Post on 27-Apr-2023
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: amaro4baya@yahoo.com
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
37
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
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
38
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)
39
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
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].
40
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
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.
41
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
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)
42
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
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
43
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
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)
The result of the Hotelling T-squared test statistic for 10,000 permutations was obtained as given:
Test stat: 1.9366
Numerator df: 5
Denominator df: 14
Permuation P-value: 0.0747
Number of permutations : 10000
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
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
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>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 statistic for 10,000 randomizations was obtained as given:
Monte-Carlo test
Call: as.randtest(sim = res[-1], obs = obs)
Observation: 0.3603113
Based on 10000 replicates
Simulated p-value: 0.749625
Alternative hypothesis: greater
Std.Obs Expectation Variance
-0.80701939 0.46562060 0.01702808
In putting data generated using Cauchy 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. The function hotelling.test wasemployed to call the Hotelling T-Squared test for 10, 000 permutations.
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,
45
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
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)
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: 0.94862
Numerator df: 5
Denominator df: 34
Permuation P-value: 0.4808
Number of permutations : 10000
In putting data generated using Cauchy 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 method for RVcoefficient for 10, 000 randomizations.
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
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
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)
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: 0.1040866
Based on 10000 replicates
Simulated p-value: 0.4914509
Alternative hypothesis: greater
47
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
Std.Obs Expectation Variance
-0.4505670 0.2651695 0.1278144
In putting data generated using Cauchy distribution for sample size 30 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(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)
The result of the Hotelling T-squared test statistic for 10,000 permutations was obtained as given:
Test stat: 1.3247
Numerator df: 5
Denominator df: 54
Permuation P-value: 0.2277
Number of permutations : 10000
In putting data generated using Cauchy distribution for sample size 30 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.
48
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
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)
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)
49
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
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: 0.3083813
Based on 10000 replicates
Simulated p-value: 0.2930707
Alternative hypothesis: greater
Std.Obs Expectation Variance
-0.04104809 0.32172092 0.10560875
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. The function hotelling.test wasemployed to call the Hotelling T-Squared test for 10, 000 permutations.
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
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
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: 0.44659
Numerator df: 5
Denominator df: 74
Permuation P-value: 0.9938
Number of permutations : 10000
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
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
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)
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: 0.09736166
Based on 10000 replicates
Simulated p-value: 0.630037
Alternative hypothesis: greater
Std.Obs Expectation Variance
-0.80756583 0.34875727 0.09690798
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. The function hotelling.test wasemployed to call the Hotelling T-Squared test for 10, 000 permutations.
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
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
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)
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: 0.90172
Numerator df: 5
Denominator df: 94
Permuation P-value: 0.5242
Number of permutations : 10000
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
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
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)
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: 0.6942657
Based on 10000 replicates
Simulated p-value: 0.40026
Alternative hypothesis: greater54
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
Std.Obs Expectation Variance
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)
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: 0.89616
Numerator df: 5
Denominator df: 4
Permuation P-value: 0.5597
Number of permutations : 10000
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
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
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)
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: 0.6358249
Based on 10000 replicates
Simulated p-value: 0.3487651
Alternative hypothesis: greater
Std.Obs Expectation Variance
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
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
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)
The result of the Hotelling T-squared test statistic for 10,000 permutations was obtained as given:
Test stat: 1.7077
Numerator df: 5
Denominator df: 14
Permuation P-value: 0.2015
Number of permutations : 10000
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
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
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>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: 0.6793432
Based on 10000 replicates
Simulated p-value: 0.590341
Alternative hypothesis: greater
Std.Obs Expectation Variance
-0.27193770 0.70666089 0.01009138
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. The function hotelling.test wasemployed to call the Hotelling T-Squared test for 10, 000 permutations.
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
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
-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)
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: 1.0641
Numerator df: 5
Denominator df: 34
Permuation P-value: 0.3953
Number of permutations : 10000
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
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
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)
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: 0.4929715
Based on 10000 replicates
Simulated p-value: 0.5664434
Alternative hypothesis: greater
Std.Obs Expectation Variance
0.23490131 0.46477650 0.01440696
In putting data generated using Normal distribution for sample size 30 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(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
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
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)
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: 0.63577
Numerator df: 5
Denominator df: 54
Permuation P-value: 0.6712
Number of permutations : 10000
In putting data generated using Normal distribution for sample size 30 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(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
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
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)
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: 0.7426857
62
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
Based on 10000 replicates
Simulated p-value: 0.8050195
Alternative hypothesis: greater
Std.Obs Expectation Variance
-0.946962198 0.804399389 0.004247146
In putting data generated using Normal 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. The function hotelling.test wasemployed to call the Hotelling T-Squared test for 10, 000 permutations.
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)
The result of the Hotelling T-squared test statistic for 10,000 permutations was obtained as given:
Test stat: 0.68191
Numerator df: 5
Denominator df: 74
Permuation P-value: 0.633
Number of permutations : 10000
In putting data generated using Normal distribution for sample size 40 on the R 2.13.0 command
63
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
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
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
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)
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: 0.8912391
Based on 10000 replicates
Simulated p-value: 0.8854115
Alternative hypothesis: greater
Std.Obs Expectation Variance
-1.139897042 0.915506951 0.000453243
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. The function hotelling.test wasemployed to call the Hotelling T-Squared test for 10, 000 permutations.
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
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
-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)
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: 1.1849
Numerator df: 5
Denominator df: 94
Permuation P-value: 0.3092
Number of permutations : 10000
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 labeled DA andDB respectively. The function RVdist.randtest was employed to call the randomization method for RVcoefficient for 10, 000 randomizations.
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
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
-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>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)67
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
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
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: 0.9176748
Based on 10000 replicates
Simulated p-value: 0.3625637
Alternative hypothesis: greater
Std.Obs Expectation Variance
0.3587462045 0.9091964191 0.0005585318
68
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
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
In putting data generated using Normal 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. The function hotelling.test wasemployed to call the Hotelling T-Squared test for 10, 000 permutations.
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”)))
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:
69
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
Test stat: 6.0409
Numerator df: 8
Denominator df: 51
Permuation P-value: 0
Number of permutations : 10000
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)
70
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
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)
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 permutations wasobtained as given:
Monte-Carlo test
Call: as.randtest(sim = res[-1], obs = obs)
Observation: 0.9937748
Based on 10000 replicates
Simulated p-value: 9.999e-05
Alternative hypothesis: greater
Std.Obs Expectation Variance
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
71
SOP TRANSACTIONS ON STATISTICS AND ANALYSIS
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.
72
Comparison of Randomization method for RV coefficient and Permutation Method for Hotelling T-Squared
References
[1] M. J. Anderson, “Permutation tests for univariate or multivariate analysis of variance and regression,”Canadian Journal of Fisheries and Aquatic Sciences, vol. 58, no. 3, pp. 626–639, 2001.
[2] K. R. CLARKE, “Non-parametric multivariate analyses of changes in community structure,” Aus-tralian journal of ecology, vol. 18, no. 1, pp. 117–143, 1993.
[3] K. J. Gaston and B. H. McArdle, “The temporal variability of animal abundances: measures, methodsand patterns,” Philosophical Transactions of the Royal Society of London. Series B: BiologicalSciences, vol. 345, no. 1314, pp. 335–358, 1994.
[4] M. J. Anderson, “A new method for non-parametric multivariate analysis of variance,” Australecology, vol. 26, no. 1, pp. 32–46, 2001.
[5] E. S. Eddington, “Randomization tests, 1995.”[6] B. Manly and B. Randomization, “Monte carlo methods,” Randomization, bootstrap, and Monte
Carlo methods in biology, pp. 69–78, 1997.[7] P. I. Good, “Resampling methods,” 2000.[8] S. R. A. Fisher, “The coefficient of racial likeness” and the future of craniometry,” Journal of the
Royal Anthropological Institute of Great Britain and Ireland, vol. 66, pp. 57–63, 1936.[9] E. Lehmann, “Nonparametric statistical methods based on ranks. 1975.”
[10] M. D. Ernst et al., “Permutation methods: A basis for exact inference,” Statistical Science, vol. 19,no. 4, pp. 676–685, 2004.
[11] D. A. Jackson and K. M. Somers, “Are probability estimates from the permutation model of mantel’stest stable?,” Canadian Journal of Zoology, vol. 67, no. 3, pp. 766–769, 1989.
[12] C. Aronu and G. Ebuh, “Application of mantels permutation technique on asphalt production innigeria,” International Journal of Statistics and Applications, vol. 3, no. 3, pp. 81–85, 2013.
[13] C. Aronu, G. Ebuh, G. Ogbogbo, and A. Bilesanmi, “Partial mantel analysis on estimating theresemblance of students performance,” International Journal of Scientific and Engineering Research,vol. 4, no. 7, pp. 2269–2273, 2013.
[14] M. A. Greenacre, “Simple permutation test for clusteredness,” Barcelona Graduate School ofEconomics Working Paper Series, vol. 555, pp. 1–18, 2011.
[15] C. O. A. F. I. Aronu and C. A. Ofodile, “Stress management amongst traders and public civil workersin nigeria,” Advances in Economics (AE), vol. 2, no. 1, pp. 1–10, 2014.
[16] C. Aronu and A. Bilesanmi, “Assessing the knowledge, attitude and factors affecting team buildingamongst health workers in nigeria using the permutation method for hotelling t-squared analysis,”American Journal of Theoretical and Applied Statistics, vol. 2, no. 6, pp. 184–190, 2013.
[17] C. Aronu, G. Ebuh, G. Ogbogbo, and A. Bilesanmi, “Measuring the resemblance in weight of twogroup of broiler birds using the mantel test analysis,” International Journal of Agriculture andForestry, vol. 3, no. 4, pp. 145–151, 2013.
[18] B. Manly, Randomization and Monte Carlo methods in biology. Chapman & Hal, 1991.[19] Y. Escoufier, “Operator related to a data matrix: a survey,” in Compstat 2006-Proceedings in
Computational Statistics, pp. 285–297, Springer, 2006.[20] P. Robert and Y. Escoufier, “A unifying tool for linear multivariate statistical methods: the rv-
coefficient.,” Journal of the Royal Statistical Society. Series C (Applied Statistics), vol. 3, pp. 257–265, 1976.
[21] S. Dray, A.-B. Dufour, et al., “The ade4 package: implementing the duality diagram for ecologists,”Journal of statistical software, vol. 22, no. 4, pp. 1–20, 2007.
73