A basic issue in sample design is how many units should be selected
at each stage in order to efficiently estimate population values. If
strata are used, the number of units to allocate to each stratum must be
determined. In this vignette, we review some basic techniques for sample
size determination in single-stage samples using the package
PracTools (Valliant, Dever, and
Kreuter 2023) that contains specialized routines to facilitate
the calculations, most of which are not found in other packages. We
briefly summarize some of selection methods and associated formulas used
in designing single-stage samples and describe the capabilities of
PracTools. Technical background is in Valliant, Dever, and Kreuter (2018), ch. 3.
First, the package must be loaded with
Complex samples can involve any or all of stratification, clustering,
multistage sampling, and sampling with varying probabilities. Many texts
cover these topics, including Cochran
(1977), Lohr (1999), Särndal, Swensson, and Wretman (1992), and Valliant, Dever, and Kreuter (2018). Here we
discuss single-stage designs with and without stratification and
formulas that are needed for determining sample allocations. The
PracTools package will not select samples, but the R
sampling package (Tillé and Matei
2016) will select almost all the types used in practice.
Simple Random Sampling
Simple random sampling without replacement (srswor) is a
method of probability sampling in which all samples of a given size
\(n\) have the same probability of
selection. The function sample in R base (R Core Team 2020) will select simple random
samples either with or without replacement. One way of determining an
srswor sample size is to specify that a population value \(\small \theta\) be estimated with a certain
coefficient of variation (CV) which is defined as the ratio of
the standard error of the estimator, \(\small
\hat{\theta}\), to the value of the parameter: \(\small CV(\hat{\theta}) =
\sqrt{Var(\hat{\theta})}/\theta\). For example, suppose that
\(\small y_{k}\) is a value associated
with element \(\small k\), \(\small U\) denotes the set of all elements
in the universe, \(\small N\) is the
number of elements in the population, and the population parameter to be
estimated is the mean, \(\small \bar{y}_{U} =
\sum_{k \in U}y_{k}/N\). With a simple random sample, this can be
estimated by the sample mean, \(\small
\bar{y}_{s} = \sum_{k \in s}y_{k}/n\), where \(\small s\) is the set of sample elements
and \(\small n\) is the sample size.
Setting the required \(\small CV\) of
\(\small \bar{y}_{s}\) to some desired
value \(\small CV_{0}\) in an
srswor leads to a sample size of \[\small
n=\frac{\frac{S_U^{2}}{\bar{y}_{U}^{2}}}{CV_{0}^{2}
+\frac{S_U^{2}}{N\bar{y}_{U}^{2}}}\]
The R function, nCont, will compute a sample size using
either a target \(\small CV_{0}\) or a
target variance, \(\small V_{0}\), of
\(\small \bar{y}_s\) as input. The
parameters used by the function are shown below and are described in the
help page for the function:
nCont(CV0=NULL, V0=NULL, S2=NULL, ybarU=NULL, N=Inf, CVpop=NULL)
- Example 1: Sample size for a target CV. Suppose
that we estimate from a previous survey that the population CV of some
variable is 2.0. If the population is extremely large and
CV0 (the target CV ) is set to 0.05, then the call to the R
function is nCont(CV0 = 0.05, CVpop = 2). The resulting
sample size is 1,600. If the population size is N = 500, then
nCont(CV0 = 0.05,CVpop = 2, N = 500) results in a sample
size of 381 after rounding. The finite population correction
(fpc) factor has a substantial effect in the latter case.
The function nProp will perform the same computation for
estimated proportions.
- Example 2: Sample sizes for a vector of target CVs.
Often it will be useful to show a client the sample sizes for a series
of precision targets. This will be especially true when the budget is
uncertain and a researcher would like to think about options. We can ask
for the sample sizes for a vector of values of
CV0 from
0.01 to 0.21 in increments of 0.02 with:
ceiling(nCont(CV0 = seq(0.01, 0.21, 0.02), CVpop=2))
[1] 40000 4445 1600 817 494 331 237 178 139 111 91
ncont returns unrounded sample sizes having quite a few
decimal places; ceiling rounds to the next highest
integer.
Using a margin of error to find sample sizes
Many investigators prefer to think of setting a tolerance for how
close the estimate should be to the population value. If the tolerance,
sometimes called the margin of error (MOE), is \(\small e\) and the goal is to be within
\(\small e\) of the population mean
with probability \(\small 1-\alpha\),
this translates to \[\begin{equation}
\small Pr \left(\left|\bar{y}_{s} -\bar{y}_{U} \right|\le
e\right)=1-\alpha.
\end{equation}\] This is equivalent to setting the half-width of
a \(\small 100\left(1-\alpha \right)\)
two-sided confidence interval (CI) to \(\small
e=z_{1-\alpha\left/2\right.} \sqrt{V\left(\bar{y}_{s} \right)}\),
assuming that \(\small \bar{y}_{s}\)
can be treated as being normally distributed. The term \(\small z_{1-\alpha\left/2\right.}\) is the
\(\small 100\left(1-\alpha\left/2\right.
\right)\) percentile of the standard normal distribution, i.e.,
the point with \(\small
1-\alpha\left/2\right.\) of the area to its left. On the other
hand, if we require \[\begin{equation}
\small Pr \left(\left|\frac{\bar{y}_{s} -\bar{y}_{U}}{\bar{y}_{U}}
\right|\le e\right)=1-\alpha, \end{equation}\]
this corresponds to setting \(\small
e=z_{1-\alpha\left/2\right. }CV\left(\bar{y}_{s}\right)\). If we
set the MOE in (1) to \(\small e_{0}\),
then the above equation can be manipulated to give the required sample
size as
\[\begin{equation}
\small n=\frac{z_{1-\alpha\left/2\right.}^{2}
S_U^{2}}{e_{0}^{2}+z_{1-\alpha\left/2\right.}^{2}
S_U^{2}\left/N\right.}.
\end{equation}\]
Similarly, if the MOE in (2) is set to \(\small e_{0}\), we obtain \[\begin{equation}
\small n=\frac{z_{1-\alpha\left/2\right.}^{2} S_U^{2}\left/
\bar{y}_{U}^{2}\right.}{e_{0}^{2} +z_{1-\alpha \left/2\right.}^{2}
S_U^{2}\left/ \left(N\bar{y}_{U}^{2} \right)\right.}
\end{equation}\] The functions nContMoe and
nPropMoe will make the sample size calculations based on
MOEs for continuous variables and for proportions.
- Example 3: Sample sizes for proportions based on an
MOE. Suppose that we want to estimate a proportion for a
characteristic where an advance estimate is \(\small p_{U} =0.5\). The MOE is to be \(\small e\) when \(\small \alpha =0.05\). In other words, the
sample should be large enough that a normal-approximation 95% confidence
interval should be \(\small 0.50\pm e\)
as implied by (1). For example, if \(\small
e=0.03\) and the estimated proportion were actually 0.5, we want
the confidence interval to be \(\small 0.50\pm
0.03=\left[0.47,0.53\right]\). The sample size is highly
dependent on the width of the confidence interval as seen in the
following table. Sample sizes are evaluated using the formula given in
(3) with \(\small S_U^2 =
Np_U(1-p_U)/(N-1)\), \(\small p_{U}
=0.5\) and \(\small z_{0.975}
=1.96\). The command to generate the sample sizes listed in the
table below is
ceiling(nPropMoe(moe.sw=1, e=seq(0.01,0.08,0.01), alpha=0.05, pU=0.5))
#> [1] 9604 2401 1068 601 385 267 196 151
| 0.01 |
9,604 |
|
0.05 |
385 |
| 0.02 |
2,401 |
|
0.06 |
267 |
| 0.03 |
1,068 |
|
0.07 |
196 |
| 0.04 |
601 |
|
0.08 |
151 |
The parameter moe.sw=1 says to compute the sample size
based on (3). moe.sw=2 would use the MOE relative to \(\small \bar{y}_U\) in (4).
Stratified Simple Random Sampling
Simple random samples are rare in practice for several reasons. Most
surveys have multiple variables and domains for which estimates are
desired. Selecting a simple random sample runs the risk that one or more
important domains will be poorly represented or omitted entirely. In
addition, variances of survey estimates often can be reduced by using a
design that is not srswor.
A design that remedies some of the problems noted for an
srswor is referred to as stratified simple random sampling
(without replacement) or stsrswor. As the name indicates, an
srswor design is administered within each design stratum.
Strata are defined with one or more variables known for all
units and partition the entire population into mutually exclusive groups
of units. We might, for example, divide a population of business
establishments into retail trade, wholesale trade, services,
manufacturing, and other sectors. A household population could be
divided into geographic regions—north, south, east, and west. For an
stsrswor, we define the following terms:
\(\small N_{h}\) = the known number
of units in the population in stratum h (\(\small h=1,2,\ldots ,H\))
\(\small n_{h}\) = the size of the
srswor selected in stratum h
\(\small y_{hi}\) = the value of the
y variable for unit i in stratum h
\(\small
S_{Uh}^{2}=\sum_{i=1}^{N_{h}}\left(y_{hi}-\bar{y}_{U_h}\right)^{2}\left/
\left(N_{h} -1\right)\right.\), the population variance in
stratum h
\(\small U_{h}\) = set of all units
in the population from stratum h
\(\small s_{h}\) = set of \(\small n_{h}\) sample units from stratum
h
\(\small c_h\) = cost per sample
unit in stratum h
The population mean of y is \(\small \bar{y}_{U}
=\sum_{h=1}^{H}W_{h}\bar{y}_{U_h}\), where \(\small W_{h}=N_{h}\left/ N\right.\) and
\(\small \bar{y}_{U_h}\) is the
population mean in stratum h. The sample estimator of \(\small \bar{y}_{U}\) based on an
stsrswor is \[\begin{equation}
\small \bar{y}_{st} =\sum_{h=1}^{H}W_{h} \bar{y}_{s_h}, \notag
\end{equation}\] where \(\small
\bar{y}_{s_h} =\sum_{i\in s_{h}}y_{hi}\left/n_{h}\right.\). The
population sampling variance of the stratified estimator of the mean is
\[\begin{equation}
\small Var\left(\bar{y}_{st}\right) = \sum_{h=1}^{H}W_{h}^{2}
\frac{1-f_{h}}{n_{h}} S_{Uh}^{2}, \notag
\end{equation}\] where \(\small f_{h} =
n_{h}\left/N_{h}\right.\). The total cost of the sample is \[\small C = \sum_{h=1}^{H} c_h n_h.\]
There are various ways of allocating the sample to the strata,
including:
Proportional to the \(\small
N_h\) population counts
Equal allocation (all \(\small
n_h\) the same)
Cost-constrained optimal in which the allocation minimizes the
variance of \(\small \bar{y}_{st}\)
subject to a fixed budget
Variance-constrained optimal in which the allocation minimizes
the total cost subject to a fixed variance target for \(\small \bar{y}_{st}\)
Neyman allocation, which minimizes the variance of the estimated
mean disregarding the \(\small c_h\)
costs
The R function, strAlloc, will compute the proportional,
Neyman, cost-constrained, and variance-constrained allocations. The
parameters accepted by the function are shown below.
n.tot = fixed total sample size
Nh = vector of pop stratum sizes or pop stratum proportions (required parameter)
Sh = stratum unit standard deviations, required unless alloc = "prop"
cost = total variable cost
ch = vector of costs per unit in strata
V0 = fixed variance target for estimated mean
CV0 = fixed CV target for estimated mean
ybarU = pop mean of y
alloc = type of allocation, must be one of "prop", "neyman", "totcost", "totvar"
If the stratum standard deviations are unknown (as would usually be
the case), estimates can be used.
The parameters can only be used in certain combinations, which are
checked at the beginning of the function. Basically, given an
allocation, only the parameters required for the allocation are allowed
and no more. For example, the Neyman allocation requires
Nh, Sh, and n.tot. The function
returns a list with three components—the allocation type, the vector of
sample sizes, and the vector of sample proportions allocated to each
stratum. Three examples of allocations are Neyman, cost constrained, and
variance constrained (via a target CV):
# Neyman allocation
Nh <- c(215, 65, 252, 50, 149, 144)
Sh <- c(26787207, 10645109, 6909676, 11085034, 9817762, 44553355)
strAlloc(n.tot = 100, Nh = Nh, Sh = Sh, alloc = "neyman")
#>
#>
#>
#> allocation = neyman
#> Nh = 215, 65, 252, 50, 149, 144
#> Sh = 26787207, 10645109, 6909676, 11085034, 9817762, 44553355
#> nh = 34.641683, 4.161947, 10.473487, 3.333804, 8.798970, 38.590108
#> nh/n = 0.34641683, 0.04161947, 0.10473487, 0.03333804, 0.08798970, 0.38590108
#> anticipated SE of estimated mean = 1727173
# cost constrained allocation
ch <- c(1400, 200, 300, 600, 450, 1000)
strAlloc(Nh = Nh, Sh = Sh, cost = 100000, ch = ch, alloc = "totcost")
#>
#>
#>
#> allocation = totcost
#> Nh = 215, 65, 252, 50, 149, 144
#> Sh = 26787207, 10645109, 6909676, 11085034, 9817762, 44553355
#> nh = 30.605403, 9.728474, 19.989127, 4.499121, 13.711619, 40.340301
#> nh/n = 0.2574608, 0.0818385, 0.1681538, 0.0378478, 0.1153458, 0.3393533
#> anticipated SE of estimated mean = 1636053
# allocation with CV target of 0.05
strAlloc(Nh = Nh, Sh = Sh, CV0 = 0.05, ch = ch, ybarU = 11664181, alloc = "totvar")
#>
#>
#>
#> allocation = totvar
#> Nh = 215, 65, 252, 50, 149, 144
#> Sh = 26787207, 10645109, 6909676, 11085034, 9817762, 44553355
#> nh = 104.54922, 33.23283, 68.28362, 15.36917, 46.83941, 137.80400
#> nh/n = 0.2574608, 0.0818385, 0.1681538, 0.0378478, 0.1153458, 0.3393533
#> anticipated SE of estimated mean = 583209.1
The output of strAlloc is a list with components:
allocation (the type of allocation), Nh,
Sh, nh, nh/n, and
anticipated SE of estimated mean. If the results are
assigned to an object, e.g.,
neyman <- strAlloc(n.tot = 100, Nh = Nh, Sh = Sh, alloc = "neyman"),
the components in the list can be accessed with syntax like
neyman$nh.
There are many variations on how to allocate a sample to strata. In
most practical applications, there are multiple variables for which
estimates are needed. This complicates the allocation problem because
each variable may have a different optimal allocation. This type of
multicriteria problem can be solved using mathematical programming as
discussed in Valliant, Dever, and Kreuter
(2018), ch. 5.
Probability Proportional to Size Sampling
Probability proportional to size (pps), single-stage
sampling is used in situations where an auxiliary variable (i.e., a
covariate) is available on the frame that is related to the variable(s)
to be collected in a survey. For example, the number of employees in a
business establishment one year ago is probably related to the number of
employees the establishment has in the current time period.
The variance formula for an estimated mean in a pps sample
selected without replacement is too complex to be useful in determining
a sample size. Thus, a standard workaround is to use the
with-replacement (ppswr) variance formula to calculate a sample
size. The result may be somewhat larger than needed to hit a precision
target, but if the sample is reduced by nonresponse, beginning with a
larger sample is prudent anyway. The simplest estimator of the mean that
is usually studied with ppswr sampling is called
``p-expanded with replacement’’ or pwr (see Särndal, Swensson, and Wretman (1992), ch.2) and
is defined as \[\begin{equation}
\small \hat{\bar{y}}_{\textit{pwr}} =\frac{1}{Nn} \sum_{i \in
s}\frac{y_{i}}{p_{i}} \notag
\end{equation}\] where \(\small
p_i\) is the probability that unit \(\small i\) would be selected in a sample of
size 1. The variance of \(\small
\hat{\bar{y}}_{\textit{pwr}}\) in ppswr sampling is
\[\begin{equation} \label{eq:vpwr}
\small Var\left(\hat{\bar{y}}_{pwr} \right)=\frac{1}{N^{2} n}
\sum_{U}p_{i} \left(\frac{y_{i}}{p_{i}} - t_U\right)^{2} \equiv
\frac{V_{1}}{N^{2}n}
\end{equation}\] where \(\small
t_U\) is the population total of \(\small y\).
If the desired coefficient of variation is \(\small CV_{0}\), \(\small \eqref{eq:vpwr}\) can be solved to
give the sample size as \[\begin{equation}
\label{eq:n.pwr}
\small n=\frac{V_{1}}{N^{2}} \frac{1}{\bar{y}_{U}^{2} CV_{0}^{2}}\;.
\end{equation}\]
Figure 1. Plot of expenditures vs. beds in
hospital population
- Example 4: Sample size in a pps sample. We
use
smho.N874, which is one of the example populations in
the PracTools package, to illustrate a pps sample
size calculation. Figure 1 plots annual expenditures per hospital versus
number of beds for the 670 hospitals that have inpatient beds. Although
the relationship is fairly diffuse, the correlation of beds and
expenditures is 0.70 so that pps sampling with beds as a
measure of size could be efficient. The code below evaluates \(\small \eqref{eq:n.pwr}\) giving a
pps sample of \(\small n =
57\), which will produce an anticipated CV of 0.149. In
contrast, an srs of \(\small n =
82\) would be necessary to obtain the same size CV.
require(PracTools)
data("smho.N874")
y <- smho.N874[,"EXPTOTAL"]
x <- smho.N874[, "BEDS"]
y <- y[x>0]
x <- x[x>0]
ybarU <- mean(y)
(N <- length(x))
#> [1] 670
CV0 <- 0.15
# calculate V1 based on pp(x) sample
pik <- x/sum(x)
T <- sum(y)
(V1 <- sum( pik*(y/pik - T)^2))
#> [1] 9.53703e+19
n <- V1 / (N*ybarU*CV0)^2
(n <- ceiling(n))
#> [1] 57
# Anticipated SE for the pps sample
(cv.pps <- sqrt(V1/(N^2*n)) / ybarU)
#> [1] 0.1495183
# sample size for an srs to produce the same SE
ceiling(nCont(CV0 = cv.pps, S2 = var(y), ybarU = ybarU, N = N))
#> [1] 82
The PracTools package includes a variety of other
functions relevant to the design of single-stage samples that are not
discussed in this vignette:
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gammaFit |
Iteratively computes estimate of \(\small \gamma\) in a model with \(\small E_M(y) =
\mathbf{x}^T\boldsymbol{\beta}\) and \(\small \sigma^2 \mathbf{x}^\gamma\). This
is useful in determining a measure of size for pps
sampling. |
nDep2sam |
Compute a simple random sample size for estimating the
difference in means when samples overlap |
nDomain |
Compute a simple random sample size using either a
target coefficient of variation or target variance for an estimated mean
or total for a domain |
nLogOdds |
Calculate the simple random sample size for estimating
a proportion using the log-odds transformation |
nProp |
Compute the simple random sample size for estimating a
proportion based on different precision requirements |
nProp2sam |
Compute a simple random sample size for estimating the
difference in proportions when samples overlap |
nWilson |
Calculate a simple random sample size for estimating a
proportion using the Wilson method |