Download ACTM – Statistics Regional Exam March 5, 2016

ACTM–StatisticsRegionalExamMarch5,2016
ForQuestions1through25recordyouranswerontheanswersheetprovided.Besuretouseapencil
tomarkyouranswers.Aftercompletingthemultiplechoicesection,gototheconstructedresponse
items.Theseareusedastiebreakeritemsonlyanddonotcountintheoverallstudentscore.Students
shouldanswertheconstructedresponseitemsinnumericalorder.Besurethatyournameisoneach
sheet.
1.Themeansalaryforcollegestudentsis$45,000.Isthisvalueastatisticoraparameter?
a. Thevalueisastatisticbecauseitisanumericalmeasurementdescribingsomecharacteristicofa
population.
b. Thevalueisaparameterbecauseitisanumericalmeasurementdescribingsomecharacteristic
ofasample.
c. Thevalueisaparameterbecauseitisanumericalmeasurementdescribingsomecharacteristic
ofapopulation.
d. Thevalueisastatisticbecauseitisanumericalmeasurementdescribingsomecharacteristicofa
sample.
2.Atownobtainscurrentemploymentdatabypolling10,000ofitscitizensthismonth.Identifythe
typeofobservationalstudy.
a. Cross-sectional
b. Prospective
c. Retrospective
d. Noneofthese
3.ThefrequencydistributionbelowsummarizesthehomesalepricesinthecityofSummerhillforthe
monthofJune.Determinethewidthofeachclass.
Salespriceinthousand$ Frequency
80.0-110.9
2
111.0-141.9
5
142.0-172.9
7
173.0-203.9
10
204.0-234.9
3
235.0-265.9
1
a. 30
b. 28
c. 61
d. 31
ACTM–StatisticsRegionalExamMarch5,2016
4.Anursemeasuredthebloodpressureofeachpersonwhovisitedherclinic.Belowisarelativefrequencybargraphforthesystolicbloodpressurereadingsforthosepeopleagedbetween25and40.
Thebloodpressurereadingsweregiventothenearestwholenumber.Identifythecenterofthethird
class.
Rela]veFrequency
a. 130
b. 124
0.4
c. 120
0.3
d. 125
0.2
0.1
0
100-110 110-120 120-130 130-140 140-150 150-160
SystolicBloodPressure(mmHg)
5.Themanagerofabankrecordedtheamountoftimeeachcustomerspentwaitinginlineduringpeak
businesshoursoneMonday.Thefrequencydistributionbelowsummarizestheresults.Estimatethe
meanwaitingtime.Roundtoonedecimalplace.
WaitingTime(inminutes) Numberofcustomers
0-3
9
4-7
9
8-11
13
12-15
6
16-19
6
20-23
1
24-27
2
a. 6.6minutes
b. 9.9minutes
c. 9.7minutes
d. 45minutes
6.Michaelgetstestgradesof71,76,80,and86.Hegetsa90onhisfinalexam.Findtheweighted
meanifthetestseachcountfor10%andthefinalexamcountsfor60%ofthefinalgrade.Roundtoone
decimalplace.
a. 85.3
b. -71.2
c. 241.8
d. 80.6
ACTM–StatisticsRegionalExamMarch5,2016
7.Theagesofthemembersofagymhaveameanof46yearsandastandarddeviationof12years.
WhatcanyouconcludefromChebyshev’stheoremaboutthepercentageofgymmembersaged
between14.8and77.2?
a. Thepercentageisatmost85.2%
b. Thepercentageisatleast85.2%
c. Thepercentageisatleast61.5%
d. Thepercentageisapproximately61.5%
8.Mario’sweeklypokerwinningshaveameanof$301andastandarddeviationof$60.Lastweekhe
won$182.Howmanystandarddeviationsfromthemeanisthat?
a. 0.99standarddeviationsbelowthemean
b. 1.98standarddeviationsbelowthemean
c. 1.98standarddeviationsabovethemean
d. 0.99standarddeviationsabovethemean
9.Thetablesummarizestheresultsoftestingforacertaindisease.Ifoneoftheresultsisrandomly
selected,whatistheprobabilitythatitisafalsenegative(testindicatesthepersondoesnothavethe
diseasewheninfacttheydo)?Whatdoestheprobabilitysuggestaboutthetest?
PositiveTest NegativeTest
Result
Result
Subjecthasthedisease
115
5
Subjectdoesnothavethedisease 11
180
a. 0.595;Theprobabilityofthiserrorishighsothetestisnotveryaccurate.
b. 0.035;Theprobabilityofthiserrorislowsothetestisfairlyaccurate.
c. 0.016;Theprobabilityofthiserrorislowsothetestisfairlyaccurate.
d. 0.042;Theprobabilityofthiserrorislowsothetestisfairlyaccurate.
10.Findtheprobabilitythatfourrandomlyselectedpeopleallhavedifferentbirthdays.Assumeonly
365possiblebirthdays.
a. 0.9891
b. 0.9918
c. 0.9729
d. 0.9836
11.Inabatchof8,000clockradios,5%aredefective.Asampleof15clockradiosisrandomlyselected
withoutreplacementfromthe8,000andtested.Theentirebatchwillberejectedifatleastoneof
thosetestedisdefective.Findtheprobabilitythattheentirebatchwillberejected.
a.0.537b.0.050 c.0.067
d.0.463
ACTM–StatisticsRegionalExamMarch5,2016
12.Eightbasketballplayersaretobeselectedtoplayinaspecialgame.Theplayerswillbeselected
fromalistof27players.Iftheplayersareselectedrandomly,whatistheprobabilitythattheeight
tallestplayerswillbeselected?
a. 1/2,220,075
b. 1/40,320
c. 1/213,127,200
d. 8/27
13.Refertothetable,whichsummarizestheresultsofasample.Asubjectisrandomlyselected,what
istheprobabilitythesubjecthasbrownhairgiventhattheyhaveblueeyes?
GreenEyes BlueEyes
BrownHair 82
7
BlondHair 26
308
a. 0.079
b. 0.978
c. 0.022
d. 0.210
14.Anarcherisabletohitthebull’seye57%ofthetime.Ifsheshootseightarrows,whatisthe
probabilitythatshegetsexactlyfourbull’s-eyes?Assumeeachshotisindependentoftheothers.
a. 0.152
b. 0.106
c. 0.253
d. 0.004
15.Theincomesoftraineesatalocalmillarenormallydistributedwithameanof$1,100andastandard
deviationof$150.Whatpercentageoftraineesearnlessthan$900amonth?
a. 40.82%
b. 90.82%
c. 35.31%
d. 9.12%
16.SupposethattheIQscoresofadultsarenormallydistributedwithameanof100andastandard
deviationof15.FindtheIQscorethatwouldbethecutoffscoreforhavinganIQinthetop10%ofthe
population.
a. 100.5
b.119.2
c.108.1
d.80.8
ACTM–StatisticsRegionalExamMarch5,2016
17.Inapopulationof150women,theheightsofthewomenarenormallydistributedwithameanof
64.1inchesandastandarddeviationof3.10inches.If49womenareselectedatrandom,findthemean
andstandarddeviationofthepopulationofsamplemeans.Assumethatthesamplingisdonewithout
replacementanduseafinitepopulationcorrectionfactor.
a. 64.1inches,3.10inches
b. 64.1inches,2.26inches
c. 64.1inches,0.36inches
d. 52.8inches,2.55inches
18.Anewspaperarticleabouttheresultsofapollstates:“Intheory,theresultsofsuchapoll,in99
casesoutof100shoulddifferbynomorethan3percentagepointsineachdirectionfromwhatwould
havebeenobtainedbyinterviewingallvotersintheUnitedStates.”Findthesamplesizesuggestedby
thisstatement.
a. 1,068
b. 1843
c. 1509
d. 74
19.Asurveyof300unionmembersinNewYorkStaterevealsthat112favortheRepublicancandidate
forgovernor.Constructa98%confidenceintervalforthetruepopulationproportionpofallNewYork
StateunionmemberswhofavortheRepublicancandidate.Usethenormalapproximationmethod.
a. 0.304<p<0.442
b. 0.316<p<0.430
c. 0.308<p<0.438
d. Noneofthese
20.Agroupof67randomlyselectedstudentshaveameanscoreof21.9onaplacementtest.The
populationstandarddeviationis 𝜎 = 5.Whatisthe90%confidenceintervalforthemeanscoreµofall
studentstakingthetest?
a. 20.5<µ<23.3
b. 20.3<µ<23.5
c. 20.7<µ<23.1
d. 20.9<µ<22.9
21.Findthep-valueforatwo-tailedtestofmeanswithn=29andateststatistict=2.743.
a. 0.0105
b. 0.0210
c. 0.0053
d. Noneofthese
ACTM–StatisticsRegionalExamMarch5,2016
22.Theaccuracyofverbalresponsesistestedinanexperimentinwhichindividualsreporttheirheights
andthenaremeasured.Thedataconsistofthereportedheightandmeasuredheightforeach
individual.Whattypeofsampleisthis?
a. Independentsample
b. Dependentsample
c. Codependentsample
d. Clustersample
23.Thepaireddatabelowconsistofthetemperaturesonrandomlychosendaysandtheamounta
certainkindofplantgrew(inmillimeters).Findthevalueofthelinearcorrelationcoefficientr.
Temp. 62 76 50 51 71 46 51 44 79
Growth 36 39 50 13 33 33 17 6 16
a. -0.210
b. 0
c. 0.196
d. 0.256
24.Sixpairsofdatayieldr=0.444andtheregressionequation𝑦 = 5𝑥 + 2.Also,𝑦 = 18.3.Whatisthe
bestpredictedvalueofyforx=5?Useasignificancelevelof0.05andreferencethetableofcriticalr
valuesfortestingHo:𝜌 = 0.
Samplesizen Criticalrvaluesforα=0.05
4
0.950
5
0.878
6
0.811
7
0.754
a. 93.5
b. 4.22
c. 18.3
d. 27
25.Completethesentence.Thetimeittakesforthebustoarriveatthebusstopis________________.
a.adiscreterandomvariable.
b.acontinuousrandomvariable.
c.notarandomvariable.
d.arandomvariable,butitisnotdiscreteorcontinuous.
ACTM–StatisticsRegionalExamMarch5,2016
Name:_________________________________________
TIEBREAKERS
1. Fourteenpeoplewhoseteethwerethoroughlycleanedandpolishedwererandomlyassigned
totwogroupsofsevensubjectseach.Bothgroupswereassignedtouseoralrinses(no
brushing)fora2-weekperiod.Group1usedarinsethatcontainedanantiplaqueagent.Group
2,thecontrolgroup,receivedasimilarrinseexceptthat,unknowntothesubjects,therinse
containednoantiplaqueagent.Aplaqueindexx,ameasureofplaquebuildup,wasrecordedat
4,7and14days.Themeanandstandarddeviationforthe14dayplaquemeasurementsare
showninthetableforthetwogroups.
Control
Antiplaque
Group
SampleSize
7
7
Mean
1.26
0.78
Standard
0.32
0.32
Deviation
a. Statethenullandalternativehypothesesthatshouldbeusedtotesttheeffectivenessof
theantiplaqueoralrinse.
b. Dothedataprovideevidencetoindicatethattheoralantiplaquerinseiseffective?Test
using𝛼 = 0.05.
ACTM–StatisticsRegionalExamMarch5,2016
Name:_____________________________________________
2. Supposeyouwishtocomparethemeansoffourpopulationsbasedonindependentrandom
samples,eachofwhichcontainssixobservations.
A. CompletetheANOVAtablebelow.
Source
df
SS
MS
F
Treatments
339.8
Error
-
Total
473.2
-
-
B. HowmanydegreesoffreedomareassociatedwiththeFstatisticfortesting𝐻! : 𝜇! = 𝜇! =
𝜇! = 𝜇! ?
C. Dothedataprovidesufficientevidencetoindicatedifferencesamongthepopulation
means?(Use𝛼 = 0.05.)
3. PreviousgeologicalstudiesindicatethatexploratoryoilwellsdrilledintheFayettevilleshale
regionhavea40%chanceofstrikingoil.Assumethewellsbeingdrilledareindependentofone
another.LetXbethenumberofwellsdrilleduptoandincludingthewellwherestrikingoilfirst
occurs.Usingthisinformation,answerthequestionsbelow.
A. WhatistheprobabilityfunctionforX?(Specifythedistributionandtherangeofpossiblevalues
ofX.)
B. Whatistheprobabilityofstrikingoilforthefirsttimeafterthesecondwellisdrilled?
ACTM–StatisticsRegionalExamMarch5,2016
ACTMStatisticsExamSolutions
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TieBreakers
1. a.
Iftheantiplaquerinseiseffective,theplaquebuildupshouldbelessforthegroupusing
therinse.Thus𝐻! : 𝜇! − 𝜇! = 0, 𝐻! : 𝜇! − 𝜇! > 0Where1isfromthecontrolgroupand2is
fromtheantiplaquegroup.
b.
TestStatistic:𝑡 = 2.806
CriticalValue:𝑡! = 1.782(Reject𝐻! )
p-value=0.0079
Reject𝐻! .Thereisevidencetoindicatetherinseiseffective.
2. a.
Source
Treatments
Error
Total
df
3
20
23
SS
339.8
133.4
473.2
MS
113.2667
6.67
F
16.981
ACTM–StatisticsRegionalExamMarch5,2016
b.Thedegreesoffreedomforthistestare𝑑𝑓! = 3and𝑑𝑓! = 20.
c.Therejectionregionforthistestis𝐹!" > 3.10Yes,Theteststatistic,16.981fallsintherejection
regionsoreject𝐻! .Thereisevidencetoconcludetherearedifferencesinthepopulationmeans.
3. a.P(X=x)=f(x)=0.4(0.6)x-1forx≥1.
B.P(X>2)=1-P(X≤2)=1-[f(1)+f(2)]=1-[0.4+0.4*0.6]=0.36or36%chance