, 1.39, -0.8, -1.39, 0.88, -0.77, 0.83, 0, -1.91, -0.5, -0.13, -0.91)

R >A5=c(-0.2, -1.36, -0.29, -0.8, -0.94, 0.18, -1.23, 1.72, -0.8, 0.02, -0.43, -0.37, -2.53, -0.41, 0.55,0.91, -1.5, 0.47, -0.45, 1.2, 2.05, 0.66, 0.41, 0.26, 0.04, -0.84, 0.4, 2.02, -0.03, 0.02)

R >B1=c(0.28, -0.97, -0.61, -1.8, 1.02, 0.2, -0.89, -0.22, -0.03, 0.5, 1.29, 0.66, -0.62, 1.84, 1.01, 1,0.85, -0.61, 1.24, 2.03, 1.21, 2.69, -2.21, -1.83, -0.98, 0.1, -1.33, 0.38, 0.45, -0.15)

R >B2=c(-0.81, -1.22, 0.74, -1.16, -0.33, -1.17, -0.25, -1.24, 1.29, -1.43, 0.91, -1.75, 0.62, -0.89, -0.5,0.66, -2.96, 0.6, 0.07, -0.07, 0.45, -0.54, 1.47, 0.73, 0.25, -0.03, -0.48, -0.65, -0.87, -0.48)

R >B3=c(-1.83, 2.66, 0, -1.09, -0.07, -0.55, -2.16, -1.21, -0.03, -0.63, -0.72, -0.27, 0.2, 2.66, -0.66,-1.04, -0.3, 1.43, -1.11, -0.51, 0.23, -0.62, -0.83, -0.57, -1.51, -1.03, 0.2, 0.3, -1.07, -1.41)

R >B4=c(-2.35, 0.96, 0.41, -0.24, -2.31, 0.89, 0.16, -0.36, 1.38, -0.38, -0.95, -0.76, -1.44, 0.53, 1, 0.85,-0.49, 0.32, -0.19, -1.65, 0.75, -0.25, -0.52, 1.16, 0.75, -0.1, 1.29, -0.46, -0.76, -1.16)

R >B5=c(-0.37, 0.68, -0.19, 0.37, -1.46, 0.42, 0.81, 1.62, 0.4, -0.19, -0.37, -0.38, -0.11, -0.61, 0.56,0.43, -1.43, 0.15, 0.21, 1.47, -1.55, 0.71, -0.8, -0.45, 0.01, 0.87, 1.51, -1.43, 0.11, 1.16)

R >A=matrix(c(A1, A2, A3, A4, A5), nrow=5, byrow=TRUE)

R >B=matrix(c(B1, B2, B3, B4, B5), nrow=5, byrow=TRUE)

R >DA<-dist.quant(A, method = 1)

R >DB<-dist.quant(B, method = 1)

R >RVdist.randtest(DA, DB, nrepet = 10000)


Monte-Carlo test



62






-0.946962198 0.804399389 0.004247146


R >A=matrix(c(0.53, -1.19, -0.78, 0.8, 0.41, 0.95, -0.91, 0.84, -2.21, 0.47, -0.58, 1.06, -0.61, -1.75,2.86, -0.69, 0.22, 1.71, 0.33, -0.01, 0.61, -0.86, -2.03, 1.26, -1.54, 1.29, -0.62, -0.29, -0.68, 0.02, -1.4,-1.64, -0.2, 0.35, 1.97, 0.68, -1.01, 1.09, -1.43, 0.04, -1.64, -0.01, 0.01, 0.07, -0.84, -1.45, 0.95, -0.26, 1.21,-0.23, -0.83, 1.84, 0.68, 0.3, -0.99, 0.36, 1.96, 1.94, -1.95, -0.35, -0.09, -0.54, 0.76, -0.26, 1.09, 1.48, 0.42,0.85, -0.42, 1.6, -0.87, -0.2, -0.96, -0.62, 0.41, 1.86, 0.61, 0.11, 0.41, 0.55, -0.64, -0.32, -0.92, 0.53, 0.26,0.19, 0.98, -0.38, -0.63, -0.93, 1.01, 1.73, 1.73, -0.76, -0.32, 0.37, -0.32, 0.54, 0.22, -0.05, 2.52, -1.31,0.42, 0.3, 0.41, -0.71, 1.01, -0.12, -1.13, 0.79, 1.01, 0.36, 0.92, -0.17, 0.25, -1.17, -0.2, -0.71, 0.98, 1.13,-2.08, -0.37, 1.95, -0.46, -0.43, -1.53, 1.48, -0.16, 1.9, -0.62, 0.15, 0.12, -1.13, -0.71, 2.03, -1.68, -0.2,1.27, 0.48, 0.87, 0.11, -0.17, -1.83, -0.37, 0.08, -1.88, -1.05, 0.58, 1.72, 0.55, 2.36, -0.15, 0.11, -0.79,-0.66, -0.19, 1.52, -0.47, -1.39, -1, 0.31, -1.03, -0.46, -1.3, 1.84, 0.34, -0.23, -1.56, 0.7, -0.4, 1.03, 0.16,0.9, -2.08, -0.19, -0.35, 0.52, -0.68, -2.75, -0.98, 0.74, 0.29, -0.74, 3.13, -0.42, 0.23, -0.21, 1.45, 0.8, -0.18,-0.72, -0.12, 0.75, 1.49, 0.07, 0.49, -0.06, -0.51, -0.49, -0.52), nrow=40, byrow=TRUE)

R >B=matrix(c(-0.14, 1.47, 1.08, 1.18, 0.81, -0.5, -0.95, -0.25, 0.3, 1.05, -0.08, -0.17, -0.93, -1.66,0.53, -0.88, 0.54, -1.88, 0.77, -1.43, -0.55, 0.57, -1.61, -0.59, -1.24, 0.14, 0.85, -1.95, -0.03, 2.13, -0.37,-1.31, 0, 0.55, 0.7, 0.2, 0.57, -1.08, 2.38, -0.58, -1.01, 0.1, -1.27, -1.34, -0.49, 0.33, 1.06, 0.78, -0.64, -0.45,-0.72, 1.83, -0.03, 0.54, -1.17, -0.35, -0.28, -0.17, 0, -0.06, -0.93, -0.16, 0.56, -0.14, 1.37, 0.47, 1.51, 0.1,0.05, 0.64, -0.09, 2.2, -1.7, -0.49, 0.71, 1.28, -0.38, 0.5, 1.74, -0.69, -0.73, 0.33, -0.25, 1.97, 0.2, 1.21,-0.6, -0.38, 0.36, -0.36, 0.89, -1.04, -0.53, 0.55, -0.87, 0.57, 0.75, 0.76, 1.05, 0.87, 3.13, -1.1, 2.16, 0.84,-0.6, 0.18, 0.07, -0.78, -0.36, 0.47, 1.07, -0.62, -0.94, -0.51, 0.14, 0.28, 0.29, -0.2, -1, -0.24, -0.25, -0.06,1.12, 0.03, 0.9, -0.57, 0.31, -0.7, 1.37, 0.61, 0.07, -0.02, 0.19, 0.79, 0.13, 0.26, 0.7, -1.5, -1.73, 0.98, -0.84,0.08, -0.82, 0.31, -0.27, 1.11, -0.04, 0.4, 0.07, -0.33, 1.31, 0.76, -0.45, 0.84, -0.45, 0.4, -0.8, -0.57, -1.02,0.6, -1.01, 1.3, 1.03, 0.54, -0.64, 0.07, -1.03, 0.6, -1.73, -1.44, -0.64, -1.43, -0.38, 1.19, -0.09, -0.34, -1.44,0.48, 1.26, 0.31, 1.07, -0.1, 1.63, 0.31, -1.29, 0.27, 0.47, 0.05, -0.47, -0.71, 1.22, -1.09, -1.14, 0.24, -0.09,-1.87, -0.79, 0.25, -3.13, 0.39), nrow=40, byrow=TRUE)

R > Test=hotelling.test(A, B, shrinkage = FALSE, perm = TRUE, B= 10000)


Test stat: 0.68191

Numerator df: 5

Denominator df: 74



In putting data generated using Normal distribution for sample size 40 on the R 2.13.0 command

63


window, where matrix A and B are the objects or samples to be measured. Also, it should be noted thatobjects A1, A2, A3, A4 and A5 are objects of matrix A while objects B1, B2, B3, B4,and B5 are objectsof matrix B. The class distance of matrices A and B based on the canonical measure is labelled DA andDB respectively. The function RVdist.randtest was employed to call the randomization method for RVcoefficient for 10, 000 randomizations.

R>A1=c(0.53, 0.95, -0.58, -0.69, 0.61, 1.29, -1.4, 0.68, -1.64, -1.45, -0.83, 0.36, -0.09, 1.48, -0.87,1.86, -0.64, 0.19, 1.01, 0.37, 2.52, -0.71, 1.01, -1.17, -2.08, -1.53, 0.15, -1.68, 0.11, -1.88, 2.36, -0.19,0.31, 0.34, 1.03, -0.35, 0.74, 0.23, -0.72, 0.49)

R>A2=c(-1.19, -0.91, 1.06, 0.22, -0.86, -0.62, -1.64, -1.01, -0.01, 0.95, 1.84, 1.96, -0.54, 0.42, -0.2,0.61, -0.32, 0.98, 1.73, -0.32, -1.31, 1.01, 0.36, -0.2, -0.37, 1.48, 0.12, -0.2, -0.17, -1.05, -0.15, 1.52, -1.03,-0.23, 0.16, 0.52, 0.29, -0.21, -0.12, -0.6)

R>A3=c(-0.78, 0.84, -0.61, 1.71, -2.03, -0.29, -0.2, 1.09, 0.01, -0.26, 0.68, 1.94, 0.76, 0.85, -0.96, 0.11,-0.92, -0.38, 1.73, 0.54, 0.42, -0.12, 0.92, -0.71, 1.95, -0.16, -1.13, 1.27, -1.83, 0.58, 0.11, -0.47, -0.46,-1.56, 0.9, -0.68, -0.74, 1.45, 0.75, -0.51)

R>A4=c(0.8, -2.12, -1.75, 0.33, 1.26, -0.68, 0.35, -1.43, 0.07, 1.21, 0.3, -1.95, -0.26, -0.42, -0.62, 0.41,0.53, -0.63, -0.76, 0.22, 0.3, -1.13, -0.17, 0.98, -0.46, 1.9, -0.71, 0.48, -0.37, 1.72, -0.79, -1.39, -1.3, 0.7,-2.08, -2.75, 3.13, 0.8, 1.49, -0.49)

R>A5=c(0.41, 0.47, 2.86, -0.01, -1.54, 0.02, 1.97, 0.04, -0.84, -0.23, -0.99, -0.35, 1.09, 1.6, 0.41, 0.55,0.26, -0.93, -0.32, -0.05, 0.41, 0.79, 0.25, 1.13, -0.43, -0.62, 2.03, 0.87, 0.08, 0.55, -0.66, -1, 1.84, -0.4,-0.19, -0.98, -0.42, -0.18, 0.07, -0.52)

R>B1=c(-0.14, -0.5, -0.08, -0.88, -0.55, 0.14, -0.37, 0.2, -1.01, 0.33, -0.72, -0.35, -0.93, 0.47, -0.09,1.28, -0.73, 1.21, 0.89, 0.57, 3.13, 0.18, 1.07, 0.28, -0.25, -0.57, 0.07, 0.26, -0.84, 1.11, 1.31, 0.4, -0.10,0.07, -0.64, -0.34, 1.07, 0.27, 1.22, -1.87)

R>B2=c(1.47, -0.95, -0.17, 0.54, 0.57, 0.85, -1.31, 0.57, 0.1, 1.06, 1.83, -0.28, -0.16, 1.51, 2.2, -0.38,0.33, -0.6, -1.04, 0.75, -1.1, 0.07, -0.62, 0.29, -0.06, 0.31, -0.02, 0.7, 0.08, -0.04, 0.76, -0.8, 1.3, -1.03,-1.43, -1.44, -0.1, 0.47, -1.09, -0.79)

R>B3=c(1.08, -0.25, -0.93, -1.88, -1.61, -1.95, 0, -1.08, -1.27, 0.78, -0.03, -0.17, 0.56, 0.1, -1.7, 0.5,-0.25, -0.38, -0.53, 0.76, 2.16, -0.78, -0.94, -0.2, 1.12, -0.77, 0.19, -1.5, -0.82, 0.4, -0.45, -0.57, 1.03, 0.6,-0.38, 0.48, 1.63, 0.05, -1.14, 0.25)

R>B4=c(1.18, 0.3, -1.66, 0.77, -0.59, -0.03, 0.55, 2.38, -1.34, -0.64, 0.54, 0, -0.14, 0.05, -0.49, 1.74,1.97, 0.36, 0.55, 1.05, 0.84, -0.36, -0.51, -1, 0.03, 1.37, 0.79, -1.73, 0.31, 0.07, 0.84, -1.02, 0.54, -1.73,

64


1.19, 1.26, 0.31, -0.47, 0.24, -3.13)

R >B5=c(0.81, 1.05, 0.53, -1.43, -1.24, 2.11, 0.7, -0.58, -0.49, -0.45, -1.17, -0.06, 1.37, 0.64, 0.71,-0.69, 0.2, -0.36, -0.87, 0.87, -0.6, 0.47, 0.14, -0.24, 0.9, 0.61, 0.13, 0.98, -0.27, -0.33, -0.45, 0.6, -0.64,-1.44, -0.09, 0.31, -1.29, -0.71, -0.09, 0.39)







Monte-Carlo test







-1.139897042 0.915506951 0.000453243


R>A=matrix(c(1.36, 0.82, 0.86, -0.5, -0.96, -1.72, 0.73, -1.8, -0.98, -1.19, -0.44, -1.46, 0.05, -0.56,-0.11, 0.69, -0.67, -0.15, -0.75, 0.09, -1.39, 0.68, -2.34, -0.08, -1.96, -0.28, 0.82, 0.95, 0.13, 1.61, 0.82,0.55, 0.01, 0.45, 1.2, 0.5, 0.82, 0.64, -1.85, 0.41, -0.7, 0.75, -0.31, 0.49, -1.02, -0.72, 1.25, 0.91, 0.52,1.03, -0.7, -0.08, 0.29, -0.92, 1.75, -0.57, -0.36, 1.62, -0.42, 0.28, -0.91, 0.69, -0.57, 0.86, 0.52, 1.42, 0.36,0.51, 0.99, -2.09, -0.14, -0.5, 0.48, -0.44, 1.46, -0.56, 1.9, 0.71, -0.28, -1.29, 0.86, -0.08, -0.24, -1.13,-0.04, -0.76, 0.18, 2.26, -0.56, -2.03, 1.92, -0.77, 0, 0.07, -0.48, -1.73, -1.9, 1.37, -1.4, 0.21, 1.32, 1.1,-0.31, -0.66, 0.67, -3.16, -0.52, -0.6, 0.6, 0.17, -0.05, 1.92, 0.87, -0.15, -0.28, 2.24, -1.35, 0.43, 0.85, 0.39,-0.82, -1.07, -1.88, -1.35, 0.79, -1, 1.37, -1.09, 0.18, 0.59, -0.31, -0.34, -1.25, -2.15, 0.32, -1.72, -0.91,0.09, -0.41, 1.04, -0.23, 0.73, -0.5, 1.67, -0.04, 0.32, -0.57, 1.11, 0.9, 0.59, 0.87, -0.61, 1.07, 1.16, 0.98,

65


-0.07, -0.6, -0.52, 0.45, 0.91, 0.23, 2.34, 0.39, -0.22, -0.13, -0.12, -0.61, -0.57, -1.14, 1.37, 0.06, -0.73,-0.25, -0.39, -1.19, 0.87, -1.08, 0.86, 1.78, 1.48, -0.32, 0.01, 0.79, -1.56, -0.37, 0.04, 0.39, -0.5, -0.06,0.56, -0.18, 0.55, -0.03, 0.96, 0.91, 0.29, -0.11, -0.17, -1.02, -0.11, 0.64, -1.92, -0.3, 2.18, 0.26, -0.05,-0.2, -0.73, -0.1, -0.59, -0.89, 2.06, 0.63, 2.17, -0.62, 0.16, -0.53, 1.02, -0.38, 1.1, -0.64, 0.69, -0.39, -0.62,-0.96, -0.76, 0.7, -0.86, -0.03, 0.74, -1.05, 0.96, 0.02, -0.16, -1.88, 0.09, -0.45, 0.32, -0.9, -0.75, 0.84,-0.68, 0.65, 0.18, 1.43, -0.2, -0.67, 2.05, 0.17, 1.16), nrow=50, byrow=TRUE)

R>B=matrix(c(0.16, 0.99, 0.98, -1.86, 1.11, 1.06, -0.95, -0.18, -1.82, 0.08, 0.45, -0.37, 1, 0.8, 0.4,-0.42, -1.47, -1.51, 0.7, 0.21, 0.67, -0.43, -1.41, -0.66, -0.48, -0.75, 0.33, 0.89, 1.61, -0.89, 1.42, 0.94,-0.57, -0.44, -0.44, -1.45, 0.67, -0.51, 0.83, 0.49, 1.04, -0.47, -1.99, 2.13, -1.91, -0.58, 0.06, -1.2, -0.41,1.36, 1.44, -2.65, 0.23, 1.52, -0.53, 1.02, 0.18, -1.47, 0.18, -1.17, 0.04, -1.79, 0.6, 0.03, -1.06, -1.56, -1.35,1.04, 1.67, 1.08, -0.49, 1.68, -1.05, 0.68, -1.92, -0.49, 0.85, 0.2, -1.51, -0.85, 0.26, 0.08, 1.76, -0.19, -0.93,-0.67, -1.55, -0.54, 0.38, 1.2, 0, 1.13, 0.5, -0.13, 0.26, 0.83, 1.48, 0, -0.49, 0.06, -0.52, -0.8, -0.6, 0.29,-0.8, 1.36, -0.39, -0.81, -1.67, 0.71, 0.86, -0.78, -0.8, 2.27, 0.86, 0.51, 0.65, 0.57, 0.18, -0.38, -1.58, -0.28,-0.36, 0.73, -2.23, -0.35, 0.34, -0.25, 0.21, -0.59, -0.51, 1.22, -0.07, -0.84, -1.34, 1.51, -0.81, -1.25, -0.03,-0.3, 1.16, 1.37, -0.7, 0.1, -0.38, 0.38, 0.3, 0.18, -0.1, -2.34, 1.6, -1.46, -1, 2.12, -0.57, 0.01, 1.04, 0.51,-0.17, -1.9, 1.03, 0.94, -0.06, 0.59, -0.48, 1.74, 1.09, -1.39, 2.39, -1.59, -1, 0.25, 1.84, -0.32, 0.85, 0.1,2.68, -0.59, 1.81, 0.61, -0.27, 2.2, 0.4, 1.51, 1.92, 0.58, -1.58, -0.88, -0.13, -0.17, -0.47, 0.75, -1.2, -0.75,-1.75, -0.71, -0.56, 1.28, -0.39, 0.09, -1.49, 0.96, 2.94, -0.03, 0.98, 1.23, -0.57, -0.98, 0.19, -0.98, 1.46,-1.45, -0.54, 0.91, 1.07, 0.72, 0.18, -0.76, -0.57, -0.41, -1.75, -0.34, 0.29, 0.4, -0.38, -0.75, -0.32, 0.08,-0.15, 1.44, -1.09, 0.12, -0.28, -1.36, 0.05, 0.99, 0.56, 0.3, -0.43, -1.47, -2.25, 0.42, 0.48, -0.37, 0.79, -0.43,-1.43, 2.58, 0.8, 0.73), nrow=50, byrow=TRUE)



Test stat: 1.1849

Numerator df: 5

Denominator df: 94




R>A1=c(1.36, -1.72, -0.44, 0.69, -1.39, -0.28, 0.82, 0.5, -0.7, -0.72, -0.7, -0.57, -0.91, 1.42, -0.14,-0.56, 0.86, -0.76, 1.92, -1.73, 1.32, -3.16, -0.05, 2.24, -0.82, -1, -0.31, -1.72, -0.23, 0.32, 0.87, -0.07, 0.23,-0.12, 0.06, 0.87, -0.32, 0.04, -0.18, 0.29, 0.64, -0.05, -0.89, 0.16, -0.64, -0.76, -1.05, 0.09, 0.84, -0.2)

R>A2=c(0.82, 0.73, -1.46, -0.67, 0.68, 0.82, 0.55, 0.82, 0.75, 1.25, -0.08, -0.36, 0.69, 0.36, -0.5, 1.9,-0.08, 0.18, -0.77, -1.9, 1.1, -0.52, 1.92, -1.35, -1.07, 1.37, -0.34, -0.91, 0.73, -0.57, -0.61, 0.6, 2.34, -0.61,

66


-0.73, -1.08, 0.01, 0.39, 0.55, -0.11, -1.92, -0.2, 2.06, -0.53, 0.69, -0.7, 0.96, -0.45, -0.68, -0.67)

R>A3=c(0.86, -1.8, 0.05, -0.15, -2.34, 0.96, 0.01, 0.64, -0.31, 0.91, 0.29, 1.62, -0.57, 0.51, 0.48, 0.71,-0.24, 2.26, 0, 1.37, -0.31, -0.6, 0.87, 0.43, -1.88, -1.09, -1.25, 0.09, -0.5, 1.11, 1.07, -0.52, 0.39, -0.57,-0.25, 0.86, 0.79, -0.5, -0.03, -0.17, -0.3, -0.73, 0.63, 1.02, -0.39, -0.86, 0.02, 0.32, 0.65, 2.05)

R>A4=c(-0.5, -0.98, -0.56, -0.75, -0.08, 0.13, 0.45, -1.85, 0.49, 0.52, -0.92, -0.42, 0.86, 0.99, -0.44,-0.28, -1.13, -0.56, 0.07, -1.4, -0.66, 0.6, -0.15, 0.85, -1.35, 0.18, -2.15, -0.41, 1.67, 0.9, 1.16, 0.45, -0.22,-1.14, -0.39, 1.78, -1.56, -0.06, 0.96, -1.02, 2.18, -0.1, 2.17, -0.38, -0.62, -0.03, -0.16, -0.9, 0.18, 0.17)

R>A5=c(-0.96, -1.19, -0.11, 0.09, -1.96, 1.61, 1.2, 0.41, -1.02, 1.03, 1.75, 0.28, 0.52, -2.09, 1.46, -1.29,-0.04, -2.03, -0.48, 0.21, 0.67, 0.17, -0.28, 0.39, 0.79, 0.59, 0.32, 1.04, -0.04, 0.59, 0.98, 0.91, -0.13, 1.37,-1.19, 1.49, -0.37, 0.56, 0.91, -0.11, 0.26, -0.59, -0.62, 1.1, -0.96, 0.74, -1.88, -0.75, 1.43, 1.16)

R>B1=c(0.16, 1.06, 0.45, -0.42, 0.67, -0.75, 1.42, -1.45, 1.04, -0.58, 1.44, 1.02, 0.04, -1.56, -0.49,-0.49, 0.26, -0.67, 0, 0.83, -0.52, 1.36, 0.86, 0.51, -1.58, -0.35, -0.51, 1.51, 1.16, 0.38, 1.6, 0.01, 1.03, 1.74,-1, 0.1, -0.27, 0.58, -0.47, -0.71, -1.49, 1.23, 1.46, 0.72, -1.75, -0.75, -1.09, 0.99, -2.25, -0.43)

R>B2=c(0.99, -0.95, -0.37, -1.47, -0.43, 0.33, 0.94, 0.67, -0.47, 0.06, -2.65, 0.18, -1.79, -1.35, 1.68,0.85, 0.08, -1.55, 1.13, 1.48, -0.8, -0.39, -0.78, 0.65, -0.28, 0.34, 1.22, -0.81, 1.37, 0.3, -1.46, 1.04, 0.94,1.09, 0.25, 2.68, 2.2, -1.58, 0.75, -0.56, 0.96, -0.57, -1.45, 0.18, -0.34, -0.32, 0.12, 0.56, 0.42, -1.43)

R>B3=c(0.98, -0.18, 1, -1.51, -1.41, 0.89, -0.57, -0.51, -1.99, -1.2, 0.23, -1.47, 0.6, 1.04, -1.05, 0.2,1.76, -0.54, 0.5, 0, -0.6, - 0.81, -0.8, 0.57, -0.36, -0.25, -0.07, -1.25, -0.7, 0.18, -1, 0.51, -0.06, -1.39, 1.84,-0.59, 0.4, -0.88, -1.2, 1.28, 2.94, -0.98, -0.54, -0.76, 0.29, 0.08, -0.28, 0.3, 0.48, 2.58)

R>B4=c(-1.86, -1.82, 0.8, 0.7, -0.66, 1.61, -0.44, 0.83, 2.13, -0.41, 1.62, 0.18, 0.03, 1.67, 0.68, -1.51,-0.19, 0.38, -0.13, -0.49, 0.29, -1.67, 2.27, 0.18, 0.73, 0.21, -0.84, -0.03, 0.1, -0.1, 2.12, -0.17, 0.59, 2.39,-0.32, 1.81, 1.51, -0.13, -0.75, -0.39, -0.03, 0.19, 0.91, -0.57, 0.4, -0.15, -1.36, -0.43, -0.37, 0.8)

R >B5=c(1.11, 0.08, 0.4, 0.21, -0.48, -0.89, -0.44, 0.49, -1.91, 1.36, -0.53, -1.17, -1.06, 1.08, -1.92,-0.85, -0.93, 1.2, 0.26, 0.06, -0.8, 0.71, 0.86, -0.38, -2.23, -0.59, -1.34, -0.3, -0.38, -2.34, -0.57, -1.9, -0.48,-1.59, 0.85, 0.61, 1.92, -0.17, -1.75, 0.09, 0.98, -0.98, 1.07, -0.41, -0.38, 1.44, 0.05, -1.47, 0.79, 0.73)



R>DA<-dist.quant(A, method = 1)67


Table 1. SUMMARY OF RESULT FOR PERMUTATION METHOD FOR HOTELLING T-SQUARED ANDRANDOMIZATION METHOD USING CAUCHY DISTRIBUTED DATA

Sample SizePermutation Method forHotelling T-Squared

Randomization Method forRV coefficient

Test Statistic Value P-valueTest Statistic value(Observation)

P-vale

5 3.50 0.15 0.44 0.71

10 1.94 0.07 0.36 0.75

20 0.95 0.48 0.10 0.49

30 1.32 0.23 0.31 0.29

40 0.45 0.99 0.10 0.63

50 0.9 0.52 0.69 0.40

STANDARD DEVIATION 1.09 0.22

Table 2. SUMMARY OF RESULT FOR PERMUTATION METHOD FOR HOTELLING T-SQUARED ANDRANDOMIZATION METHOD USING NORMALLY DISTRIBUTED DATA

Sample SizePermutation Method forHotelling T-Squared

Randomization Method forRV coefficient

Test Statistic Value P-valueTest Statistic value(Observation)

P-vale

5 0.90 0.56 0.64 0.35

10 1.71 0.20 0.68 0.59

20 1.06 0.40 0.49 0.57

30 0.64 0.67 0.74 0.81

40 0.68 0.63 0.89 0.89

50 1.18 0.31 0.92 0.36

STANDARD DEVIATION 0.39 0.16




Monte-Carlo test







0.3587462045 0.9091964191 0.0005585318

68


3.3 Application of the Methodology

In this section we will apply the methodology using a real experimental example. Consider measuringthe equality in weight of two groups of broiler birds (see [16]). Sixty broiler birds were divided intotwo groups, one group A was fed with the first feed while group B was fed with another type of feed.The weights of the birds were recorded weekly in kilograms for a period of the eight weeks. The dataextracted from [21] was presented at the Appendix.

The hypothesis is stated as

H0 : There is no difference in the weight of the two groups of broiler birds

H1 : There is difference in the weight of the two groups of broiler birds


R>A=matrix(c(0.3, 0.55, 0.9, 1.15, 1.4, 1.7, 1.8, 1.65, 0.2, 0.6, 0.85, 1.05, 1.55, 2, 2.25, 2.65, 0.25, 0.6,0.9, 1.38, 1.85, 2.3, 2.7, 3.1, 0.3, 0.75, 1.1, 1.6, 2.1, 2.6, 3.3, 3.5, 0.4, 0.3, 1.15, 1.55, 2.05, 2.55, 3.15, 3.3,0.2, 0.6, 0.85, 1.05, 1.25, 1.5, 1.95, 2.3, 0.3, 0.6, 0.85, 1.25, 1.65, 2, 2.3, 2.5, 0.35, 0.65, 1, 1.55, 2.1, 2.5,3.1, 3.5, 0.15, 0.45, 0.65, 1.05, 1.45, 1.9, 2.35, 2.65, 0.25, 0.65, 0.9, 1.4, 1.9, 2.45, 2.85, 3.4, 0.2, 0.55,0.85, 1.2, 1.55, 1.7, 1.95, 2.2, 0.25, 0.6, 0.9, 1.25, 1.55, 2, 2.4, 2.9, 0.25, 0.55, 0.8, 1.1, 1.25, 1.5, 1.55, 1.7,0.25, 0.5, 0.9, 1.35, 1.8, 2.1, 2.5, 3.1, 0.2, 0.5, 0.75, 1.15, 1.55, 2.1, 2.55, 2.8, 0.25, 0.65, 0.9, 1.38, 1.85,2.2, 2.6, 2.85, 0.3, 0.65, 0.85, 1.3, 1.75, 2, 2.55, 2.75, 0.2, 0.55, 0.85, 1.2, 1.55, 1.8, 2.1, 2.1, 0.15, 0.5, 0.9,1.18, 1.45, 1.95, 2.35, 2.65, 0.15, 0.5, 0.75, 1.15, 1.55, 1.9, 2.2, 2.5, 0.25, 0.5, 0.8, 1.18, 1.55, 2, 2.45, 2.7,0.2, 0.5, 0.75, 1.03, 1.3, 1.8, 1.95, 2.5, 0.1, 0.23, 0.3, 0.35, 0.5, 0.65, 1.15, 1.6, 0.2, 0.6, 0.85, 1.05, 1.55, 2,2.25, 2.65, 0.2, 0.65, 0.85, 1.2, 1.55, 2, 2.25, 2.65, 0.25, 0.50, 0.75, 0.95, 1.15, 1.45, 1.75, 1.8, 0.05, 0.3,0.35, 0.45, 0.5, 0.55, 0.6, 0.75, 0.2, 0.6, 0.8, 1.2, 1.65, 2, 2.35, 2.7, 0.15, 0.5, 0.9, 1.18, 1.45, 1.95, 2.35,2.65, 0.15, 0.5, 0.75, 1.15, 1.55, 1.9, 2.2, 2.5), nrow=30, byrow=T, dimnames = list(1:30, c(”week1A”,”week2A”, ”week3A”, ”week4A”, ”week5A”, ”week6A”, ”week7A”, ”week8A”)))

R>B=matrix(c(0.2, 0.45, 0.6, 0.98, 1.35, 1.7, 1.95, 1.8, 0.2, 0.45, 0.55, 0.83, 1.1, 1.6, 2, 2.5, 0.2, 0.4,0.7, 1.03, 1.35, 2, 2.5, 3.1, 0.3, 0.4, 0.5, 0.85, 1.2, 1.55, 1.8, 2.6, 0.2, 0.35, 0.55, 0.83, 1.1, 1.05, 1.25, 1.7,0.25, 0.55, 0.85, 1.15, 1.45, 2.15, 2.8, 3.55, 0.25, 0.45, 0.75, 1.13, 1.5, 2.2, 2.95, 3.55, 0.2, 0.4, 0.65, 0.78,0.9, 1.2, 1.35, 1.55, 0.15, 0.4, 0.6, 0.9, 1.2, 1.5, 1.65, 1.85, 0.25, 0.45, 0.7, 0.85, 1, 1.2, 1.3, 1.2, 0.2, 0.4,0.6, 1, 1.4, 2, 2.5, 2.75, 0.2, 0.4, 0.55, 0.88, 1.2, 1.35, 1.6, 2, 0.2, 0.45, 0.65, 0.93, 1.2, 1.5, 1.8, 1.8, 0.3,0.55, 0.85, 1.08, 1.3, 2.4, 2.6, 2.95, 0.1, 0.25, 0.4, 0.6, 0.8, 1.15, 1.35, 1.75, 0.25, 0.5, 0.75, 1.05, 1.35, 1.9,2.45, 2.8, 0.15, 0.4, 0.6, 0.9, 1.2, 1.5, 1.65, 1.85, 0.25, 0.45, 0.7, 0.85, 1, 1.2, 1.3, 1.2, 0.25, 0.5, 0.8, 1.08,1.35, 1.9, 2.25, 2.6, 0.1, 0.25, 0.6, 0.8, 1, 1.35, 1.7, 2.05, 0.2, 0.45, 0.75, 1.03, 1.3, 1.6, 1.8, 2.05, 0.3, 0.55,0.75, 1.08, 1.4, 2.1, 2.25, 2.3, 0.1, 0.3, 0.75, 0.88, 1, 1.45, 1.85, 2.2, 0.2, 0.4, 0.6, 0.98, 1.35, 1.75, 2.25,2.6, 0.2, 0.5, 0.6, 0.93, 1.25, 1.55, 1.9, 2.25, 0.2, 0.55, 0.7, 1.13, 1.55, 2, 2.45, 2.75, 0.15, 0.35, 0.55, 0.9,1.25, 1.7, 1.9, 2.1, 0.2, 0.5, 0.7, 1, 1.45, 1.7, 2.1, 2.4, 0.3, 0.4, 0.5, 0.85, 1.2, 1.55, 1.8, 2.6, 0.2, 0.35, 0.55,0.83, 1.1, 1.05, 1.25, 1.7), nrow=30, byrow=T, dimnames = list(1:30, c(”week1B”, ”week2B”, ”week3B”,”week4B”, ”week5B”, ”week6B”, ”week7B”, ”week8B”)))



69


Test stat: 6.0409

Numerator df: 8

Denominator df: 51

Permuation P-value: 0


In putting data generated using Normal distribution for sample size 50 on the R 2.13.0 commandwindow, where matrix A and B are the objects or samples to be measured. Also, it should be noted thatobjects A1, A2, A3, A4 and A5 are objects of matrix A while objects B1, B2, B3, B4,and B5 are objectsof matrix B. The class distance of matrices A and B based on the canonical measure is labelled DA andDB respectively. The function RVdist.randtest was employed to call the randomization method for RVcoefficient statistic for 10, 000 randomizations.

Inputting the data in Table 1 and Table 2 on R 2.13.0 command window; [10], where week1A, week2A,week3A, week4A, week5A, week6A, week7A and week8A are objects of matrix A; that is, broiler birdsfed with feed A while week1B, week2B, week3B, week4B, week5B, week6B, week7B and week8B areobjects of matrix B; that is, broiler birds fed with feed B.

R>week1A=c(0.30, 0.20, 0.25, 0.30, 0.40, 0.20, 0.30, 0.35, 0.15, 0.25, 0.20, 0.25, 0.25, 0.25, 0.20,0.25, 0.30, 0.20, 0.15, 0.15, 0.25,0.20, 0.10, 0.20, 0.20, 0.25, 0.05, 0.20, 0.15, 0.15)

R>week2A=c(0.55, 0.60, 0.60, 0.75, 0.80, 0.60, 0.60, 0.65, 0.45, 0.65, 0.55, 0.60, 0.55, 0.50, 0.50,0.65, 0.65, 0.55, 0.50, 0.50, 0.50, 0.50, 0.25, 0.60, 0.65, 0.50, 0.30, 0.60, 0.50, 0.50)

R>week3A=c(0.90, 0.85, 0.90, 1.10, 1.15, 0.85, 0.85, 1.00, 0.65, 0.90, 0.85, 0.90, 0.80, 0.90, 0.75,0.90, 0.85, 0.85, 0.90, 0.75, 0.80, 0.75, 0.30, 0.85, 0.85, 0.75, 0.35, 0.80, 0.90, 0.75)

R>week4A=c(1.15, 1.05, 1.38, 1.60, 1.55, 1.05, 1.25, 1.55, 1.05, 1.40, 1.20, 1.25, 1.10, 1.35, 1.15,1.38, 1.30, 1.20, 1.18, 1.15, 1.18, 1.03, 0.35, 1.05, 1.20, 0.95, 0.45, 1.20, 1.18, 1.15)

R>week5A=c(1.40, 1.55, 1.85, 2.10, 2.05, 1.25, 1.65, 2.10, 1.45, 1.90, 1.55, 1.55, 1.25, 1.80, 1.55,1.85, 1.75, 1.55, 1.45, 1.55, 1.55, 1.30, 0.50, 1.55, 1.55, 1.15, 0.50, 1.65, 1.45, 1.55)

R>week6A=c(1.70, 2.00, 2.30, 2.60, 2.55, 1.50, 2.00, 2.50, 1.90, 2.45, 1.70, 2.00, 1.50, 2.10, 2.10,2.20, 2.00, 1.80, 1.95, 1.90, 2.00, 1.80, 0.65, 2.00, 2.00, 1.45, 0.55, 2.00, 1.95, 1.90)

R>week7A=c(1.80, 2.25, 2.70, 3.30, 3.15, 1.95, 2.30, 3.10, 2.35, 2.85, 1.95, 2.40, 1.55, 2.50, 2.55,2.60, 2.55, 2.10, 2.35, 2.20, 2.45, 1.95, 1.15, 2.45, 2.25, 1.75, 0.60, 2.35, 2.35, 2.20)

R>week8A=c(1.65, 2.65, 3.10, 3.50, 3.30, 2.30,2.50, 3.50, 2.65, 3.40, 2.20, 2.90, 1.70, 3.10, 2.80, 2.85,2.75, 2.10, 2.65, 2.50, 2.70, 2.50, 1.60, 2.70, 2.65, 1.80, 0.75, 2.70, 2.65, 2.65)

R>week1B=c(0.20, 0.20, 0.20, 0.30, 0.20, 0.25, 0.25, 0.20, 0.15, 0.25, 0.20, 0.20, 0.20, 0.30, 0.10,0.25, 0.15, 0.25, 0.25, 0.10, 0.20, 0.30, 0.10, 0.20, 0.20, 0.20, 0.15, 0.20, 0.30, 0.20)

R>week2B=c(0.45, 0.45, 0.40, 0.40, 0.35, 0.55, 0.45, 0.40, 0.40, 0.45, 0.40, 0.40, 0.45, 0.55, 0.25,0.50, 0.40, 0.45, 0.50, 0.25, 0.45, 0.55, 0.30, 0.40, 0.50, 0.55, 0.35, 0.50, 0.40, 0.35)

R>week3B=c(0.60, 0.55, 0.70, 0.50, 0.55, 0.85, 0.75, 0.65, 0.60, 0.70, 0.60, 0.55, 0.65, 0.85, 0.40,0.75, 0.60, 0.70, 0.80, 0.60, 0.75, 0.75, 0.75, 0.60, 0.60, 0.70, 0.55, 0.70, 0.50, 0.55)

R>week4B=c(0.98, 0.83, 1.03, 0.85, 0.83, 1.15, 1.13, 0.78, 0.90, 0.85, 1.00, 0.88, 0.93, 1.08, 0.60,1.05, 0.90, 0.85, 1.05, 0.80, 1.03, 1.08, 0.88, 0.98, 0.93, 1.13, 0.90, 1.08, 0.85, 0.83)

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R>week5B=c(1.35, 1.10, 1.35, 1.20, 1.10, 1.45, 1.50, 0.90, 1.20, 1.00, 1.40, 1.20, 1.20, 1.30, 0.80,1.35, 1.20, 1.00, 1.35, 1.00, 1.30, 1.40, 1.00, 1.35, 1.25, 1.55, 1.25, 1.45, 1.20, 1.10)

R>week6B=c(1.70, 1.60, 2.00, 1.55, 1.05, 2.15, 2.20, 1.20, 1.50, 1.20, 2.00, 1.35, 1.50, 2.40, 1.15,1.90, 1.00, 1.20, 1.90, 1.35, 1.60, 2.10, 1.45, 1.75, 1.55, 2.00, 1.70, 1.70, 1.55, 1.05)

R>week7B=c(1.95, 2.00, 2.50, 1.80, 1.25, 2.80, 2.95, 1.35, 1.65, 1.30, 2.50, 1.60, 1.80, 2.60, 1.35,2.45, 1.65, 1.30, 2.35, 1.70, 1.80, 2.25, 1.85, 2.25, 1.90, 2.45, 1.90, 2.10, 1.80, 1.20)

R>week8B=c(1.80, 2.50, 3.10, 2.60, 1.70, 3.55, 3.50, 1.55, 1.85, 1.20, 2.75, 2.00, 1.80, 2.95, 1.75,2.80, 1.85, 0.20, 2.60, 2.05, 2.05, 2.30, 2.20, 2.60, 2.25, 2.75, 2.10, 2.40, 2.60, 1.70)

R>A<-matrix(c(week1A, week2A, week3A, week4A, week5A, week6A, week7A, week8A), nrow =8, byrow = TRUE)

R>B<-matrix(c(week1B, week2B, week3B, week4B, week5B, week6B, week7B, week8B), nrow = 8,byrow = TRUE)

It should be of interest to note that the class distance of matrices A and B as defined above are based oncanonical measure (method = 1), labelled as DA and DB respectively.

R>DA <-dist.quant(A, method = 1)



The result of the Randomization method for RV coefficient statistic for 10,000 permutations wasobtained as given:

Monte-Carlo test




Simulated p-value: 9.999e-05



4.99551224 0.14711300 0.02872499

4. DISCUSSION

The summary of the analysis using Cauchy distributed data as displayed in Table I revealed that therandomization method for RV coefficient performed better than the permutation method for HotellingT-Squared test in terms of relative efficiency of the test statistic measure for Cauchy distributed data setsince the standard deviation of the test statistic measure for randomization method for RV coefficientobtained as 0.22 was less than the standard deviation of the test statistic measure for the permutationmethod for Hotelling T 2 obtained as 1.09. This result implies that the variation on the test statistic measurefor the randomization method for RV coefficient was smaller than that of the permutation method for

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Hotelling T-Squared across the sample sizes using Cauchy distributed data.

Also, the summary of the analysis using Normally distributed data set as displayed in Table II showedthat the randomization method for RV coefficient performed better than the permutation method forHotelling T-Squared test in terms of relative efficiency of the test statistic measure for Normally distributeddata set since the standard deviation of the test statistic measure for randomization method for RVcoefficient obtained as 0.16 was less than the standard deviation of the test statistic measure for thepermutation method for Hotelling T 2 obtained as 0.39. This result implies that the variation on the teststatistic measure for the randomization method for RV coefficient was smaller than that of the permutationmethod for Hotelling T-Squared across the sample sizes using Normally distributed data.

The result of the permutation method for Hotelling T-squared analysis for measuring the equality inweight of two groups of broiler birds showed the reference value of the test statistic measure T 2 to be6.04 and a corresponding P-value of 0.00 which falls on the rejection region of the hypothesis assuminga 95% confidence level. This result implies that there exists evidence of statistical significance in theweight of the two groups of broiler birds. Also, the result of the randomization method for RV coefficientfor measuring the equality in weight of two groups of broiler birds revealed the test statistic value as0.99 and a corresponding p-value of 0.00 which fall on the rejection region of the hypothesis assuminga 95% confidence level. This result implies that there exists evidence of statistical significance in theweight of the two groups of broiler birds. Hence, it was found that the conclusion of rejecting the nullhypothesis as revealed by the result obtained from the permutation method for Hotelling T-Squared andthe randomization method concur with result obtained by [21], where they employed the Mantel teststatistic using the same set of data.

5. CONCLUSION

This study compared the permutation method for Hotelling T-Squared and the randomization method forRV coefficient in terms of relative efficiency of their test statistic measure using the Cauchy distributionand the Normal distribution for sample size 5, 10, 20, 30, 40, and 50. The findings of this studyrevealed that the randomization method for RV coefficient statistic performed better than the permutationmethod for Hotelling T-Squared in terms of the relative efficiency of their test statistic value for the twodistributions discussed in this study. This result implies that the variation in the test statistic measure ofthe randomization method for RV coefficient statistic was lesser than that of the permutation method forHotelling T-Squared across the sample sizes. Also, the result of applying the randomization method forRV coefficient and the permutation method for Hotelling T-Squared for measuring the equality in weightof two groups of broiler birds showed that there exists significant difference on the weight of the twogroups of birds. It was observed that this result is in line with the result obtained by [21]; where theyused Mantel test analysis using the same data set. However, we conclude that the randomization methodfor RV coefficient was a better method than the permutation method for Hotelling T-Squared in termsof relative efficiency of the test statistic measure. Based on the finding of this study we recommend forfuture research comparison of the methods discussed in this study in terms of the probability of obtaininga more extreme value than the observed value of the test statistic in some reference distribution under aspecified null hypothesis using other distributions not discussed in this study.

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COMPARISON OF RANDOMIZATION METHOD FOR RV COEFFICIENT AND PERMUTATION METHOD FOR HOTELLING T-SQUARE

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