CHAPTER 28: SAMPLING in This Chapter, Students

8/14/2019 CHAPTER 28: SAMPLING in This Chapter, Students

1/20

CHAPTER 28: SAMPLINGIn this chapter, students will:. Ieam conccpts ofpopulation and saiuple. undcrstand thc concepts ofrandom. stratified, systematic and quota examplcs. leam the advantages and disadvantages ofthe various sampling methods. understand the distribution ofsanple means from a normal population. usc Central Limit Theorem to treat samplc means as having normal distributio[ whenthe sanple sizc is sufficiently large. calculatc unbiased estimates ofthe population mean and variance from a sample. solve problems ilvolving thc sarnpling distribution

SAMPLE AND POPUI,ATIONIn many^practical statisticar investigatio'sj we arc intercsted in gathering infomation about a largegroup ofindividuals or objects (known as lpopulation'). It shoujd be notid that in any investigatio"n,the population nceds to be clcarly defined. Howc-1 in many cases, it is not possiblc to obtaininformation about all meinbers ofa population tbr the lbllowing ieasons:1. The colle


2/20

gcneral comments before the details ofeach method :finit 3 t)?es .LIe rcferred to as probabilistic sarnpling in which everv member of the poDulationan equal chance of being selectcd. The sample obtained is called a random sample- ln each ofsallpling methods, a samDlins frame (see section trelow) is necded.sampling is a tjpe of nonprobabilistic sampling and the sample obtained is non-random. Iis not nee(led in quota sampling.are more likely to represent the population well through probability sampling. In gerleral,prefer probability sampling methods and consider them to be more accuratc and rigorous.

l.lof the following is a random sample?John picks out 2 identical balls liom a bag of2l, without replacment.Mark picks out 2 identical balls from a bag of2l, with replacement.Mattlew picks out 2 identical balls ftom 2 bags, one from a bag of l0 and one from a bag of11.Luke picks out 3 identical balls liom 3 bags of7, one Aom each bag.

a sampling method, if every member ofthe pop,rlatioihas an equal chance of beins selected, t!7ensampl( oblaincd is a random sarnple.The probability ofselecting the first ball is l/21- Sincc the lirst ball is not rcplaced before thcsecond ball is drawn, the probabiiity of selecting the second ball is l/20_ Tlus, we see thateach member of the population does 4qt have an equal chance l)1. being selectcd and weco{clude that the sample is non-random.Since the lirst ball is replaced before the second ball is drawn, the probabiljty ofselecting thesecond ball is l/21. Thus, we see that each member ofthc population has an equal chance oftreing selccted and we conclude that thc sample is random.The probability ofselecting a ball liom rhc bag of 10 is 1/10 and the probability ofselecting aball from the bag of 1l is l/11. Thus, we see that oach member of the population does !!)thave aa equal chance ofbeing selected and we concludc that the sample is non-random.The probability ofselecting each ball from a bag of7 is l/7. 'lhus, we see that each member ofthe population has an equal chance of being selectcd and we conclude that the sample israndom,

1-2 (Do it yourself)from IJC Prelim trxam 2007 P2 Q5)kindcrgarten has nine Year One classes and six Ycar Two classes with differcnt class sizos. Theof the kindergarlen wishcs to take a sample of 30 children to moet up with a ministryfor a short conversation. She chooses 2 children at random from each of the l5 classes. State,

a reason, whether this gives a random sample of30 children from tho kindergaden.

C2B 2


3/20

The Sampling FrameThe sampling Jrame is a complete list (to be idcal), in some form, of qtz thc members in thepopulation, from which a sample can bc selectcd.For cxamplc, in a sr.rvey of employees of a company commissioned by its managemcnt, it isrcasonable to cxpect to be able to oblain I comprellensive, up_l(, date and accuraic list of allenrployees, with sufticient classifi catory infomation.other cxanrples include population registers or electorar registers of countries, registeis of schooisand records ofscrial numbers in mallufactured goods-Howcver, in many situations, it rnay bc diflicult to maintirin a complcte and up-to_date list of thepolulation- Even if such a list cxists, duo to the gencral trend towaris greater formal protection forprivacy ofthe individual, irfomtation may be withheld.I samplingframe is neetled ifi any sampling ethod that is ran.lonlDue to difficulties in obtairing a varid sampring frane, thcre is a noed rbr us to study .onprobabilisticsampling metlrods su(h n, quote \i nllrnS.(l) Sirnple Rarrdom Sampling

ln simple random sampling, cach ildividual ir thc saniplc is chosen frorn the population purelyby chance. Each nenlbcr of thc population has an cqual chalce of bcing sclectecl_To obtain a random sample ofsize r? from a population ofsizc { weI ) obtain a sampling liame of the population,2) number the lrembers ofthe populatioq say frorn 1 to ly',3) make a rcndon selection ofr? ofthcse numbers, alrd4) idcntily the mcmbers ofthc population .cpresentcd by thc numbcrs sclcclcd.Note that random selection of the numbers can bc


4/20

influence the responses signilicantly. In such a case, it will be usetirl to ensu.e that the samplercprcsents the proportions ofmutually exclusive subgroups (ty age), in the population.Stratifi od Sampling involves1) dividing the population into mutually exclusive subgroups/strata, and2) randohilv selectins(using the sampling frame) a sanple from each stratum, withsample size proportional to the relative size of the stratum.

Suppose that in a company thl: proportions ofstaffin different age-groups are as follows:Aged undcr 40 Aged between 40 and 60 Aged over 6038v" 40yo 22vo

To obtain a sample of 200 staff, we draw random samples from the age-groups with sample sizein thc same proportion as the sizc ofeach age-group-Age-group tlnder 40 Between 40 and 60 Over 60

Sample size 38% o1200(=76) 4O"/o of 200(:80) 22o/o of 2OO(=44)Advantaqes(t It is more likely to give a good represeotative sample ofthe population.(i0 The data obtai ed liom cach stmtum can be anall,zed separately, and this usually givesmore accurate estimatcs of the population pammcters, compared to simple randomsampling.Disadvantages

(i) lt nlay be hard to obtaio the sampling lizune needed(i0 More diflicult to conduct as compared with (simple) random sampling. l-here may notbe sufficient inlbrmation about the population to delineate the subgroups.(iii) The strata may not bl- clcarly dehned.(iv) ll is rclativelr morc rinre consurning.

SystematicSamplingsampling is done through the tbllowing steps:

I ) Obtain a sampling fiamc of the population.2) Number the mlrmbcrs of the population, say ftom I to /y'.3) To draw a samplc ofsize n liom a population ofsize-l/,let tr:Nr,.4) Randomly selecr an intrgcr bclwccn I ro l.5) Then take e\,(ry 1" inrcgrr afier that.6) Identify the mcmbcrs ofthe populalion rqrresented by the numbers slected.cxample, to draw a sanple (]1250 tickcts front a population of 5300 tickets, we may take I 205100- ::j--1. In this casc, rl e selcct c!cr) 20rl' ticket cornmencing with, say the 8tl' ticket, (The numbcr25(,is randomly detennined.) In other words, we select the 8d', 28'l', 48'1', 68n', -.....

c28- 4 ,


5/20

Advantages1ii It is easier to conduct as compared with othcr t)?es ofsampling(iD lt is more cvenly spr-ead over the population. The use ofwide-spacing ofk individuals guardsagainst the list consistir4 ofclustcrs ofsiinilar individuals.Disadvantages(D It is not always possible 1o oblain a valid sampling ftame of thc popuiation and to numbc.each individual.(iD The sample obtaincd can be biased when thc members of the population have a periodic orcyclic pattem of occurrence in the sampling framc. As an ei:rmple, if a sampiirrg frarn"consists of a list of married couples in the order of husband, wife, husband, rviie, h'usband,wife..-.etc, then if every tcnth pcrson is selected alier choosing thc first of the husbands, thenthe sarnple will likely to consist only ofmales!(4) Quota SamplingQuota sampling is similar to stratified satnpliog. [n quota sanpring, the populatior is also divicied intomufually exclusive subgroups, but the sample is non random.

For example, in a market survey, an interviewer is given an assignmelt to interview 20 people in thestreet based upon quotas thrt specify or define the sample to be


6/20

lt is not a good representative ofthe population as compared with other types ofsampling .It is non-randomIt is biased as the inteNiewer may simply select those who are easiest to intewiewQuota sampling is vc.y uselll and convenient though it is non-random in nature. Info.mationgathered Aom this twe of sample should be treated with caution.

1.3sample of 5 is to be selected liom a class of 15 boys and l0 girls, Desclibe how you would chooseconlmittee using(i) random sampling(ii) stratifi ed sampling(iii)quota sampling(iv)systematic samplingTo fonn a sampling frame, assign a number to each sfudent in the class. Write the numbers Ito 25 on strips of papers, put thcm in a contaiDer, mix them up well and draw 5 numbers out-Select the 5 students assigned to thesc 5 selected numberc-Since the proportions ofboys and girls in the class rre l5/25 and I U/25 respectively, u,e needto randonly select (i5l25)x5:3 6ot. urtd,l0/25)rS = 29.1". We assign the nurnbels I to 15to the boys and randomly select (as in part (i)) 3 numbers. Wc sclcct the 3 boys that areassigned these 3 numbcrs. Similarly, we assign thc numbers i to 10 to the grls cnd rardomselcct 2 numbers bcfore selccting the corresponding girls.We decide on the propodions ofboys and girls ifl the sanple of 5. They need mrt be l5/25and10/25. Supposc we decided kr selcct 3 girls and 2 boys, we can selecl any 3 girls, fnrm l0 inthe class, and any 2 boys, liom l5i[ tho class, that arc most cool]erative ot convenient to us-They are not randoinly selected within their subgroups (unlikc stratified sampling).Number the students in the class fiom 1 to 25 (forming the sampling framc). Since samplesize is 5 and population siz,e is 25, k : 25/5: 5. Select any number from lto tr (:5), say weselect 3. Then the students to bc selected are those numbered 3,3+k,3+2k, .....i.e. thosenumbered 3,5,13, 18, and 23.

1.4employnent agoncy wants to estimate the number of unemployed people in a HDB new town,so decide to obtain infomation Aom a sample ofits adult who is ofworking age. Give oncwhy it would not be appropriate to obtain the sample by stopping adults at the town centre'sduring onc working day.

misses out peoplc who do not shof at that supennarket (biascd sample).

c28 6


7/20

Race Number of residents!!11',":sMalay 144g3lnoran 65C)thers 2aTotal 930

Example 1.5In a city, the following infonnation is availabie from the registry officc-

A sample of40 residents is to bc chosen for an intervicw. Discuss the following sainpling methods:(i) Stratilied Sampling(ii) Quota SamplingState the advantages and disadvantages ofcach mcthod.Solution(i) In stratified sampling, wc rrec(l to rdndomh) seleclMalays, (65/910)x40 Indians and (28/930)x40 Others,from 744 Chinesc, 4 Malays from 93 Malays, 3 Ildianspeople ofother raccs.

Advantages:(1) The sample rcpresents the population betfer than samples obtained through other sampiingmethods.(2) TLe data lronr each race group ca be analyzed sepantcly to llivc more detcils and thc resultscan bc more represcntalivc ofthc popuiation.Disadvantages:(l) It is more diflicult to conduct. The sampling frameimpossible to have random sclection within each racc(2) It is morc time consuoliLig.(ii) In quota sampling, the proportioN ofthc diflirent raccs ur the salnple need not be thc same asthose in the population. Supposc we docide to select 10 people from each subgroup. The 10pcople in each subgroup nced not be rundomly selcctett. wc can sclcct thosc that are mostavailablc.Advantages:(1) Data can be collected iastcr .(2) Sampling liame is nor required.Disadvantagcs:(1)1'he sample obtained does rrot represcot fhe populationother methods(2) l hc sarnlle is nnt randoml) selcct(,1.

('7441930\\40 Chinese, (93/930)r{0-fiius, we randomlv select 32 Chinescliom 65 lndi:rns and 1 person liom 28

may not bc availablc and thus it may bcltroup.

as well as samples obtained through

c28 7-


8/20

o{unbiased estimates ofthe populdtion mean a dvariance from a sdnple-ESTIMATION OF POPULATION PARAMETERSparamctcr is any numerical value describing a characteristic of a populafion. For cxamples,

mean p arird population variance o2,sanple statistic is any numerical value describirg a characte.istic ofa samplc. It is calculated f'rom

obscrvations in the sample. For cxamplc, sample meanare used to estimate thcse paftlmeters-rcn population parameters ate u known, sample statistics

a statistic is called at estirnator-

are sample mcan and sample variance?we have a sample, ofsizc 5, with the following 5sanple mean isjust thc average ofthese values. Thus,

2 { l+4+31 5

rt.;\'valuc ot sampl vifiJncc = : !(-., ;)'.(", -;)'.(,,, ;)'.(,. -;)' .(,, ;)'

5(2 3)'+0-3)'?+(a 3)'+(3-3)'"r(5 3)'5

[,STIMATEa populatior with an uDknown paramefer 0. If I is a statistic derived fiom a randorn sample

the population, thcn T is at unbiased estitnator of 0 if E(1) = 0 .do we cstimate p and o2 ?

values: ,tl: 2, 12: I, tq = .1, r.1= 3. 15= 5.the samplc mean

Ulknown PopulationParametersPopulation mean p

r Population variance)o

Sample Statisticso Sample meanX. Samplevariance estimates

estimates

c28- 8 ,


9/20

Unbiased Estimators of the Population Mean u and Variance o2From a population with unloown mean p and unknown variance 62, take a random sarnple ofsize nand let the sample mean bc- x, I r, +............-r._ I += ' I a _ -\ ,-n n?'and the sarnple variance be:6 ;I r(",)' / \,---\'{/nnThen the unfriased estimator of F, written u" it , i"V ,

I c 1, = X .lleeITo show F is an unbiased cstimator lbr F.-,., -/ r. +l. +-....... r I I -Flxl r.f : - ) ,' r.(y,rrF{y ) r...........Fr,\,)-l

l.--lnpl- pnTherclbre -Y is an unbiascd cstinator ldr /.The unbiascd esrimator of o2, written as s4 i" ;!(samlle variance)i." t :fi(samplevariance)

)'x

;IrI=,_1l -F)'l

Note:Since each ,l is drawn from the distribution of-y.E(.X)=Eol= lr

(I(";))'l, ] rin Mrr)r. .....( r'

=*[r''''c28 9-


10/20

I (", il'n-l(in MFI 5)............. ..................(2)

thc sample data given is in the form I("-") -a | (" - c)t , *lrere c is a constant, thenI(-' .)11 - lc andnEG_dI

(3)

n = 200,n = 50,

lhcr (J) and (l) arc srmrlar in lbrm.2.1unbiased estimates of thc populetron mern and varianc lrorn each of the lollowiigdrawn:19.30, r9.61, 18.27, 18.90, 19.t4,19.90,18.76, 19.10;n=50, Ir:1150, [r'?=28990n 12. r = 2.1.5. l(r .)' aa.z:

Number, x 17 1g 22 25 3l 33Frequency. / 3l 13 I 4 3 8

)(x-:oo)=20t2, l(r 3oo)'z =525262f r = 1s00, l(r - zo)' :52520

c28- 10 -


11/20

i) Using GC, enter lhe valucs as a list undcr I_r.Soln:L1 31IE19.61lE.at1E.91g-tq19.S18"76Lrfil=1 t.3Ncxt, u-sing,,


12/20

GC, entcl values ofr under Lr and their corresponding frequcncies under L217Ls2Z2Sl1

=

3t131rt3

, ,,StatsJt-LEr. oo,.1,J,.1,J,J,J-4J( L --CD/Jl5x=5.914988421sx=5.865411798.11=68

S"'? = 34.986 r:35.0 (3s.f.)n-200, !{"-roo)= 2012, 1Q-300)2 =525262.){"-:oo), [r,x ct2 (f{*-cl)2 1n-tl' n

I- ' lrrrro, 2ol2']lleel 200 I=ffi


13/20

(vi)

xn-50,I" l.r = 1s00,1500= =1050 )(x-zo)'=s2s2o50

(|(.',r-zo))'50

of the lbllowing

s'=A[tt"-zor'-onl"'- t lrrrrn too'l4el 50 I:969.796

Examplc 2.2 (Do it yourselgObtain the uobiased cstimates of thc population mean and variance from eachsamples drawn:

n - 70, lt : 2244, > tr : 728rj9

1ii1n=15, i-zs.s, I(" i)'-so.ro

r,, iz = 100, l(r-zOo)=+ozz, l(;r-zo0)' - 165590(vi) r-100, l.;r-1200, l(r-ro)'=rzlo

- I("-zotx-a' ' +)O5030- t(x 20) +2050l(:r-zo): soo

(i) 30.40, 29.61,28.21 . 31_90, 32.t4,29.90,28.76,29.1O

Numbcr of umbrellas sold ). 3 4 5 6 10Number of days 25 30 20 8 7 10

c28, 13 -


14/20

Distribulion ofsample means from a normal population.Use of the Central Ltmit meoretu to treat sampL rned6 as having rormel distributioawhen the sample .r{ze is sufrcientl! large.SAMPLE MEA}I AS A RANDOM VARIABLEX, Xz, ..., X,is a random sample ofsize n taken from any infinite population (for finite population

satrpling with replacernent) with mean p and variance d, then the sample mcan 7 is a randomwith _o): p and Var(x) = -nX, +)(^ +...-....-...X \

,)!1E8,\ + E(x, ) +...........z(x, )ln(,ri- on

+ X, +.........--.X, ),)1-,lVar( X,l+Var( X,l+.............Var( X,lfn'I r " . ^r, lo' + cr' + .......... ..o' In1 r '\ rVo')n2o

n

xl

3.1population consists ofall numbers liom 0 to 99- Samples of5 numbers are selected as follorvs:5l ,'/7 , 2'7 , 46, 40:42.33,12,90,4446,62,16,28,9893,58,20,41,8619, 64, 8, 70, 56

the m()ms ofthese sanlples and thc inan and standard deviation ofthcsc sample means.

samplesampleSarnrrlesarnplesample

c28 14


15/20

Soln:Using43.4

t? From a non normally distributed populationIf \. X2. ..., )1, is a randonr sample of sizi r taken(discrete or continuous) with mean p and variance o2 ,Limit 'I'hcorem, the distribution of the total srm X,,4 cte rr,proximatclv normal such thst

_rr + 12 +............-r,n , tho rcspcctive sample means arc 48.2, 44.2,50.0, 59.6 and

Using GC, the mear (ofall these sarnple rneans) is 49-0g and the sfandard deviation is 5.g0.(Nnte: we use Oi-l tnstead of S because in GC, 6 is the variance oftho data entered into the GC,which is requircd here)Can lou see the led of a distribution of sample means here?3.1 DISTRIBUTION Ot- THE SAMpt,E MEAN(l) From^a normaUy dislributrd populationIlix - N(p. o'1andx. , x1 , . . ., x, is a rardom sarnpre of z independent obscrvations of ,t then thcsarnple mean X has a nonnal distributlon such thatI ^ N(u. - )nX, r X, r............X,, N(n1r.no')

Aorn a not nomally distributed populationthen, lbr large', (, > 50), by rhe CentralI X ) -............X,,and the s.lrnplc mran

0)Xt+X2+..-.........X,-N(np,no'),opproximatelybycentralLimit.l.trcorem.6'(2) X - N(p, - ), approximarety by Ccotral Limir.I.hcorem.

Note:. The approximatior gets better as ,? gets larger_oG is known as thc slazzlard error ofthe sample mean_

cr28 t5 -


16/20

problems involving the sampling distributio

APPLICATION OT TIIE SAMPLING DISTRIBUTION4.1heights of a particular species of plant fbllow a normal distribution with mear 2l cm and90 cm. A random sample of 10 plants is taken and the mean height calculatcd.

the probability that this sdmple mean lies between l8 crn and 27 cm.-tr denotes the hcight ofa plant in cm.N(21, 90)

10. So X N{ 2l .q)c\actly< x < 27)

(3 sis. lis.)

4-2 (Do it yoursel0masscs of guavas are lormzrlly distributed with mean 3809 and standard deviation tog- Thcare packcd into boxes of l2 each. What is thc probability that the mean mass ofthe guavas inbox is more tha0 383g ?

noFFi-3lcdf ( 14" 27,?1,3) . a1a59467S4I

large number ofrandom samples ofsize z are taken from the distribution ofxwhcrcN(74, 36) and thc sample meais are calculatcd.X >'12) = 0.854, find, expressed to thc nearest integeE the value ofr.

a2R t6-


17/20

Soln:Mcthod I-)r- N(7,1, 36).- - N1z+,i9;ynUsing CC : Y,. normalcdltlZ. tgSJq,*l^lx

we car real lrom llre tdble drrt lor t\i lZ1, g g54,nxl0t4111Z1l1t{15

-lqt]\.s5q0E.8655q-a75Sq.ES5t9.893Ell.5{lE5Method 2 t6x-N(74,-:)n11F'22.y:6.s5a

r)rl . - 72 ,70 i=o,ss+l6lI J;'i.e. whcre Z-N(0,1)/ r'plz


18/20

ofc for which P

4.4(Do it yourself)raodom sample of 16 observatioos is to be drawn from a Nomal distribution having mean l l and

deviation 3- L"t X denote the sample mean, Find, corcct to three decimal places, thr:

4.5a random sample of size 50 is taken from each of the following distributions, find, for each case,probability that the sample ntean exceeds 5.x-'Po(4,5),

n.- B(9, 0.5),f '' Po(4,5).

E(-l) = 4.5, Var(X) :4.5sample size is large, by Central Limit Theo.em,7- Nrq.s. '5 )507"' N1a.s, o.oo;cC, P(.> 5):0.0478x.' B(e, 0_s): (eX0.5) :4.5: (9X0.5X0.5) : 2-2s

rlorFlsIr.df (5' E99'4.5,4.3).8477983384

is large, by Central Limit Theorun,) )


19/20

Using GC, P( X > 5) : 0.00921Note: Clontinuitv correctiou is NOT needed as \,vc arc dealing witb sample rnean (whiclr is cotltinuousraldom va ablc)-Ngllg: For questions involving sample mean results, look for thc phrase .,ptqlillliqll3!|lll9_qqlpbmean / arithmetic mean / average valuc".Example 4-6(Do it yourscloA random sample ofsize 60 is taken Aom each of the lbllowing distributioN:(a) x-B(10,0.a)(t) X- Po(a.3).

Dxamplc .1.7'fhc discretc random variable Xis such that E(-\,) : 4.5 and Var(, : 0.75_ Sixty obscrvations of Xarelaken and 7 is the totai sunt ofthe observations, I.ind tho snallest value ofl such that p(f < /) > 0.95Soln:T - Xt+ Xr+ X.+.........X0uSince n = 60 is large, by CLT, f - N(60x4.5,60x0.75)z - N(270,45)P(Z< r) > 0.95From GC, P(z < 281.034) = 0.95'fherefore,t >281.O34

c28- r 9


20/20

Use GC table to cherk answerY= normalc df (- Eqg,X,27O,.,{qSllset table start value to 281..034(since welready have found it earlier)

4.8(l)o it yourself)mcan time spent by childrel ofagc twclve or watching telcvision programmcs inwith a standard deviation of 0,7 hours. What will be the probability that the total100 children on watching television programmes on a Sunday excccds 365 hours?

onc day is 3.5timc spcrt by

.9...ffskHskrL=281.861ffi*5l.s1=9nl:nd!tbFe

BbTdP

TFTIn

i. Introducins Statistics(2"'r Edition) by Craham(Jpton/Ian Cook.2. Advanced Modular Mafhematics. Statistics 2 . for A and AS level 6), Gerltld Westot'er

5.

Statistics For Real Lilb Srunple Surveys by Scrgey Dorofeev / Peter GrantH2 Mathematics. A ComDrehensivo Guide for'A' Level by f'ederick IIo, David Khor, Yui-P'ng Lam, B.S. Ong.South Dast Public llcalth ObscNatory !]!>lIb-Sq4!LIqqiL!L b|, Sf.PHOi [nJbrmarion hyDesign,URl,: lltp: //wvn',.ItJcs ty

3.4.

c28 20 ,

Documents

CHAPTER 28: SAMPLING in This Chapter, Students