Measures of centrality 6mV∑=g - Amazon S3

Topic:Statisticsofcentraltendency,dispersionandinferential&relationaltechniques3.4.2.4Statisticalskills

© Tutor2u Limited 2016 www.tutor2u.net

Whatyouneedtoknow:Measuresofcentraltendency:mean,modeandmedianMeasuresofdispersion:range,inter-quartilerangeandstandarddeviationInferentialandrelationalstatisticaltechniquestoincludeSpearman’srankcorrelationandChi-squaretestandtheapplicationofsignificancetests

Measuresofcentrality

Themean,medianandmodeareknownasmeasuresofcentrality:anaimtoidentifythemidpointinadatasetthroughstatisticalmeans.Eachdoesthisinaslightlydifferentwayandmaygiveadifferentanswerifthedatasetisaskewed(asymmetrical)distribution(seediagrambelow).

Mean:ThesumofallthedatadividedbythenumberofdatavaluesExample:8+7+3+9+11+4=42÷6=Meanof7.0

Median:ThemiddatapointinadataseriesorganisedinsequenceExample:257811141821222529(fivedatavalueseitherside)

Mode:ThemostfrequentlyoccurringdatavalueinaseriesExample:2244479999121213(‘9’occursfourtimes,soisthe‘mode’)

Whyaredifferentmethodsused?Thechoiceofmeasureofcentralitydependsonthetaskbeingcarriedout.Whilethemeandailytemperatureduringaparticularmonthmayindicatetheaverageforeaseofcomparisonwithothermonths,themodewouldindicatethemostfrequentlyexperienceddailytemperaturethroughoutthemonth,whichmightbemoreusefulinformationforfarmers.Whenstudyingmeasuresofwealthinacountry’spopulation,usingthemedianismorelikelytorepresentthemajorityofpeopleratherthanthemean–whichmaybehigherthanmostpeopleexperienceifasmallproportionofpeopleearnanexceedinglyhighincome.

InmeasuresofdispersionthemedianisusedwhencalculatingInterquartileRange,whilethemeanisusedwhencalculatingStandardDeviation.

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MeasuresofdispersionThereareanumberofwaystodescribethedegreeofspreadofdata.Therange,andinterquartilerangearemeasuresof‘spread’inacollectionofdata.Theyarethemoststraightforwardofthe‘measuresofdispersion’.Tworiversmayhavethesame‘mean’depth,butonevariesconsiderablyovertheyearfromverylowtoveryhighlevels,whiletheotherhaslittlevariation.Thesemeasureshelpdescribethedegreeofdatavariance.RangeThesimplestdescriptionofvariationisastraightforwardmeasureofthedifferencebetweenthelargestandsmallestdatavalues(thedifferencebetweenthe‘highest’and‘lowest’levelofwaterinariver,forexample).Theproblemwithusingrangeisthattherecanbesomehighlyunusualresultsatthetwoextremes,knownasoutliers.Themajorityofreadingsmaybeclosetothemean,butthenaveryunusualeventmaycausethespreadtobemuchgreater.If,forinstanceaone-in-a-hundred-yearfloodraisesariverlevelsohighthatitsetsanewrecord,the‘range’wouldhavetoincludethisastheupperdatameasure.Butthesubsequentrangeofriverlevelwouldnotaccuratelydescribetheusualconditionsofflowlevelintheriver.Forthisreasonwemoreoftenuseameasurethatexcludespossibleoutliersatthetwoextremesandlookatthefiftypercentofresultseithersideofthemedian.ThisstatisticistheInterquartileRange.InterquartilerangeWhenweneedtodescribedatacollectedfromanareatocomparewithdatafromanotherarea,wemayusesomesortof‘average’tosummariseit.Wemayuse,forexample,the‘mean’pebblesizewehavemeasuredonabeachtocomparewiththemeanofanotherbeach.Butifwefindthetwomeansaresamethiscangiveaninaccurateinterpretationifwethenassumethepebblesonthetwobeachesaresimilar;the‘spread’ofpebblesononebeach,fromverysmalltoverylargemay,infact,bequitedifferentfromanotherbeachwherethepebblesizesareallveryclosetothemean.WhatistheInterquartileRange?Thisstatisticalmeasureusestheconceptofthe‘median’ratherthanthemean–themiddle-rankingvalueinarangeofdatarankedfromlargesttosmallest.Itthenfindsthemedianoftheupperhalf(UpperQuartile)andsubtractsthemedianofthelowerhalf(LowerQuartile)toproducethedifferencebetweenthequarterandthree-quartersvalue–knownastheInterquartileRange.Thisgivesanindicationofthe‘spread’ofthedataeithersideofthemedian.HowistheInterquartileRangecalculated?

• Itcanbecalculatedmanuallybycountingoutthe‘half-way’point(median),andthenthe‘halfwaypointoftheupperhalf(UQ)andthehalfwaypointofthelowerhalf(LQ)andsubtractingtheLQvaluefromtheUQvalue:

Inthissimplifiedillustration,imaginewemeasured11pebblestakenfromabeachincm:

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InterquartileRangecalculation:=UQ–LQ=19–8IQR=11Interpretation:Thereare11cmbetweenthesizeofpebblesatthequarter,andthree-quartersdispersionaroundthemedianpebblesizeonthisbeach.

• Itcanbecalculatedusingthreesimpleformulas.Theseidentifytheplaceintherankingofvalueswhereyoucanlocatethemedian,UQandLQvalues.

Forthemedian:(n+1)÷2FortheUQ:(n+1)÷4FortheLQ:3(n+1)÷4where‘n’isthenumberofvaluesinthedataset

Finalcalculation:UQ–LQ(remembertosubtractthe‘values’notthe‘rank’)=25–8IQR=17cm

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UsesoftheInterquartileRangeInterquartileRangeismostusefulwhencomparingtwoormoredatasets.Forexample,youmayhavecollectedpebblesizesfromanumberofbeachesalongacoast.Whilsttheymayhaveasimilar‘median’pebblesize,youmaynoticethatonebeachhasmuchreduced‘spread’ofpebblesizesasithasasmallerinterquartilerangethantheotherbeaches.Youmaythenwanttofocusyourfieldworkonthisbeachtotrytoworkouttheprocessescausingthistooccur.StandarddeviationThismeasureusesthemean,ratherthanthemedian,fromwhichtodescribethespreadofdata.Itmeasuresthedifferencebetweeneachdatavalueandthemean(both‘largerthan’and‘smallerthan’)andcalculatestheaverage(mean)variation.Bytakingallthevaluesintoaccountitisliabletobeinginfluencedbyextremeoutliermeasurements.Buttheimpactofthesecanbereducedbyhavingalargersamplesize.Itiscalculatedusingthefollowingequation:

𝑺𝑫 = 𝚺 𝒙 − 𝒙 𝟐

𝒏

Where:xiseachmeasurement 𝑥isthemeanofallthemeasurements ∑isthesumofthevalues nisthenumberofvaluesobtainedForexample……………Value(unit)x 𝒙 𝒙 − 𝒙 𝒙 − 𝒙 𝟐

18 19.8 -1.8 3.2422 19.8 2.2 4.8416 19.8 -3.8 14.4420 19.8 0.2 0.0413 19.8 -6.8 46.2414 19.8 -5.8 33.6427 19.8 7.2 51.8431 19.8 11.2 125.4425 19.8 5.2 27.0419 19.8 -0.8 0.6411 19.8 -8.8 77.4414 19.8 -5.8 33.6426 19.8 6.2 38.4421 19.8 1.2 1.44

∑ 458.36

SD=+,-./01+

= 32.74=5.72So,theaverage(mean)variationofmeasurementseithersideofthetotalmeanis5.72(units)

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UsesofStandarddeviationInasimilarwaytointerquartilerange,thestandarddeviationtellsyousomethingaboutaspreadofdatameasurements,andputsaprecisevalueonthespreadratherthanarange.ButitismoreusefulifcomparedwiththeSDofanothersetofmeasurements.Itdoeshavesomelimitations:itshouldbeusedwithdatathathasanormaldistribution(anequallikelihoodofvaluesbeingspreadequallyeithersideofamean).So,forexample,itwouldbevalidforthedistributionofpebblesizesonabeach,butnotthedistancepeopletraveltoshopinasuperstore,whereyouwouldexpectmorepeopletravellingtheclosertheyliveandfewerthefurtherawaytheylive(askeweddistribution).InferentialandrelationalstatisticsArelationalstatisticaimstodescribethestrengthofassociationbetweentwodistinctvariablestoseeifthereisapossiblerelationship–perhapsevenacorrelation–betweenthem.Aninferentialtechnique‘infers’fromasampleofdatawhatmightbetakingplaceintherealworldbyindicatingwhethersomethingobservedisjustavariationwithinageneralrandomness,orhasavalidbasisinbeingworthyofinvestigationassomethingofsubstance.Spearman’srankcorrelationtestThistestmeasureswhetherthereisastatisticallyreliableassociationbetweentwovariablesandthedirectionofthatrelationshipifthereisone(positive:asonevaluebecomeslarger,sodoestheother;negative:asonevaluebecomeslarger,theotherbecomessmaller).Forvalidity,itrequiresthatbothcategoriesofdataarenumeric(numbers),capableofbeingranked(putinsizeorder),andbetween10and30pairsofdata.

Rs=1 −0 78

9:;9

Where:nisthenumberofdatapairs disthedifferenceintherankbetweenthepairedvalues ∑isthesumofthedifferencesbetweenallthepairsVariable1 Rank1 Variable2 Rank2 R1-R2d d2

2.3 8 66 3 5 255.7 1 85 1 0 03.0 7 52 6 1 14.3 4 43 7 -3 91.1 11 19 11 0 03.8 6 59 5 1 14.1 5 42 8 -3 91.3 10 23 10 0 05.2 2 78 2 0 04.7 3 60 4 -1 11.9 9 32 9 0 0

∑ 46

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a)Rs=1 − 0<+01//1;11

b)Rs=1 − =>01/=?

c)Rs=1 − 0.209 d)Rs=0.791Sowhatdoes0.791signify?Thefactthatitiscloserto1.0than0suggeststhereisanassociationbetweenthetwosetsofdata.Thecloserthefinalvalueistowards1.0thestrongertherelationship.Andthefactthatitisapositivenumbertellsusitisapositiverelationship(asonevaluegetsbigger,sodoestheother).Anegativeresultisjustasvalid,aslongasitliesbetween0and-1.0andastrongnegativerelationshipsuggeststhatanincreaseinonevalueisaccompaniedbyacorrespondingdecreaseintheother.Therearetwothingsthatwecan’tbeconfidentabout:a)whethertherelationshipisacorrelation,orwhetherathirdfactor(ormore)isinfluencingbothvaluesseparately;andb)thechancethatarandomsampleproducesanapparentrelationship.Tocheckthisseconddoubt,theRsresultneedstobecheckedagainstasignificancetablethatindicatestheprobabilityofarelationshipoccurringbychance.TestsofsignificanceGeographystatisticsrequiringasignificancetestusuallyoperateatthe95%(0.05)or99%(0.01)levelofconfidencethattheresultsarenotduetoachancesample.Theparticularconfidencelevelshouldbedecidedbeforethetestiscarriedout.A95%confidencelevelstatesthatonlyin5samplesoutofevery100wouldarandomassociationoccurinthevaluesunderconsideration.TablesofcriticalvaluesareavailableagainstwhichtocheckanRscalculationresult.Lookalongtherowthatmatchesthenumberofsamplepairs(n)andidentifytherelevantcriticalvalue.Ignoringanyminussign,iftheRsvalueislargerthanthecriticalvalue,youcanassumethatitisnottheresultofachancesamplesetandthereisavalidassociationbetweenthevalues.

n 0.05(95%) 0.01(99%)11 0.618 0.755

ThefigureobtainedintheRscalculationaboveislargerthanbothcriticalvalues,sotherecanbe99%confidencethatthereisavalidrelationshipbetweenthetwopairedvariables,andjusta1%probabilitythatithasoccurredbychance.Chi-squaretestTheChi-squaretestisastatisticaltestthatisoftencarriedoutatthestartofan intendedgeographicalinvestigation.Wemayhavenoticedapattern,distributionoranomalyinafeatureofthehumanorphysicalworldandhaveahunchthat‘somethingisgoingon’toproduceit.TheChi-squaretesttellsuswhetherour‘hunch’isstatisticallysignificant–i.e.that–yes,wehavenoticedavalidgeographicalassociationbetweentwoormorevariablesthatdeservesfurtherinvestigationaspartofageographicalenquiry.Alternatively,itcanindicatethatwhatweinitiallythinkisarelatedassociationisactuallyjustarandomvariationinthefeaturewe’venoticed,anddoesn’twarrantfurtherinvestigationorresearchalongthoselinesofenquiry.

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Theequationcompareswhathasbeenmeasured(Observed)intheoccurrenceofthefeature,againstwhatmaybeanticipated(Expected)‘if’thefeaturewasrandomlyoccurring.(Note:thereshouldbenofewerthan5observationsinanyofthecategoriestousethistechnique.)First,establishaHypothesisandthenconvertittoaNullHypothesis.(NullHypothesis:whydoweneedit?Well,intheinvestigativeprocessit’snotpossibleto‘prove’somethingwith100%certainty–weonlygettoseeandexperienceapartofthewholeworld,soitmaybethatwhatwethinkwe’ve‘proved’inoneplaceis‘disproved’inanother.Butwecan‘disprove’assumptionscompletely-byfindingacontradictoryoccurrenceofit.Wecannever‘prove’ahypothesisfully,butwecanfully‘disprove’itsconverse–thenullhypothesis.Ifourstatisticaltestsallowustodisprovethenullhypothesis,thenwecan‘accept’thatourhypothesishasvalidity.Butonlytotheextentthatwecanhaveconfidencethatoursampleislargeenoughandvalid.Thisleadsontotheconceptsof‘confidencelevels’and‘criticalvalues’).

TheChi-squareequation

𝝌𝟐 = 𝑶 − 𝑬 𝟐

𝑬

Chi-squaretestexampleonwoodlanddistributionoveraregionalarea

Supposeyouhaveplacedanoverlaymapofnaturalvegetationonabasemapofsurfacegeologyforaregionandnoticedtheareasofdeciduousandmixedwoodlandseemtofeatureoncertainrocktypesmorethanothers.Youwanttoseeifthishunchiscorrectsocalculatethecombinedwoodlandcoverageinhectaresthatgrowsoneachrocktype.Hypothesis:Theareaofwoodland(deciduousandmixed)isrelatedtorocktype.Nullhypothesis:Thereisnosignificantrelationshipbetweenrocktypeandwoodlandcoverage.

Geology O(Observed)ha. E(Expected)ha. O–E (O–E)2Alluvium 8 14 -6 36Boulderclay 14 14 0 0Chalk 24 14 10 100Sandstone 11 14 -3 9Limestone 6 14 -8 64Shale 21 14 7 49 ∑ 258

1. Step1:putinthefiguresrecordedintheObservedcolumn(O)2. Step2:workouttheaverage(mean)figureforO(addupthecolumn&dividebynumberof

datasets)3. Step3:putthe‘average’intothe‘Expected’column(E)4. Step4:workoutO-Eandputintothenextcolumn5. Step5:workoutO-Esquaredandputintothenextcolumnandtotalupthecolumn6. Step6:thatisthetoppartoftheformula–nowdividebythe‘E’valuetogetyourchi-square

number

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𝜒= = 25814

=18.43OnitsowntheChi-squarestatistichaslittlemeaning–itneedsvalidatingagainstcriticalvalues.Thesearefoundintablesorongraphsthathavebeencalculatedbystatisticalexperts.Considerwhat‘confidencelevel’youwishtouse.Themostcommonlevelsingeographyare95%and/or99%.Thesemeanthat95outofevery100timesyoucarriedoutthesemeasurements(or99outof100)youwouldgetasimilarresult,buton5occasions(or1)youmaygeta‘chance’result.Theymaybeexpressedinarangeofways:

Thesecondfactor,afterConfidenceLevelisthe‘DegreesofFreedom’(df)touse.Thisisusuallycalculatedasn-1(numberofdatasetsminus1)whichinthisexampleis6(rocktypes):6-1=5.Soweusethedf5rowtolookupour‘CriticalValue’.Thetableshows‘CriticalValues’thathavebeencalculatedbystatisticalexpertsthatwejudgeourChisquareresultagainst.Ifourresultislargerthanthecriticalvalue–wehavegotavalidresultinourdatathatletsusrejectthenullhypothesisandacceptouroriginalhypothesis.Ifourresultissmallerthanthecriticalvalue,wehavetoacceptthenullhypothesis–thatthereisnokeygeographicalinferenceobservableinthisdataset.

df 0.05(95%) 0.01(99%)5 11.07 15.09

Step7:LookingattheCriticalValuestableatdf5,wecanseethatourChisquareresultof18.43islargerthanthecriticalvaluesatboththe95%and99%confidencelevel,sowecanrejectthenullhypothesisandacceptthehypothesis.Isthattheendofit?No,becausealltheChi-squaretestdoesissignifyareliablevariationindatafromwhatmightbeexpectedifwoodlandweredistributedregularlyacrossthelandscape,irrespectiveofgeology.Lookingatthedata,itappearsthatwoodlandismorelikelytogrowontwokindsofrock:chalkandshale.Aninvestigationwouldneedtoconsiderwhythisisthecase.Isitthatwoodlandhasbeenclearedofftheothergeology,possiblyforagriculture?Orarethesetworocktypesparticularlysuitablefortree-growth?Thestatisticshouldresultinfurtherquestions.

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Geographystatisticaltechniques


Coregeographicalskills3.4.2.4SelectingandusingappropriatestatisticaltestsindataanalysisAstudentisstudyingtherateofcoastalerosiononarapidly-erodingcoastlineinEastAnglia.SecondarydatahasbeenobtainedfromtheLocalEnvironmentofficeofannualcliffretreatoverthelast15yearsforaclifflinetwokilometresdown-coastofsomecoastaldefences.Yearsago15 14 13 12 11 10 9 8 7 6 5 4 3 2 184 112 73 93 142 183 165 202 189 178 221 143 89 122 196

Annualcliffretreat(cm)Calculate:

a) Themeanannualcliffretreatb) Thetotalrangeofannualcliffretreatoverthe15yearsc) Theinterquartilerangeofannualcliffretreatoverthe15years

Thestudentconsidersobtainingthesamedatafortheclifftwokilometresup-coastofthesamecoastaldefences.Longshoredriftoperatesfromnorthtosouthalongthecoastandthestudentwantstoexaminetheeffectofthecoastaldefencesonrelativeratesofcliffretreat.Estimate:

d) Themagnitudeofthemean,thesizeoftherangeandthevalueoftheinterquartilerangeforthecliffup-coast.Willtheywill‘larger’,‘smaller’,or‘similar’comparedwiththevaluesabove?

AnAlevelclasshasinput94responsestoaquestionnairesurveyintoanExceldatabase.Inanalysingthedataastudentnoticesthatthereseemstobeapatterninhowfarfromthecoastaresidentlivesandtheirattitudetothe‘effectiveness’ofthelocalcoastalprotectionmeasures–whichinvolvedtraditionalwoodengroynesandaconcreteseawall.AtfirstthestudentconsidersaSpearman’sRankcorrelationtesttoseeif‘positive’and‘negative’responsesrelatedtodistancefromthecoast.However,thisisrejectedasitdoesnotprovidenormallydistributed,butskeweddata.Thestudent,therefore,decidestoconductaChi-squaretesttoseeif‘negative’viewsaredisproportionatelyoccurringincertainpartsofthetown.Taggingthemultiplecategoryresponsesasessentially‘positive’or‘negative’thestudenttabulatesthenumberofnegativeresponsesaccordingtohousecategorieswithequalnumbersindistance-bandsfromthecoast.Respondentdistancefromcoast(m.)

‘Negative’views‘O’ E O-E (O-E)2

1-50 18 51-100 14 101-150 7 151-200 9 201-250 6 Over250 6

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Calculate:e) TheChi-squarestatistic

f) Checkitagainstthechi-squarecriticalvaluesofdf=5(n-1):0.05(11.070)and0.01(15.086)

g) Theresultcausesthestudenttoconsiderafollow-upinterviewstudywithsixresidents.Whichareasofthetownshouldbeselected,andwhatlineofenquiryshouldtheinterviewquestionsfollow?

Inanalysingthedatafurther,thestudentnoticesanapparentpatternintheratesofcliffretreat,andthemeteorologicalrecordsshowingthenumberofdaysperyeartheprevailingwindsblewfromtheeast.ItisthoughtappropriatetoconductaSpearman’srankcorrelationexercisetotestthishunchtoseeifthereisarelationshipbetweenthetwovariables:

Annualcliffretreat(cm)

Rank Daysperyearofeastwinds

Rank d d2

196 64 122 47 89 38 143 59 221 78 178 52 189 61 202 72 165 37 183 49 142 39 93 30 73 38 112 43 84 29

Conclude:

h) CalculateandinterprettheRsresultifthecriticalvaluesare:

i) Ifthereisarelationship,howstrongisit?Isitpositiveornegative?Howmightyouexplainit?

Exam-stylequestions:1. Criticallyexaminethedifferentmethodsofmeasuringthedispersionofadataset(4marks)2. Whenshould,andwhenshouldn’tyouuseaChi-squaretestondata?(4marks)ExamstylequestionsonstatisticalskillsrelatetotheAScourse.Theycanaidunderstandingpriortocompletion/submissionoftheindividualfieldinvestigationinfulfilmentoftheAcourse.

n=15 0.05 0.52 0.01 0.65

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1. Criticallyexaminethedifferentmethodsofmeasuringthedispersionofadataset(4marks)Thisanswerrequiresstudentstobeawareofmeasuresofdispersionandtobeabletocommentontheirrelativeusesandlimitations.Forfullmarksthereshouldbereferencetoatleasttwomeasuresandacommentontheirstrengthsandweaknesses.Answersmayreferto:• Range:themoststraightforwardtocalculateandasimplesubtractionofthesmallestfromthe

largestvalue.But,canbeaffectedbyoutliervaluesateitherextremewhichmaymaketheresultunrepresentativeoftheusualdataspread.

• Interquartilerange:canbecalculatedmanuallyorbyuseofasimpleformula.Benefitsby

excludingextremevalueandconsidersthemiddlefiftypercentofvaluesaroundthemedian.MostusefulwhencomparingtheIQRofonedatasetwiththatofanotherlocation/time.

• Standarddeviation:ameasurethatincludesallthevaluesbutaveragestheirdeviationfromthe

meantoprovideaprecisemeasureofspread.Usefulfordatawithanormaldistributionandforidentifyingdatawithin1(68%),2(95%),and3(99%)StandardDeviationsofthemean.ButIQRmaybemoreeffectiveatidentifyingunevenspreadinskeweddistributionswheretheIQvaluescanberelatedtothemedian.

2. Whenshould,andwhenshouldn’tyouuseaChi-squaretestondata?(4marks)

ThisanswerrequiresstudentstoshowabroadunderstandingofthevalueofaChi-squaretest,itsplaceintheenquirysequenceandtherestrictionsthatguideitsaccurateuse.Answersmayreferto:• Chi-squaregivesastatisticalvaluetothedifferencebetweenobservedvaluesandthose

expectedaccordingtotheory.Assuchitcanprovideanindicationthatageneralimpressionofunequaloccurrenceofavariableisstatisticallyvalid.Thisprovidesabasisforfurtherinvestigationandthegenerationofanenquiryquestionorhypothesisifavariancebetweenwhatisobservedandwhattheorywouldleadonetoexpect,isshowntobevalid.

• Itsstrengthsarethatisdoesnotrequireanormaldistributionofdata.Andcounted(frequency)

datacanbeinnominal(namedcategories)ratherthannumeric(number).But,thedatacategoriesmustbemutually-exclusive;anddatashouldnotbeabletofitintomorethanonecategory.

• Thelimitationsofthetestarethatnoobservedorexpectedvaluesshouldbelessthan5andthe

totalobservedshouldbemorethan20.Itshouldusecountfrequenciesandnotpercentagesorproportions.TheChi-squarestatistichasnoinherentmeaningotherthantoindicatethelikelihoodoftheobservedfeaturesoccurringbychance.Ifarelationshipissuggestedbyuseofthetechnique,itsaysnothingaboutthestrength,directionorcausesoftherelationship.

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Statisticaltechniques3.4.2.4


Q1 TrueorFalse?A Themedianisthevalueatthemiddleofarankofvaluesfromlargesttosmallest B Spearman’srankcorrelationtestshouldgivearesultbetween0and1.0 C Standarddeviationcalculatesaveragedivergenceofeachvaluefromthemean D Theusualconfidencelevelsusedingeographyare90%and95% E TheChi-squaretestshouldnotbeusedifanObservedcategoryislessthan5

Q2 Decidewhichstatisticaltest/techniquethefollowingdescriptionsapplytoA Itmaysignifyapositiveornegativerelationshipthatmaybeconsidered

strongorweak.Theresultneedsatestofsignificancetovalidateit.

B Themostcommonlyoccurringvalueinadataset.Itismoreameasureoffrequencythancentrality,particularlyinaskeweddistribution.

C Ameasureofdispersionthatincludesallthevaluesandinvolvescalculatinghowmuchtheydifferfromthemean.

D Atableofnumbersagainstwhicharesultfromastatisticalcalculationischeckedandifitislargerthanthetablenumber,hassignificance.

E Thevaluethatresultsfromthiscalculationmeanslittleonitsown,butcanindicateifthereisanunexpectedfeatureworthyoffurtherenquiry

StandardDeviationChi-squaretestModeSpearman’sRanktestCriticalValues

Q3 Matchthesymbolorfunctiontoitsmeaning,useorpurposeA Ameansofremovingnegative(-)valuesfromacalculation

B Ameansofreversingthemagnitudeeffectoftheprocessin‘A’

C Atotallingofallthevalues

D CalculatingtheInterquartileRange

E Calculatingtheaverage(mean)ofasetofvalues

F Cubingavalue/multiplyingbyitself,andtheresultbyitselfagain

𝒙𝒏𝟑UQ-LQ∑𝒙𝟐 𝒙

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Statisticaltechniques3.4.2.4


Q4 CalculatethefollowingfromthevaluesgivenInterquartileRange:4691215162426376465

StandardDeviation:

SD=%&'(()

Spearman’srank:

Rs=1 − , -.

/01/when∑d2is685andnis13

Chi-square:

𝝌𝟐 = 𝑶 − 𝑬 𝟐

𝑬 WhentheindividualO-Evaluesare:14-75-68AndEis17

Q5 Whatarethecommonerrorsmadewhencalculatingthefollowing:Interquartilerange:Spearman’srank:Chi-squaretest:

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Statisticaltechniques3.4.2.4ANSWERS


Q1 TrueorFalse?A Themedianisthevalueatthemiddleofarankofvaluesfromlargesttosmallest TrueB Spearman’srankcorrelationtestshouldgivearesultbetween0and1.0

Itcanalsogivearesultbetween0and-1.0False

C Standarddeviationcalculatesaveragedivergenceofeachvaluefromthemean TrueD Theusualconfidencelevelsusedingeographyare90%and95%95%and99% FalseE TheChi-squaretestshouldnotbeusedifanObservedcategoryislessthan5 True

Q2 Decidewhichstatisticaltest/techniquethefollowingdescriptionsapplytoA Itmaysignifyapositiveornegativerelationshipthatmaybeconsidered

strongorweak.Theresultneedsatestofsignificancetovalidateit.Spearman’sRank

testB Themostcommonlyoccurringvalueinadataset.Itismoreameasure

offrequencythancentrality,particularlyinaskeweddistribution.Mode

C Ameasureofdispersionthatincludesallthevaluesandinvolvescalculatinghowmuchtheydifferfromthemean.

Standarddeviation

D Atableofnumbersagainstwhicharesultfromastatisticalcalculationischeckedandifitislargerthanthetablenumber,hassignificance.

CriticalValues

E Thevaluethatresultsfromthiscalculationmeanslittleonitsown,butcanindicateifthereisanunexpectedfeatureworthyoffurtherenquiry

Chi-squaretest

StandardDeviationChi-squaretestModeSpearman’sRanktestCriticalValues

Q3 Matchthesymbolorfunctiontoitsmeaning,useorpurposeA Ameansofremovingnegative(-)valuesfromacalculation

𝒙𝟐

B Ameansofreversingthemagnitudeeffectoftheprocessin‘A’

𝒙C Atotallingofallthevalues

∑

D CalculatingtheInterquartileRange

UQ-LQ

E Calculatingtheaverage(mean)ofasetofvalues

𝒙

F Cubingavalue/multiplyingbyitself,andtheresultbyitselfagain

𝒏𝟑

𝒙𝒏𝟑UQ-LQ∑𝒙𝟐 𝒙

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Statisticaltechniques3.4.2.4ANSWERS


Q4 CalculatethefollowingfromthevaluesgivenInterquartileRange:4691215162426376465

UQ:37LQ:937-9=28(units)

StandardDeviation:

SD=%&'(()

7.28(2dp)

Spearman’srank:

Rs=1 − , -.

/01/when∑d2is685andnis13

1 −6x6852184

1- 1.88

=-0.88

Chi-square:

𝝌𝟐 = 𝑶 − 𝑬 𝟐

𝑬 WhentheindividualO-Evaluesare:14-75-68AndEis17

21.76

Q5 Whatarethecommonerrorsmadewhencalculatingthefollowing:Interquartilerange:• Notputtingthedatavaluesinrankorderoflargesttosmallest• Calculatingthefullrangeratherthanthequarterandthree-quartervalues• Calculatingthedifferencebetweentheranknumbersratherthantherankvalues• ForgettingtoincludetheunitsofmeasurementSpearman’srank:• Calculatingn2-nratherthann3-n• Subtracting1fromtheequationnumber,ratherthantheequationnumberfromone• NotnoticingifthefinalvalueisanegativenumberonacalculatorChi-squaretest:• NotsquaringthesumofO-E,butjustsummingtheO-Evalues.• Dividingthe(O-E)2valuebythenumberofcategories(n)ratherthantheExpectedvalue• StillusingthetestevenifoneoftheOorEvaluesislessthan5.

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Measures of centrality 6mV∑=g - Amazon S3