Name:

Class:

Date:

Chapter11-InferencesAboutPopulationVariances

CopyrightCengageLearning.PoweredbyCognero.

Page

PAGE

MultipleChoice

1.

Thesymbolusedforthevarianceofthepopulationis

a.

σ.

b.

σ2.

c.

s.

d.

s2.

ANSWER:

b

2.

Thesymbolusedforthevarianceofthesampleis

a.

σ.

b.

σ2.

c.

s.

d.

s2.

ANSWER:

d

3.

Thesamplingdistributionusedwhenmakinginferencesaboutasinglepopulation'svarianceis

a.

anFdistribution.

b.

atdistribution.

c.

achi-squaredistribution.

d.

anormaldistribution.

ANSWER:

c

4.

Whichofthefollowinghasachi-squaredistribution?

a.

(n-1)s2/σ2.

b.

(n-1)σ/s.

c.

(n-1)s/σ.

d.

(n-1)s2/σ2.

ANSWER:

a

5.

Asampleof61elementsisselectedtoestimatea95%confidenceintervalforthevarianceofthepopulation.Thechi-squarevaluestobeusedforthisintervalestimationare

a.

40.482and83.298.

b.

32.357and71.420.

c.

34.764and67.505.

d.

43.188and79.082.

ANSWER:

a

6.

Weareinterestedintestingwhetherthevarianceofapopulationissignificantlylessthan1.44.Thenullhypothesisforthistestis

a.

H0:σ2<1.44.

b.

H0:s2≥1.44.

c.

H0:σ<1.20.

d.

H0:σ2≥1.44.

ANSWER:

d

7.

Asampleof51observationsyieldedasamplestandarddeviationof5.IfwewanttotestH0:σ2=20,theteststatisticis

a.

62.5.

b.

12.5.

c.

51.25.

d.

50.

ANSWER:

a

8.

Asampleof31observationsyieldedasamplevarianceof16.IfwewanttotestH0:σ2=16,theteststatisticis

a.

20.

b.

21.

c.

30.

d.

7.5.

ANSWER:

c

9.

WhichofthefollowinghasanFdistribution?

a.

(n-1)s/σ.

b.

s1/s2.

c.

(n-1)s1/s2.

d.

.

ANSWER:

d

10.

ThevalueofF.05with8numeratorand19denominatordegreesoffreedomis

a.

2.48.

b.

2.58.

c.

3.63.

d.

2.96.

ANSWER:

a

11.

ToavoidtheproblemofnothavingaccesstoTablesofFdistributionwhenFvaluesareneededforthelowertail,thenumeratoroftheteststatisticforatwo-tailedtestshould

betheonewith

a.

thelargersamplesize.

b.

thesmallersamplesize.

c.

thelargersamplevariance.

d.

thesmallersamplevariance.

ANSWER:

c

12.

Therandomvariableforachi-squaredistributionmayassume

a.

anyvaluebetween-1to1.

b.

anyvaluebetween-∞to+∞.

c.

anynegativevalue.

d.

anyvaluegreaterthanzero.

ANSWER:

d

13.

Asampleofnobservationsistakenfromanormalpopulation.Theappropriatechi-squaredistributionhas

a.

ndegreesoffreedom.

b.

n-1degreesoffreedom.

c.

n-2degreesoffreedom.

d.

n-3degreesoffreedom.

ANSWER:

b

14.

ForanFdistribution,thenumberofdegreesoffreedomforthenumerator

a.

mustbelargerthanthenumberofdegreesoffreedomforthedenominator.

b.

mustbesmallerthanthenumberofdegreesoffreedomforthedenominator.

c.

mustbeequaltothenumberofdegreesoffreedomforthedenominator.

d.

canbelarger,smaller,orequaltothenumberofdegreesoffreedomforthedenominator.

ANSWER:

d

15.

Thebottlerofacertainsoftdrinkclaimstheirequipmenttobeaccurateandthatthevarianceofallfilledbottlesisatmost.05.Thenullhypothesisinatesttoconfirmtheclaimwouldbewrittenas

a.

H0:σ2≥

.05.

b.

H0:s2≥.05.

c.

H0:s2≤.05.

d.

H0:σ2≤

.05.

ANSWER:

d

16.

Asampleof20cansoftomatojuiceshowedastandarddeviationof.4ounces.A95%confidenceintervalestimateofthevarianceforthepopulationis

a.

.2313to.8533.

b.

.2224to.7924.

c.

.3042to.5843.

d.

.0925to.3413.

ANSWER:

d

17.

Themanageroftheservicedepartmentofalocalcardealershiphasnotedthattheservicetimesofasampleof16newautomobileshasastandarddeviationof5minutes.A95%confidenceintervalestimateforthevarianceofservicetimesforalltheirnewautomobilesis

a.

13.642to59.885.

b.

9.46to34.09.

c.

2.144to9.948.

d.

2.728to11.977.

ANSWER:

a

18.

Themanageroftheservicedepartmentofalocalcardealershiphasnotedthattheservicetimesofasampleof30newautomobileshasastandarddeviationof5minutes.A95%confidenceintervalestimateforthestandarddeviationoftheservicetimes(inminutes)foralltheirnewautomobilesis

a.

16.047to45.722.

b.

15.857to45.180.

c.

3.982to6.722.

d.

22.833to65.059.

ANSWER:

c

19.

Theproducerofacertainmedicineclaimsthattheirbottlingequipmentisveryaccurateandthatthestandarddeviationofalltheirfilledbottlesis.2ouncesorless.Asampleof20bottlesshowedastandarddeviationof.12ounces.Theteststatistictotesttheclaimis

a.

2.3.

b.

22.99.

c.

6.84.

d.

1.368.

ANSWER:

c

20.

Theproducerofacertainbottlingequipmentclaimsthatthevarianceofalltheirfilledbottlesis.027orless.Asampleof30bottlesshowedastandarddeviationof.2ounces.Thep-valueforthetestis

a.

between.025to.05.

b.

between.05to.10.

c.

.05.

d.

.025.

ANSWER:

a

21.

Weareinterestedintestingtoseeifthevarianceofapopulationislessthan7.Thecorrectnullhypothesisis

a.

σ<7.

b.

σ2≥7.

c.

σ

<49.

d.

σ2

≥49.

ANSWER:

b

22.

Arandomsampleof31salescharge

showedasamplestandarddeviationof$50.A90%confidenceintervalestimateofthepopulationstandarddeviationis

a.

1715.101to4055.589.

b.

1596.458to4466.679.

c.

39.956to66.833.

d.

41.393to63.684.

ANSWER:

d

23.

Forasamplesizeof21at95%confidence,thechi-squarevaluesneededforintervalestimationare

a.

9.591and34.170.

b.

2.700and19.023.

c.

8.260and37.566.

d.

10.283and35.479.

ANSWER:

a

24.

Thechi-squarevalueforaone-tailed(uppertail)hypothesistestata5%levelofsignificanceandasamplesizeof25is

a.

33.196.

b.

36.415.

c.

39.364.

d.

37.652.

ANSWER:

b

25.

Thechi-squarevalueforaone-tailed(lowertail)test

whenthelevelofsignificanceis.1andthesamplesizeis15is

a.

21.064.

b.

23.685.

c.

7.790.

d.

6.571.

ANSWER:

c

26.

ThecriticalvalueofF

using

α=.05

whenthereisasamplesizeof21forthesamplewiththesmallervariance,andthereisasamplesizeof9forthesamplewiththelargersamplevarianceis

a.

2.45.

b.

2.94.

c.

2.37.

d.

2.10.

ANSWER:

a

27.

Thesamplingdistributionofthequantity(n

-1)s2/σ2isthe

a.

chi-squaredistribution.

b.

normaldistribution.

c.

Fdistribution.

d.

tdistribution.

ANSWER:

a

28.

Thesamplingdistributionoftheratioofindependentsamplevariancesextractedfromtwonormalpopulationswithequalvariancesisthe

a.

chi-squaredistribution.

b.

normaldistribution.

c.

Fdistribution.

d.

tdistribution.

ANSWER:

c

29.

The95%confidenceintervalestimateofapopulationvariancewhenasamplevarianceof324isobtainedfromasampleof81itemsis

a.

14.14to174.94.

b.

243.086to453.520.

c.

16.42to194.35.

d.

254.419to429.203.

ANSWER:

b

30.

The99%confidenceintervalestimateforapopulationvariancewhenasamplestandarddeviationof12isobtainedfromasampleof10itemsis

a.

4.589to62.253.

b.

46.538to422.171.

c.

54.941to746.974.

d.

62.042to562.895.

ANSWER:

c

31.

The90%confidenceintervalestimateofapopulationstandarddeviationwhenasamplevarianceof50isobtainedfromasampleof15itemsis

a.

26.8to124.356.

b.

5.177to11.152.

c.

5.436to10.321.

d.

29.555to106.529.

ANSWER:

c

32.

Asampleof60itemsfrompopulation1hasasamplevarianceof12whileasampleof50itemsfrompopulation2hasasamplevarianceof15.Ifwewanttotestwhetherthevariancesofthetwopopulationsareequal,theteststatisticwillhaveavalueof

a.

.8.

b.

1.56.

c.

1.5.

d.

1.25.

ANSWER:

d

33.

ThevalueofF.01with9numeratorand20denominatordegreesoffreedomis

a.

2.39.

b.

2.94.

c.

2.91.

d.

3.46.

ANSWER:

d

34.

Asampleof21elementsisselectedtoestimatea90%confidenceintervalforthevarianceofthepopulation.Thechi-squarevalue(s)tobeusedforthisintervalestimationis(are)

a.

31.410.

b.

12.443.

c.

10.851and31.410.

d.

12.443and28.412.

ANSWER:

c

35.

Weareinterestedintestingwhetherthevarianceofapopulationissignificantlymorethan625.Thenullhypothesisforthistestis

a.

H0:σ2>625.

b.

H0:σ2<625.

c.

H0:σ2>625.

d.

H0:σ2<25.

ANSWER:

b

36.

Asampleof61observationsyieldedasamplestandarddeviationof6.IfwewanttotestH0:

σ2

=

40,theteststatisticis

a.

54.

b.

9.15.

c.

54.90.

d.

9.

ANSWER:

a

37.

Thecontentsofasampleof26cansofapplejuiceshowedastandarddeviationof.06ounces.Weareinterestedintestingwhetherthevarianceofthepopulationissignificantlymorethan.003.Thenullhypothesisis

a.

s2>.003.

b.

s2≤

.003.

c.

σ2>.003.

d.

σ2≤

.003.

ANSWER:

d

38.

Thecontentsofasampleof26cansofapplejuiceshowedastandarddeviationof.06ounces.Weareinterestedintestingwhetherthevarianceofthepopulationissignificantlymorethan.003.Theteststatisticis

a.

1.2.

b.

31.2.

c.

30.

d.

500.

ANSWER:

c

39.

Thecontentsofasampleof26cansofapplejuiceshowedastandarddeviationof.06ounces.Weareinterestedintestingwhetherthevarianceofthepopulationissignificantlymorethan.003.Thep-valueforthistestis

a.

.05.

b.

greaterthan.10.

c.

lessthan.10.

d.

zero.

ANSWER:

b

40.

Thecontentsofasampleof26cansofapplejuiceshowedastandarddeviationof.06ounces.Weareinterestedintestingwhetherthevarianceofthepopulationissignificantlymorethan.003.Atthe.05levelofsignificance,thenullhypothesis

a.

shouldberejected.

b.

shouldnotberejected.

c.

shouldberevised.

d.

shouldnotbetested.

ANSWER:

b

41.

Basedonthesampleevidencebelow,wewanttotestthehypothesisthatpopulationAhasalargervariancethanpopulationB.

SampleA

SampleB

n

11

10

s2

12.1

5

?

Theteststatisticforthisproblemequals

a.

.4132.

b.

1.1.

c.

2.42.

d.

2.

ANSWER:

c

42.

Basedonthesampleevidencebelow,wewanttotestthehypothesisthatpopulationAhasalargervariancethanpopulationB.

?

SampleA

SampleB

n

11

10

s2

12.1

5

?

Thep-valueisapproximately

a.

.10.

b.

.05.

c.

.025.

d.

.01.

ANSWER:

a

43.

Considerthefollowinghypothesisproblem.

n=23

s2=60

H0:σ2≤66

Ha:σ2>66

?

Theteststatistichasavalueof

a.

20.91.

b.

24.20.

c.

24.00.

d.

20.00.

ANSWER:

d

44.

Considerthefollowinghypothesisproblem.

?

n=23

s2=60

H0:σ2≤66

Ha:σ2>66

?

Ifthetestistobeperformedatthe5%level,thecriticalvalue(s)fromthechi-squaredistribution

tableis(are)

a.

10.982and36.781.

b.

12.338and33.924.

c.

12.338.

d.

33.924.

ANSWER:

d

45.

Considerthefollowinghypothesisproblem.

?

n=23

s2=60

H0:σ2≤66

Ha:σ2>66

?

Thep-valueis

a.

lessthan.025.

b.

between.025and.05.

c.

between.05and.10.

d.

greaterthan.10.

ANSWER:

d

46.

Considerthefollowinghypothesisproblem.

?

n=23

s2=60

H0:σ2≤66

Ha:σ2>66

?

Ifthetestistobeperformedatthe.05levelofsignificance,thenullhypothesis

a.

shouldberejected.

b.

shouldnotberejected.

c.

shouldberevised.

d.

shouldnotbetested.

ANSWER:

b

47.

?Inpractice,themostfrequentlyencounteredhypothesistestaboutapopulationvarianceisa

a.

?one-tailedtest,withrejectionregioninthelowertail.

b.

?one-tailedtest,withrejectionregionintheuppertail.

c.

?two-tailedtest,withequal-sizerejectionregions.

d.

?two-tailedtest,withunequal-sizerejectionregions.

ANSWER:

b

48.

ToavoidtheproblemofnothavingaccesstotablesoftheFdistributionwhenaone-tailedtestisrequiredandwithFvaluesgivenforthelowertail,let?the

a.

smallersamplevariancebethenumeratoroftheteststatistic.

b.

largersamplevariancebethenumeratoroftheteststatistic.

c.

samplevariancefromthepopulationwiththesmallerhypothesizedvariancebethenumeratoroftheteststatistic.

d.

samplevariancefromthepopulationwiththelargerhypothesizedvariancebethenumeratoroftheteststatistic.

ANSWER:

c

49.

Asampleof28elementsisselectedtoestimatea95%confidenceintervalforthevarianceofthepopulation.Thechi-squarevaluestobeusedforthisintervalestimationare?

a.

11.808and49.645.

b.

?14.573and43.195.

c.

?16.151and40.113.

d.

?15.308and44.461.

ANSWER:

b

50.

=8.231indicatesthat?

a.

?97.5%ofthechi-squarevaluesarelessthan8.231.

b.

?97.5%ofthechi-squarevaluesaregreaterthan8.231.

c.

?2.5%ofthechi-squarevaluesaregreaterthan8.231.

d.

5%ofthechi-squarevaluesareequalto8.231.

ANSWER:

b

51.

Thereisa.90probabilityofobtainingavaluesuchthat?

a.

?

<<?.

b.

≤≤?.

c.

?

≤≤?.

d.

?

<<?.

ANSWER:

c

52.

?Considerthefollowinghypothesisproblem.

n=30

H0:σ2=500

s2=625

Ha:σ2≠500

?

?Theteststatisticequals

a.

?23.20.

b.

?24.00.

c.

?36.25.

d.

?37.50.

ANSWER:

c

53.

?Considerthefollowinghypothesisproblem.

n=30

H0:σ2=500

s2=625

Ha:σ2≠500

?

?Thenullhypothesisistobetestedatthe5%levelofsignificance.Thecriticalvalue(s)fromthechi-squaredistribution

tableis(are)

a.

?42.557.

b.

?43.773.

c.

?16.047and45.722.

d.

?16.791and46.979.

ANSWER:

c

54.

?Considerthefollowinghypothesisproblem.

n=30

H0:σ2=500

s2=625

Ha:σ2≠500

?

Atthe5%levelofsignificance,the

nullhypothesis

a.

?shouldberejected.

b.

?shouldnotberejected.

c.

?shouldberevised.

d.

shouldnotbetested.

ANSWER:

b

55.

?Considerthefollowinghypothesisproblem.

n=14

H0:σ2<410

s=20

Ha:σ2>410

?

?Theteststatisticequals

a.

.63.

b.

?12.68.

c.

?13.33.

d.

?13.68.

ANSWER:

b

56.

?Considerthefollowinghypothesisproblem.

n=14

H0:σ2<410

s=20

Ha:σ2>410

?

?Thenullhypothesisistobetestedatthe5%levelofsignificance.Thecriticalvalue(s)fromthechi-squaredistribution

tableis(are)

a.

22.362.

b.

?23.685.

c.

?5.009and24.736.

d.

?5.629and26.119.

ANSWER:

a

57.

?Considerthefollowinghypothesisproblem.

n=14

H0:σ2<410

s=20

Ha:σ2>410

?

Atthe5%levelofsignificance,thenullhypothesis

a.

shouldberejected?.

b.

?shouldnotberejected?.

c.

?shouldberevised.

d.

shouldnotbetested.

ANSWER:

b

58.

?Considerthefollowingsample

information

fromPopulationAandPopulationB.

?

?

SampleA

SampleB

n

24

16

s2

32

38

??

Wewanttotestthehypothesisthatthepopulationvariancesareequal.Theteststatisticforthisproblemequals

a.

?.67.

b.

.84.

c.

?1.19.

d.

?1.50.

ANSWER:

c

59.

?ConsiderthefollowingsampleinformationfromPopulationAandPopulationB.

?

?

SampleA

SampleB

n

24

16

s2

32

38

??

Wewanttotestthehypothesisthatthepopulationvariancesareequal.Thenullhypothesisistobetestedatthe10%levelofsignificance.ThecriticalvaluefromtheFdistribution

tableis

a.

2.11.

b.

?2.13.

c.

?2.24.

d.

?2.29.

ANSWER:

b

60.

?ConsiderthefollowingsampleinformationfromPopulationAandPopulationB.

?

?

SampleA

SampleB

n

24

16

s2

32

38

??

Wewanttotestthehypothesisthatthepopulationvariancesareequal.Atthe10%levelofsignificance,thenullhypothesis

a.

shouldberejected.

b.

?shouldnotberejected.

c.

?shouldberevised.

d.

shouldnotbetested.

ANSWER:

b

61.

?Whichofthefollowingrejectionrulesisproper?

a.

?RejectH0ifp-value≤Fα.

b.

?RejectH0ifF

≤

Fα/2.

c.

?RejectH0ifp-value

≥α/2.

d.

?RejectH0ifF

≥

Fα.

ANSWER:

d

SubjectiveShortAnswer

62.

Asampleof14itemsprovidesasamplemeanof20andasamplevarianceof18.Computea95%confidenceintervalestimateforthestandarddeviationofthepopulation.

ANSWER:

3.076to6.835

63.

Asampleof30itemsprovidedasamplemeanof28andasamplestandarddeviationof6.Testthefollowinghypothesesusingα=.05.Whatisyourconclusion?

H0:σ2≤25

Ha:σ2>25

ANSWER:

?

χ2

=41.76<42.557;donotrejectH0;thereisnotsufficientsampleevidencetoconcludethatthepopulationvarianceismorethan25.

64.

Weareinterestedindeterminingwhetherornotthevariancesofthesalesareequalfortwosmallgrocerystores.Asampleof16daysofsalesateachstoreindicatedthefollowing.

StoreA

StoreB

n1=16

n2=16

s1=$130

s2=$100

?

Arethevariancesofthepopulations(fromwhichthesesamplescame)equal?

Useα=.05.

ANSWER:

?

F=1.69<2.86

p-value>.10;donotrejectH0;itcouldnotbeconcludedthatthevariancesofthetwopopulationsaredifferent.

65.

Arandomsampleof22employeesofalocalutilityfirmshowedthattheirmonthlyincomeshadasamplestandarddeviationof$110.Providea90%confidenceintervalestimateforthestandarddeviationoftheincomesofallemployeesofthefirm.

ANSWER:

88.190to148.061

66.

Arandomsampleof41scoresofstudentstakingtheACTtestshowedastandarddeviationof8points.Providea98%confidenceintervalestimateforthestandarddeviationofalltheACTtestscores.

ANSWER:

6.34to10.75(rounded)

67.

Alumbercompanyhasclaimedthatthestandarddeviationforthelengthsoftheirsix-footboardsis.4inchesorless.Totesttheirclaim,arandomsampleof16six-footboardsisselected;anditisdeterminedthatthevarianceofthesampleis.3.Dotheresultsofthesamplesupportthecompany'sclaim?Useα=.1.

ANSWER:

Sinceχ2=28.125>22.307,rejectH0;

no,thesampleresultsdonotsupportthecompany'sclaim.

68.

AneggpackingcompanyhasstatedthatthestandarddeviationoftheweightsoftheirgradeAlargeeggsis.07ouncesorless.Thesamplevariancefor51eggswas.0065ounces.Canthissampleresultconfirmthecompany'sclaim?

Useα=.1.

ANSWER:

?

Sinceχ2=66.33>63.167(p-valueisbetween.05and.10),rejectH0;no,thesampleresultcannotconfirmthecompany'sclaim.

69.

ThestandarddeviationofthedailytemperaturesinHonolululastyearwas4degreesFahrenheit.Arandomsampleof25daysresultedinastandarddeviationof5.8degreesFahrenheit.Hastherebeenasignificantincreaseinthevarianceofthetemperatures?Useα=.05.

ANSWER:

Sinceχ2=50.46>36.415(p-valueislessthan.005),rejectH0;therehasbeenasignificantincreaseinthevarianceofthetemperaturesinHonolulu.

?

70.

Atα=.1,testtoseeifthepopulationvariancesfromwhichthefollowingsamplesweredrawnareequal.

Sample1

Sample2

n1=21

n2=19

s1=18

s2=16

ANSWER:

SinceF=1.27<2.19(p-valueisgreaterthan.1),donotrejectH0.

71.

Thestandarddeviationoftheagesofasampleof16executivesfromthenorthernstateswas8.2years;whilethestandarddeviationoftheagesofasampleof25executivesfromthesouthernstateswas12.8years.Atα=.1,testtoseeifthereisanydifferencebetweenthevariancesoftheagesofallthenorthernandallthesouthernexecutives.

ANSWER:

?

SinceF=2.44>2.28(p-valueisbetween.05and.1),rejectH0.

72.

Studentadvisorsareinterestedindeterminingifthevariancesofthescoresofdaystudentsandnightstudentsarethesame.Thefollowingsamplesaredrawn:

Day

Night

n1=25

n2=31

s1=9.8

s2=14.7

?

Testtheequalityofthevariancesofthepopulations.Let

α=.05.

ANSWER:

SinceF=2.25>2.21(p-valueisbetween.02and.05),

rejectH0.

?

73.

Thescoresofasampleof5students,selectedfromalargepopulation,aregivenbelow.

Score

70

80

60

90

75

?

a.

Determineapointestimateforthevarianceofthepopulation.

b.

Determinea98%confidenceintervalforthevarianceofthepopulation.

c.

Atthe5%levelofsignificance,testtodetermineifthevarianceofthepopulationislessthanorequalto50.

ANSWER:

?

a.

125

b.

37.659to1683.502

c.

H0:σ2≤50

Ha:σ2>50

χ2=10>9.488(p-valueisbetween.025and.05);

rejectH0

74.

Amachineproducespipesusedinairplanes.Theaveragelengthofthepipeis16inches.Theacceptablevarianceforthelengthis.3inches.Asampleof25pipeswastaken.Theaveragelengthinthesamplewas15.95incheswithavarianceof.4inches.

a.

Constructa95%confidenceintervalforthepopulationvariance.

b.

Statethenullandalternativehypothesestotestwhetherornotthepopulationvarianceequals.3inches.

c.

Computetheteststatistic.

d.

Thenullhypothesisistobetestedatthe5%levelofsignificance.Statethedecisionruleforthetestusingthecriticalvalueapproach.

e.

Whatdoyouconcludeaboutthepopulationvariance?

ANSWER:

?

a.

.244to.774

b.

H0:σ2=.3

Ha:σ2≠

.3

c.

32

d.

DonotrejectH0if12.401<chi-square<39.364

RejectH0ifchi-square≥39.364orifchi-square≤12.401

e.

DonotrejectH0;noevidencetoconcludethatthepopulationvarianceisdifferentfrom.3inches.

75.

Youaregiventhefollowingresultsfromasampleof6observations.

4

6

3

4

3

10

?

a.

Determinethemeanandthestandarddeviationofthissample.

b.

Constructa98%confidenceintervalforthepopulationvariance.

ANSWER:

?

a.

5and2.68

b.

2.386to64.982

76.

Itiscrucialthatthevariance(measuredinmin2)ofaproductionprocessbelessthanorequalto25.Asampleof22istaken.Thesamplevarianceequaled26.

a.

Constructa90%confidenceintervalforthepopulationvariance.

b.

Constructa90%confidenceintervalforthepopulationstandarddeviation.

c.

Statethenullandalternativehypothesestobetested.

d.

Computetheteststatistic.

e.

Thenullhypothesisistobetestedatthe10%levelofsignificance.Usingthecriticalvalueapproach,statethedecisionruleforthetest.

f.

Whatdoyouconcludeaboutthepopulationvariance?

ANSWER:

?

a.

16.712to47.106

b.

4.088to6.863

c.

H0:σ2≤25

Ha:σ2>25

d.

21.84

e.

RejectH0ifchi-square≥29.615

f.

DonotrejectH0;noevidencetoconcludethatthepopulationvarianceoftheproductionprocessismorethan25.

77.

Ithasbeensuggestedthatnight-shiftworkersshowmorevariabilityintheiroutputlevels(units/week)thandayworkers.Below,youaregiventheresultsoftwoindependentrandomsamples.

NightShift

DayShift

SampleSize

9

8

SampleMean

520

540

SampleVariance

38

20

?

a.

Statethenullandalternativehypothesestobetested.

b.

Computetheteststatistic.

c.

Determinethep-value.

d.

Atthe.05levelofsignificance,whatdoyouconclude?

ANSWER:

?

a.

H0:

≤

Ha:

>

wherepopulation1isthenight-shiftgroup

b.

1.9

c.

Thep-valueisgreaterthan.1.

d.

Donotrejectthenullhypothesis.

Thereisnotsufficientevidencetoconcludethatthevariabilityismorewithnight-shiftworkers.

78.

Thepresidentofabankbelievesthatthevarianceofthedepositsofsuburbancustomersismorethanthevarianceofcitycustomers.Belowyouaregiventheresultsofsamplestakenfromsuburbanandcitycustomers.

?

SuburbanCustomers

CityCustomers

n

21

19

s

$150

$90

?

a.

Statethenullandalternativehypothesestobetested.

b.

Computetheteststatistic.

c.

Determinethep-value.

d.

Usingα=.05,whatdoyouconclude?

ANSWER:

?

a.

H0:

≤

Ha:

>

b.

2.78

c.

p-valueisbetween.01and.025

d.

RejectH0;thevariabilityinthedepositsismoreforsuburbancustomers.

79.

Arandomsampleof21checkingaccountsatabankshowedanaveragedailybalanceof$430withastandarddeviationof$50.

a.

Providea95%confidenceintervalestimateforthevarianceofthepopulationofthecheckingaccounts.

b.

Providea95%confidenceintervalestimateforthestandarddeviationofthepopulationofthecheckingaccounts.

ANSWER:

?

a.

1463.272to5213.221

b.

38.253to72.203

80.

Alargecompanyhasclaimedthatthestandarddeviationofthemonthlyincomesoftheiremployeesislessthanorequalto$120.Totesttheirclaim,arandomsampleof76employeesofthecompanywastaken;anditwasdeterminedthatthestandarddeviationoftheirincomeswas$135.Usingα=.10,testthecompany'sclaim.

ANSWER:

?

H0:σ2≤14400

Ha:σ2>14400

chi-square=94.92>91.061;p-valueisbetween.05and.1;rejectH0

81.

Asampleof16studentsshowedthatthevarianceinthenumberofhourstheyspendstudyingis25.Atthe5%levelofsignificance,testtoseeifthevarianceofthepopulationissignificantlydifferentfrom30.Usethecriticalvalueapproach.

ANSWER:

?

H0:σ2=30

Ha:σ2≠30

chi-square=12.5;criticalvalues:6.262and27.488;

donotrejectH0

82.

Weareinterestedindeterminingwhetherornotthevariancesofthestartingsalariesofaccountingmajorsissignificantlydifferentfrommanagementmajors.Thefollowinginformationwasgatheredfromtwosamples.

Accounting

Management

SampleSize

21

18

AverageMonthlyIncome

$3600

$3500

Variance

$900

$400

?

Atthe.1levelofsignificance,testtodeterminewhetherornotthevariancesareequal.

ANSWER:

?

H0:

=

Ha:

≠

F=2.25>2.23;rejectH0;p-valueisbetween.05and.1

83.

Theaveragescoreofasampleof30studentsontheirsecondstatisticsexaminationwas85withastandarddeviationof11.Isthevarianceofthepopulationsignificantlymorethan80?Usea.05levelofsignificance.

ANSWER:

?

H0:σ2≤80

Ha:σ2>80

chi-square=43.86>42.557(p-valueisbetween.025and.05);

rejectH0

Thepopulationvarianceissignificantlymorethan80.

84.

Acompanyclaimsthatthestandarddeviationintheirdeliverytimesislessthan5days.Asampleof27pastcustomerswastaken.Theaveragedeliverytimeinthesamplewas14dayswithastandarddeviationof4.5days.Atthe5%levelofsignificance,testthecompany'sclaim.Usethecriticalvalueapproach.

ANSWER:

?

H0:σ2≥25

Ha:σ2<25

chi-square=21.06>15.379;donotrejectH0

85.

Asampleof2