Inferences in multivariate-univariate calibration

TheStatistician (2003) 52, Part1, pp.15-39

Inferences in multivariate-univariatecalibration problems DeniseBentonand K.Krishnamoorthy Universityof Louisianaat Lafayette,USA

andThomasMathew Universityof MarylandBaltimoreCounty,Baltimore,USA [ReceivedOctober2001. RevisedAugust2002] vectorresponsevariableis considered,relatedto a scalar Summary.A normallydistributed variablethrougha linearregressionmodel.Thecalibration data,i.e. dataobtained explanatory on the responsevariablecorresponding to knownvalues of the explanatoryvariable,are to be used for makinginferencesconcerningunknownvalues of the explanatory variable.The of intervalestimationand hypothesistesting purposeof the paperis a detailedinvestigation problemsthatariseinthiscontext.Suchproblemsareaddressedinthe scenarioof bothsingle use and multipleuse of the calibration data. Inthe single-usesituation,the calibration data are used to makeinferencesconcerninga single unknownvalueof the explanatory variable. Inthe multiple-usescenario,the calibration data are used to makeinferencesconcerninga variable.A one-sidedhypothesistestingprobsequence of unknownvaluesof the explanatory lem is addressedand test proceduresare developedin the contextof bothsingle use and data.Since the test statistichas a distribution thatdependson multipleuse of the calibration some unknownparameters,a parametric bootstrapprocedureis advocatedto carryout the test. The parametric bootstrapprocedurecan be adoptedforintervalestimationas well.The of the parametric performance bootstrapprocedureis numerically investigatedandis foundto be quitesatisfactory. Some examplesare used to motivatethe problemsand to illustrate the of ourresults.The overallconclusionis thatthe parametric applicability bootstrapis a simple andsatisfactoryapproachformakinginferencesinthe calibration problem. Calibration confidenceregion;Multiple-use data;Multiple-use Keywords:Bootstrap; Percentilebootstrap; hypothesistesting;Parametric bootstrap; Single-useconfidenceregion; Single-usehypothesistesting

1. Introduction 1. 1. Statement of the problem

The multivariate-univariate calibrationprobleminvolvesa p x 1 responsevariabley related to a scalarexplanatoryvariablex, wherey is assumedto be random.Thispaperconsidersthe scenariowherex is a non-randomquantityand the relationshipbetweeny and x is a normal linear regression model. Let (yi, xi), i = 1,2, ...., n, denote n independent observations yi on

the responsevariable,correspondingto the valuexi of the explanatoryvariable.Ourassumed modelis Yi " Np(a + /3xi, E'),

(1.1)

Addressfor correspondence:Thomas Mathew, Department of Mathematicsand Statistics, University of MarylandBaltimoreCounty, 1000Hilltop Circle,Baltimore,MD 21250, USA.

E-mail:[email protected]

@2003 Royal StatisticalSociety

0039-0526/03/52015

16

andT Mathew D.Benton,K.Krishnamoorthy

i = 1, 2, ..., n, where a and / are unknown p x 1 parameter vectors and E is an unknown p x p positivedefinitematrix.Now let yo be a valueof the responsevariablecorrespondingto an unknownvalue0 of the explanatoryvariable.Similarlyto model(1.1),we assumethat Yo

~

Np(a + P0, Z),

(1.2)

whereyo is assumedto be independentof the yis in model (1.1). The problemof calibration or inverseregressionconsistsof statisticalinferenceconcerning0. The (yi, xi)s in model (1.1) are referredto as the calibrationdata. In the contextof models(1.1) and (1.2), the problemis calibrationproblemsincethe responsevariableis a vector knownas the multivariate-univariate the parameterof interest)is a scalar.The problems hence variable the and (and explanatory of confidenceregionsfor 0 and hypothethe construction deal with in this addressed paper are several below 0. Given sis tests concerning exampleswheresuch multivariate-univariate models(1.1) and (1.2)maybe used. where in and arise calibrationproblems practice 1.2. Some examples 1.2.1.

Example 1: measuringfat content of meat using near infra-redabsorbancemeasurements

Samplesof meatcan be analysedfor theirfat contentby usingnearinfra-redabsorbancemeasurements.The data that we shallconsiderconsistof 214 samplesof meat and measurements for each sampleconsist of a 100-channelspectrumof absorbances,and the contentsof moisture, fat and protein, as determinedby analyticalchemistry.(Details of the data are availdatawereobtainedby usinga ableat http: / /lib. stat. cmu. edu/datasets/tecator; TecatorInfratecFood and FeedAnalyzer.)Theprovidersof the data suggestusingthe first172 samplesto fit a model and usingthe first 13 principalcomponents,scaledto unit variance,to predictthe fat content.In fittinga linearmodel,only threeprincipalcomponentsweresignificant. Supposethat y = (yl, y2,y3)' representsthe vectorof thesethreeprincipalcomponents, and let yi denotethe valueof y for the ith sample,with a knownvaluexi for the fat content(as a percentage).We assumemodel(1.1), wherep = 3. Furthermore,if yo denotesthe valueof y for a meatsamplewith unknownfat content0, we havemodel(1.2). As an exampleof hypothesistesting,consider Ho : 0 < 10% versusHi : 0 > 10%,

(1.3)

whichis of interestin determiningwhethera sampleof groundmeat containsless than 10% fat. Currentmethodsof estimatingthe fat contentof a batchof groundmeatinvolvetakinga sampleof meat and then fryingit in a pan. The amountof fat that dripsoff is then compared with the weightof the pattyto estimatethe percentageof fat. The processobviouslybecomes considerablysimplerif the fat content(i.e. 0 in model (1.2)) can be estimatedby usingthe yis in model (1.1) and yo in model (1.2). Confidenceintervalsand tests (e.g. of the type (1.3)) are certainlyof interestin this context. The problemof estimatingfat contentin groundmeatis also discussedin Martensand Naes (1989),pages 108-112. 1.2.2.

Example 2: thepaint finish data

The paint finishdata first appearedin Brown(1982) and werelater analysedby Brownand Sundberg(1987),Oman(1988)and Mathewand Zha (1996).In this example,x is a scalarrepresentingthe viscosityof the paint samplesand y is a bivariateobservationvectorconsisting of two measurementson certainopticalpropertiesof the samples.The viscosityof each paint

Problems Calibration Multivariate-Univariate

17

samplewasone of threedifferentvalues,codedas -1, 0 and 1in BrownandSundberg(1987).27 observationswereavailableforcalibrationand,if yi denotesthe observationcorrespondingto a knownviscosityxi, the modelusedby Brownand Sundberg(1987)is model(1.1).Theproblem is to make inferenceson the unknownviscosity,say 0, correspondingto a measurementyo, wheremodel(1.2) applies. 1.2.3.

Example 3: the wheat data

The wheatdata set also appearedin Brown(1982)and was laterused by Fox (1989)to exhibit conditionalmultivariatecalibration.The data set consistsof measurementson 21 samplesof hardwheat.Accuratepercentagesof waterand proteinwereobtainedalong with fourderived infra-redreflectancemeasurements.Our interestis in makinginferenceon the percentageof water in a wheat sample by using the reflectancemeasurements.It turns out that only the thirdand fourthmeasurementsare usefulin predictingthe percentageof water.Thus,if y is a bivariatevectordenotingthe thirdand fourthreflectancemeasurements,and if x denotesthe percentageof water,we havemodelssimilarto models(1.1) and (1.2). The problemwill be to makeinferencesconcerning0, the unknownpercentageof water. 1.2.4.

Example 4: chlorophenolexample

Chlorophenolsare chemicalsthat are often used in herbicidesand wood preservatives,but they havea high toxicityand are consideredto be dangerousto the environment.Methodsof detectionare expensiveand time consumingbecausethey are based on separationanalytical meatechniques.da Silvaand Laquipai(1998)discussedusingultravioletspectrophotometric surementsto predictthe concentrationof a particularchlorophenolin a solution,resultingin a quickmethodto screenfor chlorophenols.In practice,if the concentrationis abovea certain calibrationmodel,wherewe level,actionmustbe taken.Thuswe havea multivariate-univariate are interestedin testinghypothesesconcerning0, an unknownlevel of chlorophenol,by using measurements. an observedvectorof spectrophotometric 1.2.5.

Example 5: olefin example

Near infra-redspectroscopyandmultivariatecalibrationmethodsareusedto estimatethe concentrationsof olefinsin the processingof polyethylene.As a partof the on-lineprocesscontrol, it is of interestto determinewhetherthe olefinconcentrationis aboveor belowa specification limit.This allowsthe plant to makea consistentproduct,minimizingthe amountof offgrade produced.For details,see Seasholtz(1999). 1.3. Literaturereviewand summaryof the results calibrationprobIt shouldbe clearfromthe aboveexamplesthat the multivariate-univariate lem arises in many applicationsand the problemsof point and intervalestimationas well as hypothesistestingare relevantin these applications.This paperdeals with the problemsof intervalestimationandhypothesistesting.In thecontextof models(1.1)and(1.2),thefollowing hypothesistestingproblemsareconsidered: Ho : 0 = c versus H1 : 0 c, Ho: 00 c versus H1:0 > c, Ho : 0 > c versus H1: 0 < c,

wherec is known(forexample,c = 10%in problem(1.3)).

(1.4a) (1.4b) (1.4c)

18

and T Mathew D. Benton,K.Krishnamoorthy

The literatureon calibrationdistinguishesbetweentwo scenariosregardingthe use of the calibrationdata (i.e. the (yi, xi)s in model (1.1)), for the purposeof statisticalinferenceconcerning0 in model (1.2). The firstis the singleuse of the calibrationdata, in whichthe set of (yi, xi)s in model(1.1) is usedfor obtainingconfidenceregionsandhypothesistestsfor a single parameter0 correspondingto a singleobservationyo in model(1.2). This resultsin single-use confidenceregionsandsingle-usehypothesistests.Wealso havemultiple-useconfidenceregions and multiple-usehypothesistests, wherethe calibrationdata are used repeatedlyto construct confidenceregionsandhypothesistestsfora sequenceof parametervalues,similarto 0 in model (1.2),correspondingto a sequenceof observations,similarto yo in model(1.2).The distinction betweensingleuse andmultipleuse of the calibrationdatais clarifiedandexplainedin Mee and Eberhardt(1996)for the univariatecase.Theyhavealso givena cleardiscussionof the criteria to be usedfortheconstructionof single-useandmultiple-useconfidenceregions.Single-useand et al. (2001),once againfor the multiple-usehypothesistests arediscussedin Krishnamoorthy univariatecase.This paperconsidersboth single-useand multiple-useprocedures.The criteria to be usedfor thesepurposesaregivenbelow. A naturalprocedurefor obtaininga test statisticfor testingthe hypothesesin expressions (1.4) is first to assumethat a, p and E are knownin model (1.2). Whenthis is the case, the classicalestimatorof 0 can be obtainedon the basis of model (1.2) and a test statisticcan be easilyobtainedfortestingthehypothesesin expressions(1.4).Theunknownquantitiesa, ,3 and E in the test statisticcan now be replacedby estimatesobtainedby usingthe calibrationdata, i.e. by using the (yi, xi)s in model (1.1). The samecan be done for obtaininga pivot statistic for constructinga confidenceregionfor 0. Thoughthis approachis intuitivelyappealing,the difficultythat we face is that the resultingtest statistic(or pivot statistic)has a distribution that dependson nuisanceparameters.In the context of confidenceregions,this difficultyis overcomeby constructingconservativeconfidenceregions;see Mathewand Zha (1996, 1997, 1998)and Mathewet al. (1998).It shouldbe pointedout, however,that in the univariatecase (i.e. p = 1) there are test statisticsand pivot statisticswhich have distributionsfree of any unknownparameters,and Fieller'stheoremprovidesan exactsingle-useconfidenceregion;see et al. (2001)wheremultiple-usehypothesistestingis addressedin theunivariate Krishnamoorthy case. Confidenceregionscan clearlybe used for testinghypothesis(1.4a). The exact confidence regionderivedin Mathewand Kasala(1994)and the conservativeconfidenceregionderivedin MathewandZha(1996)canbothbe usedforderivingsingle-usetwo-sidedhypothesistests.Our numericalstudyindicatesthatthe test basedon the exactconfidenceregionderivedin Mathew and Kasala (1994) is biased and should be used only in some specialsituations(see Section 3.3). A pivot statisticpresentedin the presentpaperis similarto thatusedby MathewandZha (1996)to constructconfidenceregions.Specifically,the squareof ourpivotis equivalentto that of MathewandZha(1996).Thusourpivotstatisticcanbe usedforone-sidedhypothesistesting. Furthermore,we haveusedthe parametricbootstrap(PB)to takecareof the dependenceof the distributionof the pivot statisticon the unknownvectorof slopeparameters. Severalresearchershavealreadyappliednonparametric bootstraptechniquesto the calibration problem;see Jonesand Rocke(1999)for detailsand furtherreferences.Althoughmost of the referencesin this contextdeal with the univariatecase, Jonesand Rocke(1999)also considereda multivariatenon-linearcalibrationproblem.In general,the bootstrapmethodcan be appliedin both parametricand nonparametricmodes.Efronand Tibshirani(1993)devoteda few sectionson parametricbootstrappingand illustratedthe PB methodto computethe standarderrorof the samplecorrelationcoefficientin the bivariatenormaldistribution.Lee (1994) versionprovidedthat concludedthata PB resultmaybe moreaccuratethanits nonparametric

CalibrationProblems Multivariate-Univariate

19

correct.In ourset-up,the bootstrappingis done the modelassumptionis at leastapproximately i.e. undermodels(1.1) and (1.2). Section2 providesresultsalongthis direction parametrically, for obtainingconfidenceregions.Section3 dealswiththe PB procedurefor obtainingsingle-use hypothesistests. In Section4, we illustrateour resultsand, in particular,the computational procedures,by using some examples.For this, we haveused data sets based on the firstthree examplesin Section1.2. Section5 dealswithmultiple-usehypothesistesting.Someconcluding remarksappearin Section6. The majorthrustof the presentpaperis to explorethe role of the PB in constructingconcalibrationproblem.Wedescribethe PB fidenceregionsand tests in a multivariate-univariate ThenumericalresultsshowthatthePB procedureis andalsonumericallystudyits performance. indeedsatisfactory.Furthermore,the procedureis applicableforone-sidedhypothesistestingas well.As expected,somecomputationaleffortis requiredto computethePB confidenceintervals and hypothesistests.However,the requiredcomputationaleffortis muchless comparedwith what is requiredto computethe conservativeconfidenceregionsin Mathewand Zha (1996, 1997, 1998)and Mathewet al. (1998).Ourfinalconclusionsare that the PB methodis simple to use and worksquitewell in the calibrationproblem.

2. Single-use confidence regions In this section,we shallconstructa single-useconfidenceregionfor 0 in model(1.2), basedon the (yi, xi)s in model(1.1) andyo in model(1.2). Define x =

-

E

n i=l1

xi,

n

n y1 i=

(2.1)

n

sxx i=1

(xi-

S =nA(yi - a - Oxi) (Yi - a - 3xi) , i=l1

where& and 3 denotethe least squaresestimatorsof a and /3 basedon model (1.1). Clearly, yo, , /3 and S are independently distributed with y ~ Np a + ,3x, n- E

(

'

( S

(2.2)

1)

Sxx

- 2, E),

Wp(n

whereWp(n- 2, E) denotesthep-dimensionalWishartdistributionwithn - 2 degreesof freedom and scalematrixE. 2. 1. A pivot statistic

If /3and E areknown,then the least squaresestimatorof 0, say 0, is givenby x

+

/'3-' (yo - Y) E

20

D. Benton,K.Krishnamoorthy and T Mathew

Consequently, Zo

=

(1

=

)1/

- 0)

(2

1+ 1/n

(0-)

3(0-- 2)} 'c-'I {yo- - y y - #(0 {(1 + 1/n))'Y-1)3}1/2

N(0, 1).

(2.3)

The quantityZo in equation(2.3)can be usedas a pivot for obtaininga confidenceregionfor 0 when 3 andE areknown.Since 3 andE areactuallyunknown,we replacethemby theestimates ,3 and 1 =

n-21 S.

Theresultingpivot statistic,denotedby Q, is - y 'S-{-yo -/3(01 2)} + 'S}1/2 {(1 1/n)

wherewe haveignoredthe constant1/V/(n- 2) in Q. Let

'(2.4)

u=) sxx ,

O1= 3V/Sxx, Yo-

=

=

Then

v/(1+ /n)'

(2.5)

+ 1/n)}

-V/{sxx(1

u ~ Np(31, E),

vo

~

Np

(•l01,

E),

(2.6)

and Q in equation(2.4) can be expressedas Q=

u'/S-1(v0

- uO1)

(uSu)/2

(2.7)

A slightmodificationof Qis necessary(seeAppendixA) to studyits distribution.Themodified pivot statistic,say To,is givenby

To=

(n - p - 1)1/2u'S-'(vo - u01) {(1 + r)u'S-lu}l/2 - (voS-u)2(2.8) u'S-lu in termsof the(Vus-lu variablesin models

r=

vbS-lvo

Note that Tocan be easilyexpressed original (2.1)and (2.2). Note also thatit is enoughto constructa confidenceregionfor01, sincea confidenceregionfor 0 can thenbe obtainedby usingthe relationshipbetween01and0, givenin equations(2.5).The pivots Q2 and T2 (or otherequivalentforms)havebeenconsideredby Williams(1959),Wood (1982),FujikoshiandNishii(1984),Davis and Hayakawa(1987)andMathewandZha (1996).


21

The asymptotic results in Fujikoshi and Nishii (1984) and Davis and Hayakawa (1987) and the conservative confidence regions derived by Mathew and Zha (1996) are all based on the pivot statistic T2.

2.2. The distributionof To

From the distributional results in expressions (2.2) and (2.6), and from the expression for Toin equation (2.8), it follows that the distribution of Tois invariant under non-singular transformations, i.e. it is invariant under the transformation u -* Pu, vo -* Pvo and S -- PSP', where P is any non-singular matrix. Hence, without loss of generality, we shall assume throughout this paper that E = Ip. The proof of the following theorem is given in Appendix A. Theorem1. Let Toand r be as defined in equation (2.8) and assume that E = Ip. Then, conditionally given u and r, To is distributed as tn-p-1(60), the non-central t-distribution with n - p - 1 degrees of freedom and non-centrality parameter '(/

- u)1(2.9)

{u'u(1+ r)}1/2 The distribution given in theorem 1 is not directly useful for obtaining confidence regions owing to the dependence on the unknown slope parameter il. Following the arguments of Mathew and Zha (1996), a conservative confidence region can be obtained by using a larger non-centrality parameter than 60 in equation (2.9). In particular, using the inequality u'(l31 V - u)O1//(u'u) {02(/1 - u)'(/3 - u)}1/2 -= /(02 V), we have o < (02V). Noticing that follows a central X2-distributionwithp degrees of freedom (under our assumption E = Ip), and replacing 60 by '/(02 V), a conservative confidence region can be obtained, as in Mathew and Zha (1996). Since developing exact inferential procedures based on Tois not possible, and simple approximate inferential procedures based on Toseem to be difficult, we resort to the PB method.

2.3. Theparametricbootstrapmethod forconfidence regions

The PB method can be briefly explained as follows. Let X1,..., X, be a sample of independent observations from a population with distribution function F(x; r7),where rqis an unknown parameter. Suppose that we want to estimate a real-valued function g(7). Let "~be an estimator of q based on the sample. Then, F(x; ^i) is called the PB estimate of the population distribution function F(x; rl). The samples Xil,..., Xi,, i = 1,..., m, generated from F(x; ^) are called PB samples. The sampling distribution of g('^) can be approximated by the empirical distribution of g(ii)s, where ^i is the estimate of i7 based on the ith PB sample from F(x; ^). In many situations, is can be generated directly from the sampling distribution of i. The sampling distribution may be used to make inferences about g(r). For example, when m is sufficiently large, (g(71)(am/2), 9(ij)((1-a/2)m)),

where g(7)(k) denotes the kth smallest of the g(i~i)s, serves as

an approximate 100(1 - a)% confidence interval for g(i). This is referred to as the percentile bootstrap. We can also check the performance of a PB method for a specific model. For example, let 'qo be a value chosen arbitrarilyfrom the parameterspace of q. Let i1, .. ., im be estimates based on and let m samples generated from F(x; 0o).Let ^li, ... , ^ki be estimates generated from F(x; ^1i), = of m. If the i interval for PB confidence denote the 1, ..., proportion lis containing g(7r), Ii is approximately equal to the confidence level specified, then we conclude that the PB g(ro0) method will yield satisfactory results for the model.

andT Mathew D.Benton,K.Krishnamoorthy The PB confidenceintervalbasedon the pivot statisticTocan be obtainedas follows.

22

(a) For a givenconfidencelevel 1 - a, let kI(01)and k2(01)be suchthat P {ki(01) - To - k2(01)} = 1 -

•

(b) By invertingthe inequalitiesin step (a) for 01,we obtainthe lowerand upperconfidence limits,say 011and 01urespectively,as 1

1 + ki (01)

011 =

u = + k2 = +

{

1/2

+r

(n - p - )uS-lu l+ r (

1/2'

01 k2(01) (n - p- 1)uf'S-lu

1u

where S=

u'S-' vo

(2.10)

u-lu

(c) To estimatethe valuesof k1(01)andk2(02), computeu and 01usingcalibrationdata and yo. For i = 1, 100000, generate ui - Np(u, Ip), voi ~ Np(uOl,Ip) and Si Wp(n-2, Ip).

Set

Toi

- + (n - p{(1 1)1/2U S'(voi

- ui,1)

ri)uS.lui}1/2

'

whereri denotesthe valueof r in equation(2.8) computedby usingui, voiand Si respectivelyin placeof u, vo and S. (d) Let To(a)denotethe 100athpercentileof the Toisin step(c). Then,kI(01) To(a/2) and k2(01)

"

0 - a/2). T(1

Note thatwe can also constructa confidenceintervalfor01by usingthe percentilebootstrap, i.e. based on the PB sampling distribution of 01 in equation (2.10). The procedure is as follows.

(a) For i = 1, 100000,generateui ~ Np(u,Ip), voi ~ Np(uO1,Ip) and Si - Wp(n- 2, Ip). Let 0Oi=

'voi/uSi

ui.

be the 100athpercentileof the Olisgeneratedin step (a). Then, for a given (b) Let 01(a)uiS confidence level 1 - a, (01(a/2), 01(1 - a2)) is an approximate 100(1 - a)% confidence

intervalfor 01.

2.4. Performanceof the parametricbootstrapmethod To understandthe validityof the PB procedure,andto comparethe PB resultsbasedon 01 and the pivot statisticTo,we estimatedthe coverageprobabilitiesof the PB confidenceintervalsby using the evaluationmethoddescribedin Section2.3. For example,for a given and 01, the 11 coverageprobabilityof the confidenceintervalbasedon Tocan be estimatedas follows. 2.4.1. Estimation of the coverageprobabilities of the parametric bootstrap confidenceintervals based on To (a) For i = 1, 1000, generateui - Np(31, Ip) and voi ^ Np(0 101, Ip) and Si - Wp(n- 2, Ip). (b) ComputeToiby formula(2.8), using ui,voi and Si in place of u, vo and S respectively.

Considerthis as an observedvalueof the test statisticTo.


23

, (c) For j = 1, 1000, generate uij - Np(ui, Ip), voij Np(ui81, Ip) and Sj -- Wp(n - 2, Ip). (d) ComputeToijby formula(2.8), usinguij, voijand Sj in placeof u, vo and S respectively.

Note that Toijalso dependson i. (Endthej-loop.) (e) Fora fixedi (i = 1,2...., 1000),let Ii denotethePB confidenceintervalfor 01,obtained by usingthe percentilesof the Toijs,j = 1, 2....., 1000. (f) The proportionof lis thatcontain01providesan estimateof the coverageprobabilityof the PB confidenceintervalbasedon To. The coverageof the confidenceintervalbasedon the percentilebootstrap(i.e. basedon the bootstrapdistributionof 01) can be similarlyestimated.Table 1 gives the estimatedcoverage probabilitiesand expectedlengthsof the PB confidenceintervalsbased on Toand 01, for /31= (1, 2, 3)',n = 10,20, 30 andvariousvaluesof 01.WeobservefromTable1thatthecoverage probabilitiesof the PB confidenceintervalsbasedon Toand i1arealmostequalto the specified levelwhen1011 1. In most applications,the valuesof 1811shouldnot exceed1; see remark1 below.Furthermore,for 101i1 1, the expectedlengthsof the PB intervalsbasedon Toand 01 arevirtuallyequal. Remark1. It appearsreasonableto assumethat 1011< 1 in most practicalapplications.This can be explained as follows.Note that v2 = sxx/(n - 1) is the varianceamongthexis, wheresxx is definedin expression(2.1). In carefullydesignedexperiments,the rangeof the xis will cover the rangeof practicalinterest,and 0 will be withinthis range.At the veryleast,we can assume that 10- -I1< 3vx. From the expressionfor 01 in equations(2.5), it now followsthat, when 0 --xl 3vx, we have1Oil 1 for n >, 10. If we assumethat 0 - x-i < 2vx, then 1011< 1 for n > 5.

3. Single-use hypothesis tests Sincethe single-useconfidenceregionin the previoussectioncan also providea test procedure in the case of two-sidedalternatives,i.e. hypotheses(1.4a),in this sectionwe shallconsidertests Table 1. Estimated coverage probabilitiesand expected lengths of PB confidence intervalsbased on Toand 91 forp = 3 and 3' = (1,2, 3)t Coverageprobabilitiesand lengthsfor thefollowingvaluesof n:

O1

To

-2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0

0.93 0.93 0.95 0.94 0.96 0.95 0.94 0.93 0.92

(2.3) (1.9) (1.4) (1.1) (0.9) (1.2) (1.4) (1.9) (2.4)

11

0.93 0.94 0.95 0.95 0.96 0.96 0.94 0.94 0.93

n = 30

n = 20

n = 10

(1.9) (1.6) (1.2) (1.0) (1.0) (1.0) (1.2) (1.6) (1.9)

01 TO

0.92 0.93 0.94 0.94 0.96 0.95 0.95 0.92 0.91

(1.8) (1.5) (1.2) (0.9) (0.8) (0.9) (1.2) (1.5) (1.9)

in are "Expectedlengths given parentheses.

TO

0.92 0.93 0.95 0.96 0.96 0.94 0.94 0.92 0.90

(1.7) (1.4) (1.1) (0.9) (0.8) (0.9) (1.1) (1.4) (1.7)

0.93 0.93 0.94 0.95 0.94 0.93 0.95 0.92 0.90

(1.7) (1.4) (1.1) (0.9) (0.8) (0.9) (1.1) (1.4) (1.7)

01

0.93 0.92 0.95 0.96 0.96 0.94 0.94 0.92 0.91

(1.7) (1.3) (1.1) (0.9) (0.8) (0.9) (1.1) (1.3) (1.7)

24


only for one-sided alternatives, i.e. the hypotheses (1.4b). Hypotheses (1.4c) can be handled similarly. Define

c-x

cl =

/{sxx(1

+ l/n)}

(3.1)

Hypotheses (1.4b) can now be expressed as Ho : 01 ( cl versus H1 : 01 > cl.

(3.2)

3. 1. The tests Let 3'E-1 {yo - y - 3(x - c)}

1 Z1 =

J(1 + 1/n)

(P/'E-13)1/2

(n - p - 1)1/2u'S-1(vo - ucl) {(1 + r)u'S-lu}l/2 S

(3.3)

where the various quantities in equations (3.3) are as defined in Section 2. We note that Z1 in equations (3.3) is the quantity Zo in equation (2.3) evaluated at 0 = c. Similarly, T1in equations (3.3) is obtained by replacing 01 by cl in the expression for Toin equation (2.8). The statistic T1 in equations (3.3) is a possible candidate for testing hypotheses (3.2). The use of TI is motivated by the same arguments that justified the use of Toin equation (2.8) for constructing confidence intervals. That is, Z1 in equations (3.3) is the test statistic to be used for testing hypotheses (1.4b), i.e. hypotheses (3.2), when 3 and E are known, and TI is obtained from Z1 by replacing 3 and E by their estimates. The use of Z1 in equations (3.3) is meaningful for testing hypotheses (3.2), since the distribution of Z1 is stochastically increasing in 01, as can be easily verified. However, we cannot claim a similar property for TI. Hence the performance of the test based on T1 must be numerically evaluated. From theorem 1, it follows that when 01 = and conditionally given u and r, TIis cl, distributed as tn-p-1(61), with non-centrality parameter u'(

1 - u)cI

S{(1 + r)u'u}1/2

The test basedon T1rejectsHoin hypotheses(3.2) for largevaluesof T1. A differenttest statisticcan be obtainedby usingthe followingargument.When01 = cl, P1 can be estimatedby usingu and voin equations(2.6). The resultingestimator31c is givenby U+

Al

01cl

1 + c2

and E with S/(n - 2) in equation (2.3), and after some simplifications, Replacing 31 with c1, we obtain a statistic that is proportional to the quantity u'~S-.vo* UU

S-l*=2


25

where *

=u + vocl,

= (vo - ucl)/(/(1

+ c).

vo0

Note thatthe expressionfor Q*is similarto the expressionfor Qin equation(2.7).To arriveat the distributionof Q*, we shallconsiderthe followingmodifiedstatisticT2(see AppendixB): T2=

(n -

p - 1)1/2u',S-luo

ulS-mu,

Note that T2(and also T1in equations(3.3)) can be easilyexpressedin termsof the original variablesin expressions(2.1) and (2.2).A pivot statisticequivalentto T2has beensuggestedby Mathewand Kasala(1994)for obtainingsingle-useconfidenceregions.The distributionof T2 is givenin the followingtheorem.The proofof theorem2 is givenin AppendixB. Theorem2. Let T2and be as definedin equations(3.5) and assumethat E = Ip. Then, r, T2is distributedas t,-p-1(62), with non-centralityparameter conditionallygiven and u, r,, 62 =

u',

(9

0I

1 - ci)

{u'*u*(1+ r*)(1 + c)}1/2( 2)1/2

(3.6)

Now consider the test that rejects Ho in hypothesis(3.2) for large values of T2. Since u*, Np{(1 + 1Ocl)1, (1 + c2)Ip)}, U* is more likely fall in the direction of 31 (providedthat01cl > -1), andhenceu',31 is expectedto be positive.If this happens,thenfromthe expressionfor the non-centralityparameterin equation(3.6)we can concludethatthe test that rejectsHoin hypothesis(3.2)forlargevaluesof T2attainsits maximumsizewhen01 = cl. From equation(3.6), it followsthat, when01 = Cl,T2followsa centralt-distributionwith n - p - 1 degreesof freedom.Thus,for a givenlevela, the test thatrejectsthe nullhypothesis(3.2)whenis expected to have size at most a. Here tn-p-1,1- is the 100(1 - a)th ever T2 > tn-p-1,1-a percentileof a centralt-distributionwith n - p - 1 degreesof freedom. Theseargumentsshow that the test basedon T2can be easilycarriedout by using an exact centralt-distribution.However,this is not so for the test basedon T1,sincethe distributionof T1dependson ,1. Thuswe shalluse the PB to carryout the test basedon T1,just as we did for the confidenceinterval.Note that,if we intendto employthe PB, 01in equation(2.10)can also be used as a test statistic,with largevaluesof 01providingthe rejectionregion.The PB tests basedon T1and 01can be carriedout as follows. 3.1.1.

Parametricbootstrap test based on the test statistic T1

The PB testingprocedurecan be describedas follows. (a) Using the yis andyo in models(1.1) and (1.2), computethe test statistic S

(n - p - 1)1/2u'S-l(vo - ucl) {(1 + r)u'S-lu}1/2

wherethe quantitiesS, u, vo,r andcl aregivenrespectivelyin equations(2.1),(2.5),(2.8) and (3.1). Denote the computedvalueby Tlo.

26


(b) For i = 1, 100000, generate ui -- Np(u, Ip), voi - Np(u01, Ip) and Si - Wp(n - 2, Ip).

Computethe test statisticby usingthe formulafor T1in step(a), by replacingS, u andvo by Si, ui and voirespectively,and call it Tli. (c) For a givenlevela, if the proportionof Tlisthataregreaterthan T1ois less thana, reject the nullhypothesis(3.2). HereT1ois the computedvalueof T1in step(a). The PB test basedon 01 can be carriedout similarly. 3.2. Sizes and powers of the tests based on T1 For testinghypotheses(3.2), the size and powerof the PB test basedon T1can be evaluatedas follows. (a) For i = 1, 1000, generate ui - N,(31, Ip), voi - Np(30 1, Ip) and Si ^ Wp(n - 2, Ip).

Computethe valueof T1by usingthe formulagivenabove(see step(a) in Section3.1.1), Denote the computedvalueby Tli. by replacingS, u andvoby Si, ui and voirespectively.

(b) Forj = 1, 1000, generate uij "- Np(ui, Ip), voij - Np(uicl, Ip) and Sj -^~ Wp(n - 2, Ip).

Computethe valueof T1againusinguij, voijand Sj in placeof u, vo and S respectively, and denotethe computedvalueby Tlij.Note thatTlij alsodependson i. (Endthej-loop.) (c) Computethe proportionof Tlijsgreaterthan Tli.If thisproportionis less thanthe specifieda, set Pi = 1. (Endthe i-loop.) (d) If 01 = cl, then E1o Pi/1000 is the estimatedsize of the PB test basedon T1.If 01 > Cl, then ~E1 Pi/1000 is the estimatedpowerof the PB test basedon T1. The size and powerof the PB test based on 01 can be estimatedsimilarly.Table2 givesthe estimatedsizesandpowersof the PB testsbasedon T1and01for testinghypotheses(3.2).Table 3 givesthe samewhenthe hypothesesof interestare H0 : 01 > cl versusH1 : 01 < cl.

(3.7)

From the numericalresultsin Table2 and Table3, we see that the PB test based on T1is satisfactoryfor 1011< 1.5,whereasthe PB test basedon 01is satisfactoryonly for smallvalues of 101. In particular,the sizesof the PB test based on 01 is smallerthan the specifiedlevelfor 01 > 0.5, and largerthan the specifiedlevel for 01 < -0.5. Ourlimitedsimulationsshow that 01 should not be used for testinghypotheses(3.2) and (3.7). This is in agreementwith what is known about the percentilebootstrap(i.e. based on the bootstrapdistributionof 01). That is, the use of a pivot statisticshouldbe preferredto the use of 01 for constructingconfidence intervalsand tests;see Hall andWilson(1991)and Hall (1992). Othernumericalstudiesfor some othervaluesof a, p = 4 and 01 yieldedsimilarconclusions;see Benton(2000)for some additionaltables.Wenote fromthesenumericalstudiesthat the powerof the PB tests increasesas

II3111

i=1

increases. 3.3. Powercomparisonof the tests based on T1and T2 Table4 gives the simulatedpowers(100000 runs) of the exact test based on T2in equation (3.5) and the PB test basedon TI,for testinghypotheses(3.2) and (3.7). The powerof the PB

Multivariate-Univariate CalibrationProblems

27

Table2. Estimatedpowersof the PB tests based on T1 and 01 for testing hypotheses (3.2) for a = 0.05, p = 3 and 3 = (2, 1,5) cl

01

Powersfor thefollowingvaluesof n: n = 10

1.5

1.0

0.5

0.0

-0.5

-1.0

-1.5

1.5 2.0 2.5 3.0 1.0 1.5 2.0 2.3 0.5 1.0 1.5 2.0 0.0 0.5 1.0 2.0 -0.5 0.0 0.5 1.0 -1.0 -0.5 0.0 0.5 -1.5 -1.0 -0.5 0.0

n = 30

n =20

T1

01

TI

01

T1

01

0.08 0.26 0.48 0.72 0.07 0.30 0.66 0.79 0.06 0.40 0.80 0.93 0.05 0.44 0.86 0.97 0.06 0.35 0.75 0.92 0.06 0.29 0.62 0.87 0.05 0.26 0.56 0.83

0.04 0.16 0.43 0.74 0.03 0.22 0.63 0.81 0.04 0.34 0.83 0.96 0.05 0.48 0.90 0.99 0.06 0.48 0.88 0.98 0.10 0.41 0.81 0.97 0.13 0.41 0.74 0.94

0.05 0.24 0.52 0.75 0.07 0.31 0.65 0.81 0.06 0.42 0.84 0.96 0.05 0.48 0.90 0.99 0.06 0.43 0.84 0.98 0.06 0.32 0.75 0.95 0.06 0.28 0.60 0.88

0.02 0.16 0.46 0.75 0.04 0.23 0.61 0.83 0.04 0.32 0.86 0.97 0.05 0.50 0.94 0.99 0.08 0.54 0.92 0.99 0.12 0.47 0.88 0.99 0.12 0.41 0.75 0.95

0.06 0.27 0.54 0.80 0.06 0.32 0.72 0.86 0.05 0.43 0.88 0.97 0.04 0.52 0.93 0.99 0.05 0.43 0.88 0.99 0.05 0.35 0.76 0.95 0.05 0.29 0.61 0.89

0.03 0.17 0.45 0.76 0.04 0.24 0.67 0.84 0.04 0.36 0.86 0.98 0.05 0.51 0.95 0.99 0.07 0.53 0.93 1.00 0.10 0.50 0.87 0.99 0.10 0.45 0.80 0.96

test basedon Ti was evaluatedfollowingthe procedureoutlinedin Section3.2. Regardingthe powerof the exacttestbasedon T2in equation(3.5),note that andvo*in equations(3.4)are u, + independentwithu* - Np{(1+01cl)I3, (1 + c2)Ip} andvo*, Np{1 (01 - Cl), Ip}. It is easy to see thatthe expectedvalueof the numeratorof T2is proportionalto - cl) = (01 - cl) + Olcl (1 (01 - cl)(1 + 01cl).

Therefore,if 01Cl < - 1, thenthe test maynot be ableto detectthe differenceevenif 01is much largerthancl. Forexample,forthehypothesesHo : 01 < -0.2 versusH1 : 01 > -0.2, thepower of the test basedon T2declinesfor 01 > 1.5 (see Table4). A similarobservationalso holds for the hypothesesHo : 01 > 1 versusH1 : 01 < 1. As a consequence,the test may incorrectly acceptthe nullhypothesis.Further,the confidenceintervalbasedon T2is likelyto containfalse valuesfor 01, a fact alreadyobservedby Mathewand Kasala(1994)in theirnumericalstudies. Nevertheless,it is a powerfultest if both 01 and Cl arepositiveand H1 : 01 > cl, or if they are both negativeand H1 : 01 < cl. Otherwise,the test basedon T1shouldbe preferred.

28

D. Benton,K.Krishnamoorthy and T Mathew Table 3. Estimated powers of the PB tests based on T1 and 01 for testing hypotheses (3.7) for a = 0.05, p = 3 and P' =(2, 1,5) Cl

Powersfor thefollowingvaluesof n:

01

n = 10

1.5

1.0

0.5

0.0

-0.5

-1.0

-1.5

1.5 1.0 0.5 0.0 1.0 0.5 0.0 -0.5 0.5 0.0 -0.5 -1.0 0.0 -0.5 -1.0 -1.5 -0.5 -1.0 -1.5 -2.0 -1.0 -1.5 -2.0 -2.5 -1.5 -1.0 -0.5 0.0

n =20

n =30

T1

01

T1

01

T1

01

0.06 0.22 0.51 0.79 0.05 0.29 0.67 0.87 0.07 0.32 0.72 0.89 0.04 0.37 0.78 0.90 0.06 0.34 0.70 0.84 0.07 0.27 0.57 0.74 0.06 0.21 0.43 0.63

0.11 0.34 0.72 0.92 0.09 0.41 0.84 0.97 0.09 0.45 0.88 0.99 0.05 0.42 0.88 0.98 0.03 0.30 0.77 0.94 0.05 0.22 0.56 0.83 0.04 0.15 0.38 0.68

0.05 0.26 0.61 0.88 0.05 0.32 0.73 0.94 0.05 0.44 0.86 0.98 0.06 0.48 0.90 0.98 0.06 0.38 0.81 0.95 0.06 0.34 0.67 0.89 0.07 0.26 0.53 0.77

0.11 0.41 0.79 0.96 0.10 0.47 0.87 0.99 0.08 0.56 0.93 0.99 0.06 0.50 0.93 0.99 0.04 0.32 0.81 0.98 0.03 0.25 0.63 0.91 0.04 0.17 0.46 0.75

0.07 0.27 0.61 0.87 0.06 0.36 0.77 0.97 0.06 0.44 0.90 0.98 0.05 0.52 0.92 0.99 0.06 0.44 0.87 0.98 0.07 0.35 0.73 0.93 0.07 0.27 0.57 0.79

0.13 0.42 0.79 0.95 0.11 0.53 0.88 0.99 0.08 0.55 0.95 0.99 0.06 0.52 0.94 0.99 0.03 0.37 0.85 0.99 0.04 0.24 0.67 0.94 0.04 0.19 0.48 0.77

Table 4. Estimated powers of the PB tests based on T1 and the exact test based on T2 for n = 15, a = 0.05, p = 3 and P' = (2, 1,5) H1

01

T1

T2

01 ( 1.0

01 > 1.0

1.0 1.5 2.0 2.5

0.04 0.30 0.65 0.86

0.05 0.30 0.72 0.95

01 ? -0.2

01 > -0.2

0.5 1.0 1.5 2.0 3.0

0.66 0.93 0.97 0.99 1.0

0.62 0.87 0.90 0.88 0.74

H0

H1

01

T1

T2

-0.5

01 < -0.5

01 g>1.0

01 < 1.0

-0.5 -1.0 -1.5 -2.0 -2.5 1.0 0.5 0 -0.5 -1.0

0.03 0.34 0.79 0.96 0.99 0.06 0.31 0.70 0.93 0.97

0.05 0.40 0.86 0.99 1.0 0.05 0.28 0.59 0.57 0.25

H0 01 O


29

4. Some examples We shall now illustrate the computation of the confidence regions and tests, developed in Sections 2 and 3, by using three data sets: the meat data, the paint finish data and the wheat data. For background information on these data sets and the corresponding inference problems, see Section 1.2. 4. 1. The meat data Recall that the hypotheses of interest in the meat data problem are given in expression (1.3), where 0 represents the fat content in a sample of meat. Among the data available on 214 meat samples, 172 samples were used to fit model (1.1). This leaves 42 samples which can be used for illustrating the PB procedure for testing hypotheses. Note that for these 42 samples the value of 0 is known. We have selected six of the 42 samples for testing purposes; these six samples have some values of fat content below 10 and some above. On the basis of the 172 samples used to fit the models, the various quantities have values = (0.0379, -0.0479, -0.0407) = 27 501.17, i = 18.38, y' = (-0.0167, 0.0073, 0.0113), xx and 133.78134

61.48432

36.68384

61.48432 122.48032 -60.26106 36.68384

-60.26106

127.53256

Hence 31 = 3./sxx = (6.285, - 7.943, - 6.749). Table 5 gives the six values of yo, the true values of 0 and the p-values for testing hypotheses (1.3). These p-values are based on the PB test using T1and are computed on the basis of 100000 iterations (see Section 3.1.1). We note that, among the results reported in Table 5, the wrong decisions are reached in two cases (corresponding to the third and fourth yo-values), if we use a significance level of 0.05. 4.2. The paint finish data To demonstrate how the proposed PB methods compare with other methods, we apply them to the paint data described in example 2 of Section 1.2. In this example, xi representsviscosity coded -1, 0 and 1, and yi comprises two measurements of optical properties of the surface covered with the paint, corresponding to viscosity xi, i = 1, .. .,27. Among the 27 xi-values, there were nine values corresponding to each of the coded viscosity values -1, 0 and 1. The calibration data set consists of the 27 observations used by Brown and Sundberg (1987) Table 5. Simulated p-values of the test based on T1 for testing the hypotheses Ho : 10.0

YO (-1.183, -0.523, 1.042) (-0.078, -0.036, 1.592) (1.254, 0.070, 1.448) (2.180, 2.827, -1.143) (0.207, 0.135, 0.306) (1.155, -1.041, 1.525)

0

p-value

4.8 8.6 16.0 17.0 19.4 24.8

0.74 0.77 0.10 0.06 0.01 0.00

30

D.Benton,K.Krishnamoorthy andT Mathew

and Mathew and Zha (1996). Computationsbased on the calibrationdata gave sxx = 18, x=O, & = y = (1.7478,37.9363)', = (-0.1278, -1.6922) and S-1

(6.86285

0.03052)

0.03052 0.02299" In Table6, we providethe PB confidenceintervalsbased on Toin equation(2.8) and based on 01, along with the other intervalsgiven in Mathewand Zha (1996).The PB intervalsare calculatedby using the methodoutlinedin Section2. We firstcomputedPB intervalsfor 01, which were then modifiedto obtain intervalsfor 0. The PB intervalestimatesare based on 100000PB samplesand are denotedby PB(To)and PB(O)in Table6. Table6 also gives the exactconfidenceregionfromBrown(1982),the conservativeconfidenceregionfromMathew and Zha (1996)and the likelihood-basedconfidenceregionfromBrownand Sundberg(1987), denotedby 'likelihood'in Table6. It shouldbe recalledthat Mathewand Zha's(1996)conservativeconfidenceintervalis also basedon the statisticTo.Forthisexample,theirconfidenceintervalsarein good agreementwith the PB(To)intervals.Thisconclusionis quitesimilarto thatin Jonesand Rocke(1999),namely that,in the univariatecase,bootstrappingan appropriatet-statisticproducesresultsthatarein agreementwith the standardmethod.An advantageof the PB methodis thatit is simpleto use in comparisonwith Mathewand Zha'smethodwhichis computationallyquite involved.We also noticefromTable6 thatthe PB intervalsbasedon the samplingdistributionof 0 areclose to Brown'sintervals.It shouldalso be noted that,eventhoughthe likelihood-basedconfidence regiondue to Brownand Sundberg(1987)appearsto be the narrowestin Table6, the coverage of this intervalcan be below95%;see the discussionin MathewandZha (1996),page 719. 4.3. The wheat data For the wheatdata, the yis are bivariateinfra-redreflectancemeasurementson 21 samplesof hardwheatand the xis representthe percentageof waterin the wheatsamples.Weuse the first 15 observationsto fit the model.The PB methodsaredemonstratedon the remainingobservations with the exceptionof observation20, for whichthe valuesof the variablesof interestare identicalwith those of observation19. The relevantestimatesand othersummarystatistics,basedon the first 15 observations,are = = = sxx 3.4013,i 9.52,y' (81.933,231.500),3 = (-24.918, -23.601), S

(120.910 22.075 22.075 45.072

Table6. 95%confidenceintervalsfor0 forthe paintfinishdata Yo

0

Confidenceintervalsfrom thefollowingmethods: Brown(1982)

(1.78, 38.73) (1.79, 39.83) (1.52, 35.65) (1.94, 34.09)

-1 0 1 Unknown

(-1.65, 0.92) (-1.93, 0.57) (0.39, 3.12) Empty set

Likelihood MathewandZha (1996) (-1.27, 0.57) (-1.60, 0.27) (0.71, 2.68) (-1.34, 1.16)

(-1.41, 0.67) (-1.76, 0.37) (0.61, 2.97) (-1.39, 1.23)

PB(To)

PB(O)

(-1.35, 0.68) (-1.64, 0.37) (0.60, 2.70) (-1.38, 1.16)

(-1.59, 0.89) (-1.90, 0.57) (0.39, 3.03) (-1.32, 1.16)


31

and i3 = P3Jsxx = (-45.955, -43.526). In Table 7, we provide the PB single-use confidence intervalsand the p-valuesfor testinghypothesesbased on five values of yo, followingmodel (1.2), and where0 is known.For illustration,a test of Ho : 0 ( 9.25 versusH1 : 0 > 9.25 is conducted.The single-useconfidenceintervalsin Table7 werecomputedin the samewayas for the PB(To)intervalsin the previousexample.Thep-valuesin Table7 correspondto the PB test basedon T1and arecomputedon the basisof 100000bootstrapsamples. Wenote that the test procedureresultsin correctdecisionsin fourof the fivecasesif we use a 5%significancelevel.At the 10%level,the test resultsin correctdecisionsin all cases. Thethreeexamplesgivenaboveillustratethe applicabilityandusefulnessof thePBprocedure for obtainingconfidenceintervalsand tests.

5. Multiple-useone-sided hypothesis testing Recallthat,in the multiple-usecase,the calibrationdata setis usedrepeatedlyto testa sequence of hypotheses,one at a time, correspondingto a sequenceof unknown0-values,afterobserving the correspondingvalue of yo in model (1.2). Thus, considera sequenceof observations {Y0oj},j = 1, 2, 3....,

following the model yoj

-

Np(a + P0j, E),

wherethe yojs are also independent.Problemsof confidenceregionsand hypothesistests are considerablymore difficultto solve in the multiple-usecase, as opposedto the single-usesituation. The difficultyarisesfromthe fact that two typesof uncertaintyneed to be taken into accountin the multiple-usecase-the uncertaintyin the calibrationdata and the uncertainty in the sequence{Yoj}.The problemof derivingmultiple-useconfidenceregionshas been attemptedby severalresearchers:see Mee et al. (1991) and Mee and Eberhardt(1996)for the univariatecase,andMathewandZha (1997)andMathewet al. (1998)forthe multivariate case. The multiple-useconfidenceregionsin Mathewet al. (1998)can be used for testingtwo-sided hypotheses,i.e. hypothesesof the form Hoj : Ooj= cj versusHlj : Ooj= cj,

wherethe cjs areknownconstants.Herewe shallonly considerone-sidedhypotheses,i.e. Hoj : Ooj( c versusHlj : 00j > c,

(5.1)

wherec is known.Evenin the contextof test (5.1), a moregeneralproblemis one wheredifferent constantscj areusedinsteadof a commonconstantc. Recallthat,for the meatexamplein Section1.2, it is indeedmeaningfuland practicallyusefulto test the sequenceof hypothesesin Table7. 95%confidenceintervalsfor9, andp-valuesfortesting Ho: 9 ( 9.25 versusH1: 0 > 9.25, forthe wheatdata yO (101.0, 248.0) (85.0, 235.0) (88.0, 231.0) (69.0, 222.0) (63.0, 216.0)

9

0

8.86 9.34 9.46 10.00 10.12

8.81 9.38 9.48 9.95 10.20

Confidenceinterval p-value (8.61, 9.00) (9.19, 9.56) (9.27, 9.69) (9.76, 10.14) (10.00, 10.40)

0.99 0.08 0.01 0.00 0.00

D.Benton,K.Krishnamoorthy andT Mathew expression(5.1), with c = 10%;see model(1.3). Motivatedby the definitionof Zo in equation (2.3) and Toand T1in equations(2.8) and (3.3) respectively,let 32

T3j =

(n - 2)1/23'S-1(yoj - & - 3c) S/)1(5.2)

where& and 3 denotethe least squaresestimatorsof a and 3 basedon model (1.1), and S is definedin equations(2.1). Supposethatwe rejectHojwhen T3j > k(c),

wherek(c) is to be determined. 5. 1. The criterionformultiple-usehypothesis testing Weshallnow describethe conditionto be usedfor the computationof k(c). Let if T3j> = 1, k(c), (5.3) 0 otherwise. , If a sequenceof N tests is carriedout for j = 1, 2,..., N, then the proportionof Hojsthat are rejectedis 1iN

j=Z1 The quantity k(c) is to be determined so that, for all values of Oj under Hoj (j = 1, 2, ..., N),

this proportionis smallwith a highprobability.Let

(5.4) pj = P(T3j> k(c)l&,3, S). Note that,conditionallygiven&,/• and S, the Ajs in equation(5.3) areindependentBernoulli randomvariableswith successprobabilitypj givenin equation(5.4). By the stronglawof large numbers,we obtain NS N Pj Aj for sufficientlylargeN. Therefore,we needto determinek(c) subjectto the condition -

- E max P&3,s Ho1, HoN pN ..., jN1=l

a

= 7,

(5.5)

wherea is chosento be smalland y is chosento be large(e.g.a = 0.05 and 7 = 0.95). Weshallfurthersimplifyequation(5.5). For this, note that,conditionallygiven&, 3 and S, T3j

/(n - 2)

+ 30j -

?

&-

3c))

(/o'S-10)1/2

S S-2,S

S-1

wherewe haveusedthe assumptionE = Ip. Hencepj in equation(5.4)can be expressedas

pj

= P

Z>

S (/S-S1 { 'S-23

1/2 k(c) ,IV(n - 2)

- c) /'S-1 (a + O -a & (/'S-2A)1/2

,

(5.6)

33

Problems Calibration Multivariate-Univariate where Z - N(O, 1). From equation (5.6), it follows that the maximum value of N1 Nj=1

under Hoj (i.e. under the condition j (c, j = 1, 2,...., N) is attained at 0j = c, provided that C. ,3'S-1/3 > 0. It can be argued that PlS-1,3 is very often likely to be positive; see Appendix obtain we be this to so, immediately Assuming max ^()3S-131/2'k(c) N

max H01,..., HON

- P

p

1

Z>

/ (n - 2)

B'S-2

S- -2(a + -1/2(o3'S-2)3) 1/2J ',s3c Thus, assuming that 1'S- l

> 0, condition (5.5) can be written as

1/2

(u'S-lu

u'S- (hl- h2c*)

ki(c*) (n - 2)

u'S-2u )

(u'S-2u)1/2

(5.7)

-

where zl-a is the 100(1 - a)th percentile of the standard normal distribution and

31= 13Sxx,

u= 3V/sxx, hi=

N

a+,3-y~

0,1

(hn

h2 = u - (1

~

Ip

N(0, Ip),

= (c- W?)//Sxx. c,

instead of k(c). Thus, it follows from equation (5.7) that Also, we have used the notation kI (c,) that such determined is to be (c*) kl + u'S-1 (hi - h2c*)} (n - 2)1/2z1-a(u'S-2)1/2 (u'Su)1/2c) (uS-)/ Since it is difficult to obtain kI

(c,)

=.

(5.8)

analytically, we have used the PB for computing kl

(c,).

5.2. Parametricbootstrapmethod forcomputingkl (c*)

The computational procedure for the PB is as follows. (a) Compute )3,

u = 3,/sxx.

i and sxx on the basis of the calibration data. Set c* = (c -

(b) For i = 1, 100000, generate ui -- Np(u, Ip), h1i / h2i = ui - 1 and compute Xi

(n - 2)1/2{Z, i=

and

N(0, Ip/n) and Si -~ Wp(n- 2, Ip). Set

(uS2Ui)1/2u1 + S '(hli -IU(5.9) (u!S lui)l/2

2)/,/sxx

- h2ic*)}

34

and T Mathew D. Benton,K.Krishnamoorthy Table 8. PB estimates of kl satisfying = 0.95, p = 3 condition (5.8) for a = 0.05, -y (c,) and •1 = (2, -3, 5) Estimatesof k1(c*) for thefollowing valuesof n:

c,

-2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0

n = 15

n =20

n =25

n =30

7.44 6.28 5.05 3.97 3.28 3.49 4.15 4.90 5.70

6.79 5.65 4.55 3.50 2.82 3.07 3.70 4.43 5.18

6.48 5.39 4.29 3.27 2.58 2.86 3.51 4.20 4.92

6.26 5.22 4.15 3.14 2.44 2.76 3.38 4.06 4.76

Table9. PB estimatesof the left-handside of equation (5.8) for a = 0.05,y = 0.95,p = 3

andp' = (2, -3, 5)

Estimatesfor thefollowingvaluesof n:

c,

-2.0 -1.5

n = 15

n =20

n =25

n = 30

0.94 0.94

0.95 0.95

0.94 0.96

0.95 0.94

-1.0

0.95

0.96

0.95

0.95

-0.5

0.95

0.95

0.95

0.95

0.0

0.95

0.95

0.94

0.95

0.5

0.94

0.95

0.95

0.94

1.0

0.95

0.94

0.96

0.96

1.5 2.0

0.94 0.95

0.95 0.94

0.95 0.96

0.95 0.95

(c) The 100ythpercentileof the Xis in step (b) is a PB estimateof kl (c*). For illustration,we providePB estimates(based on 100000 runs) of kl(c*) when 11 = (2, - 3, 5) and for variousvaluesof andn in Table8. c, PB estimatesof (c*), we conducteda numericalstudy To evaluatethe performanceof the kl to verifythe probabilityrequirementin equation(5.8).The numericalstudywas carriedout in the followingmanner.Weassumedthat 31 is knownand the calibrationdata arestandardized so that = 0 andy = 0. As a result&= 0. Furthermore,we set /3' = (2, -3, 5), a = 0.05 and 7 = 0.95. Wethen estimatedthe left-handside of equation(5.8) for severalvaluesof n and c*, and the resultsarepresentedin Table9. WeobservefromTable9 thatthe estimatesarein good agreementwith the specifiedvalueof y, and hencewe concludethatthe PB estimatesof kI(c*) areverysatisfactoryfor practicaluse.


35

Table 10. Valuesof the multiple-usetest statisticT3i,fortestingHoi : O < 9.25 versus Hlj : j > 9.25,forthewheatdatat Yo (101.0, 248.0) (85.0, 235.0) (88.0, 231.0)

(69.0,222.0) (63.0, 216.0)

0

T3j

8.86 9.34 9.46

-5.98 1.71 3.15

10.00

9.41

10.12

12.79

tco = 0.05 and y = 0.95.

5.3. The wheat data (continued) In Section4.3, we analysedthe wheatdataandconsideredsingle-usehypothesistestsfortesting the hypotheses Ho : 0 ( 9.25 versus H1 : 0 > 9.25. Now suppose that the calibration data

(consistingof 15 samples)will be usedto test a sequenceof suchhypotheses,i.e. we wishto test Hoj : Oj ( 9.25 versusH j : Oj > 9.25, j = 1, 2, 3, ..., where the Ojs represent the percentage of waterin a sequenceof hardwheatsamples.Forthe wheatdata, the valuesof x, y, sxx, 3 and S are given in Section 4.3, and

= (c - i-)/ sxx = (9.25 - 9.52)/VJ3.4013 = -0.1464. Table

c, 10 givesthe valuesof T3j(computed by usingequation(5.2))for the fiveyo-valuesin Table7. We computedthe value of k1(c*) followingthe PB proceduregivenin Section5.2, and the

= 2.95 when a = 0.05 and y = 0.95. In Table 10, if a value of value turned out to be k1 (c,) = then exceeds k1 (c*) 2.95, Hoj : Oj ( 9.25 is rejected for the corresponding sample. Note T3j

thatthe correctdecisionis reachedin all casesexceptfor the secondobservationin Table10. If we use a = 0.10 and 7 = 0.95, the value of ki(c*) is 2.64. Note that at this level an incorrect decisionis also madefor the secondobservation.In comparison,the single-usetest (seeSection 4.3, Table7) resultsin correctdecisionsin all the cases at the 10%level.This is not surprising becausea multiple-usehypothesistest has a wideracceptanceregionthan its corresponding single-usetest (see the descriptionin Section1.3). Remark2. Mathewet al. (1998)havederiveda conservativemultiple-useconfidenceregionin themultivariatecalibrationproblem,basedon a pivotthatis equivalentto T3jin equation(5.2). Thoughnot reportedhere,it is possibleto applythe PB procedureand to derivemultiple-use confidenceregions.

6. Concludingremarks Theworkof Brown(1982)is one of the earliestpapersdealingwiththe multivariatecalibration problem.It has broughtout the significanceof multivariatecalibrationin appliedwork,and it generatedconsiderableinterestamongresearchers,as evidencedby the subsequentpublished workon the topic;seeBrown(1993)fora review.Thepresentpaperdealswiththe specialcaseof calibrationproblems,and we haveinvestigatedthe applicabilityof the multivariate-univariate PB for derivingconfidenceregionsand hypothesistests.Numericalresultssuggestthatthe PB is an attractiveoption for makinginferencesin suchcalibrationproblems.Wehaveextensively investigatedvariousaspectsof both intervalestimationand one-sidedhypothesistesting.Even though some proceduresare availablefor constructingconfidenceregionsin the calibration problem,andhenceforcarryingout two-sidedhypothesistests,it is not clearhow to handlethe

36


one-sided hypothesis testing problem if we want to avoid using the bootstrap. Yet, the one-sided testing problem is relevant in many practical applications, as should be clear from the several examples in this paper. For obtaining confidence regions, the PB is easily adapted to the general multivariate calibration problem, where the parameter of interest is a vector. Indeed, Mathew and Zha (1996, 1997, 1998) and Mathew et al. (1998) have studied a general multivariate calibration problem in the context of a multivariate linear model. Parametric bootstrapping can be used to obtain single-use and multiple-use confidence regions based on the pivot statistics in these references. The basic ideas and the necessary computations are similar to what we have done in this paper. As noted by Jones and Rocke (1999), the bootstrap procedure is applicable to non-linear calibration problems as well, both univariateand multivariate.Our overall conclusion is that, under the models assumed, parametric bootstrapping is an excellent option for drawing inferences in the calibration problem.

Acknowledgement The authors are grateful to two reviewers for providing valuable comments and suggestions.

AppendixA: Proofof theorem1 Theproofof theorem1 is basedon technicalargumentsthataresimilarto thosein Srivastavaand Khatri (1979), section4.4. Similartechniqueshave also been used in Fujikoshiand Nishii (1984), Davis and Hayakawa(1987),Mathewand Kasala(1994)and Mathewand Zha (1996). Recallthatwe areassumingE = Ip,and u andvohavethe distributionsgivenin expression(2.6). Let

F=

F Ilul,•l

be a p x p orthogonalmatrixso thatu' = Ilull(1,0). Hereandelsewhere denotesthe Euclideannorm. Write II1'] r~sr =wW21 FSE= W=

ll

,

W12 W22 )

Wp(n

2,

Ip),

wherewll = u'S-'u/llull2is a scalar.Sinceu andvoareindependent,we seethatconditionallygivenF(i.e. conditionallygivenu)

v = 'v =

N(,

I) = N

(A

wherevl = u'vo/llullandV1 = u',101/llulI. For computing,1, we haveused the fact that E(vo)= ll01; see expression(2.6).Let Q be as definedin equation(2.7).Usingthe orthogonaltransformation r, Q can be writtenas u'FT('SP)-'(F'vo- 'uO, )

Ilull(1,O)W-1 H

{uTr(r/sr)-lru}l1/2

UI2(1,

Ilul

0)W-1

(

(01)}1/ 1/2 )

Using the inverseof a partitionedmatrix(see Rao (1973),page33), and simplifying,we obtain S=

vl -

Ilull01-Wll.2

w12W221V2

.2

(A.2)

CalibrationProblems Multivariate-Univariate whereW1l.2= givenu,

-

12WIw21.

-

Note that

Np(0, Ip-l),

W21/2w21

-

vi -IluIllO N(

37

independentof W22 and W11l.2.Also,

IlullOi,1),

where is givenabove.Writing •1l W12 W2?lV2

W2-1/2V2 -- Wl21/2

we see that,conditionallygivenW-1/2v2,

N(0,

W122 2

Hence,conditionallygivenu and v'

v•

W~21v2)

W•1v2,

IluIO - W12W221v2 N(1l - IIullI, 1 + v'2W221v2) " l Consequently,conditionallygivenu and v' W-1v2, vI

vi -

lull11O W2Wj2v Ni

-

(1 +

(A.3)

llulll

vWjv2)1/2'

(A.3)

V'Wjlv2)1/2 with n - p - 1(1+• degrees of freedom. In view of expression (A.3), Also, wll.2 follows a central x2-distribution

we modifyQin equation(A.2) and define (n - p

Ilull91-

- 1)'1/2(v +

{Wll.2(1

W12W2?lV2)

V'W?2lV2)}1/2

withn - p-1 It thenfollowsthat,conditionallygivenu andv' V2, TOfollowsa non-centralt-distribution W•' given by degreesof freedomandnon-centralityparameter60o, 60=

-Iul(A.5)

(1 + v/W21lv2)1/2

The proof is completeonce we show that Toin equation(A.4) coincideswith the expressionfor Toin equations(2.8),and 60in equation(A.5) is the samequantityas in equation(2.9).Towardsthis,note that = - lullIOl (u'30Oi- u'ul)/llull, 0l = v' W2WIv2 V•PI(r SFri)-lF'Ivo

(vS-'u)2

= voS- vo 0

U'S-lu

= r,

where r is the quantity given in equations (2.8). This completes the proof.

Appendix B: Proof of theorem 2 The proof of theorem 2 is very similar to that of theorem 1, and we shall give a very brief outline. Assume and vo* be as defined in expression (3.4). Let without loss of generality that E = I, and let u, I

(

1u

u

be a p x p orthogonal matrix so that u'Fr* = Ilu*I(1, 0). Write F', S,

Let

= W

-

12*

(Wll* W21,*W22,

=

L~~vo -

V2,

(V2* ) S

W,(n-

2, I,).

38


where v1l, is a scalar. Following the arguments in Appendix A, it can be shown that T2 in equations (3.5)

can be expressedas

(n -

T2

(1

where wll.2that, given

+

{W11.2"

=W1*

- W12* W-21,v2)

p-

W .*(1/2 1)1/2(Vl,

W22, V2,

1/2 v2?)

As in theproofof theorem1 inAppendix A, it cannowbeverified

-

= W22*W21i*.T2 follows a non-central t-distribution with n - p - 1 degrees of and r. W12, u.non-centrality freedomand v. parameter W2lv2*,,

u',1 (01- cl) ' Ilu*11{(1+ c2)(1 + r*)}1/2 It can also be verifiedthatr* coincideswiththe quantitygivenin equations(3.5).

AppendixC: An approximatelowerboundfor P(3I'S-'3 > 0) Once again, we assume that E = I,. Note that, conditionally given S, 3'S-

Therefore,

P('S-i1,

> 0) = Es{P(o3S-', = Es[P

- N(/'S-13, O'S-2/Sx,,).

> OlS)} s (xx 3'S--2 1/2 1S

Z>

PIZ >-,Jsxx

()-2)3/2}

(C.1)

whereZ ~ N(0, 1). To evaluatethe expectationin expression(C.1),let r2 be an orthogonalmatrixsuch that '2)3= 11311 (1,0..., 0)'.Thenr' SrF2 W,(n- 2, I). Let '2SF2 = (Sll s2)(C.2) (21 S22) (C.2) where sil is a scalar. Since F'2SF2 W,(n - 2, Ip,),we also have S22" - 2, Ip-_). The expectation Wp-,_(n in expression (C. 1) can be written as

)'S

2S

2

2=E[ (13'S-213p)/2J 2 Sr2)-2r,)p1}/2 { 2'r(r,

loll =E{ (1 + s12S2$S21)1/2 111311 {1 +

(C.3)

To arriveat inequality(C.3),we haveusedthe factthatE(sI2S222s21)}1/2 F'23 = 11,11(1, 0,..., 0)' andwe havealso usedthe expressionfor the inverseof the partitionedmatrixin expression(C.2).Inequality(C.3)followsfromthe fact that 1/V/(1+ x) is a convex function.Now we use the followingproperties:S2 / s21 Np-l(0, Ip-l) and is independent of S22,and

E(Sj)

=

1

n-p-2 IpIP (see lemma 7.7.1 in Anderson (1984)). Using these results, we obtain E(sl2SP2s21) = tr{

E(S2•S-1/2s2,S12SI2/22 = tr{E(S j1) E(S2I1/2521S12S21/2)} n-p-2 ( 1n

2

)

n-p-2 p-1


39

Fromexpressions(C.1),(C.3)and (C.4),we obtain P(3 S-1/3 > 0) > P Z > -( sxx I{x11+(p approximately.For large n, s,,xx= (C.5) will be large.

E•n

(x

_i- )2

-

-1)1(n

)/(n - p- 2)/2 -p

-2)11/21

(C.5)

tends to be large, and hence the right-hand side of inequality

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