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MATH 225N MATH Statistic Final, Exams of Nursing

A.T. Still University of Health Sciences (ATSU)Nursing

MATH 225N MATH Statistic Final MATH 225N MATH Statistic Final MATH 225N MATH Statistic Final

Typology: Exams

2023/2024

Available from 06/25/2024

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Download MATH 225N MATH Statistic Final and more Exams Nursing in PDF only on Docsity!

1 / 1 POINTS

Afitnesscenterclaimsthatthemeanamountoftimethatapersonspendsatthegym

pervisitis 33 minutes.Identifythenulhypothesis,H 0 ,andthealternative

hypothesis,Ha,intermsoftheparameterμ.

H 0 :μ≠ 33 ;Ha:μ= 33 H 0 :μ= 33 ;Ha:μ≠ 33 H 0 :μ≥ 33 ;Ha:μ< 33 H 0 :μ≤ 33 ;Ha:μ> 33

AnswerExplanation

Correctanswer:

H 0 :μ= 33 ;Ha:μ≠ 33

Lettheparameterμ beusedtorepresentthemean.Thenulhypothesisisalways statedwithsomeform ofequality:equal(=),greaterthanorequalto(≥),orlessthan orequalto(≤).Therefore,inthiscase,thenulhypothesisH 0 isμ= 33 .The alternativehypothesisiscontradictorytothenulhypothesis,soHa isμ ≠ 33.

QUESTION 21 / 1 POINTS

Theanswerchoicesbelowrepresentdifferenthypothesistests.Whichofthechoices areright-tailedtests?Selectalcorrectanswers.

H 0 : X≥17.1,Ha: X < 17.

H 0 :X≤ 3. 8 ,Ha:X> 3.

H 0 :X≤ 7. 4 ,Ha:X> 7.

H 0 : X = 14.4,Ha: X≠14.

H 0 : X≤3.8,Ha: X > 3.

H 0 : X≤7.4,Ha: X > 7.

H 0 : X = 3.3,Ha: X≠3.

AnswerExplanation

Correctanswer:

Thebuildinginspectorthinksthatnomorethan 15 % ofthestructuresinthecounty

werebuiltwithoutpermitswhen,infact,morethan 15 % ofthestructureswerebuilt

withoutpermits. Sointhiscase,theright-tailedtestsare:

• H 0 : X≤7.4,Ha: X > 7.

• H 0 : X≤3.8,Ha: X > 3.

QUESTION 30 / 1 POINTS

FindtheTypeIIerrorgiventhatthenulhypothesis,H 0 ,is:abuildinginspectorclaims

thatnomorethan 15 % ofstructuresinthecountywerebuiltwithoutpermits.

Thebuildinginspectorthinksthatnomorethan 15 % ofthestructuresinthecounty werebuiltwithoutpermitswhen,infact,nomorethan 15 % ofthestructuresrealywere builtwithoutpermits. Thebuildinginspectorthinksthatmorethan 15 % ofthestructuresinthecountywere builtwithoutpermitswhen,infact,morethan 15 % ofthestructuresrealywerebuilt withoutpermits. Thebuildinginspectorthinksthatmorethan 15 % ofthestructuresinthecountywere builtwithoutpermitswhen,infact,atmost 15 % ofthestructureswerebuiltwithout permits. Thebuildinginspectorthinksthatnomorethan 15 % ofthestructuresinthecounty werebuiltwithoutpermitswhen,infact,morethan 15 % ofthestructureswerebuilt withoutpermits.

AnswerExplanation

Correctanswer: ATypeIIerroristhedecisionnottorejectthenulhypothesiswhen,infact,itisfalse.In thiscase,theTypeIIerroriswhenthebuildinginspectorthinksthatnomore

Thebuildinginspectorthinksthatmorethan 15 % ofthestructuresinthecountywere

builtwithoutpermitswhen,infact,atmost 15 % ofthestructureswerebuiltwithout

permits. $$Test statistic = −2.

  • $\text{Test statistic = }-2.24$Test statistic = −2.

than 15 % ofthestructureswerebuiltwithoutpermitswhen,infact,more

than 15 % ofthestructureswerebuiltwithoutpermits.

Youranswer: Content attribution- Opens a dialog

QUESTION 41 / 1 POINTS

Supposeachefclaimsthathermeatbalweightislessthan 4 ounces,onaverage.

Severalofhercustomersdonotbelieveher,sothechefdecidestodoahypothesistest,

ata 10 % significancelevel,topersuadethem.Shecooks 14 meatbals.Themean

weightofthesamplemeatbalsis3.7 ounces.Thechefknowsfrom experiencethatthe standarddeviationforhermeatbalweightis0.5 ounces. • H 0 :μ ≥ 4 ;Ha:μ< 4 • α=0.1 (significancelevel) Whatistheteststatistic(z-score)ofthisone-meanhypothesistest,roundedtotwo

decimalplaces?

AnswerExplanation

Correctanswers: Thehypotheseswerechosen,andthesignificancelevelwasdecidedon,sothenext stepinhypothesistestingistocomputetheteststatistic.Inthisscenario,thesample

meanweight,x ¯= 3.7.Thesamplethechefusesis 14 meatbals,son = 14 .She

knowsthestandarddeviationofthemeatbals,σ=0.5.Lastly,thechefiscomparing thepopulationmeanweightto 4 ounces.So,thisvalue(foundinthenulandalternative

Astandardnormalcurvewithtwopointslabeledonthehorizontalaxis.Themeanis labeledat 0. 00 andanobservedvalueof 1. 74 islabeled.Theareaunderthecurveand totherightoftheobservedvalueisshaded.

UsingtheStandardNormalTable,wecanseethatthep-valueisequalto0.959,

whichistheareatotheleftofz=1.74.(StandardNormalTablesgiveareastothe left.)So,thep-valuewe'relookingforisp= 1 −0.959 =0.041.

QUESTION 61 / 1 POINTS

Kenneth,acompetitorincupstacking,claimsthathisaveragestackingtime

is8. 2 seconds.Duringapracticesession,Kennethhasasamplestackingtimemean

of7. 8 secondsbasedon 11 trials.Atthe 4 % significancelevel,doesthedata

providesufficientevidencetoconcludethatKenneth'smeanstackingtimeisless

than8.2 seconds?Acceptorrejectthehypothesisgiventhesampledatabelow.

• H 0 :μ=8.2 se c o nds;Ha:μ< 8.2 seconds • α=0.04 (significancelevel) • z 0 =−1. • p=0.

Donotrejectthenulhypothesisbecausethep-value 0. 0401 isgreaterthanthe

significancelevelα= 0. 04.

Donotrejectthenulhypothesisbecausethep - value0.0401 isgreaterthanthe significancelevelα=0.. Rejectthenulhypothesisbecausethep - value0.0401 isgreaterthanthesignificance levelα= 0.. Rejectthenulhypothesisbecausethevalueofz isnegative. Rejectthenulhypothesisbecause|−1.75| > 0.04. Donotrejectthenulhypothesisbecause|−1.75 |>0.04.

AnswerExplanation

Correctanswer:

InmakingthedecisiontorejectornotrejectH 0 ,ifα>p-value,rejectH 0 becausethe

resultsofthesampledataaresignificant.Thereissufficientevidencetoconclude

thatH 0 isanincorrectbeliefandthatthealternativehypothesis,Ha,maybecorrect.

Ifα≤p-value,donotrejectH 0 .Theresultsofthesampledataarenotsignificant,so thereisnotsufficientevidencetoconcludethatthealternativehypothesis,Ha,maybe correct.Inthiscase,α= 0.04 islessthanorequaltop = 0.0401,sothedecisionis

tonotrejectthenulhypothesis.

QUESTION 71 / 1 POINTS

Arecentstudysuggestedthat 81 % ofseniorcitizenstakeatleastoneprescription

medication.Ameliaisanurseatalargehospitalwhowouldliketoknowwhetherthe percentageisthesameforseniorcitizenpatientswhogotoherhospital.Sherandomly

selects 59 seniorcitizenspatientswhoweretreatedatthehospitalandfindsthat 49 of

them takeatleastoneprescriptionmedication.Whatarethenulandalternative hypothesesforthishypothesistest?

$$Test_Statistic=−0.

  • $\text{Test_Statistic}=-0.53$Test_Statistic=−0.

Findtheteststatisticforthishypothesistestforaproportion.Roundyouranswer

to 2 decimalplaces.

AnswerExplanation

Correctanswers:

Theproportionofsuccessesisp ^= 951000 =0.095.

Theteststatisticiscalculatedasfolows:

z=p^−p0p 0 ⋅(1−p 0 )n−−−−−−√

z=0.095−0.100.10⋅(1−0.10)1000−−−−−−−−√ z≈−0.

QUESTION 91 / 1 POINTS

Amedicalresearcherclaimsthattheproportionofpeopletakingacertainmedication thatdevelopserioussideeffectsis 12 %.Totestthisclaim,arandom sampleof 900 peopletakingthemedicationistakenanditisdeterminedthat 9 3peoplehave experiencedserioussideeffects.. Thefolowingisthesetupforthishypothesistest: H 0 : p = 0. 12

Ha : p ≠ 0. 12

Findthep-valueforthishypothesistestforaproportionandroundyouranswerto 3 decimalplaces.

$$P-value=0.

  • $\text{P-value=}0.124$P-value=0.

ThefolowingtablecanbeutilizedwhichprovidesareasundertheStandardNormal Curve: z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.

  • 1.8 0.036 0.035 0.034 0.034 0.033 0.032 0.031 0.031 0.030 0.
  • 1.7 0.045 0.044 0.043 0.042 0.041 0.040 0.039 0.038 0.038 0.
  • 1.6 0.055 0.054 0.053 0.052 0.051 0.049 0.048 0.047 0.046 0.
  • 1.5 0.067 0.066 0.064 0.063 0.062 0.061 0.059 0.058 0.057 0.
  • 1.4 0.081 0.079 0.078 0.076 0.075 0.074 0.072 0.071 0.069 0.

AnswerExplanation

Correctanswers: Herearethestepsneededtocalculatethep-valueforahypothesistestforaproportion: 1 .Determineifthehypothesistestislefttailed,righttailed,ortwotailed. 2 .Computethevalueoftheteststatistic. 3 .Ifthehypothesistestislefttailed,thep-valuewilbetheareaunderthestandard normalcurvetotheleftoftheteststatistic z 0 Ifthetestisrighttailed,thep-valuewilbetheareaunderthestandardnormal curvetotherightoftheteststatistic z 0 Ifthetestistwotailed,thep-valuewilbetheareatotheleftof−| z 0 |plusthe areatotherightof| z 0 |underthestandardnormalcurve

Thusthep-valueistheareaundertheStandardNormalcurvetotheleftofaz-scoreof

    1. 54 ,plustheareaundertheStandardNormalcurvetotherightofaz-scoreof 1. 54.

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.

  • 1.8 0.036 0.035 0.034 0.034 (^) 0.033 0.032 0.031 0.031 0.030 0.
  • 1.7 0.045 0.044 0.043 0.042 0.041 0.040 0.039 0.038 0.038 0.
  • 1.6 0.055 0.054 0.053 0.052 0.051 0.049 0.048 0.047 0.046 0. - 1.5 0.067 0.066 0.064 0.063 0.062 0.061 0.059 0.058 0.057 0.
  • 1.4 0.081 0.079 0.078 0.076 0.075 0.074 0.072 0.071 0.069 0.

From alookuptableoftheareaundertheStandardNormalcurve,thecorresponding areaisthen 2 ( 0. 062 )= 0. 124.

QUESTION 101 / 1 POINTS

Aneconomistclaimsthattheproportionofpeoplewhoplantopurchaseafulyelectric

vehicleastheirnextcarisgreaterthan 65 %.

Totestthisclaim,arandom sampleof 750 peopleareaskediftheyplantopurchasea fulyelectricvehicleastheirnextcar Ofthese 750 people, 513 indicatethattheydo

plantopurchaseanelectricvehicle. Thefolowingisthesetupforthishypothesistest:

H 0 :p=0.65 Ha:p>0.

Inthisexample,thep-valuewasdeterminedtobe0.026.

Cometoaconclusionandinterprettheresultsforthishypothesistestforaproportion

(useasignificancelevelof 5 %.)

ThedecisionistorejecttheNulHypothesis. Theconclusionisthatthereisenoughevidencetosupporttheclaim. ThedecisionistorejecttheNulHypothesis. Theconclusionisthatthereisenoughevidencetosupporttheclaim. ThedecisionistofailtorejecttheNulHypothesis. Theconclusionisthatthereisnotenoughevidencetosupporttheclaim.

AnswerExplanation

Correctanswer: Tocometoaconclusionandinterprettheresultsforahypothesistestforproportion usingtheP-ValueApproach,thefirststepistocomparethep-valuefrom thesample datawiththelevelofsignificance. Thedecisioncriteriaisthenasfolows:

Ifthep-valueislessthanorequaltothegivensignificancelevel,thenthenul

hypothesisshouldberejected.

So,ifp≤α,rejectH 0 ;otherwisefailtorejectH 0.

Whenwehavemadeadecisionaboutthenulhypothesis,itisimportanttowritea thoughtfulconclusionaboutthehypothesesintermsofthegivenproblem'sscenario. Assumingtheclaim isthenulhypothesis,theconclusionisthenoneofthefolowing:

  • ifthedecisionistorejectthenulhypothesis,thentheconclusionisthatthereis

enoughevidencetorejecttheclaim.

  • ifthedecisionistofailtorejectthenulhypothesis,thentheconclusionisthat

thereisnotenoughevidencetorejecttheclaim. Assumingtheclaim isthealternativehypothesis,theconclusionisthenoneofthe folowing:

H 0 : p = 0.5 ;Ha: p≠0.5 ,whichisatwo-tailedtest.

AnswerExplanation

Correctanswer:

H 0 :p= 0. 5 ;Ha:p≠ 0. 5 ,whichisatwo-tailedtest.

Thenulhypothesisshouldbetrueproportion:H 0 :p=0.5.Beckywantstoknowif thetrueproportionofheadsisdifferentfrom 0.5.Thismeansthatwejustwanttotestif theproportionisnot0.5.So,thealternativehypothesisisHa:p≠0.5,whichisatwo-

tailedtest.

QUESTION 121 / 1 POINTS

Johnownsacomputerrepairservice.Foreachcomputer,he

charges$50 plus$45 perhourofwork.Alinearequationthatexpressesthetotal

amountofmoneyJohnearnspercomputerisy= 50 + 45 x.Whatarethe independentanddependentvariables?Whatisthey-interceptandtheslope?

Theindependentvariable(x)istheamountoftimeJohnfixesacomputer.The dependentvariable(y)istheamount,indolars,Johnearnsforacomputer. Johnchargesaone-timefeeof$50 (thisiswhenx= 0 ),sothey-interceptis 50 .John earns$45 foreachhourheworks,sotheslopeis 45. Theindependentvariable(x)istheamount,indolars,Johnearnsforacomputer.The dependentvariable(y )istheamountoftimeJohnfixesacomputer. Johnchargesaone-timefeeof$45 (thisiswhenx= 0 ),sothey-interceptis 45 .John earns$50 foreachhourheworks,sotheslopeis 50. Theindependentvariable(x)istheamount,indolars,Johnearnsforacomputer.The dependentvariable(y )istheamountoftimeJohnfixesacomputer.

Theindependentvariable(x)istheamountoftimeJohnfixesacomputer.The

dependentvariable(y)istheamount,indolars,Johnearnsforacomputer. Johnchargesaone-timefeeof$ 50 (thisiswhenx= 0 ),sothey-interceptis 50. Johnearns$ 45 foreachhourheworks,sotheslopeis 45.

Johnchargesaone-timefeeof$50 (thisiswhenx= 0 ),sothey-interceptis 50 .John earns$45 foreachhourheworks,sotheslopeis 45. Theindependentvariable(x)istheamountoftimeJohnfixesacomputer.The dependentvariable(y)istheamount,indolars,Johnearnsforacomputer. Johnchargesaone-timefeeof$45 (thisiswhenx= 0 ),sothey-interceptis 45 .John earns$50 foreachhourheworks,sotheslopeis 50.

AnswerExplanation

Correctanswer:

Theindependentvariable(x)istheamountoftimeJohnfixesacomputerbecauseitis

thevaluethatchanges.Hemayworkdifferentamountspercomputer,andhisearnings aredependentonhow manyhoursheworks.Thisiswhytheamount,indolars,John

earnsforacomputeristhedependentvariable(y).

They-interceptis 50 (b= 50 ).Thisishisone-timefee.Theslopeis 45 (a = 4 5 ).

Thisistheincreaseforeachhourheworks.

QUESTION 131 / 1 POINTS

Arianakeepstrackoftheamountoftimeshestudiesandthescoreshegetsonher quizzes.Thedataareshowninthetablebelow.Whichofthescatterplotsbelow accuratelyrecordsthedata? Hoursstudying Quizscore 1 5 2 5 3 7

AscatterplothasahorizontalaxislabeledHoursstudyingfrom 0 to 10 inincrementsof 2 andaverticalaxislabeledQuizscorefrom 0 to 6 inincrementsof 1 .Thefolowing pointsareplotted:left-parenthesis 5 comma 1 right-parentheses;left-parenthesis 5 comma 2 right-parentheses;left-parenthesis 7 comma 3 right-parentheses;left- parenthesis 9 comma 4 right-parentheses;left-parenthesis 9 comma 5 right- parentheses.Alvaluesareapproximate.

AscatterplothasahorizontalaxislabeledHoursstudyingfrom 0 to 6 inincrementsof 1 andaverticalaxislabeledQuizscorefrom 0 to 9 inincrementsof 1 .Thefolowing pointsareplotted:left-parenthesis 1 comma 5 right-parentheses;left-parenthesis 2 comma 5 right-parentheses;left-parenthesis 3 comma 7 right-parentheses;left- parenthesis 4 comma 8 right-parentheses;left-parenthesis 5 comma 8 right- parentheses. AscatterplothasahorizontalaxislabeledHoursstudyingfrom 0 to 6 inincrementsof 1 andaverticalaxislabeledQuizscorefrom 0 to 12 inincrementsof 2 .Thefolowing pointsareplotted:left-parenthesis 1 comma 5 right-parentheses;left-parenthesis 2 comma 5 right-parentheses;left-parenthesis 3 comma 8 right-parentheses;left- parenthesis 4 comma 8 right-parentheses;left-parenthesis 5 comma 11 right- parentheses.Alvaluesareapproximate.

AnswerExplanation

Correctanswer: