Effective Survey Methods: Solutions to Common Statistical
School
The Chinese University of Hong Kong**We aren't endorsed by this school
Course
STAT 3003
Subject
Statistics
Date
Dec 12, 2024
Pages
2
Uploaded by PrivateHawkMaster4447
THE CHINESE UNIVERSITY OF HONG KONGDepartment of StatisticsSTAT3003: Survey MethodsProblem Sheet 2 - Solutions1. We are asked to estimatep. The design used is stratified random sampling, with proportional allocation. Theappropriate formula forμis¯Yst=1NLXi=1Ni¯YidV ar(¯Yst) =1N2LXi=1N2i(1-niNi)ˆσ2ini.Sincepis just a special case of a population mean with binary response, we note¯Yibecomes ˆpiand ˆσ2i=nini-1ˆpi(1-ˆpi), henceˆpst=1NLXi=1NiˆpidV ar(ˆpst) =1N2LXi=1N2i(1-niNi)ˆpi(1-ˆpi)ni-1.Plugging in the numbers we find ˆpst= 0.3393 anddV ar(ˆpst) = 0.003772 = 0.0614202. In order to provide a95% CI, we need to choose a percentile from thet-distribution. The sample sizes in each stratum are below30, so we use Satterthwaite’s approximation:df=⇣PLh=1khs2h⌘2PLh=1(khs2h)2nh-1=⇣PLh=1khnhnh-1ˆph(1-ˆph)⌘2PLh=1(khnhnh-1ˆph(1-ˆph)2nh-1wherekh=N2hN2(1-nhNh)1nh. Plugging in the numbers givesdf= 45.38, so we takedf= 45. Thet-percentileist45,0.975= 2.014, so our 95% CI is given by 0.3393±2.014⇥0.0614 or (0.216,0.463).2. Given the information it makes sense to apply optimal allocation.We are toldc-c0=c= 500 andN= 112 + 68 + 39 = 219. The appropriate formula isni=(c-c0)NiSi/pciPLi=1NiSipciwhereSi=qNiNi-1σi, the corrected standard deviation. This givesn1= 17.465,n2= 7.657,n3= 4.205 fora total of 29.328.We need to turn these into integers, while making sure the total cost is below 500.Forexample, choosingn1= 17,n2= 8,n3= 4 would cost 497 for a total sample size of 29 which is in budget.The corporation will be happy if the variance of the estimator is below 0.1. The variance is given byV ar(ˆYst) =1N2LXi=1N2iσ2iniNi-niNi-1and we plug into this our integer valuesn1= 17,n2= 8,n3= 4 (not the optimal values). The result is avariance of 0.0955, below the target of 0.1. The corporation has a survey that is on budget and on target forthe variance - they should be happy.1
3. The problem is of the optimization with a constraint.Minimize:V ar(¯Yst) =1N2LXi=1N2iσ2iniNi-niNi-1Subject to:c=c0+LXi=1cini.The Lagrangian isL(n1, . . . , nL,λ) =1N2LXi=1N2iσ2iniNi-niNi-1-λ(c-c0-LXi=1cini).The partial derivatives are@L@ni=-N3iNi-11N2σ2in2i+λci@L@λ=LXi=1cini+c0-cfori= 1, . . . , L. Setting these to zero tells usλcin2i=N3iNi-11N2σ2i)cini=pcipλrNiNi-1NiNσifori= 1, . . . , L. Satisfying the constant suggestsLXi=1cini=c-c0)1pλ=c-c0PLi=1qNiNi-1NiNpciσi.Henceni=(qNiNi-1NiN1pciσi)(c-c0)PLk=1qNkNk-1NkNpckσk=(NiN1pciSi)(c-c0)PLk=1NkNpckSkusing the corrected population varianceS2.d-d±⇥2