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In stratified random sampling with a linear cost function

\[ C = C_{0} + \sum_{h=1}^{L} C_{h}n_{h}, \]

the variance of estimated mean \((\bar{Y}_{st})\) is minimum for a specified cost C and the cost is minimum for a specified variance of \(\bar{Y}_{st}\) when :

  • A.\(n_{h} \propto \left\{\dfrac{w_{h}s_{h}}{\sqrt{C_{h}}}\right\}\)
  • B.\(n_{h} \propto \left\{\dfrac{\sum_{h=1}^{L} w_{h}s_{h}}{C_{h}}\right\}\)
  • C.\(n_{h} \propto \left\{\dfrac{N_{h}s_{h}}{s_{h}}\right\}\)
  • D.\(n_{h} \propto \left\{\dfrac{n\, s_{h}}{\sqrt{C_{h}}}\right\}\)

Answer: A

The optimum allocation is nh ∝ WhSh/√Ch.

Minimising \(V(\bar{Y}_{st}) = \sum W_{h}^{2}S_{h}^{2}/n_{h}\) (ignoring the finite population correction terms, which do not depend on \(n_{h}\)) subject to \(\sum C_{h}n_{h}\) fixed, or the cost subject to a fixed variance, leads by the Cauchy-Schwarz inequality or a Lagrange multiplier to

\[ n_{h} \propto \frac{W_{h}S_{h}}{\sqrt{C_{h}}} \]

So a stratum gets a larger sample when it is larger, more variable or cheaper to sample. Option B puts a sum over strata in the numerator and \(C_{h}\) without a square root; option C cancels to \(N_{h}\), which is proportional allocation; and option D drops the stratum weight.

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