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Recall that allocation concerns specifying the sample sizes in a stratified sampling design with \(L\) strata — i.e., \(n_1, n_2, \dots, n_L\). A couple of things we can take into consideration are (a) the bound of the error of estimation and (b) the cost of the survey.
Recall that under stratified random sampling the variances of \(\hat\mu\) and \(\hat\tau\) are \[ V(\hat\mu) = \frac{1}{N^2}\sum_{j=1}^L N_j^2\left(1-\frac{n_j}{N_j}\right)\frac{\sigma_j^2}{n_j} \ \ \ \text{and} \ \ \ V(\hat\tau) = \sum_{j=1}^L N_j^2\left(1-\frac{n_j}{N_j}\right)\frac{\sigma_j^2}{n_j}, \] respectively, where \(\sigma_j^2\) is the variance of the observations of the elements in the \(i\)-th stratum. The bounds on the error of estimation for \(\hat\mu\) and \(\hat\tau\) are then \[ B = 2\sqrt{V(\hat\mu)} \ \ \ \text{and} \ \ \ B = 2\sqrt{V(\hat\tau)}, \] respectively.
Assume that the cost of the survey can be computed using \[ C = c_0 + \sum_{j=1}^L n_j c_j, \] where \(c_0\) is the overhead cost and \(c_j\) is the cost-per-element in the \(j\)-th stratum.
We will consider two different approaches to optimum allocation which depend on our objective.
For a fixed bound, how do we allocate to minimize the cost?
For a fixed cost, how do we allocate to minimize the bound?
These are constrained optimization problems, but the solutions are (relatively) simple as these kinds of problems go.
First we determine how to divide \(n\) into \(n_1, n_2, \dots, n_L\). Regardless of which goal we have, it can be shown that \[ n_j = n\left(\frac{N_j\sigma_j/\sqrt{c_j}}{N_1\sigma_1/\sqrt{c_1} + N_2\sigma_2/\sqrt{c}_2 + \cdots + N_L\sigma_L/\sqrt{c_L}}\right) = n\left(\frac{N_j\sigma_j/\sqrt{c_j}}{\sum_{k=1}^LN_k\sigma_k/\sqrt{c_k}}\right). \] Note that in practice we need a good guess of \(\sigma_1, \sigma_2, \dots, \sigma_L\). Also note that this does not yet give us \(n\) or \(n_1, n_2, \dots, n_L\). It only tells us the proportion of the total sample size that should be allocated to each stratum because \[ \frac{n_j}{n} = \frac{N_j\sigma_j/\sqrt{c_j}}{N_1\sigma_1/\sqrt{c_1} + N_2\sigma_2/\sqrt{c}_2 + \cdots + N_L\sigma_L/\sqrt{c_L}} = \frac{N_j\sigma_j/\sqrt{c_j}}{\sum_{k=1}^LN_k\sigma_k/\sqrt{c_k}}. \]
Example: Recall the sword fern survey.| Stratum | Region | \(N_j\) | \(n_j\) | \(\bar{y}_j\) | \(s_j\) |
|---|---|---|---|---|---|
| 1 | Forest | 30 | 8 | 287 | 149.1 |
| 2 | Prairie | 87 | 5 | 11.3 | 16.8 |
| 117 | 13 |
If we were doing this survey again at the same location, we might use \(s_1\) and \(s_2\) as guesses of \(\sigma_1\) and \(\sigma_2\), respectively. Assume that \(c_1\) = 4 and \(c_2\) = 1. What would be \(n_1/n\) and \(n_2/n\)?
Observe that \(n_j\) is proportional to \(N_j\sigma_j/\sqrt{c_j}\). What does this tell us about the relationship between \(n_j\) and \(N_j\), \(n_j\) and \(\sigma_j\), and \(n_j\) and \(c_j\)? To which strata do we allocate large sample sizes?
Second we compute \(n\). How we do this depends on our goal.
If our goal is to minimize cost for a fixed bound on the error of estimation, then we compute \[ n = \frac{\left(\sum_{j=1}^L N_j\sigma_j/\sqrt{c_j}\right)\left(\sum_{j=1}^L N_j\sigma_j\sqrt{c_j}\right)}{N^2V + \sum_{j=1}^LN_j\sigma_j^2}, \] where \(V = B^2/4\) if we are estimating \(\mu\), and \(V = B^2/(4N^2)\) if we are estimating \(\tau\).
Example: Suppose we are estimating \(\mu\) and we want a bound on the error of estimation of \(B\) = 20 \(g/m^2\). What is the \(n\) that will give us the least expensive survey with that bound on the error of estimation? Similarly what would we use for \(n\) if we wanted to estimate \(\tau\) with a bound on the error of estimation of \(B\) = 2000 \(g/m^2\)?
Example: Suppose we want to minimize the bound on the error of estimation subject to a total cost of \(C\) = 100 and an overhead cost of \(c_0\) = 20. What is \(n\)?
Finally we combine our results from the first two steps to compute for each stratum \[ n_j = n\left(\frac{N_j\sigma_j/\sqrt{c_j}}{\sum_{k=1}^L N_k\sigma_k/\sqrt{c_k}}\right). \] Example: Given the results from the earlier examples, if we are estimating \(\mu\) what are \(n_1\) and \(n_2\) if we want to minimize cost for a bound on the error of estimation of \(B\) = 20 \(g/m^2\). What if we want to minimize the bound for a fixed cost with \(C\) = 100 and \(c_0\) = 20 when estimating \(\mu\)?
Compute the allocation fraction \[ \frac{N_j\sigma_j/\sqrt{c_j}}{\sum_{k=1}^L N_k\sigma_k/\sqrt{c_k}} \] for each stratum.
Decide if you want to minimize cost for a fixed bound, or minimize the bound for a fixed cost, and then use the appropriate formula to compute \(n\).
Compute \(n_1, n_2, \dots, n_L\) using the allocation fractions and \(n\) you computed in the previous two steps as \[ n_j = n\left(\frac{N_j\sigma_j/\sqrt{c_j}}{\sum_{k=1}^L N_k\sigma_k/\sqrt{c_k}}\right). \]
Proportional allocation is a special case where the fraction of sampled elements in each stratum equals the fraction of population elements in that stratum. That is \[ \frac{n_j}{n} = \frac{N_j}{N}, \] which implies that \(n_j = nN_j/N\). Proportional allocation is an optimum allocation if the cost-per-element is the same for all elements and all \(\sigma_j^2\) are equal. In practice we might have approximate proportional allocation where \(n_j/n \approx N_j/n\).
Example: What would \(n_1/n\) and \(n_2/n\) be for the sword fern survey using proportional allocation?
There are some practical restrictions on an optimum allocation.
Optimum \(n\) and \(n_j\) must be non-negative integers, so usually the optimum allocation is approximate.
An optimum allocation may produce \(n_j = 0\) or \(n_j = 1\). But to estimate \(\sigma_j^2\) we need all \(n_j \ge 2\).
It is possible to have an optimum allocation of \(n_j > N_j\), which is an impossible design.
For the latter two cases, we can find an optimum allocation subject to the constraint that all \(2 \le n_j \le N_j\) if we find that some \(n_j < 2\) or \(n_j > N_j\) using the method above, but how this would be done is beyond the scope of this lecture (although see below if you are curious).
The formulas given above are an analytical solution to the optimum allocation problem. These are derived using the necessary mathematics (namely calculus and what are called Lagrange multipliers). But the optimum allocation problem can also be solved numerically by using computing power instead. I have created a short demonstration of how to do this in R. You do not need to know how to do this for this course, but I have included it for any students that might be interested.