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MATH 225N MATH Statistic Final Correct Approved, Exams of Nursing

Southern Virginia UniversityNursing

MATH 225N MATH Statistic Final Correct Approved

Typology: Exams

2022/2023

Available from 05/21/2023

moraamaranga12
moraamaranga12 🇺🇸

96 documents

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MATH 225N MATH Statistic Final Correct Approved
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Partial preview of the text

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

1 / 1 POINTS

Afitnesscenterclaimstha t hemeanamountoft

imethatapersonspendsa t hegym

pervisitis 33

minutes.Identifythenu l hypothesis,H 0 ,an

dthealternative

hypothesis,Ha,intermsoftheparameterμ.

H 0 :μ≠ 33 ;Ha:μ

H 0 :μ= 33 ;Ha:μ

H 0 :μ≥ 33 ;Ha:μ

H 0 :μ≤ 33 ;Ha:μ> 33

AnswerExplanation

Co r

ectanswer:

H 0 :μ= 33 ;Ha:μ≠ 33

Le t heparameterμ

beusedtorepresen t hemean.Thenu l hypothesisisa

lways statedwithsomeform

ofequality:equal(=),greaterthanorequalto(≥)

,orlessthan

orequalto(≤).Therefore,inthiscase,thenu l hyp

othesisH 0 isμ= 33 .The

alternativehypothesisiscontradictorytothenu

l hypothesis,soHa isμ≠ 33.

QUESTI

ON 2

1 / 1 POI

NTS

Theanswerchoicesbelowrepresentdiffere

nthypothesistests.Whichofthechoices

areright-tailedtests?

Thatisco r ect!

Thatisco r ect!

Selecta l co r ectanswers.

H 0 :X≥ 17. 1 ,Ha:X< 17.

Thebulidinginspectorthinksthatnomorethan 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.

QUESTI

ON 3

0 / 1 POI

NTS

FindtheType

I e r orgiventha t henu l hypothesis,H 0 ,is:abuildin

ginspectorclaims thatnomorethan 15 %

ofstructuresinthecountywerebuiltwithoutpermi

ts.

Thebuildinginspectorthinksthatnomorethan 15 %

ofthestructuresinthecounty

werebuiltwithoutpermitswhen,infact,nomoretha

n 15 % ofthestructuresrea l ywere

builtwithoutpermits.

Thebuildinginspectorthinksthatmorethan 1

5 % ofthestructuresinthecountywere

builtwithoutpermitswhen,infact,moretha

n 15 % ofthestructuresrea l ywerebuilt

withoutpermits.

Thebuildinginspectorthinksthatmorethan 1

5 % ofthestructuresinthecountywere

builtwithoutpermitswhen,infact,atmost 15

% ofthestructureswerebuiltwithout

permits.

Thebuildinginspectorthinksthatnomoret

han 15 % ofthestructuresinthecounty

werebuiltwithoutpermitswhen,infact,m

orethan 15 % ofthestructureswerebuilt

withoutpermits.

AnswerExplanation

Co r

ectanswer:

AType

I e r oristhedecisionno t orejec t henu l hypothesisw

hen,infact,itisfalse.In thiscase,theType

I e r oriswhenthebuildinginspectorthinksthatnom

That'snotright.

ore

meanweight,x¯= 3. 7 .Thesamplethechefusesis 14

meatba l s,son= 14 .She

knowsthestandarddeviationofthemeatba l s,σ= 0. 5.

Lastly,thechefiscomparing

thepopulationmeanweigh t o 4

ounces.So,thisvalue(foundinthenu l andalterna

tive

hypotheses)isμ 0 .Nowwewi l substitutethev

aluesintotheformulatocomputethe

teststatistic:

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

So,theteststatisticforthishypothesistestisz 0

QUESTI

ON 5

1 / 1 POI

NTS

Whatisthep-valueofaright-tailedone-

meanhypothesistest,withateststatistic

ofz 0 = 1. 74?

(Donotroundyouranswer;computeyouransw

erusingavaluefrom the tablebelow.)

z1.51.61.71.81.90.000.9330.9450.9550.9640.9710.010.9340.

AnswerExplanation

Co

r

ectanswers:

Thep-

valueistheprobabilityofanobservedvalue

ofz= 1. 74 orgreaterifthenu l

hypothesisistrue,becausethishypothesi

stestisright-

tailed.Thisprobabilityisequalto

theareaundertheStandardNormalcurvetot

herightofz= 1. 74.

Thatisco r ect!

  • p=0.

Thatisco r ect!

Donotrejec t

henu l hypothesisbecausethep-

value 0. 0401 isgreaterthanthe

significancelevelα=0. 04.

Rejec t

henu l hypothesisbecausethep-value 0. 0401

isgreaterthanthesignificance levelα=0.04.

Rejec t

henu l hypothesisbecau

sethevalueofz isnegative.

Rejec t

henu l hypothesisbecau

se|−1.75|> 0. 04.

Donotrejec t

henu l hypothesisbecause|−1. 75 |> 0. 04.

AnswerExplanation

Co r

ectanswer:

Donotrejec t henu l hypothesisbecausethep-

value 0. 0401 isgreaterthanthe

significancelevelα= 0. 04.

Inmakingthedecisiontorejectornotreject

H 0 ,ifα>p-value,rejectH 0 becausethe

resultsofthesampledataaresignificant.Th

ereissufficientevidencetoconclude thatH 0

isaninco r ectbeliefandtha t healternativ

ehypothesis,Ha,maybeco r ect. Ifα≤p-

value,donotrejectH 0 .Theresultsofthesam

pledataarenotsignificant,so

thereisnotsufficientevidencetoconcludeth

a t healternativehypothesis,Ha,maybe

co r ect.Inthiscase,α= 0. 04

islessthanorequaltop= 0. 0401 ,sothedecisio

nis tonotrejec t henu l hypothesis.

QUESTI

ON 7

1 / 1 POI

NTS

Arecentstudysu g

estedthat 81 %

ofseniorcitizenstakeatleastoneprescrip

tion

medication.Ameliaisanurseatalargehospi

talwhowouldliketoknowwhetherthe

percentageisthesameforseniorcitizenpat

ientswhogotoherhospital.Sherandomly

selects 59

Thatisco r ect!

{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

Co r

ectanswer:

{H 0 :p=0.81Ha:p≠0.

Firstverifywhethera l oftheconditionshaveb e

nmet.Letp bethepopulation

proportionfortheseniorcitizenpatientstreated

atAmelia'shospitalwhotakeatleast

oneprescriptionmedication.

  1. Sincetherearetwoindependentoutcome

sforeachtrial,theproportionfo l owsa

binomialmodel.

  1. Thequestionstatestha t hesamplewasco

l ectedrandomly.

3 .Theexpectednumberofsu c esses,np= 47. 79

,andtheexpectednumberof

failures,nq=n( 1 −p)= 11. 21 ,arebothgreaterthano

requalto 5.

SinceAmeliaistestingwhethertheproport

ionisthesame,thenu l hypothesisis thatp

isequalto 0. 81

andthealternativehypothesisisthatp

isnotequalto 0. 81.

Thenu l andalternativehypothesesareshow

nbelow.

{H 0 :p=0.81Ha:p≠0.

QUESTI

ON 8

1 / 1 POI

NTS

Aresearcherclaimstha t

heproportionofcar

swithmanualtransmissionisless

than 10 %.Totes

t

hisclaim,asurveychecked

1000 randomlyselectedcars.Of

thosecars, 95 hadamanualtransmission.

Thefo l owingisthesetupforthehypothesistest:

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

  • $\text{P-value=}0.124$P-value=0.

Thefo l owingtablecanbeutilizedwhichprovidesar easundertheStandardNormal Curve: AnswerExplanation

Co

r ectanswers:

Herearethestepsn e dedtocalculatethep- valueforahypothesistestforaproportion: 1 .Determineifthehypothesistestislef t ailed, righ t ailed,ortwotailed. 2 .Computethevalueoftheteststatistic. 3 .Ifthehypothesistestislef t ailed,the p-valuewi l betheareaunderthestandard normalcurvetotheleftoftheteststatistic z 0 Ifthetestisrigh t ailed,thep- valuewi l betheareaunderthestandardnormal curvetotherightoftheteststatistic z 0 Ifthetestistwotailed,thep- valuewi l betheareatotheleftof−| z 0 |plusthe areatotherightof| z 0 | underthestandardnormalcurve Forthisexample,thetestisatwotailedte standtheteststatistic,roundingtotwo

decimalplaces,isz= 0. 1033 − 0. 120. 12 ( 1 − 0. 12 ) 900

−−−−−−−−− −−−√≈− 1. 54.

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

$$P-value=0.

Thatisco r ect!

Thusthep- valueistheareaundertheStandardNormalcurveto theleftofaz-scoreof

  1. 54 ,plustheareaundertheStandardNormalcurvetother ightofaz-scoreof 1. 54. From al o kuptableoftheareaundertheStandardNormalcurv e,theco r esponding areaisthen 2 ( 0. 062 )= 0. 124.

QUESTI

ON 10 1 / 1 POIN TS

Aneconomistclaimstha t heproportionofpeoplewho plantopurchaseafu l yelectric

vehicleastheirnextcarisgreaterthan 65 %.

Totes t

hisclaim,arandom sampleof 750

peopleareaskediftheyplantopurchasea fu l yelectricvehicleastheirnextcar

Ofthese 750 people, 513 indicatetha

t heydo plantopurchaseanelectricvehicle. Thefo l owingisthesetupforthishypothesistest:

H 0 :p=0.

5 Ha:p>0. 5

Inthisexample,thep-

valuewasdeterminedtobe 0. 026.

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

Thedecisionistorejec t heNu l Hypothesis.

Theconclusionistha t hereisenoughevidenc

etosu p or t heclaim.

Thatisco r ect!

Thedecisionistorejec t

heNu l Hypothesis.

Theconclusionistha t

hereisenoughevidencetosu p

or t

heclaim.

Thedecisionistofailtorejec t

heNu l Hypothesis.

Theconclusionistha t

hereisnotenoughevidenceto

su p

or t

heclaim.

AnswerExplanation

Co r

ectanswer:

Tocometoaconclusionandinterpre t heres

ultsforahypothesistestforproportion

usingtheP-ValueA p

roach,thefirststepistocomparethep-

valuefrom thesample

datawiththelevelofsignificance.

Thedecisioncriteriaisthenasfo l ows:

Ifthep-

valueislessthanorequaltothegivensignificance

level,thenthenu l

hypothesisshouldberejected.

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

Whenwehavemadeadecisionabou t henu l hypothesis

,itisimportan t owritea

thoughtfulconclusionabou t hehypothesesinterms

ofthegivenproblem'sscenario.

Assumingtheclaim

isthenu l hypothesis,theconclusionisthenoneoft

hefo l owing:

  • ifthedecisionistorejec t henu l hypoth

esis,thentheconclusionistha t hereis

enoughevidencetorejec t heclaim.

  • ifthedecisionistofailtorejec t henu l

hypothesis,thentheconclusionistha

t

thereisnotenoughevidencetorejec t h

eclaim.

Assumingtheclaim

isthealternativehypothesis,theconclusi

onisthenoneofthe fo l owing: