This document provides an overview of topics related to sampling with unequal probabilities, including sampling one primary sampling unit and one-stage sampling with replacement. It discusses estimating population totals using weighted estimators when probabilities of selection vary between units. The example shown illustrates calculating weights as the inverse of selection probabilities and estimating the total for a population based on data from a single sampled store. Key points are that weighted estimators are unbiased and have lower variance when probabilities are related to size measures like number of beds or students.

1. Statistics 522: Sampling and Survey Techniques
Topic 6
Topic Overview
This topic will cover
• Sampling with unequal probabilities
• Sampling one primary sampling unit
• One-stage sampling with replacement
Unequal probabilities
• Recall πi is the probability that unit i is selected as part of the sample.
• Most designs we have studied so far have the πi equal.
• Now we consider general designs where the πi can vary with i.
• There are situations where this can give much better results.
Example 6.1
• Survey of nursing home residents in Philadelphia to determine preferences on life-
sustaining treatments
• 294 nursing homes with a total of 37,652 beds (number of residents not known at the
planning stage)
• Use cluster sampling
• Suppose we choose an SRS of the 294 nursing homes and then an SRS of 10 residents
of each selected home.
• A nursing home with 20 beds has the same probability of being sampled as a nursing
home with 1000 beds.
• 10 residents from the 20 bed home represent fewer people than 10 residents from 1000
bed home.
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2. Self-weighting
• This procedure gives a sample that is not self-weighted.
• Alternatives that are self-weighted.
– A one-stage cluster sample
– Sample a fixed percentage of the residents of each selected nursing home.
The two-stage cluster design
• The two-stage cluster design (SRS of homes, then equal proportion SRS of residents
in each selected home)
– Gives a mathematically valid estimator
SRS at first stage
Three shortcomings:
• We would expect ti to be proportional to the number of beds in nursing home i, so
estimators will have large variance (Mi ).
• Equal percentage sampling in each selected home may be difficult to administer.
• Cost is not known in advance (dont know if you will get large or small homes in sample).
The study
• They drew a sample of 57 nursing homes with probabilities proportional to the number
of beds.
• Then they took an SRS of 30 beds (and their occupants) from a list of all beds within
each selected nursing home.
Properties
• Each bed is equally likely to be in the sample (note beds vs occupants).
• The cost is known before selecting the sample.
• The same number of interviews is taken at each nursing home.
• The estimators will have smaller variance
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3. Key ideas
• When sampling with unequal probabilities, we deliberately vary the selection proba-
bilities.
• We compensate by using weights in the estimation.
• The key is that we know the selection probabilities
Notation
• The probability that psu i is in the sample is πi .
• The probability that psu i is selected on the first draw is ψi .
• We will consider an artificial situation where n = 1, so πi = ψi .
Sampling one psu
• Sample size is n = 1.
• Suppose we are interested in estimating the population total.
• ti is the total for psu i.
• To illustrate the ideas, we will assume that we know the whole population.
The Example
• N = 4 supermarkets
• Size (in square meters) varies.
• Select n = 1 with probabilities proportional to size.
• Record total sales
• Using the data from one store we want to estimate total sales for the four stores in the
population.
The population
Store Size ψi ti
A 100 1/16 11
B 200 2/16 20
C 300 3/16 24
D 1000 10/16 245
Total 1600 1 300
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4. Weights
• The weights wi are the inverses of the selection probabilities ψi .
• The weighted estimator of the population total is tψ =
ˆ wi ti .
• There are four possible samples.
• We calculate tψ for each.
ˆ
The samples
Sample ψi wi ti ˆ
tψ
A 1/16 16 11 176
B 2/16 8 20 160
C 3/16 16/3 24 128
D 10/16 16/10 245 392
ˆ
Sampling distribution of the estimate tψ
Sample ψi ˆ
tψ
1 1/16 176
2 2/16 160
3 3/16 128
4 10/16 392
ˆ
Mean of the sampling distribution of tψ
ˆ 1 2 3 10
E tψ = 176 + 160 + 128 + 392 = 300 = t
16 16 16 16
• So tψ is unbiased.
ˆ
• This will always be true.
ˆ
E tψ = ψi wi ti = ti
ˆ
Variance of the sampling distribution tψ
1 2 3 10
ˆ
Var(tψ ) = (176 − 300)2 + (160 − 300)2 + (128 − 300)2 + (392 − 300)2 = 14248
16 16 16 16
Compare with the variance for an SRS:
1 1 1 1
Var(tSRS ) = (176 − 300)2 + (160 − 300)2 + (128 − 300)2 + (392 − 300)2 = 154488
ˆ
4 4 4 4
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5. Interpretation
• Store D is the largest and we expect it to account for a large portion of the total sales.
• Therefore, we give it a higher probability of being in the sample (10/16) than it would
have with an SRS (1/4).
• If it is selected, we multiply its sales by (16/10) to estimate total sales.
One-stage sampling with replacement
• Suppose n > 1 and we sample with replacement.
• This implies πi = 1 − (1 − ψi )n .
• Probability that item i is selected on the first draw is the same as the probability that
item i is selected on any other draw.
• Sampling with replacement gives us n independent estimates of the population total,
one for each unit in sample.
• We average these n estimates.
• Estimated variance is variance of the estimates divided by n
Example 6.2
• N = 15 classes of elementary stat
• Mi students in class i (i = 1 to 15)
• Values of Mi range from 20 to 100.
• We want a sample of 5 classes.
• Each student in the selected classes will fill out a questionnaire.
• (It is possible for the same class to be selected more than once.)
Randomization
• There are a total of 647 students in these classes.
• Select 5 random numbers between 1 and 647.
• Think about ordering the students by class.
• Each random number corresponds to a student and the corresponding class will be in
the sample.
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6. This method
• This method is called the cumulative-size method.
• It is based on M1 , M1 + M2 , M1 + M2 + M3 , . . .
• An alternative is to use the cumulative sums of the ψi and select random numbers
between 0 and 1.
• For this example, ψi = Mi /647
Alternative
• Systematic sampling is often used as an alternative in this setting.
– The basic idea is the same.
– Not technically sampling with replacement
– Works well as systematic sampling works well.
– See page 186 for details.
• Lahiris method
– Involves two stages of randomization
– Rejection sampling: corresponds to classroom problem in Problem Set 2.
– Can be inefficient.
– See page 187 for details
Estimation Theory
• Let Qi be the number of times unit i occurs in the sample.
1
• Then tψ =
ˆ
n
Qi ti /ψi .
• The estimated variance of ti is
ˆ
1 ti
Qi ( − tψ )2
ˆ
n(n − 1) ψi
• The estimate and its estimated variance are both unbiased.
Choosing the selection probabilities
• We want small variance for our estimator.
– Often, ti is related to the size of the psu.
– We can take ψi proportional to Mi or some other measure of the size of psu i.
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7. PPS
• This procedure is called sampling with probability proportional to size (pps).
• The formulas for the estimate and variance can be simplified for this special case.
Mi
ψi =
K
ti
= K yi
¯
ψi
• See page 190 for details
• See Example 6.5 on pages 190-192
Two-stage sampling with replacement
• Basic ideas are very similar to one-stage sampling.
• ψi is the probability that psu i is selected on the first (or any) draw.
• We take a sample of mi ssus from each selected psu.
Sampling ssu’s
• Usually we use an SRS.
• Alternatives include
– systematic sampling
– any other probability sampling method
• Note if a psu is selected more than once, a separate independent second stage sample
is required.
Estimates and SE’s
• Weights are used to make the estimators unbiased.
• Formulas are similar to those for one-stage.
• See (6.8) and (6.9) on page 192
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8. Outline of the procedure
1. Determine the ψi .
2. Select the n psus (with replacement).
3. Select the ssus.
4. Estimate the t for each selected psu,
tψ = weight × t
ˆ ˆ
ˆ
5. The average of these is tψ .
√
6. SE is the standard error of these (sd/ n).
Unequal probability sampling without replacement
• ψi is the probability of selection on the first draw.
• The probability of selection on later draws depends on which units were selected on
earlier draws.
Estimation
• πi is called the inclusion probability. ( pop πi = n)
• πi,j is the probability that both psu i and psu j are in the sample. ( j=i πi,j = (n−1)πi )
• Weights (inverse of selection probability)
– we use πi /n in place of ψi (with replacement)
• The recommended procedure is to use the Horvitz-Thompson (HT) estimator and the
ˆ ˆ
associated SE. (tHT = sam ti /πi )
• See page 196-197 for details.
• This estimator can be generalized to other designs that do not use replacement.
Randomization Theory
Framework is
• Probability sampling without replacement for the psus for the first stage
• Sampling at the second stage is independent of sampling at the first stage
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9. Horvitz-Thompson
• Randomization theory can be used to prove the Horvitz-Thompson Theorem.
– Expected value of the estimator is t.
– Formula for the variance of the estimator
The estimator
• tHT =
ˆ ˆ
ti /πi
– where the sum is over the psu’s selected in the first stage.
• Idea behind proofs is to condition on which psus are in the sample.
• Study pages 205-210
Model
• One-way random effects anova model
Yi,j = Ai + i,j
where
2
– the Ai are random variables with mean µ and variance σA
– the i,j are random variables with mean 0 and variance σ 2 .
– the Ai and the i,j are uncorrelated
The pps estimator
• πi = nMi /K – the inclusion probability
ˆ K ˆ
TP = Ti
nMi
• We rewrite this as a weighted estimator.
ˆ Mi
ti = Yi,j
mi
ˆ
tP = wi,j Yi,j
K
where wi,j = nMi
• Take expected values to show that the estimator is unbiased.
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10. Variance
• The variance can be computed.
• See page 211
• The variance depends on which psu’s are selected through the Mi .
• The variance is smallest when psu’s with the largest Mi are chosen.
Recall
• Estimate of population total is the weighted average of the ti for the selected psus.
ˆ
• The weights wi are the inverses of the probabilities of selection.
Elephants
• A circus needed to ship its 50 elephants.
• They needed to estimate the total weight of the animals.
• It is not easy to weigh 50 elephants and they were in a hurry.
• They had data from three years ago.
Sample
• The owner wanted to base the estimate on a sample.
• Dumbo had a weight equal to the average three years ago.
• The owner wanted to weigh Dumbo and multiply by 50.
• The statistician said:
NO
• You have to use probability sampling and the Horvitz-Thompson estimator.
• They compromised:
– The probability of selecting Dumbo was set as 99/100.
– The probability of selecting each of the other elephants was 1/4900.
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11. Who was selected
• Dumbo, of course.
• The owner was happy and said now we can estimate the weight of the 50 elephants as
50 times Dumbos weight, 50y.
• The statistician said
NO
• The estimate of the total weight of the 50 elephants should be Dumbos weight divided
by his probability of selection.
• This is y/(99/100) or 100y/99.
• The theory behind this estimator is rigorous
What if
• The owner asked
– What if the randomization had selected Jumbo the largest elephant in the herd?
• The statistician replied 4900y, where y is Jumbos weight.
Conclusion
• The statistician lost his circus job and became a teacher of statistics.
• bad model; highly variable estimator
• Due to Basu (1971).
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