Research Article An Analysis of Bank Service Satisfaction

Research ArticleAn Analysis of Bank Service Satisfaction Based onQuantile Regression and Grey Relational Analysis

Wen-Tsao Pan and Yungho Leu

Department of Information Management National Taiwan University of Science and Technology Taipei 10607 Taiwan

Correspondence should be addressed to Yungho Leu yhlcsntustedutw

Received 6 January 2016 Accepted 13 March 2016

Academic Editor Feng Yang

Copyright copy 2016 W-T Pan and Y LeuThis is an open access article distributed under theCreative CommonsAttribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Bank service satisfaction is vital to the success of a bank In this paper we propose to use the grey relational analysis to gauge thelevels of service satisfaction of the banks With the grey relational analysis we compared the effects of different variables on servicesatisfaction We gave ranks to the banks according to their levels of service satisfaction We further used the quantile regressionmodel to find the variables that affected the satisfaction of a customer at a specific quantile of satisfaction level The result of thequantile regression analysis provided a bank manager with information to formulate policies to further promote satisfaction ofthe customers at different quantiles of satisfaction level We also compared the prediction accuracies of the regression models atdifferent quantiles The experiment result showed that among the seven quantile regression models the median regression modelhas the best performance in terms of RMSE RTIC and CE performance measures

1 Introduction

In response to the requirement of opening market and topromote the economic growth Taiwanese government hasactively formulated policies to promote the financial liber-alization in Taiwan As a result many mergers of differentbanks have occurred in the past few years Bankmergers maychange the perception of a customer on the service qualityof a bank Therefore a bank manager should improve theservice quality of his clerks in order to promote the customerrsquossatisfaction High customer satisfaction on a bank can attractcustomers to continue doing their business with the bankand thus establishes the customersrsquo loyalty to the bank Toeffectively improve the service quality of a bank so as topromote the customerrsquos satisfaction has become an importantissue of the banking industry today

Different from the existing studies on bank servicesatisfaction [1 2] we use the data on service satisfaction ofthe public and private banks in Taiwan in this study Thequestionnaire for data collection was designed by the LouisHarris International in 1995 In this study we first collectedthe data on service satisfaction using the questionnaireThen we used the grey relational analysis to examine thecustomer satisfaction on the service of the banks including

the public and the private banks in Taiwan With the greyrelational analysis we found the variables that affected thelevels of service satisfaction of the public banks and theprivate banks We also ranked the banks according to theirlevels of service satisfaction Finally we conducted a quantileregression analysis on the 36 banks of the dataset to find thevariables that affected the levels of satisfaction for customersat different quantiles of service satisfaction The results canbe used by a bank manager to formulate different policies tofurther promote the satisfaction of the customers at differentquantile of service satisfaction Finally we compared theperformance of different quantile regressions in terms ofthree different measures of performance evaluation

The remainder of this paper is organized as followsSection 1 introduces the research motivation and purposeSection 2 reviews the literature on grey theory and quantileregression Section 3 discusses the dataset and results ofthe empirical study Section 4 gives the conclusion and thesuggestion of this paper

2 Research Method

21 Grey Relational Analysis First proposed by Deng [3] thegrey relational analysis can be used to measure the similarity

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2016 Article ID 1475148 9 pageshttpdxdoiorg10115520161475148

2 Mathematical Problems in Engineering

Table 1 An example dataset

Sequence 119896 = 1 119896 = 2 119896 = 3 119896 = 4 119896 = 5 119896 = 61199090

1000 1100 2000 2250 3000 40001199091

1000 1166 1834 2000 2340 30001199092

1000 1125 1075 1375 1625 17501199093

1000 1000 0700 0800 0900 1200

between two sequences of factors Assume that the state of asystem can be modeled by a sequence of factors Two systemstates are said to be similar if the values of their correspondingfactors are similar In a real life application a standardsequence is selected to calculate the relational grades of aset of inspected sequences Usually the standard sequenceis composed of the maximal values of each of the factors inthe current dataset By referring to the standard sequence thegrey relational analysis gives each inspected sequence a greyrelational grade (GRG) The more an inspected sequence issimilar to the standard sequence the larger the GRG of theinspected sequence is Since the standard sequence usuallyrepresents a state of the system with maximum performancea sequence with a large GRG is therefore better than asequence with a small GRG The grey relational analysis hasbeen widely used in performance evaluation [4ndash6]

The grey relational grade of an inspected sequence iscalculated through the following steps

Step 1 (define the sequences) To perform the grey relationalanalysis we need to define the sequences for comparisonTo illustrate we use Table 1 as an example Each columnin Table 1 represents a factor of the system state while eachrow in Table 1 represents a sequence of the system stateAmong all the sequences of the system state we choose orcreate a sequence called the standard sequence to measurethe performance of the other sequences which are called theinspected sequences A sequence119909

119894is denoted by a set 119909

119894(119896) |

119896 = 1 119899 Without loss of generality we choose 1199090as the

standard sequence

Step 2 (standardization) Since different factors may havedifferent ranges of values which are not comparable we needto standardize the values of all factors so that values fromdifferent factors are comparable One way of standardizationis to divide all the values of a factor by the maximum valueof the factor By doing so values of all factors are all less thanone and greater than zero

Step 3 (calculate the grey relational coefficients) The greyrelational coefficient of the 119896th factor of sequence 119909

119894is

calculated by the following equation

120585119894(119896)

=min119894min119896

10038161003816100381610038161199090 (119896) minus 119909119894 (119896)1003816100381610038161003816 + 120588max

119894max119896

10038161003816100381610038161199090 (119896) minus 119909119894 (119896)1003816100381610038161003816

10038161003816100381610038161199090 (119896) minus 119909119894 (119896)1003816100381610038161003816 + 120588max

119894max119896

10038161003816100381610038161199090 (119896) minus 119909119894 (119896)1003816100381610038161003816

(1)

Let Δ0119894(119896) denote |119909

0(119896) minus 119909

119894(119896)| Equation (1) can be written

as

120585119894(119896) =

min119894min119896Δ0119894 (119896) + 120588max

119894max119896Δ0119894 (119896)

10038161003816100381610038161199090 (119896) minus 119909119894 (119896)1003816100381610038161003816 + 120588max

119894max119896Δ0119894(119896) (2)

Note that to find the value ofmin119894min119896Δ0119894(119896) we find for

inspected sequence 119909119894 the minimum value of the differences

between factor 119896 of 1199090and factor 119896 of 119909

119894 119896 = 1 to 119899 Then we

find the minimum value among the minimum values of allthe inspected sequencesThe value ofmax

119894max119896|1199090(119896)minus119909

119894(119896)|

can be found in a similar wayThe rationale behind (2) is thatwe use the global minimum and the global maximum of thecomponentwise differences between the standard sequenceand all the inspected sequences to rescale the differencebetween a factor of an inspected sequence and the same factorof the standard sequence And then we use the rescaleddifferences also called the grey relational coefficients toreplace the values of the original factors The value of 120588belongs to (0infin) and is usually set to 05

Step 4 (find the grey relational grade) The grey relationalgrade 119903

119894of inspected sequence 119909

119894is defined to be the average

of all its grey relational coefficients The grey relational grade119903119894of 119909119894is calculated according to the following equation

119903119894=1

119899

119899

sum

119896=1

120585119894(119896) (3)

To illustrate how to calculate the grey relational grade of aninspected sequence we use the dataset in Table 1

Step 1 (define the sequence) We assume that 1199090is the

standard sequence

Step 2 (standardization) We assume that the factors of thedataset in Table 1 are comparable No standardization isrequired

Step 3 (calculate the grey relational coefficients) The compo-nentwise differences between the standard sequence and allthe inspected sequences are shown in the following

(1) Componentwise differences between 1199090and 119909

1

Δ01(1) = 0000

Δ01 (2) = 0066

Δ01(3) = 0166

Δ01(4) = 0250

Δ01(5) = 0660

Δ01(6) = 1000

(4)

Mathematical Problems in Engineering 3


2

Δ02(1) = 0000

Δ02 (2) = 0025

Δ02(3) = 0925

Δ02(4) = 0875

Δ02(5) = 1375

Δ02(6) = 2250

(5)


3

Δ03(1) = 0000

Δ03(2) = 0100

Δ03(3) = 1300

Δ03 (4) = 1450

Δ03(5) = 2100

Δ03(6) = 2800

(6)

We find the following vectors of differences

(1) Δ01= (0000 0066 0166 0250 0660 1000)

(2) Δ02= (0000 0025 0925 0875 1375 2250)

(3) Δ03= (0000 0100 1300 1450 2100 2800)

The global minimum and the global maximum of all thedifferences are 0000 and 2800 respectively Let 120588 = 05 Wewrite

120585119894 (119896) =

0 + (05) times (28)

Δ0119894(119896) + (05) times (28)

(7)

We then calculate the grey relational coefficients of theinspected sequence 119909

1in the following

1205851 (1) = 1000

1205851(2) = 09549

1205851(3) = 08939

1205851(4) = 08484

1205851(5) = 06796

1205851 (6) = 05833

(8)

Finally we find the grey relational grade (GRG) of theinspected sequence 119909

1in the following

1199031=1

6(1 + 09549 + 08939 + 08484 + 06796

+ 05833) = 08266

(9)

The grey relational grades of sequences 1199092and 119909

3can be

found in the same way

22 Quantile Regression Koenker and Bassett [7] proposedthe quantile regression method Quantile regression is usedto explore the effects of explanatory variables (119883) on theexplained variable (119884) at different quantiles of the explainedvariable Different from the conventional linear models (suchas the least squares method) that predict the mean of theexplained variable given a specific value of each explanatoryvariable a quantile regression model predicts the value ofthe explained variable at a specific quantile of the explainedvariable giving a specific value of each explanatory variableTherefore the quantile regression method facilitates thestudy of the effects of the explanatory variables on differentquantiles of the explained variable The quantile regressionmethod has been widely used in many applications [8ndash10]

3 Empirical Analysis

31 Sample Dataset and Variables Based on a questionnairedesigned by the Louis Harris International Taiwan we con-ducted a survey on the bank service quality of Taiwan in 2013We analyzed the satisfaction of bank service in Taiwan byusing the collected questionnaire dataThe dataset consists of8 public banks and 28 private banks The variables designedfor bank service satisfaction include ldquoclerk service attituderdquo(1198831) ldquocustomer-oriented servicerdquo (1198832) ldquoflexibility in han-dling customer inquiriesrdquo (1198833) ldquoservice efficiencyrdquo (1198834)ldquointerest rate favoring customersrdquo (1198835) ldquotransaction errorsrdquo(1198836) ldquoconvenience of branch locationsrdquo (1198837) ldquoservice trafficflowrdquo (1198838) and ldquoservice counter designrdquo (1198839) The value ofeach variable of a bank is obtained by averaging the scoresof the same variable on all the questionnaires for the bankThe dataset for the public bank is shown in Table 2 Thedescriptive statistics of the variables are shown in Table 3Thesource dataset is divided into three datasets The first datasetcontains all the 36 banks including the private and publicbanks the second dataset contains only the 28 private banksthe third dataset contains the 8 public banks whose data areshown in Table 2

32 Grey Relational Analysis on Bank Service SatisfactionThis study applied the Matlab tool box for grey relationalanalysis [11] to obtain a grey relational grade (GRG) for abank to represent the level of satisfaction of the bank In thisstudy we performed two analyses using the grey relationalanalysis In the first analysis we used the grey relationalanalysis to find the satisfaction levels of different variablesTo do that we first transposed the dataset so that the samplesof the dataset become the variables and the variables becomethe samples of the transposed dataset Figures 1(a) and 1(b)show the results of grey relational analysis for the eightpublic banks and the 28 private banks respectively The bolddotted lines represent the standard sequences and the thinlines represent the inspected sequences which are samplesto be evaluated The more an inspected sequence is closeto the standard sequence the larger its satisfaction gradeis According to Figure 1(a) the top three most satisfiedvariables among the nine surveyed variables for the publicbanks are ldquoconvenience of branch locationsrdquo (GRG = 08775)


Table 2 The dataset for public banks

Bank names 1198831 1198832 1198833 1198834 1198835 1198836 1198837 1198838 1198839

Farmers Bank of China 431 363 367 389 331 452 382 37 384Land Bank of Taiwan 398 344 352 374 377 462 267 287 365First Commercial Bank 404 352 331 373 341 463 361 329 331Taiwan Business Bank 369 314 321 338 328 481 364 309 4Bank of Taiwan 386 324 344 367 316 467 308 319 395Hua Nan Bank 383 312 328 363 292 457 392 325 335Chang Hwa Bank 353 288 301 353 272 478 36 325 38Taiwan Cooperative Bank 341 285 31 303 306 446 357 312 289

Table 3 Descriptive statistics of the variables in three different groups of data

Variables 1198831 1198832 1198833 1198834 1198835 1198836 1198837 1198838 1198839

First dataset (36)

Max 465 402 406 429 412 492 398 416 463Min 310 268 296 280 272 328 267 287 289Avg 397 338 346 366 339 441 345 346 380Std 034 028 023 033 029 046 033 030 039

Second dataset (28)

Max 465 402 406 429 412 492 398 416 463Min 310 268 296 280 300 328 281 294 302Avg 401 342 350 369 344 434 344 353 385Std 034 027 022 034 027 050 032 029 037

Third dataset (8)

Max 431 363 367 389 377 481 392 370 400Min 341 285 301 303 272 446 267 287 289Avg 383 323 332 358 320 463 349 322 360Std 027 027 020 025 030 011 039 022 036

ldquoclerk service attituderdquo (GRG = 05607) and ldquoservice trafficflowrdquo (GRG = 0509) while the three least satisfied variablesare ldquocustomer-oriented servicerdquo (GRG = 04364) ldquoservicecounter designrdquo (GRG = 0434) and ldquotransaction errorsrdquo(GRG = 04338)

In contrast Figure 1(b) shows that for the 28 privatebanks the top three most satisfied variables are ldquoconve-nience of branch locationsrdquo (GRG = 094) ldquoclerk serviceattituderdquo (GRG = 07302) and ldquoservice traffic flowrdquo (GRG =06594) while the three least satisfied variables are ldquotrans-action errorsrdquo (GRG = 05266) ldquoservice efficiencyrdquo (GRG =05207) and ldquocustomer-oriented servicerdquo (GRG = 05038)Therefore in regard to their advantages both the privateand public banks enjoyed ldquobranch locationsrdquo ldquoclerk serviceattituderdquo and ldquoservice traffic flowrdquo In regard to their dis-advantages both public and private banks need to improvetheir ldquocustomer-oriented servicerdquo and ldquotransaction errorsrdquoFurthermore the ldquoservice efficiencyrdquo for the private banksand the ldquocounter designrdquo for the public banks need to beimproved too

In the second analysis we compare the service satisfac-tion of different banks Figure 2(a) shows the result of thegrey relational analysis for the eight public banks It showsthat the top three banks in terms of service satisfaction areldquoFarmers Bank of Chinardquo (GRG = 08857) ldquoLand Bank ofTaiwanrdquo (GRG=06933) and ldquoFirst Commercial Bankrdquo (GRG= 06841) the bottom three banks are ldquoHua Nan Bankrdquo (GRG= 06307) ldquoChang Hwa Bankrdquo (GRG = 05996) and ldquoTaiwan

Cooperative Bankrdquo (GRG = 04932) Figure 2(b) shows thegrey relational analysis for the 28 private banks It shows thatthe top three banks in terms of service satisfaction are ldquoESUNCommercial Bankrdquo (GRG= 08586) ldquoUnion Bank of Taiwanrdquo(GRG = 08138) and ldquoYuanta Commercial Bankrdquo (GRG =07916) the bottom three banks included ldquoBank of Taipeirdquo(GRG = 04669) ldquoJih Sun International Commercial Bankrdquo(GRG = 04521) and ldquoMega International Commercial Bankrdquo(GRG = 04342)

Lastly we perform grey relational analysis on all the 36banks including the public and private banks Figure 2(c)shows that the top three banks among both public andprivate banks are ldquoESUN Commercial Bankrdquo (GRG =08643) ldquoUnion Bank of Taiwanrdquo (GRG = 08217) andldquoYuanta Commercial Bankrdquo (GRG = 07993) Note that allof them are private banks On the other hand the bottomthree banks include one public bank and two private bankswhich are ldquoTaiwan Cooperative Bankrdquo (GRG = 04773) ldquoJihSun International Commercial Bankrdquo (GRG = 04666) andldquoMega International Commercial Bankrdquo (GRG = 04483)Therefore according to the above analysis private banks usu-ally receive better service satisfaction from their customersthan public banks do

33 Quantile Regression Analysis of Different Factors on Ser-vice Satisfaction In addition to the grey relational analysisin this study we perform quantile regression analysis toexplore the factors that influence service satisfaction for


1 2 3 4 5 6 7 825

3

35

4

45

5

Sequence

Cor

resp

ondi

ng v

alue

Standard sequenceInspected sequence

(a)

0 5 10 15 20 25 3025

3

35

4

45

5

Sequence

Cor

resp

ondi

ng v

alue


(b)

Figure 1 Results of grey relational analysis for different surveyed variables

1 2 3 4 5 6 7 8 925

3

35

4

45

5

Sequence

Cor

resp

ondi

ng v

alue


(a)

1 2 3 4 5 6 7 8 925

335

445

5

Sequence

Cor

resp

ondi

ng v

alue


(b)

1 2 3 4 5 6 7 8 925

335

445

5

Sequence

Cor

resp

ondi

ng v

alue


(c)

Figure 2 Grey relational analysis on the satisfaction of eight public banks (a) 28 private banks (b) and all 36 banks (c)

those samples with low- medium- and high-levels of servicesatisfaction To do that we discretize the values of all thevariables into five intervals Each value of a variable istransformed into its corresponding label of the interval Inthis study the top 20 percent values of a variable are given thelargest label which is 5 and so on The transformed datasetfor the public banks is shown in Table 4 Note that in Table 4GRG denotes the grey relational grade of a sample bank

In this study we use the STATA software to conduct the025 quantile median and 075 quantile regressions on thedataset of all the 36 banks By using grey relational grades ofthe above grey relational analysis as the dependent variables(119884) and nine bank service satisfaction-related variables asthe independent variables (1198831ndash1198839) for quantile regressionanalysis this study determined which questionnaire survey

question items (variables) affected the bank service satisfac-tion performance for the customers at a specific quantileof service satisfaction The quantile regression analysis isconducted mainly for three different quantiles including the025 quantile median and 075 quantile According to theanalysis results shown in Table 5 ldquobank clerk service attituderdquo(1198831) ldquocustomer-oriented servicerdquo (1198832) and ldquoflexibility inhandling customer inquiriesrdquo (1198833) had no effects on thebank service satisfaction performance at different quantilesof service satisfaction However for ldquoservice efficiencyrdquo(1198834) its coefficient is significantly different from zero at10 significance level for all different quantile regressionmodels This result indicates that the waiting time for bankservice would affect the levels of service satisfaction for cus-tomers with different levels of service satisfaction Regarding


Table 4 The transformed dataset of the public bank dataset

Bank names 1198831 1198832 1198833 1198834 1198835 1198836 1198837 1198838 1198839 GRGFarmers Bank of China 5 5 5 5 3 1 5 5 5 08857Land Bank of Taiwan 4 4 4 5 5 3 1 1 4 06933First Commercial Bank 4 5 3 5 4 3 4 3 2 06841Taiwan Business Bank 2 2 2 3 3 5 4 2 5 06525Bank of Taiwan 3 3 4 4 3 3 2 2 5 06631Hua Nan Bank 3 2 3 4 1 2 5 3 3 06307Chang Hwa Bank 1 1 1 3 1 5 4 3 5 05996Taiwan Cooperative Bank 1 1 1 1 2 4 4 2 1 04932

Table 5 Quantile regression coefficients at different quantiles

Var

ModelOLS Q025 Q05 Q075

Adj 1198772 = 09528 Pseudo 1198772 = 08246 Pseudo 1198772 = 08135 Pseudo 1198772 = 08304Coef 119879 Sig Coef 119879 Sig Coef 119879 Sig Coef 119879 Sig

1198831 minus0042 minus162 minus0038 minus075 minus0044 minus133 minus00492 minus0961198832 0072 258 lowast 0068 146 0062 152 0075 1141198833 0064 149 0102 155 0057 069 minus0003 minus0031198834 0059 25 lowast 0054 188 and 0051 201 and 0085 202 and

1198835 0085 337 lowastlowast 0073 398 lowastlowast 0071 253 lowast 0128 289 lowastlowast

1198836 0071 498 lowastlowast 0062 362 lowastlowast 0065 327 lowastlowast 0083 323 lowastlowast

1198837 0060 505 lowastlowast 0065 732 lowastlowast 0063 453 lowastlowast 0057 224 lowast

1198838 0055 262 lowast 0044 238 lowast 0041 15 0044 0941198839 0050 339 lowastlowast 0042 294 lowastlowast 0056 34 lowastlowast 0038 119Note andmeans that it is significant under 10 significance level lowastmeans that it is significant under 5 significance level lowastlowastmeans that it is significant under1 significance level

the variable of ldquointerest rate favoring customersrdquo (1198835)according to Table 5 its coefficient is different from zero at 1significance level in all the three quantile regression modelsFigure 3 shows that for the customers at the 075 quantileof service satisfaction the ordinary least squares (OLS) (thedash line on the plot) regression tends to underestimate theeffect of ldquointerest rate favoring customersrdquo (1198835) on servicesatisfaction Therefore customers with a high level of servicesatisfaction will be especially concerned about whether thebank can adjust the interest rates to their advantages Thissignifies that a bank manager needs to pay more attention tothe policy of interest rate adjustment to further promote theservice satisfaction of his customers who already have a highlevel of service satisfaction

For the variable of ldquotransaction errorsrdquo (1198836) Table 5shows that its coefficients are different from zero for all thethree different quantile regression models at 1 significancelevel Furthermore Figure 3 shows that for high quantilesof service satisfaction the ordinary least squares regressionmodel tends to underestimate the effect of 1198836 Figure 4shows the results of 025 05 and 075 quantile regressionmodels in terms of 1198836 and (1198836)2 (stands for 1198836 squared)In the plot the satisfaction levels are calculated accordingto the regressed quadratic equations of variable 1198836 Thetrends of the curves indicate that fewer transaction errorswould result in higher satisfaction levels (in terms of GRG)

Note that a large value of 1198836 in Figure 4 represents a casewith fewer transaction errors and therefore receives a highersatisfaction level from the customers Also note that for themedian regression model the level of service satisfactionincreases as the number of transaction errors decreasesHowever the trend is reversed as the number of transactionerrors is further reduced to be below a certain number acase that is counter to our intuition This represents thatthe quadric equation of 1198836 is not adequate to capture therelationship between the service satisfaction levels and1198836

Table 5 shows that the coefficient of 1198837 (convenience ofbranch location) is significantly different from zero at 1 sig-nificance level for both 025 quantile and median regressionmodels For the 075 quantile regressionmodel its coefficientis significantly different from zero at 5 significance levelThe results show that the convenience of branch location canaffect the service satisfaction regardless of different quantilesof service satisfaction In addition Figure 3 shows that theordinary least squares regression overestimates the effectof 1198837 on service satisfaction for high quantiles of servicesatisfaction As to the variable of ldquoservice traffic flowrdquo (1198838)Table 5 shows that its coefficient is different from zero onlyfor the 025 quantile regression model at 5 significancelevel In other words improving the service traffic flow canpromote the service satisfaction for customers with low levelsof service satisfaction Finally regarding ldquoservice counter

Mathematical Problems in Engineering 7X1

020

010

000

minus010

minus020

minus030

Quantile0 02 04 06 08 1

X2

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X3

050

000

minus050

Quantile0 02 04 06 08 1

X4

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X5

040

030

020

010

000

minus010

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X6

Quantile0 02 04 06 08 1

X7

030

020

010

000

minus020

minus010

Quantile0 02 04 06 08 1

X8

030

020

010

000

minus010

020

010

000

minus020

minus010

Quantile0 02 04 06 08 1

X9

Quantile0 02 04 06 08 1

Figure 3 Satisfaction performance results of all 36 banks

designrdquo (1198839) its coefficients are different from zero for the025 and 05 quantile regression at 1 significance levelTherefore customers with low or medium levels of servicesatisfaction care more about the ldquoservice counter designrdquo andto improve the design of the service counter may furtherpromote the service satisfaction of the customers with lowto medium levels of service satisfaction The comprehensiveanalysis of the effects of different variables in this study couldhelp a bankmanager to promote the service satisfaction of hiscustomers with different levels of service satisfaction

34 Performance Comparison of Different Quantile Regres-sion Models Finally by using the grey relational gradeas the dependent variables (119884) and the nine bank ser-vice satisfaction-related variables as the independent vari-ables (1198831ndash1198839) this study built seven forecasting modelsincluding Q15 Q25 Q35 Q50 Q65 Q75 and Q85 5-fold

cross-validation is used to find the values of performancemeasures of the seven models To perform 5-fold cross-validation thirty-six samples of the dataset are divided intofive groups Four groups of them are used to build a modeland the rest is used as a test dataset to calculate the values ofdifferent performance measures This procedure is repeatedfor five times each with a different group of the dataset asthe test dataset The five values of a performance measureare averaged to render the reported value of the performancemeasure The performance measures for this study includeRMSE RTIC and CE [12] The equations for these measuresare as follows

The equation for RMSE (Root Mean Squared Error) is

RMSE = radicsum119873

119905=1(119909119905minus 119909119905)2

119873

(10)


Table 6 Performance of the seven quantile regression models

Index Q15 Q25 Q35 Q50 Q65 Q75 Q85RMSE 004227 003610 012830 003494 003906 004014 008812RTIC 000519 000372 003859 000338 000412 000434 002032CE 074438 078248 012402 084030 083163 082688 018862

Satis

fact

ion

perfo

rman

ce

07

065

06

055

05

X6

3 35 4 45 5

Q25

Q50

Q75

Figure 4 Regressed curves for different quantiles in terms of1198836

The equation for RTIC (RevisionTheil Inequality Coefficient)is

RTIC = radicsum119873

119905=1(119909119905minus 119909119905)2

sum119873

119905=11199091199052 (11)

The equation for CE (coefficient of efficiency) is

CE = 1 minussum119873

119905=1(119909119905minus 119909119905)2

sum119873

119905=1(119909119905minus 119909119905)2 (12)

Note that a model with small values of RMSE and RITCperforms better than a model with large values of RMSE andRITC For CE a good forecasting model has a CE value closeto 1The performance of the seven quantile regressionmodelsis shown in Table 6

Table 6 shows that the median regression model hasthe smallest RMSE and RTIC values and the largest CEvalue Therefore the median regression model has the bestprediction accuracy among all the seven quantile regressionmodels

4 Conclusion and Suggestion

The major contribution of this paper is to explore bankservice satisfaction performance based on the 2013 servicesatisfaction survey data of 36 public and private banks inTaiwanThis paper proposes to use the grey relational analysisto gauge the service satisfaction of a customer based on ninequestion items in a questionnaire With the grey relationalanalysis we have found some variables that contribute more

on the service satisfaction of customers than the othervariables We also ranked the banks according to theirgrey relational grades of satisfaction levels Furthermore thequantile regression analysis was used to further explore thedeterminant factors of service satisfaction for the customersat different quantiles of service satisfactionsWith regressionson different quantiles the manager of a bank can find thefactors that are more concerned by their customers at aspecific quantile of service satisfaction As a result the man-ager can formulate different policies to promote the servicesatisfaction of customers at different quantiles of servicesatisfaction Finally this study examined the performance ofdifferent regression models The experimental result showedthat among the seven quantile regressionmodels themedianregressionmodel has the best performance in terms of RMSERTIC and CE performance measures

Competing Interests

The authors declare that they have no competing interests

References

[1] P Gerrard and B Cunningham ldquoBank service quality acomparison between a publicly quoted bank and a governmentbank in Singaporerdquo Journal of Financial Services Marketing vol6 no 1 pp 50ndash66 2001

[2] C T Ennew andM R Binks ldquoThe impact of service quality andservice characteristics on customer retention small businessesand their banks in the UKrdquo British Journal of Management vol7 no 3 pp 219ndash230 1996

[3] J Deng ldquoThe control problems of grey systemrdquo Systems ampControl Letters vol 1 no 5 pp 288ndash294 1982

[4] N Tosun ldquoDetermination of optimum parameters for multi-performance characteristics in drilling by using grey relationalanalysisrdquo International Journal of Advanced ManufacturingTechnology vol 28 no 5-6 pp 450ndash455 2006

[5] G Nagpal M Uddin and A Kaur ldquoA hybrid techniqueusing grey relational analysis and regression for software effortestimation using feature selectionrdquo International Journal of SoftComputing and Engineering vol 1 no 6 pp 20ndash27 2012

[6] H Hasani S A Tabatabaei and G Amiri ldquoGrey relationalanalysis to determine the optimum process parameters foropen-end spinning yarnsrdquo Journal of Engineered Fibers andFabrics vol 7 no 2 pp 81ndash86 2012

[7] R Koenker and J Bassett ldquoRegression quantilesrdquo Econometricavol 46 no 1 pp 33ndash50 1978

[8] L Meligkotsidou I D Vrontos and S D Vrontos ldquoQuantileregression analysis of hedge fund strategiesrdquo Journal of Empiri-cal Finance vol 16 no 2 pp 264ndash279 2009


[9] R Alhamzawi and K Yu ldquoVariable selection in quantile regres-sion via Gibbs samplingrdquo Journal of Applied Statistics vol 39no 4 pp 799ndash813 2012

[10] M Harding and C Lamarche ldquoA quantile regression approachfor estimating panel data models using instrumental variablesrdquoEconomics Letters vol 104 no 3 pp 133ndash135 2009

[11] K L Wen S K Chang-Chien C K Yeh CWWang and H SLin Apply MATLAB in Grey System Theory Chuan Hwa BookCo 2006

[12] W-T Pan ldquoMixed modified fruit fly optimization algorithmwith general regression neural network to build oil and goldprices forecasting modelrdquo Kybernetes vol 43 no 7 pp 1053ndash1063 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


Table 1 An example dataset

Sequence 119896 = 1 119896 = 2 119896 = 3 119896 = 4 119896 = 5 119896 = 61199090

1000 1100 2000 2250 3000 40001199091

1000 1166 1834 2000 2340 30001199092

1000 1125 1075 1375 1625 17501199093

1000 1000 0700 0800 0900 1200

between two sequences of factors Assume that the state of asystem can be modeled by a sequence of factors Two systemstates are said to be similar if the values of their correspondingfactors are similar In a real life application a standardsequence is selected to calculate the relational grades of aset of inspected sequences Usually the standard sequenceis composed of the maximal values of each of the factors inthe current dataset By referring to the standard sequence thegrey relational analysis gives each inspected sequence a greyrelational grade (GRG) The more an inspected sequence issimilar to the standard sequence the larger the GRG of theinspected sequence is Since the standard sequence usuallyrepresents a state of the system with maximum performancea sequence with a large GRG is therefore better than asequence with a small GRG The grey relational analysis hasbeen widely used in performance evaluation [4ndash6]

The grey relational grade of an inspected sequence iscalculated through the following steps

Step 1 (define the sequences) To perform the grey relationalanalysis we need to define the sequences for comparisonTo illustrate we use Table 1 as an example Each columnin Table 1 represents a factor of the system state while eachrow in Table 1 represents a sequence of the system stateAmong all the sequences of the system state we choose orcreate a sequence called the standard sequence to measurethe performance of the other sequences which are called theinspected sequences A sequence119909

119894is denoted by a set 119909

119894(119896) |

119896 = 1 119899 Without loss of generality we choose 1199090as the

standard sequence

Step 2 (standardization) Since different factors may havedifferent ranges of values which are not comparable we needto standardize the values of all factors so that values fromdifferent factors are comparable One way of standardizationis to divide all the values of a factor by the maximum valueof the factor By doing so values of all factors are all less thanone and greater than zero

Step 3 (calculate the grey relational coefficients) The greyrelational coefficient of the 119896th factor of sequence 119909

119894is

calculated by the following equation

120585119894(119896)

=min119894min119896

10038161003816100381610038161199090 (119896) minus 119909119894 (119896)1003816100381610038161003816 + 120588max

119894max119896

10038161003816100381610038161199090 (119896) minus 119909119894 (119896)1003816100381610038161003816

10038161003816100381610038161199090 (119896) minus 119909119894 (119896)1003816100381610038161003816 + 120588max

119894max119896

10038161003816100381610038161199090 (119896) minus 119909119894 (119896)1003816100381610038161003816

(1)

Let Δ0119894(119896) denote |119909

0(119896) minus 119909

119894(119896)| Equation (1) can be written

as

120585119894(119896) =

min119894min119896Δ0119894 (119896) + 120588max

119894max119896Δ0119894 (119896)

10038161003816100381610038161199090 (119896) minus 119909119894 (119896)1003816100381610038161003816 + 120588max

119894max119896Δ0119894(119896) (2)

Note that to find the value ofmin119894min119896Δ0119894(119896) we find for

inspected sequence 119909119894 the minimum value of the differences

between factor 119896 of 1199090and factor 119896 of 119909

119894 119896 = 1 to 119899 Then we

find the minimum value among the minimum values of allthe inspected sequencesThe value ofmax

119894max119896|1199090(119896)minus119909

119894(119896)|

can be found in a similar wayThe rationale behind (2) is thatwe use the global minimum and the global maximum of thecomponentwise differences between the standard sequenceand all the inspected sequences to rescale the differencebetween a factor of an inspected sequence and the same factorof the standard sequence And then we use the rescaleddifferences also called the grey relational coefficients toreplace the values of the original factors The value of 120588belongs to (0infin) and is usually set to 05

Step 4 (find the grey relational grade) The grey relationalgrade 119903

119894of inspected sequence 119909

119894is defined to be the average

of all its grey relational coefficients The grey relational grade119903119894of 119909119894is calculated according to the following equation

119903119894=1

119899

119899

sum

119896=1

120585119894(119896) (3)

To illustrate how to calculate the grey relational grade of aninspected sequence we use the dataset in Table 1

Step 1 (define the sequence) We assume that 1199090is the

standard sequence

Step 2 (standardization) We assume that the factors of thedataset in Table 1 are comparable No standardization isrequired

Step 3 (calculate the grey relational coefficients) The compo-nentwise differences between the standard sequence and allthe inspected sequences are shown in the following


1

Δ01(1) = 0000

Δ01 (2) = 0066

Δ01(3) = 0166

Δ01(4) = 0250

Δ01(5) = 0660

Δ01(6) = 1000

(4)



2

Δ02(1) = 0000

Δ02 (2) = 0025

Δ02(3) = 0925

Δ02(4) = 0875

Δ02(5) = 1375

Δ02(6) = 2250

(5)


3

Δ03(1) = 0000

Δ03(2) = 0100

Δ03(3) = 1300

Δ03 (4) = 1450

Δ03(5) = 2100

Δ03(6) = 2800

(6)


(1) Δ01= (0000 0066 0166 0250 0660 1000)

(2) Δ02= (0000 0025 0925 0875 1375 2250)

(3) Δ03= (0000 0100 1300 1450 2100 2800)


120585119894 (119896) =

0 + (05) times (28)

Δ0119894(119896) + (05) times (28)

(7)


1in the following

1205851 (1) = 1000

1205851(2) = 09549

1205851(3) = 08939

1205851(4) = 08484

1205851(5) = 06796

1205851 (6) = 05833

(8)


1in the following

1199031=1

6(1 + 09549 + 08939 + 08484 + 06796

+ 05833) = 08266

(9)


3can be








Bank names 1198831 1198832 1198833 1198834 1198835 1198836 1198837 1198838 1198839



Variables 1198831 1198832 1198833 1198834 1198835 1198836 1198837 1198838 1198839

First dataset (36)

Max 465 402 406 429 412 492 398 416 463Min 310 268 296 280 272 328 267 287 289Avg 397 338 346 366 339 441 345 346 380Std 034 028 023 033 029 046 033 030 039

Second dataset (28)

Max 465 402 406 429 412 492 398 416 463Min 310 268 296 280 300 328 281 294 302Avg 401 342 350 369 344 434 344 353 385Std 034 027 022 034 027 050 032 029 037

Third dataset (8)

Max 431 363 367 389 377 481 392 370 400Min 341 285 301 303 272 446 267 287 289Avg 383 323 332 358 320 463 349 322 360Std 027 027 020 025 030 011 039 022 036








1 2 3 4 5 6 7 825

3

35

4

45

5

Sequence

Cor

resp

ondi

ng v

alue


(a)

0 5 10 15 20 25 3025

3

35

4

45

5

Sequence

Cor

resp

ondi

ng v

alue


(b)


1 2 3 4 5 6 7 8 925

3

35

4

45

5

Sequence

Cor

resp

ondi

ng v

alue


(a)

1 2 3 4 5 6 7 8 925

335

445

5

Sequence

Cor

resp

ondi

ng v

alue


(b)

1 2 3 4 5 6 7 8 925

335

445

5

Sequence

Cor

resp

ondi

ng v

alue


(c)









Var













020

010

000

minus010

minus020

minus030

Quantile0 02 04 06 08 1

X2

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X3

050

000

minus050

Quantile0 02 04 06 08 1

X4

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X5

040

030

020

010

000

minus010

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X6

Quantile0 02 04 06 08 1

X7

030

020

010

000

minus020

minus010

Quantile0 02 04 06 08 1

X8

030

020

010

000

minus010

020

010

000

minus020

minus010

Quantile0 02 04 06 08 1

X9

Quantile0 02 04 06 08 1







119905=1(119909119905minus 119909119905)2

119873

(10)




Satis

fact

ion

perfo

rman

ce

07

065

06

055

05

X6

3 35 4 45 5

Q25

Q50

Q75




119905=1(119909119905minus 119909119905)2

sum119873

119905=11199091199052 (11)



119905=1(119909119905minus 119909119905)2

sum119873

119905=1(119909119905minus 119909119905)2 (12)






Competing Interests


References





















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











2

Δ02(1) = 0000

Δ02 (2) = 0025

Δ02(3) = 0925

Δ02(4) = 0875

Δ02(5) = 1375

Δ02(6) = 2250

(5)


3

Δ03(1) = 0000

Δ03(2) = 0100

Δ03(3) = 1300

Δ03 (4) = 1450

Δ03(5) = 2100

Δ03(6) = 2800

(6)


(1) Δ01= (0000 0066 0166 0250 0660 1000)

(2) Δ02= (0000 0025 0925 0875 1375 2250)

(3) Δ03= (0000 0100 1300 1450 2100 2800)


120585119894 (119896) =

0 + (05) times (28)

Δ0119894(119896) + (05) times (28)

(7)


1in the following

1205851 (1) = 1000

1205851(2) = 09549

1205851(3) = 08939

1205851(4) = 08484

1205851(5) = 06796

1205851 (6) = 05833

(8)


1in the following

1199031=1

6(1 + 09549 + 08939 + 08484 + 06796

+ 05833) = 08266

(9)


3can be








Bank names 1198831 1198832 1198833 1198834 1198835 1198836 1198837 1198838 1198839



Variables 1198831 1198832 1198833 1198834 1198835 1198836 1198837 1198838 1198839

First dataset (36)

Max 465 402 406 429 412 492 398 416 463Min 310 268 296 280 272 328 267 287 289Avg 397 338 346 366 339 441 345 346 380Std 034 028 023 033 029 046 033 030 039

Second dataset (28)

Max 465 402 406 429 412 492 398 416 463Min 310 268 296 280 300 328 281 294 302Avg 401 342 350 369 344 434 344 353 385Std 034 027 022 034 027 050 032 029 037

Third dataset (8)

Max 431 363 367 389 377 481 392 370 400Min 341 285 301 303 272 446 267 287 289Avg 383 323 332 358 320 463 349 322 360Std 027 027 020 025 030 011 039 022 036








1 2 3 4 5 6 7 825

3

35

4

45

5

Sequence

Cor

resp

ondi

ng v

alue


(a)

0 5 10 15 20 25 3025

3

35

4

45

5

Sequence

Cor

resp

ondi

ng v

alue


(b)


1 2 3 4 5 6 7 8 925

3

35

4

45

5

Sequence

Cor

resp

ondi

ng v

alue


(a)

1 2 3 4 5 6 7 8 925

335

445

5

Sequence

Cor

resp

ondi

ng v

alue


(b)

1 2 3 4 5 6 7 8 925

335

445

5

Sequence

Cor

resp

ondi

ng v

alue


(c)









Var













020

010

000

minus010

minus020

minus030

Quantile0 02 04 06 08 1

X2

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X3

050

000

minus050

Quantile0 02 04 06 08 1

X4

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X5

040

030

020

010

000

minus010

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X6

Quantile0 02 04 06 08 1

X7

030

020

010

000

minus020

minus010

Quantile0 02 04 06 08 1

X8

030

020

010

000

minus010

020

010

000

minus020

minus010

Quantile0 02 04 06 08 1

X9

Quantile0 02 04 06 08 1







119905=1(119909119905minus 119909119905)2

119873

(10)




Satis

fact

ion

perfo

rman

ce

07

065

06

055

05

X6

3 35 4 45 5

Q25

Q50

Q75




119905=1(119909119905minus 119909119905)2

sum119873

119905=11199091199052 (11)



119905=1(119909119905minus 119909119905)2

sum119873

119905=1(119909119905minus 119909119905)2 (12)






Competing Interests


References





















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Bank names 1198831 1198832 1198833 1198834 1198835 1198836 1198837 1198838 1198839



Variables 1198831 1198832 1198833 1198834 1198835 1198836 1198837 1198838 1198839

First dataset (36)

Max 465 402 406 429 412 492 398 416 463Min 310 268 296 280 272 328 267 287 289Avg 397 338 346 366 339 441 345 346 380Std 034 028 023 033 029 046 033 030 039

Second dataset (28)

Max 465 402 406 429 412 492 398 416 463Min 310 268 296 280 300 328 281 294 302Avg 401 342 350 369 344 434 344 353 385Std 034 027 022 034 027 050 032 029 037

Third dataset (8)

Max 431 363 367 389 377 481 392 370 400Min 341 285 301 303 272 446 267 287 289Avg 383 323 332 358 320 463 349 322 360Std 027 027 020 025 030 011 039 022 036








1 2 3 4 5 6 7 825

3

35

4

45

5

Sequence

Cor

resp

ondi

ng v

alue


(a)

0 5 10 15 20 25 3025

3

35

4

45

5

Sequence

Cor

resp

ondi

ng v

alue


(b)


1 2 3 4 5 6 7 8 925

3

35

4

45

5

Sequence

Cor

resp

ondi

ng v

alue


(a)

1 2 3 4 5 6 7 8 925

335

445

5

Sequence

Cor

resp

ondi

ng v

alue


(b)

1 2 3 4 5 6 7 8 925

335

445

5

Sequence

Cor

resp

ondi

ng v

alue


(c)









Var













020

010

000

minus010

minus020

minus030

Quantile0 02 04 06 08 1

X2

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X3

050

000

minus050

Quantile0 02 04 06 08 1

X4

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X5

040

030

020

010

000

minus010

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X6

Quantile0 02 04 06 08 1

X7

030

020

010

000

minus020

minus010

Quantile0 02 04 06 08 1

X8

030

020

010

000

minus010

020

010

000

minus020

minus010

Quantile0 02 04 06 08 1

X9

Quantile0 02 04 06 08 1







119905=1(119909119905minus 119909119905)2

119873

(10)




Satis

fact

ion

perfo

rman

ce

07

065

06

055

05

X6

3 35 4 45 5

Q25

Q50

Q75




119905=1(119909119905minus 119909119905)2

sum119873

119905=11199091199052 (11)



119905=1(119909119905minus 119909119905)2

sum119873

119905=1(119909119905minus 119909119905)2 (12)






Competing Interests


References





















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










1 2 3 4 5 6 7 825

3

35

4

45

5

Sequence

Cor

resp

ondi

ng v

alue


(a)

0 5 10 15 20 25 3025

3

35

4

45

5

Sequence

Cor

resp

ondi

ng v

alue


(b)


1 2 3 4 5 6 7 8 925

3

35

4

45

5

Sequence

Cor

resp

ondi

ng v

alue


(a)

1 2 3 4 5 6 7 8 925

335

445

5

Sequence

Cor

resp

ondi

ng v

alue


(b)

1 2 3 4 5 6 7 8 925

335

445

5

Sequence

Cor

resp

ondi

ng v

alue


(c)









Var













020

010

000

minus010

minus020

minus030

Quantile0 02 04 06 08 1

X2

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X3

050

000

minus050

Quantile0 02 04 06 08 1

X4

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X5

040

030

020

010

000

minus010

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X6

Quantile0 02 04 06 08 1

X7

030

020

010

000

minus020

minus010

Quantile0 02 04 06 08 1

X8

030

020

010

000

minus010

020

010

000

minus020

minus010

Quantile0 02 04 06 08 1

X9

Quantile0 02 04 06 08 1







119905=1(119909119905minus 119909119905)2

119873

(10)




Satis

fact

ion

perfo

rman

ce

07

065

06

055

05

X6

3 35 4 45 5

Q25

Q50

Q75




119905=1(119909119905minus 119909119905)2

sum119873

119905=11199091199052 (11)



119905=1(119909119905minus 119909119905)2

sum119873

119905=1(119909119905minus 119909119905)2 (12)






Competing Interests


References





















Volume 2014




Journal of











Journal of


Function Spaces






Algebra













Var













020

010

000

minus010

minus020

minus030

Quantile0 02 04 06 08 1

X2

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X3

050

000

minus050

Quantile0 02 04 06 08 1

X4

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X5

040

030

020

010

000

minus010

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X6

Quantile0 02 04 06 08 1

X7

030

020

010

000

minus020

minus010

Quantile0 02 04 06 08 1

X8

030

020

010

000

minus010

020

010

000

minus020

minus010

Quantile0 02 04 06 08 1

X9

Quantile0 02 04 06 08 1







119905=1(119909119905minus 119909119905)2

119873

(10)




Satis

fact

ion

perfo

rman

ce

07

065

06

055

05

X6

3 35 4 45 5

Q25

Q50

Q75




119905=1(119909119905minus 119909119905)2

sum119873

119905=11199091199052 (11)



119905=1(119909119905minus 119909119905)2

sum119873

119905=1(119909119905minus 119909119905)2 (12)






Competing Interests


References





















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










020

010

000

minus010

minus020

minus030

Quantile0 02 04 06 08 1

X2

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X3

050

000

minus050

Quantile0 02 04 06 08 1

X4

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X5

040

030

020

010

000

minus010

030

020

010

000

minus010

Quantile0 02 04 06 08 1

X6

Quantile0 02 04 06 08 1

X7

030

020

010

000

minus020

minus010

Quantile0 02 04 06 08 1

X8

030

020

010

000

minus010

020

010

000

minus020

minus010

Quantile0 02 04 06 08 1

X9

Quantile0 02 04 06 08 1







119905=1(119909119905minus 119909119905)2

119873

(10)




Satis

fact

ion

perfo

rman

ce

07

065

06

055

05

X6

3 35 4 45 5

Q25

Q50

Q75




119905=1(119909119905minus 119909119905)2

sum119873

119905=11199091199052 (11)



119905=1(119909119905minus 119909119905)2

sum119873

119905=1(119909119905minus 119909119905)2 (12)






Competing Interests


References





















Volume 2014




Journal of











Journal of


Function Spaces






Algebra












Satis

fact

ion

perfo

rman

ce

07

065

06

055

05

X6

3 35 4 45 5

Q25

Q50

Q75




119905=1(119909119905minus 119909119905)2

sum119873

119905=11199091199052 (11)



119905=1(119909119905minus 119909119905)2

sum119873

119905=1(119909119905minus 119909119905)2 (12)






Competing Interests


References





















Volume 2014




Journal of











Journal of


Function Spaces






Algebra





















Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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