3 STRATIFIED SIMPLE RANDOM SAMPLING
• Suppose the population is partitioned into disjoint sets of sampling units called strata.
If a sample is selected within each stratum, then this sampling procedure is known as
stratified sampling.
• If we can assume the strata are sampled independently across strata, then
(i) the estimator of t or y
U
can be found by combining stratum sample sums or means
using appropriate weights
(ii) the variances of estimators associated with the individual strata can be summed
to obtain the variance an estimator associated with the whole population. (Given
independence, the variance of a sum equals the sum of the individual variances.)
• (ii) implies that only within-stratum variances contribute to the variance of an estimator.
Thus, the basic motivating principle behind using stratification to produce an estimator
with small variance is to partition the population so that units within each stratum are as
similar as possible. This is known as the stratification principle.
• In ecological studies, it is common to stratify a geographical region into subregions that are
similar with respect to a known variable such as elevation, animal habitat type, vegetation
types, etc. because it is suspected that the y-values may vary greatly across strata while
they will tend to be similar within each stratum. Analogously, when sampling people, it
is common to stratify on variables such as gender, age groups, income levels, education
levels, marital status, etc.
• Sometimes strata are formed based on sampling convenience. For example, suppose a
large study region appears to be homogeneous (that is, there are no spatial patterns) and
is stratified based on the geographical proximity of sampling units. Taking a stratified
sample ensures the sample is spread throughout the study region. It may not, however,
lead to any significant reduction in the variance of an estimator.
• But, if the y-values are spatially correlated (y values tend to be similar for neighboring
units), geographically determined strata can improve estimation of population parameters.
Notation: H = the number of strata
N
h
= number of population units in stratum h h = 1, 2, . . . , H
N =
P
H
h=1
N
h
= the number of units in the population
n
h
= number of sampled units in stratum h h = 1, 2, . . . , H
n =
P
H
h=1
n
h
= the total number of units sampled
y
hj
= the y-value associated with unit j in stratum h
y
h
= the sample mean for stratum h
t
h
=
N
h
X
j=1
y
hj
= stratum h total t =
H
X
h=1
N
h
X
j=1
y
hj
=
H
X
h=1
t
h
= the population total
y
hU
=
t
h
N
h
= stratum h mean y
U
=
1
N
H
X
h=1
N
h
X
j=1
y
hj
=
t
N
= the population mean
45
• If a simple random sample (SRS) is taken within each stratum, then the sampling design
is called stratified simple random sampling.
• For stratum h, there are
N
h
n
h
possible SRSs of size n
h
. Therefore, there are
N
1
n
1
N
2
n
2
···
N
H
n
H
possible stratified SRSs for specified stratum sample sizes n
1
, ··· , n
H
.
• If S
strat
is a stratified SRS, then the probability of selecting S
strat
is
P (S
strat
) =
H
Y
h=1
1
N
h
n
h
=
1
N
1
n
1
N
2
n
2
···
N
H
n
H
• Thus, every possible stratified SRS having stratum sample sizes n
1
, ··· , n
H
has the same
probability of being selected.
3.1 Estimation of y
U
and t
• Because a SRS was taken within each stratum, we can apply the estimator formulas for
simple random sampling to each stratum. We can estimate each stratum population mean
y
hU
and each stratum population total t
h
. The formulas are:
d
y
hU
= y
h
=
1
n
h
n
h
X
j=1
y
hj
b
t
h
= N
h
y
h
= (24)
• Because each
b
t
h
is an unbiased estimator of the stratum total t
h
for i = 1, 2, . . . , k, their
sum will be an unbiased estimator of the population total t. That is,
b
t
str
=
is an unbiased estimator of t. An unbiased estimator of y
U
is a weighted average of the
stratum sample means
c
y
U
str
=
b
t
str
N
=
1
N
H
X
h=1
N
h
y
h
or, equivalently,
c
y
U
str
=
where is the weighting factor for stratum h.
• Before we can study V (
b
t
str
) and V (
c
y
U
str
), we need to look at the within-stratum variances.
• Because a SRS is taken within stratum h, we can apply the results for simple random
sampling estimators to each stratum. The variances of the stratified SRS estimators of
the mean and total are:
V (
c
y
U
h
) = V (
b
t
h
) = (25)
where S
2
h
=
1
N
h
− 1
N
h
X
j=1
(y
hj
− y
hU
)
2
is the finite population variance for stratum h.
46
• Because the simple random samples are independent across the strata, the variance of
b
t
str
is the sum of the individual stratum variances:
V (
b
t
str
) =
H
X
i=1
V (
b
t
h
) =
H
X
i=1
(26)
• Dividing by N
2
, gives the V (
c
y
U
str
):
V (
c
y
U
str
) =
1
N
2
V (
b
t
str
) =
1
N
2
H
X
i=1
N
h
(N
h
− n
h
)
S
2
h
n
h
(27)
• Because S
2
h
is unknown, we use s
2
h
to get an unbiased estimator of V (
b
t
h
):
b
V (
b
t
h
) = (28)
where s
2
h
is the sample variance of the n
h
y-values sampled from stratum h.
• Substitution of (28) into (26) and (27) produce the estimated variances of the stratified
SRS estimators:
b
V (
b
t
str
) =
H
X
h=1
N
h
(N
h
− n
h
)
s
2
h
n
h
b
V (
c
y
U
str
) =
1
N
2
H
X
h=1
N
h
(N
h
− n
h
)
s
2
h
n
h
(29)
• Taking a square root of
b
V (
b
t
str
) or
b
V (
c
y
U
str
) yields the corresponding standard error.
This will be used when generating confidence intervals for t or y
U
.
• For the estimated variances of the estimators given in (29), we are assuming that all n
h
> 1
(because s
2
h
is undefined for n
h
= 1). Cochran (1977 pages 138-140) discusses two potential
methods of dealing with the extreme case where all n
h
= 1.
Stratification Example with Strong Spatial Correlation
• Abundance counts for the population in Figures 5a and 5b show a strong diagonal spatial
correlation. The region has been gridded into a 20 ×20 grid of 10 m ×10 m quadrats. The
total abundance t = 13354. This population was stratified in two different ways:
(i) Into the four 10 × 10 strata shown in Figure 5a.
Stratum sizes are N
h
= 100 and stratum sample sizes are n
h
= 5 for h = 1, 2, 3, 4.
Stratum sample totals
P
n
h
j=1
y
hj
are 124, 158, 172, and 223 for h = 1, 2, 3, 4.
Stratum sample means y
h
are 24.8, 31.6, 34.4, and 44.6 for h = 1, 2, 3, 4.
Stratum sample variances are s
2
1
= 21.7, s
2
2
= 13.3, s
2
3
= 45.3, and s
2
4
= 41.3.
47
(ii) Into seven unequal size diagonally-oriented strata shown in Figure 5b.
Stratum sizes are N
1
= N
7
= 45, N
2
= N
6
= 60, N
3
= N
5
= 66, and N
4
= 58.
Stratum sample sizes are n
1
= n
7
= 3, n
2
= n
3
= n
5
= n
6
= 5, and n
4
= 4.
Stratum sample totals
P
n
h
j=1
y
hj
are 65, 122, 153, 143, 178, 203, and 143 for h =
1, 2, 3, 4, 5, 6, 7, respectively.
Stratum sample means y
h
are 21.6, 24.4, 30.6, 35.75, 35.6, 40.6, and 47.6 for h =
1, 2, 3, 4, 5, 6, 7, respectively.
Stratum sample variances are s
2
1
= 10.3, s
2
2
= 14.8, s
2
3
= 19.3, s
2
4
= 4.25, s
2
5
= 8.3,
s
2
6
= 10.8, and s
2
7
= 26.3, respectively.
• For the stratified SRSs in Figure 5a and Figure 5b:
– Calculate
b
t
str
,
c
y
U
str
, and their standard errors.
– Calculate 95% confidence intervals for t and y
U
.
3.1.1 Confidence Intervals for y
U
and t
• If all of the stratum sample sizes n
h
are sufficiently large (Thompson suggests n
h
≥ 30),
approximate 100(1 − α)% confidence intervals for y
U
and t are
c
y
U
str
± z
∗
q
b
V (
c
y
U
str
)
b
t
str
± z
∗
q
b
V (
b
t
str
) (30)
where z
∗
is the upper α/2 critical value from the standard normal distribution.
• For smaller sample sizes, the following confidence intervals have been recommended:
c
y
U
str
± t
∗
q
b
V (
c
y
U
str
)
b
t
str
± t
∗
q
b
V (
b
t
str
) (31)
where t
∗
is the upper α/2 critical value from the t(d) distribution. In this case, d is
Satterthwaite’s (1946) approximate degrees of freedom d where
d =
P
H
h=1
a
h
s
2
h
2
P
H
h=1
(a
h
s
2
h
)
2
/(n
h
− 1)
=
(
b
V (
b
t
str
))
2
P
H
h=1
(a
h
s
2
h
)
2
/(n
h
− 1)
(32)
where a
h
= N
h
(N
h
− n
h
)/n
h
.
• Lohr (page 79) mentions that some software packages will use n − H degrees of freedom
(instead of the approximate degrees of freedom). Both R and SAS use n−H as the default
degrees of freedom.
• If the stratum sample sizes n
h
are all equal and the stratum sizes N
h
are all equal, then
the degrees of freedom reduces to d = n − H where n =
P
n
h
is the total sample size.
• One-sided confidence intervals can by generated just like those using SRS. Just use t
∗
using
the upper α critical value from the t(d) distribution.
48
47
49
49
3.2 Using R and SAS to Analyze a Stratified SRS
Datasets used in the R code
R dataset from Figure 5a R dataset from Figure 5b
------------------------ ------------------------
count fpc stratum count fpc stratum
25 100 1 18 45 1
30 100 1 23 45 1
18 100 1 24 45 1
28 100 1 23 60 2
23 100 1 21 60 2
30 100 2 21 60 2
26 100 2 28 60 2
35 100 2 29 60 2
34 100 2 25 66 3
33 100 2 32 66 3
38 100 3 27 66 3
27 100 3 35 66 3
30 100 3 34 66 3
44 100 3 34 58 4
33 100 3 37 58 4
36 100 4 38 58 4
41 100 4 34 58 4
53 100 4 33 66 5
47 100 4 38 66 5
46 100 4 37 66 5
38 66 5
32 66 5
40 60 6
44 60 6
38 60 6
44 60 6
37 60 6
49 45 7
42 45 7
52 45 7
R code for Stratified SRS (Figure 5a)
source("c:/courses/st446/rcode/confintt.r")
# t-based confidence intervals for SRS in Figure 5a
library(survey)
strat5adat <- read.table("c:/courses/st446/rcode/fig5a.txt", header=T)
# strat5adat
strat_design <- svydesign(id=~1, fpc=~fpc, strata=~stratum, data=strat5adat)
strat_design
esttotal <- svytotal(~count,strat_design)
print(esttotal,digits=15)
confint.t(esttotal,degf(strat_design),level=.95)
confint.t(esttotal,degf(strat_design),level=.95,tails=’lower’)
confint.t(esttotal,degf(strat_design),level=.95,tails=’upper’)
estmean <- svymean(~count,strat_design)
print(estmean,digits=15)
confint.t(estmean,degf(strat_design),level=.95)
confint.t(estmean,degf(strat_design),level=.95,tails=’lower’)
confint.t(estmean,degf(strat_design),level=.95,tails=’upper’)
50
R output for Stratified SRS (Figure 5a)
(For the population total)
-------------------------------------------------------------------
mean( count ) = 13540.00000
SE( count ) = 480.66620
Two-Tailed CI for count where alpha = 0.05 with 16 df
2.5 % 97.5 %
12521.03317 14558.96683
-------------------------------------------------------------------
-------------------------------------------------------------------
mean( count ) = 13540.00000
SE( count ) = 480.66620
One-Tailed (Lower) CI for count where alpha = 0.05 with 16 df
5 % upper
12700.81272 infinity
-------------------------------------------------------------------
-------------------------------------------------------------------
mean( count ) = 13540.00000
SE( count ) = 480.66620
One-Tailed (upper) CI for count where alpha = 0.05 with 16 df
lower 95 %
-infinity 14379.18728
-------------------------------------------------------------------
(For the population mean)
-------------------------------------------------------------------
mean( count ) = 33.85000
SE( count ) = 1.20167
Two-Tailed CI for count where alpha = 0.05 with 16 df
2.5 % 97.5 %
31.30258 36.39742
-------------------------------------------------------------------
-------------------------------------------------------------------
mean( count ) = 33.85000
SE( count ) = 1.20167
One-Tailed (Lower) CI for count where alpha = 0.05 with 16 df
5 % upper
31.75203 infinity
-------------------------------------------------------------------
-------------------------------------------------------------------
mean( count ) = 33.85000
SE( count ) = 1.20167
One-Tailed (upper) CI for count where alpha = 0.05 with 16 df
lower 95 %
-infinity 35.94797
-------------------------------------------------------------------
51
R code for Stratified SRS (Figure 5b)
The R code is exactly the same as the R code for the Figure 5a data analysis except you read
in the data file fig5b.txt.
R output for Stratified SRS (Figure 5b)
(For the population total)
-------------------------------------------------------------------
mean( count ) = 13462.70000
SE( count ) = 256.02201
Two-Tailed CI for count where alpha = 0.05 with 23 df
2.5 % 97.5 %
12933.07812 13992.32188
-------------------------------------------------------------------
-------------------------------------------------------------------
mean( count ) = 13462.70000
SE( count ) = 256.02201
One-Tailed (Lower) CI for count where alpha = 0.05 with 23 df
5 % upper
13023.91117 infinity
-------------------------------------------------------------------
-------------------------------------------------------------------
mean( count ) = 13462.70000
SE( count ) = 256.02201
One-Tailed (upper) CI for count where alpha = 0.05 with 23 df
lower 95 %
-infinity 13901.48883
-------------------------------------------------------------------
(For the population mean)
-------------------------------------------------------------------
mean( count ) = 33.65675
SE( count ) = 0.64006
Two-Tailed CI for count where alpha = 0.05 with 23 df
2.5 % 97.5 %
32.33270 34.98080
-------------------------------------------------------------------
-------------------------------------------------------------------
mean( count ) = 33.65675
SE( count ) = 0.64006
One-Tailed (Lower) CI for count where alpha = 0.05 with 23 df
5 % upper
32.55978 infinity
-------------------------------------------------------------------
-------------------------------------------------------------------
mean( count ) = 33.65675
SE( count ) = 0.64006
One-Tailed (upper) CI for count where alpha = 0.05 with 23 df
lower 95 %
-infinity 34.75372
-------------------------------------------------------------------
52
Using Proc Surveymeans in SAS:
• When the stratum unit totals (N
d
) are known, you must create a variable called total
that assigns N
h
to each stratum level. It must be called total . In the following examples,
the stratum variable is called Area.
• You also need to create a weight variable which takes on the value N
h
/n
h
. In the following
examples, the weight variable is called W and it appears in the Weight statement.
• Include the option total=(dataname) in the Proc Surveymeans statement. (dataname)
is the name of the data set. In the first example, the dataname is fig 5a. In the second
example, the dataname is fig 5b.
• Include a Stratum statement that contains the stratum variable.
• In the Var statement, include the response variable y. In these examples, y is Count.
• If you want one-sided confidence intervals for y
U
or t, in the Proc Surveymeans statement
enter lclm or uclm for y
U
and lclmsum or uclmsum for t. In the second example, I
included all 4 options.
• The list option in the Stratum statement produces a table containing information about
each stratum.
Analysis of the Stratified SRS in Figure 5a
data fig5a;
input Area Count @@;
datalines;
1 18 1 23 1 28 1 25 1 30
2 35 2 30 2 26 2 33 2 34
3 33 3 27 3 30 3 44 3 38
4 47 4 36 4 41 4 53 4 46
;
data fig5a; set fig5a;
if Area = 1 then _total_= 100; *** _total_ = Nh ;
if Area = 2 then _total_= 100;
if Area = 3 then _total_= 100;
if Area = 4 then _total_= 100;
if Area=1 then W = 100/5; *** W = Nh / nh ;
if Area=2 then W = 100/5;
if Area=3 then W = 100/5;
if Area=4 then W = 100/5;
title1 ’Analysis of Stratified SRS in Figure 5a’;
proc surveymeans data=fig5a total=fig5a mean clm sum clsum df;
Stratum Area / list;
Var Count;
Weight W;
run;
===============================================================
Analysis of Stratified SRS in Figure 5a
The SURVEYMEANS Procedure
Data Summary
Number of Strata 4
Number of Observations 20
Sum of Weights 400
53
Stratum Information
Stratum Population Sampling
Index Area Total Rate N Obs Variable N
------------------------------------------------------------------------
1 1 100 5.00% 5 Count 5
2 2 100 5.00% 5 Count 5
3 3 100 5.00% 5 Count 5
4 4 100 5.00% 5 Count 5
------------------------------------------------------------------------
Statistics
Std Error
Variable DF Mean of Mean 95% CL for Mean
--------------------------------------------------------------------
Count 16 33.850000 1.201666 31.3025829 36.3974171
--------------------------------------------------------------------
Variable Sum Std Dev 95% CL for Sum
-------------------------------------------------------------------
Count 13540 480.666204 12521.0332 14558.9668
-------------------------------------------------------------------
Analysis of Stratified SRS in Figure 5b
data fig5b;
input Area Count @@;
datalines;
1 18 1 24 1 23
2 28 2 29 2 21 2 21 2 23
3 34 3 27 3 35 3 32 3 25
4 34 4 38 4 37 4 34
5 32 5 38 5 37 5 38 5 33
6 37 6 38 6 44 6 44 6 40
7 42 7 49 7 52
;
data fig5b; set fig5b;
if Area = 1 then _total_= 45; *** _total_ = Nh ;
if Area = 2 then _total_= 60;
if Area = 3 then _total_= 66;
if Area = 4 then _total_= 58;
if Area = 5 then _total_= 66;
if Area = 6 then _total_= 60;
if Area = 7 then _total_= 45;
if Area=1 then W = 45/3; *** W = Nh / nh ;
if Area=2 then W = 60/5;
if Area=3 then W = 66/5;
if Area=4 then W = 58/4;
if Area=5 then W = 66/5;
if Area=6 then W = 60/5;
if Area=7 then W = 45/3;
title1 ’Analysis of Stratified SRS in Figure 5b’;
proc surveymeans data=fig5b total=fig5b mean clm sum clsum df
lclm uclm lclmsum uclmsum ;
Stratum Area / list;
Var Count;
Weight W;
run;
54
Analysis of Stratified SRS in Figure 5b
The SURVEYMEANS Procedure
Data Summary
Number of Strata 7
Number of Observations 30
Sum of Weights 400
Stratum Information
Stratum Population Sampling
Index Area Total Rate N Obs Variable N
----------------------------------------------------------------------
1 1 45 6.67% 3 Count 3
2 2 60 8.33% 5 Count 5
3 3 66 7.58% 5 Count 5
4 4 58 6.90% 4 Count 4
5 5 66 7.58% 5 Count 5
6 6 60 8.33% 5 Count 5
7 7 45 6.67% 3 Count 3
----------------------------------------------------------------------
Statistics
Std Error
Variable DF Mean of Mean 95% CL for Mean
-----------------------------------------------------------------------
Count 23 33.656750 0.640055 32.3326953 34.9808047
-----------------------------------------------------------------------
Lower 95% Upper 95%
One-Sided CL One-Sided CL
Variable for Mean for Mean Sum Std Dev
-----------------------------------------------------------------------
Count 32.559778 34.753722 13463 256.022011
-----------------------------------------------------------------------
Lower 95% Upper 95%
One-Sided One-Sided
Variable 95% CL for Sum CL for Sum CL for Sum
-----------------------------------------------------------------
Count 12933.0781 13992.3219 13024 13901
-----------------------------------------------------------------
3.3 Efficiency of Stratified Simple Random Sampling
• Because the variance formulas for
b
t
str
and
c
y
U
str
are determined only from within-stratum
variances, the precision of the estimators can be improved by forming strata with small S
2
h
values (strata with similar y-values within each stratum). We will compare
b
V (
c
y
U
) from a
SRS to
b
V (
d
y
str
) from a stratified SRS.
• The population variance can be rewritten as the weighted sum of within-stratum and
between-stratum variabilities:
S
2
=
1
N − 1
H
X
h=1
N
h
X
j=1
(y
hj
− y
U
)
2
=
1
N − 1
"
H
X
h=1
(N
h
− 1)S
2
h
+
H
X
h=1
N
h
(y
hU
− y
U
)
2
#
55
• By substituting this alternative form of S
2
into V (
c
y
U
) and V (
c
y
U
str
), it can be shown that:
V (
c
y
U
) − V (
c
y
U
str
) =
N − n
Nn(N − 1)
"
H
X
h=1
N
h
(y
hU
− y
U
)
2
−
1
N
H
X
h=1
(N − N
h
)S
2
h
#
.
• If this difference in variances is positive, or, equivalently, if
H
X
h=1
N
h
(y
hU
− y
U
)
2
>
1
N
H
X
h=1
(N − N
h
)S
2
h
,
then we say that
c
y
U
str
is more efficient than
c
y
U
.
• A stratified SRS estimator will be more efficient than the SRS estimator of y
U
or t if
the variability between stratum means is sufficiently large relative to the within-stratum
variability. This is what happened with the stratification used in Figures 5a and 5b.
3.4 Allocation of Sampling Units
• Given that we have enough resources to allocate n units among the H strata, how do we
determine the stratum sample sizes n
h
?
• Situation 1: If all strata are the same size and no prior information is available about
the population, a reasonable choice would be to assign equal (or nearly equal) sample sizes
to the strata. That is, n
h
≈ .
– Example: Consider the stratified population in Figure 5a. Suppose there are enough
resources to take a sample of size n = 50. How many samples should be taken for
each stratum assuming Situation 1?
• Situation 2: If the strata are not all the same size and no prior information is available
about the population, a reasonable choice would be to assign sample sizes proportional to
the sizes of the strata relative to the population size N. That is, n
h
. This is
known as proportional allocation.
– Example: Consider the stratified population in Figure 5b. Suppose there are enough
resources to take a sample of size n = 50. How many samples should be taken for
each stratum assuming proportional allocation?
• Situation 3: The allocation scheme that minimizes V (
b
t
str
) is called optimum allocation
and requires
n
h
=
Because the S
2
h
values are unknown, we would need prior estimates (possibly from past
data or published studies) to attempt optimum allocation.
– Example: Consider the stratified population in Figure 5b. Suppose there are enough
resources to take a sample of size n = 50 and we have prior estimates of s
1
= 3.2,
s
2
= 3.8, s
3
= 4.4, s
4
= 2.1, s
5
= 2.9, s
6
= 3.3, and s
7
= 5.1. How many samples
should be taken for each stratum assuming optimum allocation?
56
• Situation 4: In some cases, if the cost of sampling units varies from stratum to stratum,
then the total cost of taking a stratified SRS may determine how to allocate units to strata.
– Let c
0
be the fixed (also called “overhead”) cost of the survey that does not depend
on what units are in sample. Let c
h
be the cost to sample a unit from stratum h.
The total cost C of the sample will be C = c
0
+
H
X
h=1
c
h
n
h
.
– Case I: For a fixed total cost C, the smallest variance V (
c
y
U
) or V (
b
t) is achieved
by choosing n
h
such that:
n
h
=
(C − c
0
)N
h
S
h
/
√
c
h
P
H
h=1
N
h
S
h
√
c
h
– Case II: For a fixed (specified) variance V (
c
y
U
), the smallest cost is achieved by
first determining the total sample size n such that
n =
P
H
h=1
N
h
S
h
√
c
h
P
H
h=1
N
h
S
h
/
√
c
h
N
2
V +
P
H
h=1
N
h
S
2
h
where V is the fixed variance specified by the researcher. Then, the stratum sample
size n
h
for h = 1, 2, . . . , H is
n
h
=
nN
h
S
h
/
√
c
h
P
H
h=1
N
h
S
h
/
√
c
h
For a fixed V (
b
t), use V = V (
b
t)/N
2
in the formula.
• If all of the costs (c
h
) are the same, then the total sample size formula reduces to
n =
P
H
h=1
N
h
S
h
2
N
2
V +
P
H
h=1
N
h
S
2
h
• Because the S
h
values are unknown in either Case I or Case II, we would need prior
estimates (possibly from past data or published studies) to attempt optimum allocation.
Example of Situation 4: Case I: Suppose there is a fixed total cost C = $3000 and a fixed
overhead cost of c
0
= $500. Consider the stratification used in Figure 5b. The unit sampling
costs are
c
1
= $20 per unit from stratum 1
c
2
= c
3
= $25 per unit from stratum 2 or 3
c
4
= $30 per unit from stratum 4
c
5
= c
6
= $35 per unit from stratum 5 or 6
c
7
= $40 per unit from stratum 7
57
Then, using s
2
h
as an estimate of S
2
h
: n
h
=
(C − c
0
)N
h
s
h
/
√
c
h
P
H
h=1
N
h
s
h
√
c
h
=
2500 N
h
s
h
/
√
c
h
7657.776
rounded projected
Stratum N
h
s
h
c
h
N
h
s
h
√
c
h
N
h
s
h
/
√
c
h
n
h
n
h
cost
1 45 3.215 20 647.006 32.350 10.6 11 $220
2 60 3.847 25 1154.100 46.164 15.1 15 $375
3 66 4.393 25 1449.690 57.988 18.9 19 $475
4 58 2.062 30 655.054 21.835 7.1 7 $210
5 66 2.881 35 1124.919 32.141 10.5 10 $350
6 60 3.286 35 1166.414 33.326 10.9 11 $385
7 45 5.132 40 1460.593 36.515 11.9 12 $480
The estimated total cost is $ + c
0
= + $500 = requiring sampling
units.
Example of Situation 4: Case II: Suppose there is a fixed variance of V = V (
b
t) = .35.
Consider the stratification used in Figure 5b. The costs are the same as Case I.
Then, using s
h
as an estimate of S
h
:
n =
P
H
h=1
N
h
S
h
√
c
h
P
H
h=1
N
h
S
h
/
√
c
h
N
2
V +
P
H
h=1
N
h
S
2
h
≈
(7657.776)(260.319)
(400
2
)(.35) + 5254.1
=
Then, substitution yields
n
h
=
nN
h
S
h
/
√
c
h
P
H
h=1
N
h
S
h
/
√
c
h
=
(37.433)N
h
S
h
/
√
c
h
260.319
≈ N
h
S
h
/
√
c
h
.
rounded projected
Stratum N
h
s
h
c
h
N
h
s
h
√
c
h
N
h
s
h
/
√
c
h
N
h
S
2
h
n
h
n
h
cost
1 45 3.215 20 647.006 32.350 465.0 4.65 5 $100
2 60 3.847 25 1154.100 46.164 888.0 6.64 7 $175
3 66 4.393 25 1449.690 57.988 1273.8 8.34 8 $200
4 58 2.062 30 655.054 21.835 246.5 3.14 3 $ 90
5 66 2.881 35 1124.919 32.141 547.8 4.62 5 $175
6 60 3.286 35 1166.414 33.326 648.0 4.79 5 $175
7 45 5.132 40 1460.593 36.515 1185.0 5.25 5 $200
7657.776 260.319
Thus, the minimum cost to achieve V is $ + c
0
= + $500 = requiring
a total of sampling units.
58
