Reference p.121 of Scheaffer, Mendenhall, and Ott

An advertising firm, interested in determining how much to emphasize television advertising in a certain county decides to conduct a sample survey to estimate the average number of hours each week that households within that county watch television. The county has two towns, A and B, and a rural area C. Town A is built around a factory and most households contain factory workers with school-aged children. Town B contains mainly retirees and rural area C residents are mainly farmers.

There are 155 households in town A, 62 in town B and 93 in rural area C. The firm decides to select 20 households from Town A, 8 households from Town B, and 12 households from the rural area. The results are given in the following table:

Here is the Minitab output that describes the data from each stratum: ( N in the output denotes numbers of data)

Usually, a sample is selected by some probability design from each of the L strata in the population, with selections in different strata independent of each other. The special case where from each stratum a simple random sample is drawn is called a stratified random sample.

Notation

• L = the number of strata
• N_h = number of units in each stratum h
• n_h = the number of samples taken from stratum h
• N = the total number of units in the population, i.e., N₁ + N₂ + ... + N_L

For our "Watching TV" example the following values are:

L = 3, \(N_1\) = 155, \(N_2\) = 62, \(N_3\) = 93, N = 155 + 62 + 93 = 310

Estimating the Population Total

\(\hat{\tau}_{st}=\sum\limits_{h=1}^L \hat{\tau}_h\)

The total is from each stratum added up where \(\hat{\tau}_h\) is an unbiased estimator for \(\tau_h\).

Since selections in a different strata are independent, the variance is:

\(Var(\hat{\tau}_{st})=\sum\limits_{h=1}^L Var(\hat{\tau}_h)\), and

\(\hat{V}ar(\hat{\tau}_{st})=\sum\limits_{h=1}^L \hat{V}ar(\hat{\tau}_h)\)

The formula is computed differently according to the sampling scheme within each stratum. For stratified random sampling, i.e., take a random sample within each stratum:

\(\hat{\tau}_h=N_h \bar{y}_h\)

\(\hat{V}ar(\hat{\tau}_{st})=\sum\limits_{h=1}^L N_h \cdot (N_h-n_h)\cdot \dfrac{s^2_h}{n_h}\)

\(s^2_h=\dfrac{1}{n_h-1}\sum\limits_{i=1}^{n_h}(y_{hi}-\bar{y}_h)^2\)

You can see that this turns out pretty easy to remember, and one can easily obtain the estimates for the population mean.

\(\hat{\mu}_{st}=\dfrac{\hat{\tau}_{st}}{N}\)
\(\hat{V}ar(\hat{\mu}_{st})=\dfrac{1}{N^2}\hat{V}ar(\hat{\tau}_{st})\)

For stratified random sampling:

\(\bar{y}_{st}=\dfrac{1}{N} \sum\limits_{h=1}^L N_h \bar{y}_h\)

\(\hat{V}ar(\bar{y}_{st})=\sum\limits_{h=1}^L \left(\dfrac{N_h}{N}\right)^2 \left(\dfrac{N_h-n_h}{N_h}\right) \dfrac{s^2_h}{n_h}\)

\(s_h\) is the sample standard deviation of h stratum as given in Minitab.

Confidence Intervals

When all of the stratum sizes are small, an approximate 100(1-\(\alpha\))% CI for \(\tau\) is:

\(\hat{\tau}_{st} \pm t\sqrt{\hat{V}ar(\hat{\tau}_{st})}\)

However, when the stratum sample sizes are at least 30, use z to approximate t.

What are the degrees of freedom for the t used in this formula for the confidence interval? Intuitively we would want this to be, (\(n_1-1)+(n_2-1)+...+(n_L-1)\), and this is correct when the variances of all strata are all the same. But when this is not the case and we can not pool the degrees of freedom, we will need to use the Satterwaithe approximation for the degrees of freedom as follows:

\(d=\left(\sum\limits_{h=1}^L a_h s^2_h\right)^2/\sum\limits_{h=1}^L \dfrac{(a_h s^2_h)^2}{(n_h-1)}\)

where, \(a_h=\dfrac{N_h(N_h-n_h)}{n_h}\)

In particular, when \(N_h\) are all equal, \(n_h\) are all equal and \(s^2_h\) are all equal , the d.f. = n - L.

For the TV example:

\(a_1=\dfrac{N_1(N_1-n_1)}{n_1}=\dfrac{155(155-20)}{20}=1046.25\)

\(a_2=\dfrac{N_2(N_2-n_2)}{n_2}=\dfrac{62(62-8)}{8}=418.5\)

\(a_3=\dfrac{N_3(N_3-n_3)}{n_3}=\dfrac{93(93-12)}{12}=627.75\)

\begin{align}
d&= \dfrac{(a_1s^2_1+a_2s^2_2+a_3s^2_3)^2}{\dfrac{(a_1s^2_1)^2}{n_1-1}+\dfrac{(a_2s^2_2)^2}{n_2-1}+\dfrac{(a_3s^2_3)^2}{n_3-1}}\\
&= \dfrac{(1046.5\cdot(5.95)^2+418.5\cdot(15.25)^2+627.75\cdot(9.36)^2)^2}{\dfrac{(1046.5\cdot(5.95)^2)^2}{20-1}+\dfrac{(418.5\cdot(15.25)^2)^2}{8-1}+\dfrac{(627.75\cdot(9.36)^2)^2}{12-1}}\\
&=21.09\\
\end{align}

Using R

Here is the code for R for this example:

Datafile:  TVhour.txt
R code:  Chapter6_TVhour.R.txt

A firm knows that 40% of its accounts receivable are wholesale and 60% are retail. However, to identify an account without pulling a file and looking at it is difficult. An auditor randomly sampled 100 accounts without replacement. Here are the results of his sampling:

The principal of a Prep school for boys wants to estimate the average weight of the 7th-grade boys in the school. There are 4 classes, 24 students in class 1, 36 in class 2, 30 students in class 3, and 30 in class 4.

For administrative ease, he decides to use stratified sampling with each class as a stratum. The principal has enough time and money to obtain data for 20 students, and because the cost of sampling is the same in each stratum, he decides to use proportional allocation, which gives \(n_1=4, n_2=6, n_3=5\) and \(n_4=5\). The data (in lbs.) is given in the following table:

Here is the Minitab output that describes the data from each stratum:

For a 95% CI, we need to compute Satterwaithe's formula to get the degree of freedom:

\(d=\dfrac{\left(\sum\limits_{h=1}^L a_h s^2_h \right)^2}{\sum\limits_{h=1}^L \dfrac{(a_h s^2_h)^2}{n_h-1}}\)

\(a_h=\dfrac{N_h(N_h-n_h)}{n_h}\)

Plug in the formula and we get that d = 13.7576.

Round it down to 13, to be more conservative, and use d.f. = 13.

Then, an approximate 95% CI is:

\(99.3 \pm 2.160\sqrt{2.93}\)
\(=99.3 \pm 3.697\)

Looking back at the data, if we had used simple random sampling, would our CI have been tighter or looser?

Usually, the stratified random sampling will overall perform better because we usually use stratified random sampling when the stratum is more homogeneous.

There is no reason that the classes are more homogeneous in weight, and therefore there is no reason why this stratified random sampling is any better than simple random sampling.

Since the data had been collected by stratified sampling, the above method treating it as srs is the wrong way to compute the variance for this problem. How the variance is computed depends on the method by which the sample was taken. We did the computation just to show that if hypothetically, the data was collected by s.r.s. with the data turning out to be as shown (for illustration's sake), then the margin of error will be smaller.

Moral of this example:

Stratifying on class, which is not related to weight, does not result in smaller variances within the strata. On the other hand, if stratification had other purposes such as to estimate the parameters of each subgroup, it still makes sense to stratify, though the purpose is not to get estimates with smaller variance. For this particular example, the stratification to estimate the average weight for each class may be relevant.

The advertising firm wants to estimate the proportion of households in the county that view the television show "American Idol".

\(N_1=155,N_2=62, N_3=93\). As before, we stratify by town and the sample results are:

We plug in the values and we can get the following:

The following display the estimated variance for each stratum:

\begin{align}
\hat{V}ar(\hat{p}_1)&= \left(\dfrac{N_1-n_1}{N_1}\right)\cdot \dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1-1}\\
&= \left(\dfrac{155-20}{155}\right)\cdot \dfrac{0.8(0.2)}{19}\\
&= 0.007\\
\end{align}

\begin{align}
\hat{V}ar(\hat{p}_2)&= \left(\dfrac{N_2-n_2}{N_2}\right)\cdot \dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2-1}\\
&= \left(\dfrac{62-8}{62}\right)\cdot \dfrac{0.25(0.75)}{7}\\
&= 0.024\\
\end{align}

\begin{align}
\hat{V}ar(\hat{p}_3)&= \left(\dfrac{N_3-n_3}{N_3}\right)\cdot \dfrac{\hat{p}_3(1-\hat{p}_3)}{n_3-1}\\
&= \left(\dfrac{93-12}{93}\right)\cdot \dfrac{0.5(0.5)}{11}\\
&= 0.02\\
\end{align}