3.2 Multiple-start balanced modified systematic sampling
4.2.1 Methodology
The seven cases of RMSS are given in Table 4.1. As is the case with RLSS, we first divide the population into two strata, where the first stratum, ST1, contains the first (n−r)k units and the second stratum,ST2, contains the remaining r(k+ 1) units. We next need to select (n−r) sampling units fromST1 (note that RLSS applies LSS to select (n−r) sampling units using the sampling intervalkinST1) andr sampling units fromST2 (note that RLSS applies LSS to select r sampling units using the sampling interval (k+ 1) in ST2).
Table 4.1: Possible cases of RMSS.
n k r (n−r) Case
even even even even A
even odd even even A
odd even odd even B
odd odd even odd C
odd odd odd even D
odd even even odd E
even even odd odd F
even odd odd odd G
We know that MSS is a linear trend free sampling design if the sample size is even.
Likewise, CESS is a linear trend free sampling design if the sampling interval is odd.
Thus, if we consider Table 4.1, we can easily deduct that MSS should be applied in ST1 for Cases A, B and D, as (n−r) is even, as well as inST2 for Cases A, C and E, as r is even. Similarly, CESS should be applied in ST1 for Cases C and G, as k is odd, as well as inST2 for Cases B and F, as (k+ 1) is odd. Note that if both MSS and CESS offer linear trend free sampling, then MSS is preferred since MSS is a randomized design, unlike CESS which requires no randomization. Thus, we are left with Cases E and F forST1 and Cases D and G forST2, of which we will apply MSS. Hence, we can expect RMSS to be a linear trend free sampling design for Cases A, B and C. Note that each case of RMSS
entails some form of randomized sampling. Selecting a sample of sizenfrom a population of sizeN =nk+r using RMSS, consists of the following steps:
(i) Select two random startsk1 and k2 from 1 tokand 1 to k+ 1, respectively.
(ii) The sample chosen from ST1 is given by
sk1={Uk1+jk, U(n−r)k−jk−k1+1|j= 0, ...,(n−r−2)/2}, for A, B and D
={U[(2j−1)k+1]/2|j = 1, ..., n−r}, for C and G
={Uk1+(n−r−1)k/2}
∪ {Uk1+jk, U(n−r)k−jk−k1+1|j= 0, ...,(n−r−3)/2}, for E and F.
(iii) The sample chosen from ST2 is given by
sk2={Uk2+j(k+1)+(n−r)k, UN−j(k+1)−k2+1|j= 0, ...,(r−2)/2}, for A, C and E
={Uj+k(j+n−r−1/2)|j= 1, ..., r}, for B and F
={Uk2+N−(r+1)(k+1)/2}
∪ {Uk2+j(k+1)+(n−r)k, UN−j(k+1)−k2+1|j= 0, ...,(r−3)/2}, for D and G.
(iv) The final sample of size nis given by s=sk1∪sk2.
Ifn−r = 1, then the sampling unit for Cases E and F in ST1, is obtained by randomly selecting a unit from the firstkunits. Similarly, ifr= 1, then the sampling unit for Cases D and G inST2, is obtained by randomly selecting a unit from the last k+ 1 units.
For RMSS, the first-order inclusion probability for unit Uq is given by πq = 1/k, ifq ∈ {1, ...,(n−k)r} for A, B, D, E and F
= 1/(k+ 1), ifq ∈ {(n−k)r+ 1, ..., N} for A, C, D, E and G
= 1, ifq ∈ {[(2j−1)k+ 1]/2|j= 1, ..., n−r}for C and G
= 1, ifq ∈ {j+k(j+n−r−1/2)|j= 1, ..., r}for B and F
= 0, otherwise.
Note thatπq = 0 or 1, for Cases C and G inST1 and B and F in ST2, as CESS is applied in these instances. Now, let us denote thek1th sample mean fromST1 asyk1 and thek2th sample mean from ST2 as yk2, which are estimates of the stratum means from ST1 and ST2 (denoted byY1 and Y2), respectively. Thus, the sample mean for RMSS is obtained
by using the first-order inclusion probabilities on the Horvitz-Thompson (1952) estimator, i.e.
YˆHT = (n−r)kyk1+r(k+ 1)yk2
N =yRM SS. (4.6)
Note that estimatoryRM SS is an unbiased estimator of the population mean for Cases A, D and E, since πq 6= 0 for all q, i.e. MSS is applied in both ST1 and ST2 for Cases A, D and E. Likewise, estimatoryRM SS is biased for Cases B, C, F and G, since πq = 0 for someq, i.e. CESS is applied in either ST1 orST2 for Cases B, C, F and G. However, if we consider Cases B, C, F and G, we can easily verify that under the assumption of a perfect linear trend in the population, estimatoryRM SS is unbiased for Cases B and C, as yk1 =Y1 andyk2 =Y2, i.e. for Case B, MSS is applied inST1 where (n−r) is even and CESS is applied inST2 where (k+ 1) is odd. Likewise, for Case C, CESS is applied in ST1 wherek is odd and MSS is applied in ST2 wherer is even. Thus, in each case, both designs offer linear trend free sampling and hence RMSS is a linear trend free sampling design.
Let,
sa={k1 +jk,(n−r)k−jk−k1 + 1|j= 0, ...,(n−r−2)/2}, sb ={[(2j−1)k+ 1]/2|j= 1, ..., n−r},
sc={k1 + (n−r−1)k/2}
∪ {k1 +jk,(n−r)k−jk−k1 + 1|j= 0, ...,(n−r−3)/2},
sd={k2 +j(k+ 1) + (n−r)k, N −j(k+ 1)−k2 + 1|j= 0, ...,(r−2)/2}, se={j+k(j+n−r−1/2)|j = 1, ..., r},
sf ={k2 +N−(r+ 1)(k+ 1)/2}
∪ {k2 +j(k+ 1) + (n−r)k, N −j(k+ 1)−k2 + 1|j= 0, ...,(r−3)/2}.
Thus, the second-order inclusion probabilities, πqz, for the pair of units (Uq, Uz), q, z ∈
{1, ..., N}(q6=z), are given as follows:
Case A: πqz = 1/k, ifq and z∈sa
= 1/k(k+ 1), ifq∈sa and z∈sd, orq ∈sd andz∈sa
= 1/(k+ 1), ifq and z∈sd
= 0, otherwise.
Case B: πqz = 1/k, ifq and z∈sa
= 1/k, ifq∈sa and z∈se, orq∈se and z∈sa
= 1, ifq and z∈se
= 0, otherwise.
Case C:πqz = 1/(k+ 1), ifq and z∈sd
= 1/(k+ 1), ifq∈sb and z∈sd, orq∈sdand z∈sb
= 1, ifq and z∈sb
= 0, otherwise.
Case D: πqz = 1/k, ifq and z∈sa
= 1/k(k+ 1), ifq∈sa and z∈sf, orq ∈sf and z∈sa
= 1/(k+ 1), ifq and z∈sf
= 0, otherwise.
Case E:πqz = 1/k, ifq and z∈sc
= 1/k(k+ 1), ifq ∈sc and z∈sd, orq∈sd and z∈sc
= 1/(k+ 1), ifq and z∈sd
= 0, otherwise.
Case F: πqz = 1/k, ifq and z∈sc
= 1/k, ifq ∈sc and z∈se, orq ∈se and z∈sc
= 1, ifq and z∈se
= 0, otherwise.
Case G:πqz = 1/(k+ 1), ifq and z∈sf
= 1/(k+ 1), ifq ∈sb and z∈sf, orq∈sf and z∈sb
= 1, ifq and z∈sb
= 0, otherwise.
Hence, it is impossible to obtain an unbiased estimate of the variance of estimatoryRM SS, as certain second-order inclusion probabilities,πqz, will be zero for each Case. Also, note that the variance of the sample mean is unobtainable when conducting CESS, as there is only one possible sample selected, thus the variance of estimator yRM SS is undefined for Cases B, C, F and G.