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}