Two-stage sampling plansfor clustered population

Jiahua Chen

This is prepared for Stat 344, Chapter 8

2014

Jiahua Chen Week10a

Equal-sized clusters first

Assume that the finite population is made of N clusters ofequal size M.

Consider a sampling plan that is carried out in two stages.

an SRSWOR of n clusters from N clusters.within each sampled cluster, an SRSWOR of m elements.

Observations are denoted as yij for j = 1, 2, . . . ,m, andi = 1, 2, . . . , n.

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What is new here?

Unlike the “single stage” clustered sampling plan, we do not obtainresponse values on all M elements in the sample cluster.

Instead, only a subset of this cluster are inspected with theirresponse values measured.

This is the reason why this sample plan is a “two-stage” plan.

Jiahua Chen Week10a

What is new here?

Unlike the “single stage” clustered sampling plan, we do not obtainresponse values on all M elements in the sample cluster.

Instead, only a subset of this cluster are inspected with theirresponse values measured.

This is the reason why this sample plan is a “two-stage” plan.

Jiahua Chen Week10a

Inference under two-stage sampling plan

Inference targets are population cluster level mean Y orelement level mean ¯Y = Y /M.

We estimate ¯Y by the most straight estimators:

¯y =n∑

i=1

m∑j=1

yij/(nm) = n−1n∑

i=1

yi

with yi = m−1∑m

j=1 yij .

Estimating Y by M ¯y .

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Unbiasedness

We make use of the fact E (¯y) = E{E (¯y |C )} which is a formulamany of us may not remember.

Intuitively, this formula says that the ”average value of ¯y can becomputed in two steps:

obtain the average of ¯y in each instance of C ;

take the average of the above average over instances of C .

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Unbiasedness

Let the instance of C being that the 9th cluster in the populationis sampled. If so,

E (y9|the 9th cluster) =∑M

j=1 y9j = Y9.

Any cluster in the population could be sampled and the chancesare the same, hence,

E (yi , any i) = N−1∑N

j=1 Yi = ¯Y .

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Unbiasedness

Because ¯y is the average of yi over an srswor of size n:

E (¯y) = n−1n∑

i=1

E (yi ) = ¯Y .

This gives us unbiasedness of ¯y as an estimator of ¯Y undertwo-stage cluster sampling plan with equal cluster size.

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Variance

How does the randomness of yi affect the variance Var( ¯Y )?

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Variance formula (8.4)

We plan to use Var(¯y |C ) = Var(E{¯y |C}) + E{Var(¯y |C )} toget Var(¯y).

We have interpreted E{¯y |C} as computing the conditionalexpectation of the cluster sample mean yi given cluster i .

If this i = 9, then E (y9| 9th cluster) = M−1∑M

j=1 y9j = Y9.

This consideration leads to

E{¯y |C} = E (n−1n∑

i=1

yi |C ) = n−1n∑

i=1

Yi .

Jiahua Chen Week10a

Refreshing the purpose now: try to getVariance formula (8.4)

Do not forget: usingVar(¯y |C ) = Var(E{¯y |C}) + E{Var(¯y |C )} to get Var(¯y).

The first task is to compute E{¯y |C} which has been done:

E (yi | ith cluster) = Yi .

E (¯y |C ) = n−1∑n

i=1 Yi .

We now move to

Var (E (¯y |C )) = Var(n−1∑

Yi ).

Jiahua Chen Week10a

Variance of E (¯y | clusters)

What is random in E (¯y |C ) = n−1∑

Yi?

The cluster means Yi are not random;

Which subset of clusters are included in∑n

i=1 is random.

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Variance of E (¯y | clusters)

Because the clusters in the sample are obtained based onSRSWOR, we get

Var(n−1∑

Yi ) = (1

n− 1

N)S2

1

where

S21 =

1

N − 1

N∑i=1

(Yi − ¯Y )2

and this is the population variance of the cluster mean response.

More directly, S21 defined above tells us how different the cluster

means are in this population

Jiahua Chen Week10a

Variance of n−1∑

Yi

Worth to remind us again: usingVar(¯y |C ) = Var(E{¯y |C}) + E{Var(¯y |C )} to get Var(¯y).

We have succeeded on the first step at

Var(E{¯y |C}) = Var(n−1∑

Yi ) = (1

n− 1

N)S2

1 .

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Next task

The next step is on E{Var(¯y |C )}.

Recall that,

conditioning on C is to have clusters fixed.

So Var(¯y |C ) is the variation in ¯y when the clusters in the sampleare fixed.

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Variation given a single cluster

Suppose the 9th cluster is in the sample.

How much is Var(y9)?

We note y9 is the sample mean based on m elements obtainedSRSWOR from M elements in the 9th cluster of the population.

This tells us

Var(y9) = (1

m− 1

M)S2

2,9

where S22,9 is the variation in the 9th cluster.

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When clusters are given, we have

Var(¯yn|C ) = Var(n−1n∑

i=1

yi |C )

= n−2(1

m− 1

M)

n∑i=1

S22,i .

This is Var(¯y |C ) in Var(¯y) = Var(E{¯y |C}) + E{Var(¯y |C )}.

The final step to taking average over all possible samples of nclusters.

Jiahua Chen Week10a

Taking average over all possible sample of n clusters,

E{Var(¯yn|C )} = E{n−2(1

m− 1

M)

n∑i=1

S22,i}

= n−1(1

m− 1

M)

{N−1

N∑i=1

S22,i

}

= n−1(1

m− 1

M)S2

2

where

S22 = N−1

N∑i=1

S22,i

is the population average of the within cluster variations.

Jiahua Chen Week10a

We obtained

Var(E{¯yn|C}) = (1

n− 1

N)S2

1

earlier, and from the last slide

E{Var(¯yn|C )} = n−1(1

m− 1

M)S2

2 .

The formula Var(¯y) = Var(E{¯y |C}) + E{Var(¯y |C )} leads to

Var(¯y) = (1

n− 1

N)S2

1 + n−1(1

m− 1

M)S2

2 .

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Variance estimation

We estimate these s2 by their sample versions:s21 = (n − 1)−1

∑ni=1(yi − ¯y)2 and s22 = n−1

∑ni=1 s

22i .

An unbiased estimator of Var(¯y) is given by

v(¯y) = (1

n− 1

N)s21 + n−1(

1

m− 1

M)s22 .

Jiahua Chen Week10a

Sample size determination

How do we determine the sample size required to meet someprecision/cost requirements?

The answer: solving the inequality based on whateverrequirements are posted.

Jiahua Chen Week10a

Sample size determination

How do we determine the sample size required to meet someprecision/cost requirements?

The answer: solving the inequality based on whateverrequirements are posted.

Jiahua Chen Week10a

Unequal cluster sizes

We now move to unequal cluster size situation for two-stagesampling plan.

The finite population is made of N clusters (primary samplingunits).

The clusters have different numbers of elements (ultimatesampling units).

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Two-stage sampling design

Two-stage sampling plan is as follows.

in the first stage, select an SRSWOR of n primary units;(regardless of their sizes).

in the second stage, select mi elements srswor from the ithcluster in the sample with Mi elements.

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Two-stage sampling design

We have implicitly assumed Mi is known once this cluster isselected.

We have also assumed mi is somehow pre-determined.

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Response values

The response value is again denoted as yij , for the ith clusterand jth element in that cluster.

At population level, i = 1, 2, . . . ,N and j = 1, 2, . . . ,Mi .

At sampling level, i = 1, 2, . . . , n and j = 1, 2, . . . ,mi .

Be aware the abuse of notation: y13 at the population level isnot y13 at the sampling level.

Jiahua Chen Week10a

Response values

The response value is again denoted as yij , for the ith clusterand jth element in that cluster.

At population level, i = 1, 2, . . . ,N and j = 1, 2, . . . ,Mi .

At sampling level, i = 1, 2, . . . , n and j = 1, 2, . . . ,mi .

Be aware the abuse of notation: y13 at the population level isnot y13 at the sampling level.

Jiahua Chen Week10a

Population parameters

We use Yi =∑Mi

j=1 yij for cluster total. The population total

is Y =∑N

i=1 Yi .

The average response of the ith cluster is Yi = Yi/Mi .

The population mean of the cluster average isY = N−1

∑Ni=1 Yi .

The population mean of response at element level is

¯Y =

∑Ni=1 Yi∑Ni=1Mi

.

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Population parameters

When Mi ’s are not all equal, ¯Y does not related to Yi in asimple way but

¯Y =

∑Ni=1Mi Yi∑Ni=1Mi

.

Within cluster variations remain as

S22i = (Mi − 1)−1

Mi∑j=1

(yij − Yi )2.

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Summary statistics related to populationparameters

We use

yi = m−1i

mi∑j=1

yij

for sample mean of the responses in the ith sampling unit.

We estimate population mean of elements ¯Y by

ˆYR =

∑ni=1Mi yi∑ni=1Mi

.

This estimator estimates Yi by Mi yi

It then takes average over the clusters in the sample to get anestimate of the population mean at the element level.

Jiahua Chen Week10a

Summary statistics related to populationparameters

We estimate Y =∑N

i=1 Yi naturally by

YR = (N∑i=1

Mi )ˆYR .

Notice that subscript R is used here because the estimator isa ratio type estimator.

Jiahua Chen Week10a

Variance formulas

The variance formulas are given as (8.9) and (8.10) in thetextbook.

Var(YR) ≈ N2(1/n − 1/N)N∑i=1

M2i (Yi − ¯Y )2/(N − 1)

+(N/n)N∑i=1

M2i (1/mi − 1/Mi )S

22i

v(YR) = N2(1/n − 1/N)n∑

i=1

M2i (yi −

ˆYR)2/(n − 1)

+(N/n)n∑

i=1

M2i (1/mi − 1/Mi )s

22i

First remark: no need to memorize them, not insightful.The last bonus question of this course: why the first formulais an approximation but the second one is not?

Jiahua Chen Week10a

Variance formulas

The variance formulas are given as (8.9) and (8.10) in thetextbook.

Var(YR) ≈ N2(1/n − 1/N)N∑i=1

M2i (Yi − ¯Y )2/(N − 1)

+(N/n)N∑i=1

M2i (1/mi − 1/Mi )S

22i

v(YR) = N2(1/n − 1/N)n∑

i=1

M2i (yi −

ˆYR)2/(n − 1)

+(N/n)n∑

i=1

M2i (1/mi − 1/Mi )s

22i

First remark: no need to memorize them, not insightful.The last bonus question of this course: why the first formulais an approximation but the second one is not?

Jiahua Chen Week10a

Variance formulas

The variance formulas are given as (8.9) and (8.10) in thetextbook.

Var(YR) ≈ N2(1/n − 1/N)N∑i=1

M2i (Yi − ¯Y )2/(N − 1)

+(N/n)N∑i=1

M2i (1/mi − 1/Mi )S

22i

v(YR) = N2(1/n − 1/N)n∑

i=1

M2i (yi −

ˆYR)2/(n − 1)

+(N/n)n∑

i=1

M2i (1/mi − 1/Mi )s

22i

First remark: no need to memorize them, not insightful.The last bonus question of this course: why the first formulais an approximation but the second one is not?

Jiahua Chen Week10a