1.Cochran, W.G. (1963) Sampling Techniques Survey …

1 STAT3014/3914 Applied Statistics-SamplingPreliminaryReferences 1. Cochran, (1963) Sampling Techniques , Wiley, New Kish, L. (1995) Survey Sampling , Wiley Inter. Lohr, (1999) Sampling : Design and Analysis, Duxbury McLennan, W. (1999)An Introduction to Sample Surveys, , outline1. Simple random samples and population correction factor. Sample size determination. In-ference over Stratified Ratio and regression estimators. Hartley-Ross estimator. Ratio estimator for strat-ified samples. Regression Systematic Sampling and cluster Sampling with unequal proportional to size(PPS) Sampling . The STAT3014 (2015) Second semesterDr. J. Chan1 STAT3014/3914 Applied random sample1 Simple Random Samples (SRS) PopulationWe have a finite number of elements,NwhereNisassumed population ,whereYiis a numerical value associated withi-th element.

2 We adopt the notation where capital letters refer to char-acteristics of the population; small letters are used for the correspondingcharacteristics of a Total:Y=N i=1Yi,Population Mean: = Y=YN=1NN i=1Yi,Population Variance: 2=1NN i=1(Yi Y)2andS2=NN 1 2=1N 1N i=1(Yi Y) are fixed (population) quantities, to be we have to consider two numerical values: (Yi,Xi), i= 1, ,Nan additional population quantity of interest isR=N i=1 YiN i=1Xi=YX= Y Xtheratio of STAT3014 (2015) Second semesterDr. J. Chan2 STAT3014/3914 Applied random Random SamplingFocus on the numerical valuesYi, i= 1, ,N. A random sampleof sizenis takenwithout replacement: the observed valuesy1, ,ynarerandom variablesand arestochastically dependent. Thesamplingframeis a list of the valuesYi, i= 1, , estimatorfor Yis Y=1nn i=1yi= yand hence forY=N is Y=N properties of yare complicated by the dependence of theyi variance:s2=1n 1n i=1(yi y)2 Fundamental ResultsE( y) =.

3 Var( y) =(1 nN)S2n= (1 f)S2n=(N nN 1) 2nvar( y) =(1 nN)s2nE(s2) =S2wherefis the Sampling fraction and the finite population correction( ) is 1 STAT3014 (2015) Second semesterDr. J. Chan3 STAT3014/3914 Applied random sampleProof:Lety=1n i Syi=1nN i=1yiIiwhere the sample membershipindicatorIi={1 if elementiis in the sample,0 if , we haveE(Ii) = 0 Pr(Ii= 0) + 1 Pr(Ii= 1) = i=nN,E(I2i) = 02 Pr(Ii= 0) + 12 Pr(Ii= 1) = i=nN,E(IiIj) = 0 0 Pr(Ii= 0&Ij= 0) + 0 1 Pr(Ii= 0&Ij= 1) +1 0 Pr(Ii= 1&Ij= 0) + 1 1 Pr(Ii= 1&Ij= 1)= ij=n(n 1)N(N 1)Var(Ii) =E(I2i) E2(Ii) = i(1 i) =nN(1 nN),Cov(Ii,Ij) =E(IiIj) E(Ii)E(Ij) = ij i j=n(n 1)N(N 1) (nN) (y) =1nN i=1yiE(Ii) =1nN i=1yi nN=1NN i=1yi=Y .UnbiasedE[var(y)] [(N nN)s2yn] =(N nN)S2yn Var(y) 1:show that Var(y) =(N nN) STAT3014 (2015) Second semesterDr.}

4 J. Chan4 STAT3014/3914 Applied random sampleVar(y) = Var(1nN i=1yiIi)=1n2[N i=1y2iVar(Ii) + 2 i j,i<jyiyjCov(Ii,Ij)]=1n2{N i=1y2i[nN(1 nN)]+ 2 i j,i<jyiyj[n(n 1)N(N 1) (nN)2]}=nn2{1N(1 nN)N i=1y2i+ 21N(n 1N 1 nN) i j,i<jyiyj}=1n(1 nN){1NN i=1y2i+ 21N(1 nN) 1N(n 1) n(N 1)N(N 1) i j,i<jyiyj}=1n(1 nN){1NN i=1y2i+ 21 NNN nn NN(N 1) i j,i<jyiyj}=1n(1 nN)1N 1{N 1NN i=1y2i 21N i j,i<jyiyj}=1n(1 nN)1N 1{N i=1y2i 1N(N i=1y2i+ 2 i j,i<jyiyj)}=1n(1 nN)1N 1{N i=1y2i 1N(N i=1yi)(N i=1yi)}=(1 nN)S2yn=N nNNN 1 2yn=(N nN 1) 2ynProof 2:show thatE(s2y) =S2ywhereS2y=1N 1N i=1(yi Y)2=NN 1 (s2y) =E[1n 1n i=1(yi y)2]=1n 1E{n i=1[(yi Y) (y Y)]2}SydU STAT3014 (2015) Second semesterDr. J. Chan5 STAT3014/3914 Applied random sample=1n 1E[n i=1(yi Y)2 n(y Y)2]=1n 1[n i=1E(yi Y)2 nE(y Y)2]=1n 1[n i=1 Var(yi) nVar(y)]=1n 1[n 2y n(N nN 1) 2yn]=(n N nN 1) 2yn 1=(nN n N+nN 1) 2yn 1=NN 1 2y=S2yCentral Limit PropertyFor large sample sizen(n >30 say), and small to moderatef,we havethe approximation( y )/ Var( y) N(0,1).

5 Confidence Interval for = YReplacingS2bys2,an approximate 95% for andYarerespectively y n 1 f,N( y n 1 f)Read Tutorial 10 STAT3014 (2015) Second semesterDr. J. Chan6 STAT3014/3914 Applied random sampleExample:(Industrial firm) An industrial firm is concerned about thetime spent each week by staff on certain tasks. The time-log sheets of aSRS ofn= 50 employees show the average amount of time spent on thesetasks is hours, with a sample variances2= The companyemploysN= 750 staff. Estimate the total number of man-hours usedeach week on the tasks and construct a 95% CI for the :FromN= 750 time-log sheets, a SRS ofn= 50 sheets wasobtained. The average amount of time used in the sample isy= Sincen= 50 is large, we Hence Y=Ny= 750 = hoursvar( Y) =N2(1 nN)s2n= 7502(1 50750) 23,625se( Y) = 23,625 = CI forY= (Ny se( Y), Ny+ se( Y))= ( , + )= ( , )SydU STAT3014 (2015) Second semesterDr.

6 J. Chan7 STAT3014/3914 Applied random Random Sampling for method for SRS can be applied to estimate thetotal number, orproportion(or %) of units which possess some qualitative attribute. Letthis subset of the population 1 ifi C= 0 ifi / Cand similarly foryi s .Customary notation: = Y=Pis the population proportion,Y=N i=1Yi=NPis the population total count, y=pis the sample i=1yi=np, Q= 1 Pandq= 1 Y=Np= i=1(Yi Y)2=N i=1Y2i N Y2=N Y N Y2=NP(1 P),we haveS2=NP(1 P)/(N 1) P(1 P)and similarly,s2=np(1 p)/(n 1) p(1 p) ands2n=np(1 p)n(n 1)=p(1 p)n 1 p(1 p)nSydU STAT3014 (2015) Second semesterDr. J. Chan8 STAT3014/3914 Applied random Size CalculationsTo calculate the sample size needed for samplingyet to be carried out,we want to be at least 100(1 )% sure the estimate yof is within100 % of the actual value of ( 1 = , z /2= ).

7 ThatisPr{| y | } 1 Pr(| y | Var( y) Var( y)) 1 / Var( y) z /2 (1 nN)S2n ( )2z2 /2 S2n S2N ( )2z2 /2 S2n ( )2z2 /2+S2N n S2( )2z2 /2+S2N n NS2N( )2/z2 /2+S2qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqq 0 yzNormalStandard normal + SE( y) SE( y)Area 1 z /2= -Ignoring ( takingf= 0 or Var( y) =S2/nwhenNis unknown)n z2 /2S2( )2 z2 /2s2( y)2wheres2and yare estimates from a pilot Survey ands2 p(1 p) STAT3014 (2015) Second semesterDr. J. Chan9 STAT3014/3914 Applied random sampleExample:(blood group)1. What size sample must be drawn from a population of sizeN= 800in order to estimate the proportion with a given blood group towithin ( an absolute error of 4%) with probability What sample size is needed if we know that the blood group is presentin no more than 30% of the population?

8 Solution:1. We haveN= 800,S2' p(1 p) = , = Note thatS2=p(1 p) is max atp= and thisgives the most (1 p) NS2N 2 /z2 /2+S2=800( )800( ) + NowS2=p(1 p) = ( ) = NS2N 2 /z2 /2+S2=800( )800( ) + Tutorial 10 STAT3014 (2015) Second semesterDr. J. Chan10 STAT3014/3914 Applied random over Subpopulations-PoststratificationMotivat ing example:(dentist) There are 200 children in a dentist takes a simple random sample of 20 and finds 12 childrenwith at least one decayed tooth and a total of 42 decayed teeth. Anotherdentist quickly checks all 200 children and finds 60 with no decayed the total number of decayed : 1 decayed teethC2: no decayed teeth TotalN1= 140N2= 60N= 200n1= 12n2= 8n= 20 i C1yi= 42 =n i=1y i'&$%'&$%C1n1= 12C2n2= 8N1= 140N2= 60 i C1yi= 42 From a population ofNindividuals,onesimple random sample ofnindividualsyi, i= 1, ,nis drawn.

9 Separate estimates might bewanted for one of a number of subclasses{C1,C2, }which aresubsetsof the population ( Sampling frame) :1. Unavailability of a suitable Sampling frame for each stratum eventhough the stratum sizesN1,..,NLare often obtainable from offi-cial Inability to classify population elements into an appropriate stratumwithout actual contact, personal characteristics such as educational level and politicalpreference and household characteristics such as owned/rented ac-commodation, income level and household size are unknown3. Multi-variate and multi-purpose nature of most Post-stratification is to correct the distorted sample proportion dueto STAT3014 (2015) Second semesterDr. J. Chan11 STAT3014/3914 Applied random sampleExample:POPULATION ( Sampling FRAME)SUBPOPULATIONA ustralian populationunemployed Queenslandersretailerssupermarketsthe employedthe employed working overtimeSolution:The estimate for theoverallaverage no.

10 Of decay teeth usingoverall sample meanis Y= y =1n i C1yi=4220= estimate for theoverallorconditionaltotal number of decayedteeth,Yis Ypst,1m1=N y = 200 = 420,ignoring the information ofn1= 12 andN1= 140 from the seconddentist. Using these information and condition onthose with at leastone decayed teeth, the average no. of decay teeth is Ypst,1m1=N y N1=200( )140= 3 Alternative estimate for the average no. of decayed teeth using thecon-ditional sample meanis Ypst,1m2= y1=n y n1=4212= the estimate for the total no. of decayed teeth is Ypst,1m2=N1 y1= 140( ) = 490 Read Tutorial 10 STAT3014 (2015) Second semesterDr. J. Chan12 STAT3014/3914 Applied random sampleFormulae:Denote the total number of items in classClbyNl.