MATH 225N MATH Statistic Final (NEWEST 2024), Exams of Nursing

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/ 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 = 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

H 0 :X≤ 3. 8 ,Ha:X> 3.

H 0 :X≤ 7. 4 ,Ha:X> 7.

Correctanswer:

Remembertheformsofthehypothesistests.

• Right-tailed:H 0 :X≤X 0 ,Ha:X >X 0.
• Left-tailed:H 0 :X≥X 0 ,Ha:X < X 0.
• Two-tailed:H 0 : X = X 0 ,Ha: X≠X 0.

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

$$0.

• $0.041$0.

hypotheses)isμ 0 .Nowwewilsubstitutethevaluesintotheformulatocomputethe

teststatistic:

z 0 =x¯−μ0σn√=3.7−40.514√≈−0.30.134≈−2.

So,theteststatisticforthishypothesistestisz 0 =−2.24.

QUESTION 51 / 1 POINTS

Whatisthep-valueofaright-tailedone-meanhypothesistest,withateststatistic

ofz 0 = 1.74?(Donotroundyouranswer;computeyouranswerusingavaluefrom the tablebelow.) z1.51.61.71.81.90.000.9330.9450.9550.9640.9710.010.9340. 9460.9560.9650.9720.020.9360.9470.9570.9660.9730.030. 370.9480.9580.9660.9730.040.9380.9490.9590.9670.9740. 50.9390.9510.9600.9680.9740.060.9410.9520.9610.9690. 50.070.9420.9530.9620.9690.9760.080.9430.9540.9620. 0.9760.090.9440.9540.9630.9710. AnswerExplanation Correctanswers: Thep-valueistheprobabilityofanobservedvalueofz= 1.74 orgreaterifthenul

hypothesisistrue,becausethishypothesistestisright-tailed.Thisprobabilityisequalto

theareaundertheStandardNormalcurvetotherightofz = 1.74.

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?

{H 0 :p=0.81Ha:p>0. {H 0 :p≠0.81Ha:p=0. {H 0 :p=0.81Ha:p<0. {H 0 :p=0.81Ha:p≠0.

AnswerExplanation

Correctanswer:

{H 0 :p=0.81Ha:p≠0.

Firstverifywhetheraloftheconditionshavebeenmet.Letp bethepopulation

proportionfortheseniorcitizenpatientstreatedatAmelia'shospitalwhotakeatleast oneprescriptionmedication.

1. Sincetherearetwoindependentoutcomesforeachtrial,theproportionfolowsa binomialmodel.
2. Thequestionstatesthatthesamplewascolectedrandomly.

3 .Theexpectednumberofsuccesses,np =47.79,andtheexpectednumberof

failures,nq =n( 1 −p)=11.21,arebothgreaterthanorequalto 5.

SinceAmeliaistestingwhethertheproportionisthesame,thenulhypothesisis

thatp isequalto0.81 andthealternativehypothesisisthatp isnotequalto0.81.

Thenulandalternativehypothesesareshownbelow.

{H 0 :p=0.81Ha:p≠0.

QUESTION 81 / 1 POINTS

Aresearcherclaimsthattheproportionofcarswithmanualtransmissionisless

than 10 %.Totestthisclaim,asurveychecked 1000 randomlyselectedcars.Of

thosecars, 95 hadamanualtransmission.

Thefolowingisthesetupforthehypothesistest:

{H 0 :p=0.10Ha:p<0.

$$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

Forthisexample,thetestisatwotailedtestandtheteststatistic,roundingtotwo

decimalplaces,isz =0.1033−0.120.12( 1 −0.12) 900 −−−−−−−−−

−−− √ ≈−1.54.

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:

• ifthedecisionistorejectthenulhypothesis,thentheconclusionisthatthereis

enoughevidencetosupporttheclaim.

• ifthedecisionistofailtorejectthenulhypothesis,thentheconclusionisthat

thereisnotenoughevidencetosupporttheclaim.

Inthisexample,thep-value=0.026.Wethencomparethep-valuetothelevelof

significancetocometoaconclusionforthehypothesistest.

Inthisexample,thep-valueislessthanthelevelofsignificancewhichis0.05.

Sincethep-valueisgreaterthanthelevelofsignificance,theconclusionistorejectthe nullhypothesis.

QUESTION 111 / 1 POINTS

Becky'sstatisticsteacherwasteachingtheclasshow toperform thez-testfora

proportion.Beckywasboredbecauseshehadalreadymasteredthetest,soshe decidedtoseeifthecoinshehadinherpocketwouldcomeupheadsortailsinatruly

random fashionwhenflipped.Shediscretelyflippedthecoin 30 timesandgot

heads 18 times. Beckyconductsaone-proportionhypothesistestatthe 5 % significancelevel,totest whetherthetrueproportionofheadsisdifferentfrom 50%.

Whichanswerchoiceshowsthecorrectnulandalternativehypothesesforthistest? H 0 :p =0.6;Ha: p >0.6,whichisaright-tailedtest. H 0 :p= 0.5 ;Ha: p < 0.5 ,whichisaleft-tailedtest. H 0 :p=0.6 ;Ha: p≠0.6,whichisatwo-tailedtest.

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

AscatterplothasahorizontalaxislabeledHoursstudyingfrom 0 to 6 inincrementsof 1 andaverticalaxislabeledQuizscorefrom 0 to 10 inincrementsof 2 .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 9 right-parentheses;left-parenthesis 5 comma 9 right- parentheses.Alvaluesareapproximate.

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.