\documentstyle[12pt,fullpage,fancyheadings]{article}
\pagestyle{fancy}
\lhead{Statistics 36-303}
\rhead{Spring, 2008}
\setlength{\headrulewidth}{0pt}
\begin{document}

\def\today{February 26, 2008}

\title{Notes on Stratified Sampling\\
for Statistics 36-303: Sampling, Surveys, and Society} 

\author{Stephen E. Fienberg\\
Department of Statistics\\ 
Carnegie Mellon University\\
(Revised by Brian Junker, Spring 2008)}
\maketitle
\section*{I. Why Stratify?}

        One of the simplest extensions to simple random sampling
        involves the use of stratification of the population of
        interest.  In stratified random sampling, we divide the population
        into distinct subpopulations called {\em {strata}}, and within
        each stratum we select a separate random sample.  There are
        four primary reasons for using stratification:

\begin{enumerate}
\item The subpopulations that form the strata are ``domains" of study
  or interest and thus we need to get separate estimates for each
  stratum.  For example, we may take separate samples for men and
  women because we are interested in making separate inferences for
  the subpopulations as well as comparing them.
  
\item The subpopulations or strata are such that different methods or
  sampling schemes may be suitable for the different strata. For
  example, if we are estimating the number of words in a textbook, we
  might take different samples for different sections: the front
  matter. the main text, the references, and the index.
  
\item The population is geographically diverse and thus stratification
  is a convenient way of organizing sampling and data collection.

\item By employing stratification in a careful fashion, we can get
  improved precision for our estimates of population quantities.
\end{enumerate}
        
In these notes, we explain, both by example and by formula, the
potential gains in precision due to stratification, i.e. reason 4
above.  We exploit the formal results on the calculation of
expectations and variances for linear combinations of random
variables, from the handout distributed in class.  Later in the
course, we will discuss the combining of stratification with other
sampling devices such as clustering.

\section*{II. Stratification for the Estimation of Proportions}

Here we are interested in estimating the proportion $p$ of some
attribute in the population (what is the proportion of yellow balls?
What is the proportion of people who ride PAT buses?, etc.)

{\bf {Example 1: Three Urns.}} Consider three urns, with the $i$th urn
containing $N_i$ balls of two types, $N_ip_i$ yellow ones and
$N_i(1-p_i)$ blue ones for $i=1,2,3.$ We are interested in the
attribute yellow and our goal is to estimate the overall proportion of
yellow ball in the three urns combined. The composition of the three
urns is:\\

\begin{enumerate}       
\item $N_1= 100, p_1=0.1$,\\ 
\item $N_2= 100, p_2=0.2$,\\
\item $N_3= 100, p_3=0.3$.\\ 
\end{enumerate} 



\noindent What we would like to do is compare the precision of a
stratified random sample with $n_i=10$ balls from each of the three
urns with a simple random sample of size $n= \sum^3_{i=1}n_i = 30$
balls from a pooled urn with composition $N= \sum^3_{i=1}N_i= 300$ and
$p=0.2.$\\

To compare the precision of the two different sampling schemes, we
need look at the variances of the estimates of $p$ for the two
situations or the corresponding standard deviations.\\

\subsection*{IIa. Precision for Simple Random Sampling}
        
The variance of a sample proportion, $\hat{p}$, based on an SRS
without replacement, of size $n$, from a population of size $N$, with
population proportion of attributes equal to $p$, is (Groves, p. 100):
 \begin{equation}
        Var(\hat{p})= \frac{p(1-p)}{n-1} (1-f)  ~~~ \mbox{where
          $f=n/N$}.
\end{equation}
{\bf{Example 1 (cont.):}} For our urn with $N=300$ and $p=.2$, the
variance of a simple random sample of size $n=30$ is\\
\begin{equation}       
 Var(\hat{p})= \frac{.2(1-.2)}{29}\left(1-\frac{30}{300}\right) = 0.004966
\end{equation}
and
  \begin{equation}
 s.d.(\hat{p})=0.0705.
\end{equation}
Of course if we do not know $p$, a good conservative strategy is to
pretend $p=1/2$ since that will make the variance as large as possible.
\subsection*{IIb. Precision for Stratified Random Sampling}
        
Now suppose we have a population of size $N$ divided into $H$
strata. We take a stratified random sample with $n_h$ observations in
the $i$th stratum, which has population size $N_h$ and a population
proportion of attributes equal to $p_h$. Let $\hat{p}_h$ be the
sample proportion in the $i$th stratum.  Our overall sample size is
$n= \sum_h n_h$ and the ``pooled" estimate of $p$,
$\hat{p}_{strat}$, weights the sample proportions from each
stratum according to the population sizes from the strata, i.e.,
\begin{equation}        
\hat{p}_{strat}= \sum^H_{h=1} \bigg(\frac{N_h}{\sum_k N_k} \bigg)
\hat{p}_h.
\end{equation}
Weighting the sample proportions according to the population sizes
makes the stratified estimate, $\hat{p}_{strat}$, unbiased, i.e.,
if we average over all possible such samples that are drawn without
replacement, we get the overall population proportion, $p$:

\begin{eqnarray}
E(\hat{p}_{strat}) & = & E \bigg[ \sum^H_{h=1} \bigg( \frac{N_h}{
 \sum_k N_k}\bigg) \hat{p}_h \bigg]\\
 & = & \sum^H_{h=1} \bigg( \frac{N_h}{ \sum_k N_k} \bigg)E(
 \hat{p})= \sum^H_{h=1} \bigg( \frac{N_h}{ \sum_k N_k} \bigg)p_h\nonumber\\ 
 & = & p. \nonumber
\end{eqnarray}

We can compute the variance for $\hat{p}_{strat}$ as a special
case of the formula for the variance of a linear combination of
independent random variables, $\{X_h\}$:\\
\begin{equation}
       Var\left( \sum^H_{h=1}a_h X_h\right)= \sum^H_{h=1}a^2_h Var(X_h).
\end{equation}
% (See the seperate handout on expectations and variances.)\\ 
%
Thus we have that

\begin{eqnarray}
Var( \hat{p}_{strat})& = & Var \bigg( \sum_h
\bigg(\frac{N_h}{\sum_k N_k} \bigg)
\hat{p}_h \bigg)\\
& = & \sum^H_{h=1} \bigg(\frac{N_h}{\sum_k N_k} \bigg)^2Var(
\hat{p}_h)\nonumber\\
& = & \sum^H_{h=1} \bigg(\frac{N_h}{\sum_k N_k} \bigg)^2
\frac{p_h(1-p_h)}{n_h-1} (1-f_h). \nonumber
\end{eqnarray}
       

{\bf {Example 1 (cont.):}} In our example with the 3 urns, this
formula for the variance of the stratified estimate simplifies because
they have the same sample size, 10, and population size, 100:

\begin{eqnarray}
Var( \hat{p}_{strat})& = &\left(\frac{100}{300}\right)^2\frac{1 - 10/100}{10-1}
 \sum_{h=1}^3 p_h(1-p_h)\\
& = & (0.0111) \sum^3_{h=1}p_h(1-p_h)\nonumber\\
& = & (0.0111) ( .1\times.9 + .2\times.8 + .3\times.7 )\nonumber\\
& = & 0.00511\nonumber
\end{eqnarray}  

\noindent and the standard deviation of $\hat{p}_{strat}$,
$$
s.d.( \hat{p}_{strat})=.0715,
$$ is smaller than that for the simple random sample of equivalent
overall size that we computed above.  The reduction in the standard
error is less than $2\%$.  This may not seem all that substantial but
it is a gain that we can achieve in a consistent fashion if we choose
our sample sizes for the strata with care as we have in this
example.\\

\subsection*{IIc. Gains from Proportionate Stratified Sampling}


        The improvement of the variance in our example of the urns is
        not a coincidence, but rather an illustration of an important
        result that explains the gain that results from
        stratification.  If we choose sample sizes for our strata in
        proportion to the population sizes, i.e., we let
\begin{equation}
n_i \propto N_i, 
\end{equation}
then we can rewrite $ \hat{p}_{strat}$ as
\begin{equation}
\overline {\hat{p}}= \sum_i \bigg(\frac{n_i}{\sum_jn_j} \bigg)
\hat{p}_i\\
\end{equation}
        
\noindent and the variance of the proportionate stratified estimate as\\ 
\begin{eqnarray}        
Var(\overline{\hat{p}})& = & \sum_i \bigg(\frac{n_i}{\sum_jn_j}
\bigg)^2 \frac{p_i(1-p_i)}{n_i} \frac{(N-n_i)}{(N-1)}.
\end{eqnarray}

Then we can show algebraically that
\begin{equation}
        Var(\overline {\hat{p}}) \leq Var(\hat{p}),
\end{equation}
as we saw in our example of the three urns where the sample sizes
where chosen in accord with proportionate sampling.  The gains that
result from stratification when we sample proportional to population
size are modest, however, unless the $\{p_i\}$ for the strata are
widely spread.\\

{\bf {Example 1 (cont.):}} Let us consider another version of our urn
problem but with two urns each with $N_i=150$ balls. In the first urn
there are no yellow balls and thus $p_1=0.0$ while in the second urn
there are 60 yellow balls (and 90 blue ones) so that $p_2=0.4$.  The
pooled urn again has composition $N= \sum_iN_i= 300$ and $p=0.2$.  Now
we compute the variance of $\overline{\hat{p}}$ for the two
strata situation as

\begin{eqnarray}
Var(\overline{\hat{p}}) & = & \frac{(15/30)^2}{15}
        \frac{(150-15)}{(150-1)} \sum^2_{i=1}p_i(1-p)\\
& = & \frac{\sum^2_{i=1}p_i(1-p_i)}{149}\nonumber\\
& = & \frac{9}{4} \frac{(0 \times1)+(.4 \times.6)}{149}\nonumber\\
& = & \frac{.54}{149}=.003624
\end{eqnarray}
 
       
and 

$$
s.d.(\overline{\hat{p}})=.062.
$$

In this situation there is a reduction of (1 - 0.602/0.694)= 13.3$\%$
in the standard deviation of our estimate.  Clearly by setting up an
urn with no yellow balls, and then choosing a second urn so that the
overall composition of the pooled urns remains the same, we have
illustrated the maximum possible gain for proportionate stratified
sampling.  In real life problems it is often difficult to achieve such
gains and thus, when we are interested in attributes, we typically
carry out stratification primarily for other reasons.

Proportionate random sampling has another feature worthy of note: By
choosing a sample size for each of the strata in proportion to its
population size, the method yields an equal probability of selection
for each individual element of the population.  As a consequence our
sample estimate for $p$ assigns equal weights to each observation.
This property is often referred to as {\em {self-weighting}}.  What
this means is that $\overline{\hat{p}}$ is just the overall
proportion of elements in the sample with the attribute of interest.
Of course, simple random sampling is also self-weighting.


\vspace{.25in} 
{\bf {Example 2: North Carolina Citizen Survey.}}  Henry (1990, p. 63)
describes a series of North Carolina household surveys carried out in
the late 1970s and early 1980s to examine the effectiveness of
programs in human services, transportation, and economic development.
The survey employed stratified sampling with separate samples being
drawn from two strata: (a) a list of 1975 North Carolina income tax
returns, and (b) a list of North Carolina residents eligible for
Medicaid assistance in 1975.  The initial samples for the strata were
selected in proportion to the sizes of the two lists.  Later when the
response rate from the Medicaid list was lower than that for the tax
list, the proportion from list (b) was increased and the sample was no
longer self-weighting.

\subsection*{IId. Interpreting the Gain From Stratification}
 
        One way to interpret the gain that results from proportionate
        stratification is to think in terms of the efficiency of the
        sampling design.  We ask what size simple random sample would
        produce the same precision as that achieved by our
        proportionate stratified sample?  Let  $n_{equiv}$ be
        this value. Then we could think of the savings associated with
        stratification in terms of the ratio $n/n_{equiv}$.  This
        turns out to be the same (except for the finite population
        correct) as the following definition:\\
\begin{equation}
Efficiency \hspace{.025in}of \hspace{.025in}the \hspace{.025in}design=\frac{Var(\overline{\hat{p}})}{Var(\hat{p})}.
\end{equation}
\vspace{.25in}
        {\bf {Example 1 (cont.):}}  For our three urn-example the
        efficiency of stratification was $ \frac{.004646}{.004816} = 96.5\%$
        whereas for the extreme two-urn example the efficiency was
        $\frac{.003624}{.004816} = 75.3\%$.\\

\subsection*{IIe. Stratification Versus Quotas: Beware}

        Many market research firms still use some version of quota
        sampling, in order to disguise low response rates and to
        ``fix" to sample up.  The proponents of this approach to
        sampling either directly or indirectly claim that you don't
        have to worry about call-backs to nonresponders as long as you
        get the predetermined numbers of individuals from designated
        subpopulations.  These subpopulations are then represented as
        strata and the quotas as the subpopulation sample sizes in
        stratified random sample.  While this quota approach to
        sampling was discredited man y years ago, vestiges remain and
        you need to be on the lookout for signs of its use.
        Suspicious symptoms of quota are round numbers of respondents
        in designated categories and the absence of information on how
        the sample was drawn and what was done about nonresponse.\\

{\bf {Example 3: Quarterly Survey of Allegheny County, Pennsylvania.}}
The {\em {Pittsburgh Post-Gazette}} (February 8, 1994) reported on an
a quarterly survey conducted by a market research firm of Allegheny
County residents to determine attitudes toward public policy issues
facing the Pittsburgh area. In the survey conducted during the
three-week period from October 26 to November 13, 1993, the newspaper
reported that there were 600 respondents, 300 males and 300 females,
and that the poll had a plus or minus error rate of 4 percent."  The
round numbers of men and women should make you suspicious, especially
in the absence of any further information on the sampling method. The
error rate reported in this news article sounds suspiciously like 2
times the standard deviation of some estimate of a population
proportion, corresponding to an approximate 95$\%$ confidence
interval.  Let's check out the calculation.  Suppose we had a simple
random sample of $n= 600$ with population proportion of attributes
equal to $p=.5$.  The population size of Allegheny county is in excess
of one million and thus we can ignore the finite population
correction. Thus the variance of $\hat{p}$ is
$$
        Var(\hat{p})=.5 \times\frac{.5}{600} = .0004166
$$ 
and the estimated standard deviation is .0204 or just over 2$\%$.  If
we had used stratified random sampling, with a sample of size $n_i=
300$ from separate strata for men and women, and different
proportions, $p_i$, then the variance of a estimated proportion would
be smaller, since we would have a proportionate stratified sample.
The error estimate looks like it came out of an elementary statistics
textbook and reinforces our suspicion about the possible uses of quota
sampling.

\section*{III. Stratification for Measurements}
        
The same idea of using stratification works for measurements as well
as for proportions, but now the gains due to a careful choice of
stratification may be quite substantial.  Further we can achieve even
greater gains than those offered by proportionate stratified sampling.

Suppose we wish to estimate a population quantity, $\mu=
\overline{Y}$, the mean of the measurements for the entire
population. For a simple random sample of size $n$, we use as our
estimate the sample average, $\overline{y}$.  The variance of
$\overline{y}$ is
\begin{equation}
        Var(\overline{y})= \frac{\sigma^2}{n} \frac{(N-n)}{(N-1)}.
\end{equation}

\subsection*{IIIa. Proportionate Stratification}

        Now we divide the population into $k$ subpopulations or strata
        of size $N_i$ where $ \sum N_i=N.$ For the $i$th stratum we
        estimate the stratum mean $\mu_i= \overline{Y}_i$ based on a
        sample of size $n_i$ and the stratum sample average,
        $\overline{y}_i$.  As before we take our stratified sample
        estimate to be a weighted combination of the stratum
        averages:\\
\begin{equation}
        \overline{y}_{strat}= \sum^k_{i=1} \bigg(\frac{N_i}{\sum_jN_j} \bigg)\overline{y}_i.
\end{equation}
Just as in the case with attributes, weighting the sample means in
proportion to the population sizes makes the stratified estimate,
$\overline{y}_{strat}$, unbiased, i.e., if we average over all
possible such samples that are drawn without replacement, we get the
overall population proportion, $\mu$, i.e.

\begin{eqnarray}
E(\overline{y}_{strat})=\sum^k_{i=1}\bigg(\frac{N_i}{\sum_jN_j}\bigg)\mu_i=\mu.
\end{eqnarray}

The variance of this stratified estimate is

\begin{eqnarray}
Var(\overline{y}_{strat})& = & Var \bigg( \sum^k_{i=1}
\bigg( \frac{N_i}{\sum_jN_j} \bigg)\overline{y}_i \bigg)\\
& = & \sum^k_{i=1} \bigg(\frac{N_i}{\sum_jN_j} \bigg)^2Var(\overline{y}_i)\nonumber
\end{eqnarray}
 
\noindent and by recalling that
   \begin{equation}     
Var(\overline{y}_i)= \frac{\sigma^2_i}{n_i}
\frac{(N_i-n_i)}{(N_i-1)},
\end{equation}
we can write
\begin{equation} 
        Var(\overline{y}_{strat})= \sum^k_{i=1} \bigg(\frac{N_i}{\sum_jN_j} \bigg)^2
        \frac{\sigma^2_i}{n_i} \frac{(N_i-n_i)}{(N_i-1)}.
\end{equation}
In practice we typically do not know the stratum variances and so we
substitute sample estimates.

\vspace{.25in} {\bf {Example 4: Taxpayer Incomes.}} Williams (1978)
gives a simple illustration of gains from stratification for a small
population of measurements on the income (in thousands of dollars) for
a population of $N=9$ taxpayers:

\begin{tabular}{c|c|c}
{\bf {Taxpayer No.}} & {\bf {Actual Income (\$ thousands)}} & {\bf
  {Reported Income}}\\ \hline
1 & 60 & 50\\
2 & 72 & 56\\
3 & 68 & 66\\
4 & 94 & 76\\
5 & 90 & 90\\
6 & 102 & 100\\
7 & 116 & 112\\
8 & 130 & 110\\
9 & 200 & 175\\ \hline
\end{tabular}
\vspace{.25in}

For the complete population we have $N= 9$, and for the actual income $\mu=103.56$ and $\sigma^2=
1621.14$.  The variance for a simple random sample of $n=3$ is\\
$$
        var(\overline{y})= \frac{1621.14}{3}
        \frac{(9-3)}{(9-1)}=405.285.
$$ 
Now suppose we use the values of the reported income in order to
stratify the population into two subpopulations consisting of the
first eight observations and the ninth observation:\\

       $ N_1=8, \mu_1=91.50$  and $\sigma^2_1=515.75$;
 $N_2=1, \mu_2=200$ and $\sigma^2_2=0$.

\vspace{.25in}
We now take a sample of size $n=3$, with $n_1=2$ from stratum 1 and
$n_2=N_2=1$ from stratum 2.  Then
$$
Var(\overline{y}_{strat})= \bigg(\frac{8}{9} \bigg)^2 \frac{515.75}{2}
\frac{(8-2)}{(8-1)}+0=174.65.
$$ 
The stratified estimate in this case has a variance which is less than
half that for a simple random sample. Of course, we are rarely so luck
as to be able to identify the largest element in a population and give
it it's own stratum!

\vspace{.25in}
Next, suppose we choose sample sizes for our strata in proportion to the
population sizes, i.e.,  we let
\begin{equation}
n_i \propto N_i,
 \end{equation}
and we label the resulting estimate as $\overline{\overline{y}}$,
then

\begin{equation} 
        Var(\overline{\overline{y}})= \sum_i
        \bigg(\frac{n_i}{\sum_jn_j} \bigg)^2
        \frac{\sigma^2_i}{n_i} \frac{(N_i-n_i)}{(N_i-1)}.
\end{equation}

\noindent Now some messy algebra yields that

\begin{equation}
        Var(\overline{\overline{y}}) \leq Var(\overline{y}).
\end{equation}
This is the same result we had above for sampling of
attributes. Proportionate stratification improves the precision of our
estimate relative to simple random sampling.

Just as in the case of sampling for attributes, proportionate
stratification produces an estimate $\overline{\overline{y}}$ that
assigns equal weights to the observations, i.e., if we let $y_{ij}$ be
the $j$th observation from the $i$th strata, then we can rewrite the
formula for $\overline{\overline{y}}$ as

\begin{equation}
        \overline{\overline{y}}= \frac{\sum_i\sum_jy_{ij}}{n} .
\end{equation}


This is the same self-weighting property that we discussed earlier.\\


{\bf {Example 2 (cont.):}} The North Carolina surveys measured, among
other variables, the income of households. Thus we would expect
greater precision from the stratified sample that was used than from a
simple random sample.  The nonresponse problem and the later
adjustments to the sample discussed by Henry (1990) make the realized
sample accuracy somewhat more complicated.

\vspace{.25in} {\bf {Example 4 (cont.):}} We continue with the example
of the taxpayer actual incomes, but now we have two strata organized
geographically: stratum 1 contains $N_1=6$ middle income taxpayers,
numbers 1, 2, 3, 4, 6 and 7, while stratum 2 contains $N_2=3$
taxpayers, numbers 5, 8, and 9.  For these strata we have\\
 
$\mu_1=85.33$ and $\sigma^2_1=412.79; \mu_2=140$, and
$\sigma^2_2=2066.66$.\\ 

        
We again take a stratified random sample of size $n=3$, but this time
we sample in proportion to population size, so that $n_1=2$ and
$n_2=1$, and
\begin{eqnarray*}
Var(\overline{\overline{y}})& = & \bigg( \frac{6}{9} \bigg)^2 \frac{412.79}{2}
\frac{(6-2)}{(6-1)}+ \bigg(\frac{3}{9} \bigg)^2 \frac{2066.66}{1}
\frac{(3-1)}{(3-1)}\\
& = & 73.385+229.629=303.01.
\end{eqnarray*}         
This result is an improvement upon the variance for simple random
sampling, but it isn't as good when we treated the largest taxpayer as
forming a separate stratum and included that individual in the sample
with certainty.\\

{\bf {Example 5: Test Scores.}} Henry (1990, pp. 99-106) gives a
detailed example with different examples of proportionate
stratification and achieves estimated efficiency gains of 40$\%$ as
measured by the sample estimates of
$Var(\overline{y})/Var(\overline{\overline{y}})$.


\subsection*{IIIb. Optimal Allocation}

        It turns out that we can achieve even greater gains in
        stratification by selecting our sample rates $f_i=n_i/N_i$   in
        the strata proportional to the population standard deviations,
        $\sigma_i$, i.e., by choosing

\begin{equation}
        n_i \propto N_i \sigma_i,
\end{equation} 
subject to the restriction that the stratum sample sizes add up to the
overall sample size, i.e.,

\begin{equation} 
        n= \sum^k_{i=1}n_i.
\end{equation}
This is a version of something that statisticians called {\em {optimal
    allocation}} and, if we let $\overline{y}_{opt}$ be the estimate
using these sampling rates in the strata, then some further algebra
yields that

\begin{equation} 
       Var(\overline{y}_{opt}) \leq Var(\overline{\overline{y}})
        \leq Var(\overline{y}).
\end{equation}
We can determine the efficiency of optimal allocation relative to
proportional allocation or simple random sampling by considering the
ratios of the corresponding variances.  As in the case of
proportionate stratification for attributes above, we can interpret
these ratios approximately as determining the equivalent sample sizes
required to yield the same precision.  Thus the ratio
$Var(\overline{\overline{y}})/Var(\overline{y}_{opt})$ gives,
approximately, the relative proportional increase in sample size that
would be required for proportional stratification to achieve a similar
level of precision as optimal allocation.  Note that using optimal
allocation of our sample to strata with $n_i \propto N_i \sigma_i$
makes the probability of selection for the individuals in some strata
higher than in others.  The result is that we do not have a
self-weighting sample and $\overline{y}_{opt}$ is not an estimate
which weights each observation equally.  This is a price we pay for
the increase in precision.

\vspace{.25in} 
{\bf {Example 4 (cont.):}} We continue with the example of the
taxpayer actual incomes, and with the same two strata organized
geographically of sizes $N_1=6$ and $N_2=3$ where\\


$\mu_1=85.33$ and $\sigma^2_1=412.79; \mu_2=140,$ and
         $\sigma^2_2=2066.66.$\\

We again take a stratified random sample of size $n=3$, but this time
we attempt an optimal allocation so we want to choose our sample sizes
such that

$$ 
       \frac{n_1}{n_2}= \frac{N_1 \sigma_1}{N_2 \sigma_2}.
$$
Of course, we can only do this if we know the stratum variances which
we do not.  Thus we'll need to use a ``guess" for the ratio $\sigma_1/
\sigma_2$. Suppose we take this to be 1/3 (the true value being .447).
Then
 
$$
\frac{n_1}{n_2}= \frac{6}{3} \frac{1}{3}= \frac{2}{3}.
$$
But we can't divide a sample of size 3 into two parts in the ratio 2
to 3 and so we will need to do the best we can allocation and . The
resulting estimate is not quite , and we label it with a prime as .
The variance of this approximation to the optimum allocation is
\begin{eqnarray*}
Var(\overline{y} \prime_{opt})& = & \bigg( \frac{6}{9} \bigg)^2 \frac{412.79}{1}
\frac{(6-1)}{(6-1)}+ \bigg(\frac{3}{9} \bigg)^2 \frac{2066.66}{3}
  \frac{(3-2)}{(3-1)}\\
& = & 183.46+57.407=240.87.
\end{eqnarray*}       
The approximate optimum allocation estimate has variance smaller than
that for the proportionate stratified estimate, as the theory
suggested, and we have

$$
Var(\overline{\overline{y}})/Var(\overline{y}
\prime_{opt})=\frac{303.01}{240.87}=1.26
$$    
corresponding to an efficiency gain of 26$\%$.  Again we have not
produced an estimate with precision equal to that when we treated the
largest taxpayer as forming a separate stratum and included that
individual in the sample with certainty, but we have come fairly
close.\\

You might be asking yourself why we didn't try to achieve this extra
efficiency associated with optimal allocation of our sample across
strata when we were working with attributes.  It turns out that the
result described above does work for attributes too, but it has no
practical value.  This is because the population variance in the $i$th
stratum, $p_i(1-p_i)$, is a function of the unknown stratum
proportion, $p_i$, and thus we need to know $p_i$ in order to choose
the optimal stratum sample size.

\section*{IV. Disproportionate Allocation}

        Optimal allocation is just one illustration of why we might
        opt to give up the nice self-weighting properties that go with
        proportionate sampling.  More often we choose to weigh some
        strata more than others because we wish to make inferences
        about a subpopulation and the accuracy of proportionate
        stratification or optimal allocation is not sufficient for our
        purposes.\\


{\bf {Example 6: The National Health and Nutrition Examination
    Survey.}}  The National Health and Nutrition Examination Survey
(NHANES) conducted regularly by the National Center for Health
Statistics (NCHS), assesses the health and nutritional status of the
general population of the United States.  Data are collected through
interviews, physical examinations, clinical and laboratory testing,
and other related procedures. NHANES provides data that allow NCHS to
estimate the prevalence of specific conditions and chronic diseases
and information on nutritional status and deficiencies.

Several years ago, the Congress mandated a special sample supplement
to NHANES to increase the number of Hispanics in the survey and to
enable accurate estimates for the Hispanic population.  In forthcoming
waves of NHANES, there will be disproportionate sampling for other
population groups as well such as Blacks, low-income persons,
pre-school children, and the elderly.


\vspace{.25in}
A second reason to give up the self weighting property of
proportionate sampling occurs when we are less interested in
estimating an overall proportion or mean and more interested in
comparisons among strata. Suppose, for example, that in your group
project you wish to compare attitudes about extra-curricular
activities for H\&SS students with those of students in all other
colleges. Think of H\&SS and all other colleges as two strata.  Then
instead of dividing your sample in proportion to the size of the
colleges you might choose sample sizes for your strata to make the
within stratum variances equal. Because the stratum population sizes
are unequal and we are sampling without replacement our choice of
stratum sample sizes will also not be equal, but they will be quite
different from those in proportional allocation.

\vspace{.5in}
{\bf {Historical Note:}} The formal idea of stratification appeared
early in the 20th century but it was the pioneering paper by Jerzy
Neyman in 1934 that popularized the approach and introduced the idea
of optimal allocation, which was found also in an 1923 paper by
Tschuproff.  Neyman's version of the optimal allocation incorporates
differing costs for sampling in different strata.\\

 
{\bf {Note on Proofs of Results:}} Many textbooks on sampling actually
work out the messy formulas to show the gains due to proportionate
sampling and to optimal allocation.  Good references are Cochran
(1977) or Lohr (1999).\\


\noindent {\bf {References}} \\

\noindent Cochran, W. G. (1977). {\em {Sampling Techniques.}}
3rd. Edition. Wiley: New York.\\

\noindent Neyman, J. (1934). On the two different aspects of the
representative method: the method of stratified sampling and the
method of purposive selection (with discussion). {\em {Journal of the
    Royal Statistical Society}}, {\bf {97}}, 558-625.\\

\noindent Henry, G. T. (1990). {\em {Practical Sampling.}} Sage:
Newbury Park, CA. \\

\noindent Lohr, S. (1999).  {\em Sampling: Design and
  Analysis}. Duxbury: Pacific grove, CA.\\

\noindent Williams, B. (1978). {\em {A Sampler on Sampling.}} Wiley: New York.


\end{document}

































