Chapter 8 Review - Weeblychristensen60.weebly.com/.../4/54343837/chapter_8_review.pdfChapter 8...

AP STATISTICS

Chapter 8 Review

8.1 Objectives:ü Distinguish between a statistic and a population characteristicü Understand the definition of a statisticü Understand the definition of sampling variabilityü Understand the definition of a sampling distribution

8.1 Objectives:ü Have a sense that all statistics have sampling distributions

8.2 Objectives:ü Understand the general properties of the sampling distribution of the sample meanü Understand the conditions for applying the Central Limit Theorem

8.2 Objectives:ü Be able to use the Central Limit Theorem and the normal curve probability calculations to find the the probability that the sample mean falls in a given interval when the sample size is large.ü Have a sense the the Central Limit Theorem can be applied if nexceeds 30

8.3 Objectives:ü Understand the general properties of the sampling distribution of the sample proportionü Find the mean and standard deviation of the sampling distribution of given p and the sample size.

8.3 Objectives:ü Be able to use the properties of the sampling distribution of the sample proportion and the normal curve to find the probability that the sample proportion falls in a given interval when the sample size is largeü Have a sense that the Central Limit Theorem is being applied to a rescaled binomial random variable

1998 Form A FR #1

1998 Form A FR #1 Rubric

2010 Form A FR #4b

2010 Form A FR #4b Rubric

Statistic• A number that that can be computed from

sample data

• Some statistics we will use includex – sample means – standard deviationp – sample proportion

• The observed value of the statistic depends on the particular sample selected from the population and it will vary from sample to sample.

Sampling Distributions• The probability distribution of a

statistic: The sampling distribution describes the long-run behavior of the statistic.

Sampling Distributions of X• The probability distribution of the

sample mean X based on a random sample of size n. When the population is normal or the sample size is large, the sampling distribution of X is (approximately) normal.

• The mean of the sampling distribution EQUALS the mean of the population.

• As the sample size increases, the standard deviation of the sampling distribution decreases.

µx = µ

asn σx

General Properties of Sampling Distributions of X

Rule 1:

Rule 2:

This rule is exact if the population is infinite, and is approximately correct if the population is finite and no more than 10% of the population is included in the sample

µµ =x

nxσσ =

General Properties Continued . . .Rule 3:

When the population distribution is normal, the sampling distribution of X is also normal for any sample size n.

General Properties Continued . . .

Rule 4: Central Limit TheoremWhen n is sufficiently large, the sampling distribution of X is well approximated by a normal curve, even when the population distribution is not itself normal.

CLTcansafelybeappliedifn exceeds30.

Sampling Distributions of p• The probability distribution of the

sample proportion p, based on a random sample of size n. When the sample size is sufficiently large, the sampling distribution of p is approximately normal.

General Properties for Sampling Distributions of p

Rule 1:

Rule 2:

pp =µ ˆ

npp

p)1(

ˆ−

=σ

Rule 3: When n is large and p is not too near 0 or 1, the sampling distribution of pis approximately normal.

Chapter 8 Review

Review Questions

1.AccordingtointernaltestingdonebytheGet-A-Griptirecompany,themeanlifetimeoftiressoldonnewcarsis23,000miles,withastandarddeviationof2500miles.

IftheclaimbyGet-A-Gripistrue,whatisthemeanofthesamplingdistributionof forsamplesofsizen =4?


IftheclaimbyGet-A-Gripistrue,whatisthestandarddeviationofthesamplingdistributionof forsamplesofsizen =4?

𝜎"# =𝜎"𝑛=25002

= 1250


Ifthedistributionoftirelifeisapproximatelynormal,whatistheprobabilitythatthemeanofarandomsampleofsizen =4oftirelifetimeswillbelessthan20,000miles?

2.Considersamplingfromapopulationwhoseproportionofsuccessesisp =0.1.Asthesamplesize,n,increases,somecharacteristicsofthesamplingdistributionofp change.Whichofthefollowingcharacteristicswillchangeasn increases,andwhatisthenatureofthechange?

Themeanofthesamplingdistributionof

Asthesamplesizeincreases, staysthesame.


Thestandarddeviationofthesamplingdistributionof

Asthesamplesizeincreases,getssmaller.


Theshapeofthesamplingdistributionof𝑝+?

Asthesamplesizeincreases,thesamplingdistributionof𝑝+ becomeslessskewedandmorenormal.

3. Supposeweartificiallycategorizepopulationsasapproximatelynormalornot,andsamplesaslargeorsmall.Thiscategorizationresultsin4categories:

a)Smallsamplesfromanapproximatelynormalpopulationb)Largesamplesfromanapproximatelynormalpopulationc)Smallsamplesfromapopulationthatisnotapproximatelynormald)Large samplesfromapopulationthatisnotapproximatelynormal

What,ifanything,canbesaidabouttheshape ofthesamplingdistributionof foreachofthese4situations?

a) Smallsamplesfromanapproximatelynormalpopulation

Thesamplingdistributionof𝑋# willbeapproximatelynormal.

b)Largesamplesfromanapproximatelynormalpopulation

Thesamplingdistributionof𝑋# willbeapproximatelynormal

c)Smallsamplesfromapopulationthatisnotapproximatelynormal

Thesamplingdistributionof𝑋# willresembletheshapeoftheoriginalpopulation(notapproximatelynormal).

d)Largesamplesfromapopulationthatisnotapproximatelynormal

Thesamplingdistributionof𝑋# willbeapproximatelynormal

4.Considersamplingfromaskewed population.Asthesamplesize,n,increases,somecharacteristicsofthesamplingdistributionof change.Doesanincreasingsamplesizecausechangesinthecharacteristicsofthesamplingdistributionshownbelow?Ifso,specificallyhowdoesthesamplingdistributionchange?

Themeanofthesamplingdistributionof

As the sample size increases, stays the same.



As the sample size increases, gets smaller.



As the sample size increases, will decrease.


The shapeofthesamplingdistributionof

Asthesamplesizeincreases,thedistributionof𝑋#becomeslessskewedandmorenormal.

5.Considerthefollowing"population":{2,2,4,5}.Supposethatarandomsampleofsizen =2istobeselectedwithoutreplacement fromthispopulation.Thereare6possiblesamples(sincetheorderofselectiondoesnotmatter).Computethesamplemean foreachofthesesamplesandusethatinformationtoconstructthesamplingdistributionof𝑋# .(Displayitintableform.)

The6possiblesamplesinclude2and2,2and4,2and5,2and4,2and5,4and5.

Thedistributionof 𝑋# is:

𝑋# 2 3 3.5 4.5𝑃 𝑋# 1

626

26

16

6.Somebiologistsbelievetheevolutionofhandednessislinkedtocomplexbehaviorssuchastool-use.Underthistheory,handednesswouldbegeneticallypassedonfromparentstochildren.Thatis,left-handedparentswouldbemorelikelytohaveleft-handedchildrenthanright-handedparents.Analternatetheoryassertsthathandednessshouldberandom,withleft- andright-handednessequallylikely.Inarecentstudyusingasimplerandomsampleofn =76right-handedparents,50ofthechildrenbornwereright-handed.(p =0.658.)Supposehandednessisarandomoccurrencewitheitherhandequallylikelytobedominant,implyingthattheprobabilityofaright-handedoffspringisp =0.50.

Inarecentstudyusingasimplerandomsampleofn =76right-handedparents,50ofthechildrenbornwereright-handed.(p =0.658.)Supposehandednessisarandomoccurrencewitheitherhandequallylikelytobedominant,implyingthattheprobabilityofaright-handedoffspringis p=0.50.

Showthatitisreasonabletoapproximatethesamplingdistributionof usinganormaldistribution.

Inarecentstudyusingasimplerandomsampleofn =76right-handedparents,50ofthechildrenbornwereright-handed.(p =0.658.)Supposehandednessisarandomoccurrencewitheitherhandequallylikelytobedominant,implyingthattheprobabilityofaright-handedoffspringis p=0.50.Assumingleft- andright-handedchildrenareequallylikelyfromright-handedparents,whatistheprobabilityofobservingasampleproportionofatleast𝑝+ =0.658?𝜇0+ = 𝑝 = 0.5

𝜎0+ =𝑝(1 − 𝑝)

𝑛 =0.5(0.5)76 = 0.0574

𝑃 �̂� ≥ .658 = 𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 .658,∞, . 5,0.0574 = 0.0030

7.Youwillbeaskedtosketchcurvesrepresentingthedistributionsofasetofdata,aswellasthesamplingdistributionsofthemeanunderdifferentconditions.Youneednotgetthesegraphsperfectlycorrect,butshouldclearlyindicatedifferentaspectsofthecurves,suchaslocation,variability,andshape.

TheStateFisheriesDepartmentwishestostocktheStyxRiverwithfish,andwouldlikethespeciestonotonlysurvivebutthrive.(The"substrate"ofariverisanimportantdeterminantofthequalityofspawninghabitat.)Thepebblediametersintheriverareapproximatelynormallydistributedwithameanof24mm,andastandarddeviationof8mm.TheFisheriespeoplewillselectarandomsampleofpebblesfromtheStyxRiverinanattempttojudgetheaveragepebblesize.

Onthescalebelow,sketchcurvesrepresentingthedistributionoftheoriginalpopulationandthesamplingdistributionof𝑋# forasampleofsizen =16.

the mean will remain the same , the shape will remain normal, while the standard deviation will decrease from to , making the graph of 𝑋# skinnier and taller.

Whatistheapproximateprobabilitythatasampleofn =16fromthispopulationwouldresultinasamplemeangreaterthan30mm?

𝜇"# = 𝜇" = 24

𝜎"# =𝜎"𝑛=84

𝑃 𝑋# ≥ .30 = 𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 30,∞, 24,2 = 0.0013

Anotherissuethatmustbeconsideredinstockingariverwithfishisthatnotonlynaturalpredatorsofthefish,butfishermenandfisherwomen(fisherpersons?)willbetakingfishfromtheriver,andthusdrivingthepopulationdown.ThegraphbelowisarelativefrequencyhistogramofthelengthsofthegentleHecatefishintheStyxRiver,asrecordedduringamassivefishtaggingsurvey.Themeanlengthforthesefishwas0.8feet.

Supposeweweretotakearandomsample,n =4,fromthispopulationoffishandmeasuretheirlengths.Describetheshape,center,andspreadofthesamplingdistributionof.

Theshapeofthedistributionwillbeskewedright,butnotasmuchasthepopulation.Itwillnotbeapproximatelynormalbecausethesamplesizeissmall(n =4).Themeanofthedistributionwillremainthesameasthatofthepopulation,𝜇"# = 𝜇" = 0.8.Thestandarddeviationofthedistribution,𝜎"# =

EFG= EF

H,however,willbehalfofwhatitwasinthe

originalpopulation.

8.Thefirstlarge-scalestudyofthehumansexratioinvolvedover6,000familieseachhaving12children.(Thiswasdonein19thCenturyGermany--largefamiliesweremorecommonthenandthere.)52%ofthechildrentheyobservedwereboys.Supposethat21stCenturyresearcherswishtoreplicatethisobservationalstudy.Inordertoseeifthisproportionofboysmighthavechangedintheinterveningyears,supposetheresearcherstrackdown50familieswith12children.Fromthese600children,arandomsampleof50childrenistaken.30ofthe50childrenwereboys(p =0.6.)

52%ofthechildrentheyobservedwereboys……50familieswith12children.Fromthese600children,arandomsampleof50childrenistaken.30ofthe50childrenwereboys(p =0.6.)Showthatitisreasonabletoapproximatethesamplingdistributionof usinganormaldistribution.

np =50(.52)=26 ≥10andn(1- p)=50(1- .52)=24 ≥10

52%ofthechildrentheyobservedwereboys……50familieswith12children.Fromthese600children,arandomsampleof50childrenistaken.30ofthe50childrenwereboys(p =0.6.)Ifthemoderntruepopulationproportionofnewbornboysisp =0.52,whatistheprobabilityofobservingasampleproportionofatleast=0.6?

𝜇0+ = 𝜇0 = 0.52

𝜎0+ =𝑝(1 − 𝑝)

𝑛 =0.52(0.48)

50 = 0.0707

𝑃 �̂� ≥ .6 = 𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 .6,∞, . 52,0.0707 = 0.1289

8.32Thenicotinecontentinasinglecigaretteofaparticularbrandhasadistributionwithmean0.8mgandstandarddeviation0.1mg.If100ofthesecigarettesareanalyzed,whatistheprobabilitythattheresultingsamplemeannicotinecontentwillbelessthan0.79?lessthan0.77?X=nicotinecontentofarandomlyselectedcigarette

=averagenicotinecontentof100randomlyselectedcigarettes

𝑋# ≈ 𝑁𝑜𝑟 𝜇"# = 0.8,𝜎"# =0.1100

8.32Thenicotinecontentinasinglecigaretteofaparticularbrandhasadistributionwithmean0.8mgandstandarddeviation0.1mg.If100ofthesecigarettesareanalyzed,whatistheprobabilitythattheresultingsamplemeannicotinecontentwillbelessthan0.79?lessthan0.77?

𝑋# ≈ 𝑁𝑜𝑟 𝜇"# = 0.8,𝜎"# =0.1100

8.33Letx1,x2,...,x100 denotetheactualnetweights(inpounds)of100randomlyselectedbagsoffertilizer.Supposethattheweightofarandomlyselectedbaghasadistributionwithmean50poundsandvariance1pound2.Let 𝑋# bethesamplemeanweight(n =100).

a. Describethesamplingdistributionof𝑋#.

=averageweightof100randomlyselectedbagsoffertilizer.

𝑋# ≈ 𝑁𝑜𝑟 𝜇"# = 50,𝜎"# =1100

= 0.1

8.33Letx1,x2,...,x100 denotetheactualnetweights(inpounds)of100randomlyselectedbagsoffertilizer.Supposethattheweightofarandomlyselectedbaghasadistributionwithmean50poundsandvariance1pound2.Let bethesamplemeanweight(n =100).

b.Whatistheprobabilitythatthesamplemeanisbetween49.75poundsand50.25pounds?

8.33Letx1,x2,...,x100 denotetheactualnetweights(inpounds)of100randomlyselectedbagsoffertilizer.Supposethattheweightofarandomlyselectedbaghasadistributionwithmean50poundsandvariance1pound2.Let bethesamplemeanweight(n =100).

c.Whatistheprobabilitythatthesamplemeanislessthan50pounds?

8.34Supposethat20%ofthesubscribersofacabletelevisioncompanywatchtheshoppingchannelatleastonceaweek.Thecablecompanyistryingtodecidewhethertoreplacethischannelwithanewlocalstation.Asurveyof100subscriberswillbeundertaken.Thecablecompanyhasdecidedtokeeptheshoppingchannelifthesampleproportionisgreaterthan.25.Whatistheapproximateprobabilitythatthecablecompanywillkeeptheshoppingchannel,eventhoughtheproportionofallsubscriberswhowatchitisonly.20?

8.34 Whatistheapproximateprobabilitythatthecablecompanywillkeeptheshoppingchannel,eventhoughtheproportionofallsubscriberswhowatchitisonly.20?

�̂�=sampleproportionofsubscriberswhowatch

𝑝+ ≈ 𝑁 𝜇0+ = 0.2, 𝜎0+ =0.2(0.8)100

1 − 𝑃 𝑝+ ≥ 0.25 =

1 − 𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 0.2,∞, 0.25,0.2(0.8)100

= 0.1056

8.35Waterpermeabilityofconcretecanbemeasuredbylettingwaterflowacrossthesurfaceanddeterminingtheamountlost(ininchesperhour).Supposethatthepermeabilityindexx forarandomlyselectedconcretespecimenofaparticulartypeisnormallydistributedwithmeanvalue1000andstandarddeviation150.

normallydistributedwithmeanvalue1000andstandarddeviation150.a.Howlikelyisitthatasinglerandomlyselectedspecimenwillhaveapermeabilityindexbetween850and1300?

X=Permeabilityindexforarandomlyselectedspecimen

normallydistributedwithmeanvalue1000andstandarddeviation150.b.Ifthepermeabilityindexistobedeterminedforeachspecimeninarandomsampleofsize10,howlikelyisitthatthesamplemeanpermeabilityindexwillbebetween950and1100?𝑋#=averagepermeabilityindexforarandomlyselectedsampleof10specimens

𝑋# ≈ 𝑁𝑜𝑟 𝜇"# = 1000,𝜎"# =15010

= 0.1

normallydistributedwithmeanvalue1000andstandarddeviation150.b.Ifthepermeabilityindexistobedeterminedforeachspecimeninarandomsampleofsize10,howlikelyisitthatthesamplemeanpermeabilityindexwillbebetween850and1300?𝑋#=averagepermeabilityindexforarandomlyselectedsampleof10specimens

𝑋# ≈ 𝑁𝑜𝑟 𝜇"# = 1000,𝜎"# =15010

= 0.1

8.35Waterpermeabilityofconcretecanbemeasuredbylettingwaterflowacrossthesurfaceanddeterminingtheamountlost(ininchesperhour).Supposethatthepermeabilityindexx forarandomlyselectedconcretespecimenofaparticulartypeisnormallydistributedwithmeanvalue1000andstandarddeviation150.a.Howlikelyisitthatasinglerandomlyselectedspecimenwillhaveapermeabilityindexbetween850and1300?b.Ifthepermeabilityindexistobedeterminedforeachspecimeninarandomsampleofsize10,howlikelyisitthatthesamplemeanpermeabilityindexwillbebetween950and1100?between850and1300?

8.36Newsweek(November23,1992)reportedthat40%ofallU.S.employeesparticipatein“self-insurance”healthplans(p =.40).

a.Inarandomsampleof100employees,whatistheapproximateprobabilitythatatleasthalfofthoseinthesampleparticipateinsuchaplan?

=sampleproportionofemployeesintheplan

𝑝+ ≈ 𝑁 𝜇0+ = 0.4, 𝜎0+ =0.4(0.6)100

= 0.0490

b.Supposeyouweretoldthatatleast60ofthe100employeesinasamplefromyourstateparticipatedinsuchaplan.Wouldyouthinkp =.40foryourstate?Explain.

This tells us that, if 40% of people in a state participate in a “self-insurance” plan then it is very unlikely that more than 60 people in a sample of 100 would participate in such a plan. Therefore, if more than 60 people in a sample of 100 were found to participate in this type of plan, we would strongly doubt that for that state.

8.37Theamountofmoneyspentbyacustomeratadiscountstorehasameanof$100andastandarddeviationof$30.Whatistheprobabilitythatarandomlyselectedgroupof50shopperswillspendatotalofmorethan$5300?(Hint:Thetotalwillbemorethan$5300whenthesamplemeanexceedswhatvalue?)X=amountofmoneyspentbyacustomeratadiscountstore

=averageamountofmoneyspentbya50customersatadiscountstore

𝑋# ≈ 𝑁𝑜𝑟 𝜇"# = 100,𝜎"# =3050

= 4.243

Whatistheprobabilitythatarandomlyselectedgroupof50shopperswillspendatotalofmorethan$5300?

=averageamountofmoneyspentbya50customersatadiscountstore

𝑋# ≈ 𝑁𝑜𝑟 𝜇"# = 100,𝜎"# =3050

= 4.243

Chapter 8 Review - Weeblychristensen60.weebly.com/.../4/54343837/chapter_8_review.pdfChapter 8...

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