Docsity
Docsity
Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity

Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips
Guidelines and tips
Sell on Docsity
Sell on Docsity
Docsity AI
Docsity AI
Log inSign up

Prepare for your exams

Study with the several resources on Docsity

Find documents

Prepare for your exams with the study notes shared by other students like you on Docsity

Search for your university

Find the specific documents for your university's exams

Docsity AINEW

Summarize your documents, ask them questions, convert them into quizzes and concept maps

Explore questions

Clear up your doubts by reading the answers to questions asked by your fellow students

Earn points to download

Earn points by helping other students or get them with a premium plan

Share documents
download point icon20 Points

For each uploaded document

Answer questions
download point icon5 Points

For each given answer (max 1 per day)

All the ways to get free points
Get points immediately

Choose a premium plan with all the points you need

Study Opportunities

Choose your next study program

Get in touch with the best universities in the world. Search through thousands of universities and official partners

Community

Ask the community

Ask the community for help and clear up your study doubts

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

Statistical Analysis of Test Scores, Exams of Nursing

Chamberlain College of NursingNursing

A statistical analysis of test scores, including the calculation of means, modes, and standard deviations. It also includes hypothesis testing using the t-distribution and the chi-square distribution. The data set consists of test scores for various subjects and grades.

Typology: Exams

2023/2024

Available from 06/11/2024

nclexmaster
nclexmaster 🇺🇸

5

(1)

1.9K documents

1 / 91

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
1
1
/
1
P
O
I
N
T
S
A
fi
t
n
e
ss
c
e
n
t
e
r
c
l
a
i
m
s
t
h
a
tt
h
e
m
e
a
n
a
m
o
un
t
o
f
t
i
m
e
t
h
a
t
a
p
e
r
s
o
n
s
p
e
n
d
s
a
tt
h
e
g
y
m
p
e
r
v
i
s
i
t
i
s
33
m
i
nu
t
e
s
.
I
d
e
n
t
i
f
y
t
h
e
nu
l
h
y
p
o
t
h
e
s
i
s
,
H
0
,
a
n
d
t
h
e
a
l
t
e
r
n
a
t
i
v
e
h
y
p
o
t
h
e
s
i
s
,
H
a
,
i
n
t
e
r
m
s
o
f
t
h
e
p
a
r
a
m
e
t
e
r
μ
.
H0
:
μ
≠
33
;
H
a
:
μ
=
33
H0
:
μ
=
33
;
H
a
:
μ
≠
33
H0
:
μ
≥
33
;
H
a
:
μ
<
33
H0
:
μ
≤
33
;
H
a
:
μ
>
33
A
n
s
w
e
r
E
x
p
l
ana
t
i
on
C
o
rr
e
c
t
an
s
w
e
r
:
H
0
:
μ
=
33
;
H
a
:
μ
≠
33
L
e
tt
h
e
p
a
r
a
m
e
t
e
r
μ
b
e
u
s
e
d
t
o
r
e
p
r
e
s
e
n
tt
h
e
m
e
a
n
.
T
h
e
nu
l
h
y
p
o
t
h
e
s
i
s
i
s
a
l
w
a
y
s
s
t
a
t
e
d
w
i
t
h
s
o
m
e
f
o
r
m
o
f
e
q
u
a
li
t
y
:
e
q
u
a
l
(
=
)
,
g
r
e
a
t
e
r
t
h
a
no
r
e
q
u
a
l
t
o
(
≥
)
,
o
r
l
e
ss
t
h
a
n
o
r
e
q
u
a
l
t
o
(
≤
)
.
T
h
e
r
e
f
o
r
e
,
i
n
t
h
i
s
c
a
s
e
,
t
h
e
nu
l
h
y
p
o
t
h
e
s
i
s
H
0
i
s
μ
=
33
.
T
h
e
a
l
t
e
r
n
a
t
i
v
e
h
y
p
o
t
h
e
s
i
s
i
sc
on
t
r
a
d
i
c
t
o
r
y
t
o
t
h
e
nu
l
h
y
p
o
t
h
e
s
i
s
,
s
o
H
a
i
s
μ
≠
33
.
Q
U
ES
T
I
O
N
2
1
/
1
P
O
I
N
T
S
T
h
e
a
n
s
w
e
r
c
h
o
i
c
e
s
b
e
l
o
w
r
e
p
r
e
s
e
n
t
d
i
ff
e
r
e
n
t
h
y
p
o
t
h
e
s
i
s
t
e
s
t
s
.
W
h
i
c
h
o
f
t
h
e
c
h
o
i
c
e
s
a
r
e
r
i
gh
t-
t
a
il
ed
t
e
s
t
s
?
S
e
l
e
c
t
a
l
c
o
rr
e
c
t
a
n
s
w
e
r
s
.
•
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a
pf1b
pf1c
pf1d
pf1e
pf1f
pf20
pf21
pf22
pf23
pf24
pf25
pf26
pf27
pf28
pf29
pf2a
pf2b
pf2c
pf2d
pf2e
pf2f
pf30
pf31
pf32
pf33
pf34
pf35
pf36
pf37
pf38
pf39
pf3a
pf3b
pf3c
pf3d
pf3e
pf3f
pf40
pf41
pf42
pf43
pf44
pf45
pf46
pf47
pf48
pf49
pf4a
pf4b
pf4c
pf4d
pf4e
pf4f
pf50
pf51
pf52
pf53
pf54
pf55
pf56
pf57
pf58
pf59
pf5a
pf5b

Partial preview of the text

Download Statistical Analysis of Test Scores 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.

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:

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?

{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.

Forthisexample,thetestisatwotailedtestandtheteststatistic,roundingtotwo

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

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: