Chapter One-converted business research .pdf

ADVICE TO ALL EMPLOYEES

1. Build a home earlier. Be it rural home or urban home. Building a house at 50 is not an achievement. Don't get used to government houses. This comfort is so dangerous. Let all your family have good time in your house.

2. Go home. Don't stick at work all the year. You are not the pillar of your department. If you drop dead today, you will be replaced immediately and operations will continue. Make your family a priority.

3. Don't chase promotions. Master your skills and be excellent at what you do. If they want to promote you, that's fine if they don't, stay positive to your personal.
development.

4. Avoid office or work gossip. Avoid things that tarnish your name or reputation. Don't join the bandwagon that backbites your bosses and colleagues. Stay away from negative gatherings that have only people as their agenda.

5. Don't ever compete with your bosses. You will burn your fingers. Don't compete with your colleagues, you will fry your brain.

6. Ensure you have a side business. Your salary will not sustain your needs in the long run.

7. Save some money. Let it be deducted automatically from your payslip.

8. Borrow a loan to invest in a business or to change a situation not to buy luxury. Buy luxury from your profit.

9. Keep your life,marriage and family private. Let them stay away from your work. This is very important.

10. Be loyal to yourself and believe in your work. Hanging around your boss will alienate you from your colleagues and your boss may finally dump you when he leaves.

11. Retire early. The best way to plan for your exit was when you received the employment letter. The other best time is today. By 40 to 50 be out.

12. Join work welfare and be an active member always. It will help you a lot when any eventuality occurs.

13.Take leave days utilize them by developing yr future home or projects..usually what you do during yr leave days is a reflection of how you'll live after retirement..If it means you spend it all holding a remote control watching series on Zee world, expect nothing different after retirement.

14. Start a project whilst still serving or working. Let your project run whilst at work and if it doesn't do well, start another one till it's running viably. When your project is viably running then retire to manage your business. Most people or pensioners fail in life because they retire to start a project instead of retiring to run a project.

15. Pension money is not for starting a project or buy a stand or build a house but it's money for your upkeep or to maintain yourself in good health. Pension money is not for paying school fees or marrying a young wife but to look after yourself.

16. Always remember, when you retire never be a case study for living a miserable life after retirement but be a role model for colleagues to think of retiring too.

17. Don't retire just because you are finished or you are now a burden to the company and just wait for your day t

1. Sampling and Sampling Distributions
.
1
Chapter One

2. 2
Objectives
After completing this chapter you will be able to:
▪explain the concept of sampling
▪determine the sampling distribution of the mean
▪determine the sampling distribution of the proportion

3. Sample:isasubsetofthepopulation
Sampling:istheprocessormethodofsampleselection
fromthepopulation.
Samplingdistribution:isaprobabilitydistributionforthe
possiblevaluesofasamplestatistic.
Whatissamplingandsamplingdistribution?

4. 4
Definition of terms
Parameter:characteristicormeasureobtainedfromapopulation.
Statistic:characteristicormeasureobtainedfromasample.
Sampling:theprocessormethodofsampleselectionfromthepopulation.
Samplingunit:theultimateunittobesampledorelementsofthepopulationtobesampled.
Examples:
•IfsomebodystudiesScio-economicstatusofthehouseholds,householdsarethesampling
unit.
•Ifonestudiesperformanceoffreshmanstudentsinsomecollege,thestudentisthesampling
unit.
Samplingframe:isthelistofallelementsinapopulationunderstudy.
Example:Listofhouseholds.
▪Listofstudentsintheregistraroffice.

5. 5
Errors in sample survey:
Therearetwotypesoferrors
i. Sampling error:
❖It is the discrepancy between the population value and sample value.
❖arise due to inappropriate sampling techniques applied
ii. Non sampling errors:are errors due to procedure bias
It arise due to:
❖Incorrect responses and Measurement
❖Errors at different stages in processing the data.

6. 6
The Need for Sampling
Sampleisused/neededto
➢Reducedcost
➢Greaterspeed
➢Greateraccuracy
➢Greaterscope
➢Moredetailedinformationcanbeobtained.
Typesofsampleselections
Therearetwotypesofsampleselections:
i.Sampleselectionwithreplacementand
ii.Sampleselectionwithoutreplacement

7. 7
Sampling with replacement
➢Itisthemethodofselectionofaunitfromapopulationinwhichthe
unitisreturnedtothepopulationbeforethenextselectionismade.
➢Thepopulationsizeateachselectionremainthesame(constant).
➢Eachpopulationunitcanhaveachanceofselectionmorethanonce.
➢ThereareN
n
possiblesamplesofsizen.
Note:theprobabilityateachselectionofaunitisthesame.

8. 8
Sampling without replacement
➢Aunitselectedfromapopulationisnotreturned/replacedtothe
populationforthenextselection.
➢Thepopulationsizereducesbyoneforeachselection.
➢ThereareNCnpossiblesamples.
➢Theprobabilityofselectionisnotconstant.Itis1/Nforthe1
��
unit,
1/(N-1)forthenext,1/(N-2)forthethirdandsoon
➢Note:Sampleoutcomesarestatisticallyindependentwhensamplingwith
replacementandarestatisticallydependentwhensamplingwithoutreplacement

9. 9
Example
Apopulationconsistsofthefivenumbers:2,3,6,8,11.Findallpossiblesamplesof
sizetwowhichcanbedrawn
i)withoutreplacementii)withreplacement.
Solution:(i)thereare5C2=10samplesofsizetwowhichcanbedrawnwithout
replacementnamely(2,3),(2,6),(2,8),(2,11),(3,6),(3,8),(3,11),(6,8),(6,11),(8,11),
Theselection(2,3)isconsideredthesameas(3,2).
ii)5
2
=25samplesofsizetwowhichcanbedrawnwithreplacement.Namely(2,2)
(2,3)(2,6)(2,8)(2,11)(3,2)(3,3)(3,6)(3,8)(3,11)(6,2)(6,3)(6,6)(6,8)(6,11)(8,
2)(8,3)(8,6)(8,8)(8,11)(11,2)(11,3)(11,6)(11,8)(11,11)

10. 10
Sampling Technique
➢Therearetwodifferenttypesofsamplingtechniques.RandomSamplingor
probabilitysamplingandNonrandomsamplingornon-probabilistic
sampling
1.RandomSamplingorprobabilitysampling
•Everysingleobservationofthesampledpopulationhasanon-zerochanceof
beingactuallyincludedinthesample.
•Thoseitemsfromthepopulationthataretobeincludedinthesample
cannotbeidentifiedinadvance.

11. 11
Count…
•Randomsamplescanbeonlyanalyzedbyusingstatisticalmethods.
•Thedifferenttechniquesoftakingarandomsampleare:
➢Simple random sampling
➢Stratified random sampling
➢Cluster sampling
➢Systematic sampling

12. 12
i.Simple Random Sampling:
▪All elements in the population have the same pre-assigned non-zero
probabilityto be included in to the sample.
▪Every possible sample of specific size has an equal chance of being
selected.
▪Lists of all elements are needed.
▪sampling can be done with or without replacement.
▪It can be done either using the lottery method or table of random
numbers.

13. 13
Count….
Inthelotterymethod,firstgiveauniqueidentificationcodetoeachunitof
thepopulation.Then,writedownthecodesonidenticalpiecesofpapers,mix
themupinbowlandselecttheunitswhosecodesappearontherandomly
selectedpiecesofpapers.
Inthetableofrandomnumbersmethod:toselectsampleofsizen,first
makealistofthepopulationtobesampledandgiveadistinctcodenumber
toeachunitofthepopulationthen,choosethedirectionofselectionrandomly
andfinallytakenunits.

14. 14
Stratified Random Sampling
▪Thepopulationwillbedividedintonon-overlapping(meanseachandeveryunit
inthepopulationbelongstooneandonlyonestratum)andexhaustivegroups
calledstrataanditformedinawaythatelementsinthesamestratashouldbe
moreorlesshomogeneouswhiledifferentindifferentstrata.
Inshort,Elementswithinstratashouldbehomogeneous&betweenstratashould
beheterogeneous).
▪Samplefromeachgroup/stratawillbeselectedbyusingSRS.
▪Itisappliedifthepopulationisheterogeneous.

15. 15
Count…
▪Someofthecriteriafordividingapopulationintostrataare:Sex(male,female);Age
(under18,18to28,29to39,);Occupation(blue-collar,professional,andother)
etc....
▪InstratifiedsamplingthegivenpopulationofsizeNisdividedintosay,krelatively
homogeneousstrataofsizesN
1,N
2,N
3,…,N
KrespectivelysuchthatN=σ
??????=1
??????
??????
??????.
▪Drawsimplerandomsamples(withoutreplacement)fromeachofthekstrata.
▪Letsampleofsizen,bedrawnfromthei
th
strata,(i=1,2,3,…,k)suchthat
n=σ
??????=1
??????
�
??????,wherenisthetotalsamplesizeformapopulationofsizeN.

16. 16
Example
1.supposethepresidentofuniversitywantstoknow,theexperienceofafour-year
studentsattheuniversity.Furthermore,thepresidentwishestoseeifthe
experienceofthestudentsisdifferfromyeartoyear(1
st
from2
nd
,2
nd
from3
rd
andthe3
rd
from4
th)
students.Thepresidentwilldividesthestudentsin4groups
&randomlyselectstudentsfromeachgrouptouseinthesample.
2.Apopulationconsistofmalesandfemaleswhoaresmokers&nonsmokers.The
researcherwillwanttoincludeinthesamplepeoplefromeachgroupthatis,males
whosmoke,maleswhodonotsmoke,femaleswhosmoke,andfemaleswhodonot
smoke.Toaccomplishthisselection,theresearcherdividesthepopulationintofour
subgroupsandthenselectsarandomsamplefromeachsubgroup.

17. 17
Proportional Allocation of sample size in stratified sampling
•The items are selected from each stratum in the same proportion as they exist in the
population.
•The allocation of sample sizes is termed as proportional if the sample fraction, i.e., if
the ratio of the sample size to the population size, remains the same in all the strata.
Mathematically the principle of proportional allocation gives:
??????1
??????
1
=
??????2
??????
2
=
??????3
??????
3
=…
????????????
??????
By the property of ratio and proportions, each of these ratios is equal to the ratio of the
sum of numerators to the sum of denominators,
i.e.,
??????
1
??????
1
=
??????
2
??????
2
=
??????
3
??????
3
=
??????
??????
??????
=
??????
1+??????
2+??????
3….+??????
??????
??????
1+??????
2+??????
3+⋯??????
??????
=
σ
??????=1
??????
??????
??????
σ
??????=1
??????
??????
??????
=
??????
??????
= c, (constant )
since the total sample size n , and the population size N are fixed.
Hence, �
1=??????
1(
??????
??????
), �
2= ??????
2(
??????
??????
), �
3= ??????
3(
??????
??????
), �
??????= ??????
??????(
??????
??????
), (i = 1, 2, 3,..., k )

18. 18
Example
•Astratifiedsampleofsizen=80istobetakenformapopulationofsizeN=2000,
whichconsistsoffourstrataforwhichN1=500,N2=1200,N3=200andN4=100.If
weuseproportionalallocation,howlargeasamplemustbetakenfromeachstratum?
•Solution:Inproportionalallocation,weknowthatthesamplesizefortheithstratum
isgivenby
•�
??????=??????
??????(
??????
??????
),(i=1,2,3,...,k),Then
•�
1=??????
1(
??????
??????
)=500(
80
2000
)=20mustbetakenfromthe1ststratum.
•�
2=??????
2(
??????
??????
)=1200(
80
2000
)=48,�
3=??????
3(
??????
??????
)=200(
80
2000
)=8
•�
4=??????
4(
??????
??????
)=100(
80
2000
)=4

19. 19
Cluster Sampling
➢Populationisdividedintonon-overlappinggroupscalledclusters
anditformedinawaythatelementswithinaclusterare
heterogeneous.
➢Randomlyselectssomeoftheseclustersandusesallmembersofthe
selectedclustersasthesubjectsofthesamples.
➢ClusterSamplingusedwhenthepopulationislargeanddifficultto
generateasimplerandomsampleorwhenitinvolvessubjectsexistin
inalargegeographicarea.

20. 20
Example
1.EstimatetheaverageannualHHincomeinaAA.LeteachSubcity
representacluster.Asampleofclusterscouldberandomlyselected,and
everyhouseholdwithintheseclusterscouldbeinterviewedtofindthe
averageannualHHincomeinAA.
2.ifaresearcherwantedtodoastudyinvolvingthepatientsinthe
hospitalsinAA.City.Somehospitalscouldbeselectedatrandom,and
thepatientsinthesehospitalswouldbeinterviewedinacluster.

21. 21
Systematic Sampling:
➢Thistechniqueisrecommendedifthecompletelistofthesamplingunits,is
availableandtheunitsarearrangedinsomesystematicordersuchas
alphabetical,chronological,geographicalorder,etc
➢Theprocedurestartsbydeterminingthefirstelementtobeincludedinthesample.
➢Onlythefirstsampleisselectedatrandomandtheremainingunitsareautomatically
selectedinadefinitesequence.Thentakethek
th
itemfromthesamplingframe
➢Let,??????=������������??????�,�=��������??????�,�=
??????
??????
�ℎ��ℎ����������������������.
➢Choseanynumberbetween1and�.Supposeitis�
�ℎ
,(1≤�≤�)
➢The�
�ℎ
unitisselectedatfirstandthen(�+�)
�ℎ
,(�+2�)
�ℎ
,....���untilthe
requiredsamplesizeisreached.

22. 22
Example
Supposewewanttoselect50votersformalistofvoterscontaining
1,000namesarrangedsystematically.
Solution
Heren=50andN=1,000,k=
N
n
=
1,000
50
=20.
Weselectanynumberfrom1to20atrandomandthecorresponding
voterinthelistisselected.
Lettheselectednumberis6.Then,thesystematicsamplewillconsistof
50votersinthelistatserialnumbers:6,26,46,66,…,966,986.
1
st
=6,2
nd
=6+20=26,3
rd
=26+20=46andetc.…..

23. 23
Non-Random Sampling or non-probability sampling
•Asampleisselectedatthecompletedecisionoftheinvestigator.
•Personalknowledgeandopinionareusedtoidentifyitemsfromthe
populationthataretobeincludedinthesample.
Heretherearethreemethodsoftakinganon-randomsample.
i.Judgmentsampling(purposivesampling)
ii.Conveniencesampling
iii.QuotaSampling

24. 24
1.JudgmentSampling
•Inthiscase,thepersontakingthesamplehasdirectorindirectcontrol
overwhichitemsareselectedforthesample.
2.ConvenienceSampling
•Theinvestigatorselectsasamplefromthepopulationinamannerthatis
relativelyeasyandconvenient/suitable.
3.QuotaSampling
•It requires the sample to contain a certain number of items with a given
characteristic.

25. 25
Sampling Distribution
➢ConsiderallpossiblesamplesofsizenformapopulationofsizeN(with
orwithoutreplacement).Wecancomputeanysamplevalue(statistic)such
asmean,standarddeviation,variance,etc.foreachsample.Thesestatistics
varyfromsampletosample.Therefore,theyarerandomvariables.
➢Valueofstatisticalongwithitsprobabilityofoccurrenceiscalled
samplingdistributionofthestatistic.
➢Thestandarddeviationofthesamplingdistributionofastatisticiscalled
thestandarderrorofthatstatistic.

26. 26
Sampling Distribution of the Mean
•ConsiderallsamplesofsizenfromthepopulationofsizeN.Then
samplemeanofeachsample(ത??????
??????)alongwithitsprobabilityof
occurrence(relativefrequency)iscalledthesamplingdistributionof
themean.
•itisdescribedbytwoparameters:theexpectedvalue(
??????
)=ധ??????,ormean
ofthesamplingdistributionofthemean,andthestandarddeviationof
themeanorstandarderrorofthemean??????
??????.

27. 27
Properties of the Sampling Distribution of Means
1.Themeanofthesamplingdistributionofthemeansisequaltothepopulation
mean.µ=??????
??????
=ധ??????.
2.thestandarddeviationofthesamplingdistributionofthemeans(standarderror)is
equaltothepopulationstandarddeviationdividedbythesquarerootofthesample
size:??????
??????=δ/√n.Thisholdtrueifandonlyifn<0.05NandNisverylarge.IfNis
finiteandn≥0.05N,??????
??????=
??????
??????
∗
??????−??????
??????−1
.Theexpression
??????−??????
??????−1
iscalledfinite
populationcorrectionfactor/finitepopulationmultiplier.
Inthecalculationofthestandarderrorofthemean,ifthepopulationstandard
deviationδisunknown,thestandarderrorofthemean??????
??????,canbeestimatedbyusing
thesamplestandarderrorofthemean??????
??????
whichiscalculatedasfollows:
??????
??????
=ൗ
??????
??????
&#3627408476;&#3627408479;??????
??????
=
??????
??????
∗
??????−??????
??????−1
.
3.Thesamplingdistributionofmeansisapproximatelynormalforsufficientlylarge
samplesizes(n≥30).

28. 28
Example
Apopulationconsistsofthefollowingages:10,20,30,40,and50.Arandomsampleofthreeis
tobeselectedfromthispopulationandmeancomputed.Developthesamplingdistributionof
themeanifselectioniswithoutreplacement.
Solution:
The number of simple random samples of size n that can be drawn without replacement from a
population of size N is NC
n.
With N= 5 and n = 3, 5C3 = 10 samples can be drawn from the
population as:
Sampled items
Sample means (ҧ&#3627408485;)
10, 20, 30 20.00
10, 20, 40, 23.33
10, 20, 50 26.67
10, 30, 40 26.67
10, 30, 50 30.00
10, 40, 50 33.33
20, 30, 40 30.00
20, 30, 50 33.33
20, 40, 50 36.67
30, 40, 50 40.00
total 300.00

29. 29
Count…
A systematic organization of the above figures gives the following:
??????=
σ??????
??????
=
σ??????
??????
=&#3627409361;&#3627409358;,Regardlessofthesamplesize??????=ന??????And
??????=
σ??????
??????−??????
&#3627409360;
??????
=
&#3627409359;&#3627409358;&#3627409358;&#3627409358;
&#3627409363;
=&#3627409359;&#3627409362;.&#3627409359;&#3627409362;&#3627409360;
•??????
??????
=
??????
??????
∗
??????−??????
??????−&#3627409359;
=
&#3627409359;&#3627409362;.&#3627409359;&#3627409362;&#3627409360;
&#3627409361;
∗
&#3627409363;−&#3627409361;
&#3627409363;−&#3627409359;
=&#3627409363;.????????????&#3627409362;=
σ??????
??????−??????
&#3627409360;
??????
=
&#3627409361;&#3627409361;&#3627409361;.&#3627409362;
&#3627409359;&#3627409358;
=&#3627409363;.????????????&#3627409362;
•Since averaging reduces variability ??????
??????< δ except the cases where δ = 0 and n = 1.
Sample mean (ത??????)Frequency Prob. (relative freq.) of ത??????
20.00 1 0.1
23.33 1 0.1
26.67 2 0.2
30.00 2 0.2
33.33 2 0.2
36.67 1 0.1
40.00 1 0.1
10.00 1.00
Columns 1 & 2 show frequency distribution of
sample means.
Columns 1 and 3 show sampling distribution of
the mean.

30. 30
Central Limit Theorem and the Sampling Distribution of the Mean
•TheCentralLimitTheorem(CLT)statesthat:
1.Ifthepopulationisnormallydistributed,thedistributionofsamplemeansisnormal
regardlessofthesamplesize.
2.Ifthepopulationfromwhichsamplesaretakenisnotnormal,thedistributionof
samplemeanswillbeapproximatelynormalifthesamplesize(n)issufficiently
large(n≥30).Thelargerthesamplesizeisused,thecloserthesampling
distributionistothenormalcurve
Therelationshipbetweentheshapeofthepopulationdistributionandthe
shapeofthesamplingdistributionofthemeaniscalledtheCentralLimit
Theorem.

31. 31
The significance of the Central Limit Theorem
➢itpermitstousesamplestatisticstomakeinferenceaboutpopulation
parameterswithoutknowinganythingabouttheshapeofthefrequency
distributionofthatpopulationotherthanwhatwecangetfromthesample.
➢Italsopermitstousethenormaldistribution(curveforanalyzingdistributions
whoseshapeisunknown.
➢Itcreatesthepotentialforapplyingthenormaldistributiontomanyproblems
whenthesampleissufficientlylarge.

32. 32
Example 1and 2
1.Thedistributionofannualearningsofallbanktellerswithfiveyearsof
experienceisskewednegatively.ThisdistributionhasameanofBirr15,000anda
standarddeviationofBirr2000.Ifwedrawarandomsampleof30tellers,whatis
theprobabilitythattheirearningswillaveragemorethanBirr15,750annually?
Andinterprettheresult?
2.Supposethatduringanyhourinalargedepartmentstore,theaveragenumberof
shoppersis448,withastandarddeviationof21shoppers.Whatistheprobability
ofrandomlyselecting49differentshoppinghours,countingtheshoppers,and
havingthesamplemeanfallbetween441and446shoppers,inclusive?

33. 33
Solution 1
1.Calculateµand??????
??????
µ=Birr15,000
??????
??????=δ/√n=2000/√30=Birr365.15
2.CalculateZfor??????
&#3627408461;
??????
=
??????−ധ??????
??????
??????
=
??????−??????
??????
??????
,&#3627408461;
15,750=
15,750−15,000
365
=+2.05
3.Findtheareacoveredbytheinterval
P(ത??????>15,750)=P(Z>+2.05)
=0.5-P(0to+2.05)
=0.5–0.47892
=0.02018
•4. Interpret the results
Thereisa2.02%chancethattheaverageearningbeingmorethanBirr15,750
annuallyinagroupof30tellers.

34. 34
Solution 2
1.Calculateµand??????
??????
µ=448shoppers ??????
??????=δ/√n=21/√49=3
2. Calculate Z for
??????
&#3627408461;
??????
=
??????−ന??????
??????
??????
=
??????−??????
??????
??????
&#3627408461;
441=
441−448
3
=−2.33 &#3627408461;
446=
446−448
3
=−0.67
3.Findtheareacoveredbytheinterval
P(441≤
??????
≤15,750)=P(-2.33≤Z≤-0.67)
=P(0to-2.33)-P(0to-0.67)
=0.49010–0.24857
=0.24153
4. Interpret the results
Thereisa24.153%chanceofrandomlyselecting49hourlyperiodsforwhich
thesamplemeanfallsbetween441and446shoppers.

35. 35
Example 3, 4 and 5
3.Aproductioncompany’s350hourlyemployeesaverage37.6yearofage,withastandard
deviationof8.3years.Ifarandomsampleof45hourlyemployeesistaken,whatisthe
probabilitythatthesamplewillhaveanaverageageoflessthan40years?
4.Supposethatarandomsamplesizeof36isbeingdrawnfromapopulationwithameanof
278.If86%ofthetimethesamplemeanislessthan280,whatisthepopulationstandard
deviation?
5.Ateachergivesatesttoaclasscontainingseveralhundredstudents.Itisknownthatthe
standarddeviationofthescoresisabout12points.Arandomsampleof36scoresisobtained.
a)Whatistheprobabilitythatthesamplemeanwilldifferfromthepopulationmeanby
morethan6points?
b)Whatistheprobabilitythatthesamplemeanwillbewithin6pointsofthepopulation
mean?

36. 36
Solution 3
1.Calculateµand??????
??????
µ=37.6years n/N=45/350>5%......FPCFisneeded
??????
??????=
??????
??????
∗
??????−??????
??????−1
??????
??????=
8.3
45
∗
350−45
350−1
=1.16
2. Calculate Z for
??????
&#3627408461;
??????
=
??????−ധ??????
??????
??????
=
??????−??????
??????
??????
&#3627408461;
40=
40−37.6
1.16
=+2.07
3.Findtheareacoveredbytheinterval
P (ത??????< 40) = P (Z < +2.07)
= 0.5 + P (0 to +2.07)
= 0.5 + 0.48077
= 0.98077
4. Interpret the results: There is a 98.08% chance of randomly selecting 45 hourly employees
and their mean age be less than 40 years.

37. 37
Solution 4
µ=278n=36??????=280P(??????<280)=0.86δ=?
(
??????
??????
=0.36)=+1.08
&#3627408461;
??????
=
??????−??????
??????
??????
&#3627408461;
280=
280−278
??????
??????
+1.08=
280−278
??????
??????
+1.08=
2
??????
??????
??????
??????
=
2
1.08
=1.85
??????
??????
=
??????
??????
1.85=
??????
36
1.85=
??????
6
??????=6∗1.85=11.1

38. 38
Solution 5
a.n=36δ=12??????
??????
=
??????
??????
=
12
36
=
12
6
=2P(??????>µ+6)+P(??????<µ-6)=?
&#3627408461;
??????+6=
??????+6−??????
2
=+3 &#3627408461;
??????−6=
??????−6−??????
2
=−3
P (??????> µ +6) + P (Z> µ -6) = P (Z> 3) + P (Z< -3)
=[0.5–P(0to+3)]+[0.5–P(0to-3)]
=(0.5–0.49865)+(0.5–0.49865)
=0.00135(2)=0.00270
b.n=36δ=12??????
??????
=
??????
??????
=
12
36
=
12
6
=2
P(µ-6≤??????≤µ+6)=P(-3≤Z≤3)
=P(0to3)*2
=0.49865*2
= 0.9973
Ifthepopulationstandarddeviationis12,inarandomsampleof36scoresthereisa99.73%
chanceofgettingasamplemeanscoretoliewithin6pointsofthepopulationmean.

39. 39
Sampling Distribution of Proportions (??????)
•Thepopulationproportionisobtainedbytakingtheratioofthenumberofelements
inapopulationwithaspecificcharacteristictothetotalnumberofelementsinthe
population.
•Therearenumerousproblemsinbusinessthatwanttoknowtheproportionofitems
inapopulationthatpossessacertaincharacteristic.Forexample,
-Aqualitycontrolengineermightwanttoknowwhatproportionsofproductsofan
assemblylinearedefective.
-Alaboreconomistmightwanttoknowwhatproportionofthelaborforceis
unemployed.

40. 40
count…
•The sample proportion is computed by dividing the frequency that a given
characteristic occurs in a sample by the number of items in the sample.
??????=
??????
??????
, Where ??????= sample proportions
X = number of items in a sample that possess the characteristic
n = number of items in the sample
•Samplingdistributionoftheproportionisdescribedbytwoparameters:themean
ofthesampleproportions,E(??????)andthestandarddeviationoftheproportions,??????
??????
whichiscalledthestandarderroroftheproportion.

41. 41
Properties of Sampling distribution of ??????
1.ThepopulationproportionP,isalwaysequaltothemeanofthesampleproportion,
i.e.,P=E(??????).
2.Thestandarderroroftheproportionisequalto:??????
??????
=
??????&#3627408478;
??????
,
whereP=populationproportion,q=1–Pandn=samplesize.
Or
??????
??????
=
??????&#3627408478;
??????
∗
??????−??????
??????−1
, where
??????−??????
??????−1
= finite population correction factor.
•The finite population correction factor is not needed if n < 0.05N.

42. 42
Central Limit Theorem (CLT) and Sampling distribution of ??????
TheCLTstatesthatnormaldistributionapproximatestheshapeofthedistributionof
sampleproportionsifnpandnqaregreaterthan5.Consequently,wesolveproblems
involvingsampleproportionsbyusinganormaldistributionwhosemeanandstandard
deviationare:
??????
??????
=??????,??????
??????
= ൗ
??????&#3627408478;
&#3627408475;&#3627408462;&#3627408475;&#3627408465;&#3627408461;
??????
=
??????−??????
??????
??????
NB:Thesamplingdistributionof&#3627408477;canbeapproximatedbyanormaldistribution
wheneverthesamplesizeislargei.e.,npandnq>5.

43. 43
Examples
1.Supposethat60%oftheelectricalcontractorsinaregionuseaparticularbrandofwire.
Whatistheprobabilityoftakingarandomsampleofsize120fromtheseelectrical
contractorsandfindingthat0.5orlessusethatbrandofwire?
2.If10%ofapopulationofpartsisdefective,whatistheprobabilityofrandomly
selecting80partsandfindingthat12ormorearedefective?
3.Supposethatapopulationproportionis.40andthat80%ofthetimeyoudrawarandom
samplefromthispopulation,yougetasampleproportionof0.35ormore.Howlargea
samplewereyoutaking?
4.Ifapopulationproportionis0.28andifthesamplesizeis140,30%ofthetimethe
sampleproportionwillbelessthanwhatvalueifyouaretakingrandomsamples?

44. 44
Solution 1
n = 120 P = 0.6 q = 0.4 P (&#3627408477;<0.5) =?
1.Checkthatnpandnq>5
120*0.6=72,and120*0.4=48.Botharegreaterthan5.
2.Calculate??????
??????
??????
??????
=
??????&#3627408478;
??????
==
0.6∗0.4
120
=0.0477
3. Calculate Z for &#3627408477;
&#3627408461;
&#3627408477;=
??????−??????
??????
??????
, &#3627408461;
0.5=
0.5−0.6
0.0477
=−2.24
4. Find the area covered by the interval
P(&#3627408477;<0.5)=P(Z<-2.24)
=0.5-P(0to-2.24)
=0.5–0.48745
=0.01255
5.Interprettheresults
Theprobabilityoffinding50%orlessofthe
contractorstousethisparticularbrandofwireisvery
low(1.255%)ifwetakearandomsampleof120
contractors.

45. 45
Solution 2
n = 80, P = 0.1, X = 12 P (&#3627408477;>0.15) =?
1. 80*0.1 = 8, and 80*0.9 = 72. Both are greater than 5.
2.&#3627408477;= X/n = 12/80 = 0.15 ??????
??????
=
0.10∗0.90
80
=0.0335
3. &#3627408461;
&#3627408477;=
??????−??????
??????
??????
&#3627408461;
0.15=
0.15−0.1
0.0335
=+1.49
4. P (&#3627408477;>0.15) = P (Z >+ 1.49)
= 0.5 –P(0 to + 1.49)
= 0.5 –P (0 to + 1.49)
= 0.5 –0.43189 = 0.06811
5. About 6.81% of the time, twelve or more defective parts would appear in a random sample
of eighty parts when the population proportion is 0.10.

46. 46
Solution 3
P= 0.4 P (&#3627408477;>0.35) = 0.80 n =?
1. (
??????
??????
= 0.30) = 0.84 ??????
??????
=
??????&#3627408478;
??????
; squaring both sides
2.&#3627408461;
0.35=
0.35−0.4
??????
??????
0.0595=
.4∗.6
??????
-0.84=-0.05/??????
??????
0.0595
2
=
.4∗.6
??????
2
0.0035=0.24/n
0.84??????
??????
=0.05 0.0035=0.24/n
??????
??????
=0.05/0.84 n=0.24/0.0035
=0.0595 n=68

47. 47
Solution 4
P=0.28 n=140 P(&#3627408477;<X)=0.30X=?
(
??????
??????
=0.2)=-0.52
&#3627408461;
??????
=
??????−??????
??????
??????
−0.52=
??????−0.28
0.0379
−0.0197=??????−0.28
??????=0.26
??????
??????
=
??????&#3627408478;
??????
=
0.28∗0.72
140
=0.0379