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\def\date{\today}

\begin{center}
\large\bf
36-303: Sampling Surveys and Society\\
Stratified Sampling Examples\\
\rm Brian Junker\\
132E Baker Hall\\
brian@stat.cmu.edu\\
\date
\end{center}

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\rhead{Stratified Samples}
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\tableofcontents

\clearpage

\section{Key Ideas}

\subsection{Why stratify?}

Strata are just subgroups of the target population that have some
feature in common (gender, major, region, income, \ldots)
%
Reasons to stratify include:
\begin{itemize}
\item We need to make a separate inference for each stratum (e.g. we
  want to estimate men’s and women’s incomes separately);
\item Different sampling schemes would be used in each stratum (PA
  voters in PA [telephone survey?], vs PA voters in Iraq [mail or
    email?]);
\item Population is geographically diverse (Minnesota, Illinois, Ohio,
  Pennsylvania, \ldots);
\item Reduce variance of estimates (and reduce sample size) by
  exploiting similarity among members of the same stratum.
\end{itemize}

\subsection{Notation and Basic Facts}

We assume there are $H$ strata.  Within each stratum, we will use the
following notation:
\begin{itemize}\parsep=0pt\itemsep=0pt\parskip=0pt\em
\item $N_h$ = population size in each stratum
\item $n_h$ = sample size in each stratum
\item $f_h$ = $n_h/N_h$ = sampling fraction, each stratum
\item $W_h$ = $N_h/N$ = stratum weights, each stratum
\item $N = \sum_{h=1}^H N_h$ = total population size
\item $n = \sum_{h=1}^H n_h$ = total sample size
\end{itemize}
We also assume throughout that within each stratum we are taking a
simple random sample (SRS) without replacement, of size $n_h$ from the
stratum population of size $N_h$.

Let $y$ be some attribute of interest.  $y$ could be numerical, like
a person's income, or it could be Bernoulli, like $y=1$ if the person
rides PAT buses, and $y=0$ otherwise.

\subsubsection{Means}

The population average for $y$ is
\begin{equation}
	\bar y_{pop} = \frac1N\sum_{i=1}^N y_i
	= \frac1N\sum_{h=1}^H \sum_{i=1}^{N_h} y_{hi}
	= \sum_{h=1}^H \frac{N_h}{N} \frac1{N_h}\sum_{i=1}^{N_h} y_{hi}
	= \sum_{h=1}^H \frac{N_h}{N} \bar y_{h,pop}
        = \sum_{h=1}^H W_h \bar y_{h,pop}
        \label{eq:pop-ave}
\end{equation}
where $\bar y_{h,pop} = \frac1{N_h} \sum_{i=1}^{N_h} y_i$ and 
\begin{equation}
  W_h = N_n/N . \label{eq:weights}
\end{equation}
So we can see that $\bar y_{pop}$ is a weighted sum of the population
averages in each stratum, $\bar y_{h,pop}$, where the weights $W_h$ in
\eq{eq:weights} are proportional to the size of the stratum.

To make an estimate from stratified sample data we can mimic the
idea in \eq{eq:pop-ave} with sample averages:
\begin{equation}
	\bar y_{st} = \frac1n\sum_{i=1}^n y_i
	= \sum_{h=1}^H \frac{N_h}{N} \bar y_h
        = \sum_{h=1}^H W_h \bar y_h        
        \label{eq:st.ybar}
\end{equation}
where $\bar y_h = \frac1{n_h} \sum_{i=1}^{n_h}  y_{hi}$. and
the $W_h$ are as in \eq{eq:weights}.

It is important to realize that the stratified sample average is
unbiased for the population average:
\[
 \let\osum\sum\def\sum{\displaystyle\osum}
 \begin{array}{rcl@{~~~~~~~~~}p{2.5in}}
  E[\bar y_{st}] & = & E\left[\sum_{h=1}^H W_h \bar y_h\right]  
 & (by \eq{eq:st.ybar})\\
    & = & \sum_{h=1}^H W_h E[\bar y_h] \\
    & = & \sum_{h=1}^H W_h \bar y_{h,pop} & ($E[\bar y_{h,srs}] = \bar
  y_{h,pop}$ in each stratum $h$) \\
    & = & \bar y_{pop} & (by \eq{eq:pop-ave}).
 \end{array}
\]

\subsubsection{Variances}

Within each stratum, since we are doing SRS without replacement, we
have 
\begin{equation}
	Var(\bar y_h) = (1-f_h) \frac{s^2_h}{n_h} 
	\mbox{~~~~~~where~~~~~~} s^2_h = 
	\frac1{n_h-1} \sum_{i=1}^{n_h} (y_{hi} - \bar y_h)^2
        \label{eq:within.var}
\end{equation}
as usual.  Note that for Bernoulli data, there is a simpler expression
for the within-cluster standard error:
\begin{equation}
      s_h^2/n_h = \frac{p_h(1-p_h)}{n_h-1}
      \label{eq:prop-adjust}
\end{equation}
To obtain an overall variance estimate for $\bar y_{st}$ we have:
\begin{eqnarray}
	Var(\bar y_{st})  & = & Var\left(\sum_{h=1}^H W_h \bar
        y_h\right) \nonumber \\
	& = & \sum_{h=1}^H Var(W_h \bar y_h)
	~ = ~  \sum_{h=1}^H W_h^2Var( \bar y_h) \nonumber \\
	& = &  \sum_{h=1}^H W_h^2 (1-f_h) \frac{s_h^2}{n_h}
        \label{eq:total.var}
\end{eqnarray}

\section{Self-weighting Samples, and Proportionate Sampling}

\subsection{Self-weighting samples}

In general, 
\[
   \bar y_{st} = \sum_{h=1}^H W_h \bar y_h
\]
need not be the same as
\[
   \bar y_{srs} = \frac1n \sum_{i=1}^n y_i .
\]
When we are lucky enough to have $\bar y_{srs} = \bar y_{st}$ the
stratification is said to be ``self-weighting''.

\subsection{Proportionate sampling}

One case of a self-weighing stratification is called ``proportionate
sampling''.  We get proportionate sampling when we force the sampling
fraction $f_h$ to be the same in every stratum,
\begin{equation}
   f_h \equiv f, ~\forall h
   \label{eq:proportionate}
\end{equation}
Since $f_h=n_h/N_h$ it is easy to see that if every $n_h/N_h\equiv f$
then also $n/N=f$, and so
\[
    n_h/n = (n_h/N_h)(N_h/N)(N/n) = f (N_h/N) (1/f) = N_h/N
\]
Therefore,
\begin{equation}
    \bar y_{st} = \sum_{h=1}^H W_h \bar y_h 
    = \sum_{h=1}^H (N_h/N) \bar y_h 
    = \sum_{h=1}^H (n_h/n) \bar y_h 
    = \sum_{h=1}^H (n_h/n) \frac{1}{n_h}\sum_{i=1}^{n_h}  y_{hi}
    = \frac1n\sum_{i=1}^n y_i 
    = \bar y_{srs}
    \label{eq:self-weighting}
\end{equation}

\subsection{Design Effect}

The nice part about self-weighting samples is that anybody, with no
statistical training, can calculate $\bar y_{srs}$ and it will be the
right number for $\bar y_{st}$ as well.  The only difference is that
instead of 
\begin{equation}
   Var (\bar y_{srs}) = (1-f)\frac{s^2}{n}
   \label{eq:srs.var}
\end{equation}
we will have
\begin{equation}
   Var(\bar y_{st}) = \sum_{h=1}^H W_h^2 (1-f_h)\frac{s_h^2}{n_h}
   \label{eq:st.var}
\end{equation}
The ``art'' of stratified sampling is making sure that $Var(\bar
y_{st})$ is smaller than $Var(\bar y_{sr})$.  The {\em design effect}
is defined to be 
\begin{equation}
d^2 = \frac{Var(\overline y_{st})}{Var(\overline y_{srs})}
	=\frac{\sum_{h=1}^H W_h^2 (1-f_h) \frac{s_h^2}{n_h}}
	{(1-f)\frac{s^2}{n}}
        \label{eq:deff}
\end{equation}
and we have succeeded in the art of stratified sampling  when $d^2<1$.
Cochran (1961) did some experiments with stratification and showed
that $d^2<1$ when
\begin{itemize}
\item 2--6 strata are used (more strata tend to reduce $Var(\bar
  y_{st})$ further, but there are diminishing returns as $H$ grows);
\item Elements are more similar to each other within strata than
  between (e.g., substantively meaningful strata);
\item Proportionate sampling is used.
\end{itemize}

\section{Examples}

\subsection{Confidence Intervals for SRS and Proportionate Stratified
  Samples}

\subsubsection{Proportion $p_{pop}$}

Consider the three urns we played with in class:
\begin{itemize}
\item Urn 1: $N_1=100$ balls; 10 yellow, 90 blue, $p_1=0.1$
\item Urn 2: $N_2=100$ balls; 20 yellow, 80 blue, $p_2=0.2$
\item Urn 3: $N_3=100$ balls; 30 yellow, 70 blue, $p_3=0.3$ \\
\item Combined urns: $N=300$, 60 yellow, 240 blue $p=0.2$.
\end{itemize}
We want to compare $n=30$ SRS without replacement from the combined
urns, with a stratified sample of $n_h=10$ balls each from the three
urns.
\begin{description}

\item[{\boldmath$n=30$} SRS:] Using formula \eq{eq:srs.var} (and the adjustment
  in equation \eq{eq:prop-adjust}),
\[
   Var (\hat p_{srs}) = (1-f)\frac{s^2}{n} 
   = (1-f)\frac{p(1-p)}{n-1} = (1-30/300)\frac{0.2(1-0.2)}{30-1}
   = 0.00497
\]
so, if we draw 30 balls and 8 of them are yellow, then $\hat
p=8/30=0.27$ and we get a 95\% CI 
that is
\[
    0.27 \pm (1.96)\sqrt{0.00497} = 0.27 \pm 0.1381
\]

\item[{\boldmath $n_h=10$} stratified SRS's from each urn:] Using
  formula \eq{eq:st.var}, and noting that $f_h=10/100=0.1$ for all
  strata (proportionate sampling) and $W_h=100/300=1/3$ for all strata,
\begin{eqnarray*}
   Var(\hat p_{st}) = \sum_{h=1}^H W_h^2 (1-f_h)\frac{s_h^2}{n_h}
    = \sum_{h=1}^H W_h^2 (1-f_h)\frac{p_h(1-p_h)}{n_h-1}
    = \sum_{h=1}^H (1/3)^2 (1-0.1)\frac{p_h(1-p_h)}{10-1} \\
    = \frac{(1/3)^2(0.9)}{9}\left[ (0.1)(0.9) + (0.2)(0.8) +
      (0.3)(0.7)\right]
    = 0.00511
\end{eqnarray*}
If 2 of the 10 balls from Urn 1 are yellow, 2 from Urn 2 are
yellow, and 3 from Urn 3 are yellow, we can use the fact that we are
doing {\em proportionate sampling} to obtain $\hat p=8/30=0.27$
(otherwise we have to use formula \eq{eq:st.ybar}).  We get a 95\% CI
\[
   0.27 \pm (1.96)\sqrt{0.00511} = 0.27 \pm 0.1401
\]
\end{description}
Note that in this case, {\em stratification was not as good as a plain
  old SRS!}, although the difference wasn't enormous.  The problem is
that the three strata are a little too much like the combined urn.  It
is better when the strata have much less internal variability than the
combined population would have.  For example:
\begin{description}
\item[Setup:] Same setup but now suppose the three urns have
  proportions of yellow $p_1=0.1$, $p_2=0.5$ and $p_3=0.9$.  The
  combined population will now have a proportion $p=0.5$ of yellow
  balls.  We'll continue to have $N_h=100$, $n_h=10$.

\item[SRS CI:] The margin of error for a 95\% CI is going to be
\[
   (1.96)\times\sqrt{(1-f)\frac{p(1-p)}{n-1}}
   = (1.96)\times\sqrt{(1-30/300)\frac{(0.5)(1-0.5)}{30-1}}
   = 0.1726
\]

\item[Stratified CI:] The margin of error for a 95\% CI is going to be
\begin{eqnarray*}
  &&(1.96)\times\sqrt{\sum_{h=1}^H W_h^2 (1-f_h)\frac{p_h(1-p_h)}{n_h-1}} \\
  &&= (1.96)\times\sqrt{\frac{(1/3)^2(1-10/100)}{9}\left[ (0.1)(0.9) + (0.5)(0.5) +
      (0.9)(0.1)\right]} 
  = 0.1355  
\end{eqnarray*}

\end{description}
Now the stratified CI wins, because stratum 1 and 3 have so much less
internal variability than the combined population does.  In this case
also, the design effect is
\[
    d^2 = \frac{Var(\hat p_{st})}{Var(\hat p_{srs})}
    =\frac{(0.1355/1.96)^2}{(0.1726/1.96)^2}
    = 0.616
\]
so the stratified variance for $\bar y$ is only 61\% of the SRS variance!

\subsubsection{Numerical $\bar y_{pop}$}

This example is a little contrived, in order to make all of the parts
``visible''.  But the underlying principle, that stratification
reduces variance, still holds.


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, shown in \tbl{t:taxframe}.

\begin{table}
\begin{center}
\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}
\caption{Sampling frame for Williams (1978)
  example.}\label{t:taxframe}
\end{center}
\end{table}

\begin{description}
\item[SRS:]
For the complete population we have $N=9$, and for the actual income 
 $\bar y_{pop} = 103.56$ and $s^2_{pop}=1621.14$.  The variance for an
SRS of size $n=3$ is
\[
   Var(\bar y_{srs}) = = (1-f)\frac{s^2}{n}
   = (1- 3/9)\frac{1621.14}{3} = 360.25
\]
(if we didn't know $s^2_{pop}$ to plug in for $s^2$ we would plug in a
sample variance instead), so the 95\% margin of error for estimating
$\bar y_{pop}$ would be
\[
   (1.96)\times \sqrt{360.25}=37.20 
\]

\item[Stratified:]
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:\\

     \hfil  $ N_1=8, \bar y_{1,pop}=91.50$  and $s^2_{1,pop}=515.75$;
 $N_2=1, \bar y_{2,pop}=200$ and $s^2_{2,pop}=0$.

\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 (note that this is not a proportionate
sample!).  The variance for this stratified sample is
\[
   Var(\bar y_{st}) = \sum_{h=1}^H W_h^2 (1-f_h)\frac{s_h^2}{n_h}
   = (8/9)(1-3/8)\frac{515.75}{8} + (1/9)(1-1/9)\frac{0}{1}
   = 35.82 + 0 = 35.82
\]
and the 95\% margin of error for estimating $\bar y_{pop}$ will
be 
\[
   (1.96)\times\sqrt{35.82} = 11.73
\]
The design effect in this case is
\[
    d^2 = \frac{Var(\bar y_{st})}{Var(\bar y_{srs})}
    =\frac{35.82}{360.32}
    = 0.0994
\]
so in this case the variance for the stratified sample is {\em less
  than 10\%} of the variance from an SRS.

\end{description}
Note especially in the Variance calcualtion above the expression
``35.82 + 0''.  The zero comes because there is no variability in
stratum 2, just the one observation.  Low within stratum variability
causes low variance, which causes low sample size (see next
subsection!).  This is a good thing.

\subsection{Sample Size Calculations for SRS and Proportionate
  Stratified Samples}

\subsubsection{Proportion $p_{pop}$}

We will illustrate calculating a sample size to get a 0.10 margin of
error for a 95\% CI.  We will assume a population of 300 units.
\begin{description}

\item[SRS:] We've done this a few times before.  We first calculate
  the SRS with replacement sample size
\[
   n_0 \geq \frac{(Z)^2(SD)^2}{(ME)^2} = \frac{(1.96)^2(0.5)^2}{(0.10)^2} = 96.04
\]
(where we take $SD=0.5$ as a worst case),
and then apply the correction for SRS without replacement.  
\[
     n \geq \frac{Nn_0}{N+n_0} = \frac{(300)(96.04)}{300+96.04} = 72.75
\]

\item[Stratified:] Let us suppose that our population breaks into
  three strata, with $N_1=50$, $N_2=100$, and $N_3=150$ units per
  stratum.  Here are two possible approaches to determining sample
  sizes for the three strata:
\begin{itemize}
\item {\em Trial and error with proportionate sampling.}  One might
  imagine a table like this:
\begin{center}
\begin{tabular}{ccccc}
$f$ & $n_h = f\times N_h$ & $p_h$ & ME $=$ $1.96\times\sqrt{\mbox{\em
    Eq.~\eq{eq:st.var}}}$ & $n=n_1+n_2+n_3$ \\
\hline
0.10 & (~~5, 10, 15) & (0.5, 0.5, 0.5) & 0.18 & 30 \\
0.20 & (10, 20, 30) & (0.5, 0.5, 0.5) & 0.12 & 60 \\
0.30 & (15, 30, 45) & (0.5, 0.5, 0.5) & 0.09 & 90 \\
0.25 & (13, 25, 37) & (0.5, 0.5, 0.5) & 0.10 & 75 
\end{tabular}
\end{center}
Note that here we have used just about the worst possible assumption,
namely that $p_1=p_2=p_3$, so we get no advantage (and actually
probably a slight disdadvantage) for stratifying.  

On the other hand if we knew something about $p_1$, $p_2$, $p_3$
(perhaps from a pilot study?), then things look better:
\begin{center}
\begin{tabular}{ccccc}
$f$ & $n_h = f\times N_h$ & $\hat p_{h,pilot}$ & ME $=$ $1.96\times\sqrt{\mbox{\em
    Eq.~\eq{eq:st.var}}}$ & $n=n_1+n_2+n_3$ \\
\hline
0.100 & (~~5, 10, 15) & (0.1, 0.5, 0.9) & 0.14 & 30 \\
0.200 & (10, 20, 30) & (0.1, 0.5, 0.9) & 0.09 & 60 \\
0.150 & (~~8, 15, 22) & (0.1, 0.5, 0.9) & 0.11 & 45\\
0.175 & (~~9, 17, 26) & (0.1, 0.5, 0.9) & 0.10 & 53\\
\end{tabular}
\end{center}
$n=53$ is more efficient than $n= 72$ or 73 that would be needed for
SRS.  Note that we have used the additional information in the $\hat
p_{pilot}$'s to get this better sample size!

\item {\em Optimal Allocation.} It turns out that we can achieve
  greater gains from stratification by choosing our sampling fractions
  to be proportional to the standard deviations in each stratum.  The
  idea is
\[
         n_h = c s_h
\]
where $c$ is a constant, and $s_h$ is the (population) SD for stratum
$h$.  For fixed overall sample size $n$, we also need
\[
         \sum_{h=1}^H n_h = n, ~~~~{\mbox{i.e.}}~ \sum_{h=1}^H c s_h = n
\]
and so, clearly,
\[
      c = \frac{n}{\sum_{h=1}^H s_h}
\]
and therefore
\[
         n_h = \frac{s_h}{\sum_{k=1}^H s_k}\times n
\]
We can apply this to our sample size problem for a margin of error of
0.10 with another 'trial and error table', this time for the overall
sample size $n$:
\begin{center}
\begin{tabular}{cccc}
$n$ & $n_h = s_h/\sum_k s_k$ & $f_h=n_h/N_h$ &  ME $=$
  $1.96\times\sqrt{\mbox{\em Eq.~\eq{eq:st.var}}}$ \\
\hline
33  & (11, 11, 11) & (0.22, 0.11, 0.07) & 0.18 \\
66  & (22, 22, 22) & (0.44, 0.22, 0.15) & 0.12 \\
99  & (33, 33, 33) & (0.66, 0.33, 0.22) & 0.09 \\
90  & (30, 30, 30) & (0.60, 0.30, 0.20) & 0.10
\end{tabular}
\end{center}
Note that this is {\em not} a proportionate allocation scheme; the
$f_h$'s are not all equal.

Note also that we have taken $p=0.5$ again, so that every $s_h$ is
0.5, and so we are really just doing 1/3-1/3-1/3 allocation.  This is
probably why the result is not so great ($n=90$ suggested here,
vs. the best so-far of $n=53$ above.

As before we can improve things by using pilot estimates for $p$.  Now
if we use our fictional pilot estimates $\hat p_{pilot} = (0.1, 0.5,
0.9)$, we discover that 
\[
    s_1 = \sqrt{(0.1)(0.9)} = 0.3;~ 
    s_2 = \sqrt{(0.5)(0.5)} = 0.5;~
    s_3 = \sqrt{(0.9)(0.1)} = 0.3
\]
and so the proportions of $n$ allocated to each stratum would be
\[
    n_1 = s_1/\mbox{$\sum_h$} s_h = (0.27) n ; ~
    n_2 = s_2/\mbox{$\sum_h$} s_h = (0.45) n ; ~
    n_3 = s_3/\mbox{$\sum_h$} s_h = (0.27) n ; ~
\]
Note that most of the sample is being spent on the stratum with the
largest variance; that is why this method works.  Our new trial and
error table is as follows:
\begin{center}
\begin{tabular}{cccc}
$n$ & $n_h = s_h/\sum_k s_k$ & $f_h=n_h/N_h$ &  ME $=$
  $1.96\times\sqrt{\mbox{\em Eq.~\eq{eq:st.var}}}$ \\
\hline
33  & (~~9, 15, ~~9) & (0.18, 0.15, 0.06) & 0.13 \\
66  & (18, 30, 18) & (0.36, 0.30, 0.12) & 0.09 \\
45  & (12, 21, 13) & (0.25, 0.20, 0.08) & 0.11 \\
54  & (14, 25, 15) & (0.30, 0.25, 0.10) & 0.10
\end{tabular}
\end{center}
Once again this is not a proportionate allocation scheme.  It is also
worth noting that the $N_h$'s don't seem to matter much: smaller
samples are taken in the first and third strata, because that's where
the variance, and hence the uncertainty, is the lowest -- even though
strata 1 and 3 are the smallest and largest strata.

Our answer here is comparable to the answer we got with proportionate
sampling; indeed $n=53$ also produces a 10\% margin of error here.


\end{itemize}

\end{description}

\subsubsection{Numerical $\bar y_{pop}$}

To illustrate sample size calculations for a numerical variable, I
will consider the problem of estimating average income for a
population of size 300, that can be broken up into three strata:
\begin{center}
\begin{tabular}{cccc}
Stratum ($h$) & $N_h$ & $\bar y_{h,pop}$ & $s^2_{h,pop}$ \\
\hline
1             & ~~50    & 140 & 2000\\ 
2             & 100   & ~~85 & ~~415 \\
3             & 150   & ~~50 & ~~200
\end{tabular}\\
\parbox{2.5in}{\footnotesize Note:
$\bar y_{h,pop}$ is the stratum mean income, in 1,000's of dollars.
$s^2_{h,pop}$ is the stratum variance in income.}
\end{center}
From this table one can easily calculate that 
% \[
%    \bar y_{pop} = \frac{(50)(140) + (100)(85) + (150)(50)}{300} = 76.67
% \]
% and
\[
     s^2_{pop} = \frac{(50)(2000) + (100)(415) + (150)(200)}{300} = 571.67
\]
so that the population SD is $\sqrt{572.67} = 23.91$ (if we didn't
know this we could estimate it from pilot data, for example).

We want to know the sample size needed to estimate the population mean
income with a 95\% margin of error of $\pm$ 5 thousand dollars.

\begin{description}

\item[SRS] This is our old familiar friend.  It proceeds pretty much
  as before.  We first calculate the SRS with replacement sample size
\[
   n_0 \geq \frac{(Z)^2(SD)^2}{(ME)^2} = \frac{(1.96)^2(23.91)^2}{(5)^2} = 87.85
\]
and then apply the correction for SRS without replacement.  
\[
     n \geq \frac{Nn_0}{N+n_0} = \frac{(300)(87.85)}{300+87.85} = 67.95
\]




\item[Stratified]  We try the same two approaches that we tried for
  proportions:

\begin{itemize}

\item {\em Trial and error with proportionate sampling.}  Here is our
  trial and error table:
\begin{center}
\begin{tabular}{ccccc}
$f$ & $n_h = f\times N_h$ & $s_h$ & ME $=$ $1.96\times\sqrt{\mbox{\em
    Eq.~\eq{eq:st.var}}}$ & $n=n_1+n_2+n_3$ \\
\hline
0.10 & (~~5, 10, 15) & (44.72, 12.04, 14.14) & 7.45 & 30 \\
0.20 & (10, 20, 30) & (44.72, 12.04, 14.14) & 4.96 & 60 \\
\end{tabular}
\end{center}
So, in this case, stratification with proportionate allocation
requires only a sample of size $n=60$, which is
already beating SRS's sample size requirement of $n=68$.

\item {\em Optimal Allocation.} Reacalling from above that our
  standard deviations are $s_1=44.72$, $s_2=12.04$, and $s_3=14.14$
  As in the previous subsection, the idea is to take
\[
    n_1 = s_1/\mbox{$\sum_h$} s_h = (0.63) n ; ~
    n_2 = s_2/\mbox{$\sum_h$} s_h = (0.17) n ; ~
    n_3 = s_3/\mbox{$\sum_h$} s_h = (0.20) n ; ~
\]
Again we build a trial-and-error table for various sample sizes $n$:
\begin{center}
\begin{tabular}{cccc}
$n$ & $n_h = s_h/\sum_k s_k$ & $f_h=n_h/N_h$ &  ME $=$
  $1.96\times\sqrt{\mbox{\em Eq.~\eq{eq:st.var}}}$ \\
\hline
30 & (19, 5, 6) & (0.37, 0.05, 0.04) & 7.03\\
60 & (38, 10, 12) & (0.76, 0.10, 0.08) & 4.64\\
45 & (28, ~~8, ~~9) & (0.57, 0.08, 0.06) & 5.55 \\
54 & (34, ~~9, ~~11) & (0.68, 0.09, 0.07) & 4.97
\end{tabular}
\end{center}
Once again this is {\em not} a proportionate allocation scheme; the
$f_h$'s are not all equal.  For this problem, the ``optimal'' scheme
produces a somewhat better sample size requirement, $n=54$ than the
proportionate allocation scheme, $n=60$.

\end{itemize}

\end{description}

\section{Proportionate vs. Non-Proportionate Stratified Samples}

{\em Proportionate sampling} schemes are easy to explain: the formula
for $\bar y$ is the same under proportionate stratified sampling as it
is under SRS (because proportionate sampling is self-weighting);
moreover it is easy to explain to someone that you sampled more from
the bigger strata, and that is what proportionate sampling does.

In some ways taking a bigger sample from the bigger stratum is looking
at the problem wrong.  The things that are hard to estimate are not
things in big strata, but rather things that vary a lot within their
stratum.  {\em Optimal sampling} takes account of this by allocating
sample size to each stratum, proportionate to that stratum's SD,
rather than proportionate to that stratum's size.  As we saw above,
optimal sampling can produce lower sample sizes than proportionate
sampling.  You pay for this greater efficiency by not being able to
use $\bar y_{srs}$ as your estimator anymore (instead, use the
weighted $\bar y_{st}$ from equation \eq{eq:st.ybar}).

Proportionate sampling gets you most of the benefits of stratified
sampling, in getting you a reduced sample size.  Optimal sampling
reduces the sample size a little bit more, but sometimes not much
more. 

If you are only interested in estimating overall population
quantities, then once you have assigned sample sizes to each stratum
by one of these methods, you are done.  However, you may want to
estimate some quantities separately in a stratum as well.  

\noindent{\bf Example:}

If you were doing a survey of satisfaction with sports facilities at
Carnegie Mellon, your strata might be: staff, faculty,
student-athletes, and other students.  You might be epsecially
interested in the proportion of student athletes who are satisfied.
But that is a small stratum, so proportionate sampling will allocate
only a few observations to that stratum.

You could do a separate, plain-vanilla SRS calculation in that one
stratum, to see how many student athletes you need to just estimate
their satisfaction.  This will probably lead to a larger sample size
in that stratum than proportionate sampling suggests.  If so, go ahead
and increase the sample size for the student-athletes stratum.  It
will allow you to make a better estimate for student-athletes, and it
will not wreck anything else in the survey (in fact it should help
your overall estimates just a little bit).


\section{Pro's and Con's of Stratified Samples}

Pro's:
\begin{itemize}
\item Can make your sample ``more representative'' of the population
  by making sure you get some observations from each stratum.
\item Can ``oversample'' some strata because you want really good
  estimates for those strata.
\item Can generally design a set of strata so that the margin of error
  is smaller for stratified samples than for nonstratified.
\end{itemize}

\noindent
Con's:
\begin{itemize}
\item Figuring out what strata to use, largely a trial-and-error game.
\item Figuring out the sample size for each stratum, largely a
  trial-and-error game.
\item Need to know 
\begin{itemize}
\item The population sizes of the strata, $N_h$
\item The overall population size of your sampling frame, $N$
\item Rough guesses about the variances $\hat s^2_{h,pilot}$ and/or the
  proportions $\hat p_{h,pilot}$, often from a pilot study or other
  background information about your strata.
\end{itemize}
in order to make stratification work for you to reduce sample size.
\item Once you have committed to stratified sampling, now you have to
  make the extra effort necessary to reach your target sample size
  {\em in each stratum}.  It will make for a better survey, but it is
  also more work for you.
\end{itemize}
\end{document}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\item {\em Allocate some fraction of the error to each stratum.} We
  know that we want
\[
    (ME)^2 \geq (1.96)^2 \left[W_1^2 (1-f_1) \frac{s_1^2}{n_1} + W_2^2 (1-f_2)
    \frac{s_2^2}{n_2} + W_3^2 (1-f_3) \frac{s_3^2}{n_1}\right]
\]
Plugging in $ME=0.10$ for this problem, and knowing the weights
$W_1=N_1/N=50/300=1/6$, $W_2=1/3$, $W_3=1/2$, we get 
\[
   \frac{(ME)^2}{(1.96)^2} = 0.002603 \geq 
   (1/6)^2 (1-f_1) \frac{s_1^2}{n_1} + (1/3)^2 (1-f_2)
    \frac{s_2^2}{n_2} + (1/2)^2 (1-f_3) \frac{s_3^2}{n_1}
\]
Now we could allocate, say, 1/3 of the squared ``error'',
$(0.10)^2/(1.96)^2=0.000868$ to each term on the right, so that we
would want to solve:
\begin{itemize}
\item $0.000868 \geq (1/6)^2 (1-f_1) \frac{s_1^2}{n_1}$.  This can be
  solved as if it were an SRS problem, treating 1/6 as ``$z$'':
\[
    n_{10} \geq \frac{(1/6)^2(0.5)^2}{0.000868} = 8.003333
\]
so
\[
     n_1 \geq \frac{N_1n_{10}}{N_1+n_{10}} =
     \frac{(50)(8.00333)}{58.00333} = 6.90
\]

\item Similarly, 
\[
    n_{20} \geq \frac{(1/3)^2(0.5)^2}{0.000868} = 32.01
\]
so
\[
    n_2 \geq  \frac{N_2n_{20}}{N_2+n_{20}} =
    \frac{(100)(32.01)}{132.01} = 24.25
\]

\item Finally
\[
    n_{30} \geq \frac{(1/2)^2(0.5)^2}{0.000868} = 72.03
\]
so
\[
    n_3 \geq  \frac{N_3n_{30}}{N_3+n_{30}} =
    \frac{(150)(72.03)}{150+72.03} = 48.66
\]
\end{itemize}
So, altogether, the sample size estimate is $7+25+49=81$.  This isn't
as good as 72 or 73 predicted by simple SRS.  One could improve things
by either using pilot estimates for the $p$'s or using something other
than 1/3-1/3-1/3 allocations of the ``error''.

