Decision Theoretic Approach for Sample Size Estimation.

This is problem 24 from Chapter 2 of the Sampling: Design and Analysis textbook.

In a decision theoretic approach, two functions are specified:
L(n) = Loss or “cost” of a bad estimate
C(n) = Cost of taking the sample
Suppose that for some constants $c_0$, $c_1$, and $k$,
$L(n) = kV(\bar{y}_s) = k(1 - \frac{n}{N})\frac{S^2}{n}$
$C(n) = c_o + c_1n$
What sample size n minimizes the total cost L(n) + C(n)?

So, besides $L(n)$ resembling $e =$ margin of error $= z_{a/2}\sqrt{1-\frac{n}{N}}\frac{s}{\sqrt{n}}$, and seeing that $n = \frac{C(n) - c_o}{c_1}$, I have no idea how to solve this problem.

You are asked to solve $\min_{n\in\mathbb{N}_{\geq1}}L(n)+C(n)$. Since minimizing over integers is problematic, let me deal with $\min_{n\in[0,\infty)}L(n)+C(n)$. Note that $L(n)+C(n)=\frac{kS^{2}}{n}-\frac{kS^{2}}{N}+c_{o}+c_{1}n$. The first order condition for the problem is (you should check that the objective function is convex in $n$) $-\frac{kS^{2}}{n^{2}}+c_{1}$. Hence optimal $n=\sqrt{\frac{kS^{2}}{c_{1}}}$.
Does this help? What is $S$ and $N$ in your problem?
• Ah, so finding the partial differentiation with respect to n gives us the minimum? So the min $n = \sqrt{\frac{kS^2}{c_1}}$? Why do I need to find $S$ and $N$? – JasonDor Feb 3 '17 at 4:31
• @JasonDor Well, I am asking what $S$ and $N$ are. In the optimization I treated them as parameters. If those depend on $n$, since the notation suggest $S$ might to do with variance of the sample, the solution to the problem will change. – Jan Feb 3 '17 at 6:23
• $S$ is the population standard deviation, which is just the square root of the variance? Sorry I'm not sure what you are getting at – JasonDor Feb 3 '17 at 20:48