Solved – Neyman Allocation Standard Deviation for Proportions

samplingstratification

I am currently working towards the Royal Statistical Society's Higher Diploma and have run across a strange result in one of their sample papers that I can't work out – wonder if anyone can help me.

In Q3 of this paper you are asked to work out the optimal (Neyman) allocation for strata sizes in a sample.

According to everything I've seen, the formula is:

$$n_h = n \cdot \frac{ N_h S_h}{ Σ_i N_i S_i}$$

However, in the sample solution the standard deviation seems to be calculated as √(nhph(1 – ph)) – in other words, the nh in the numerator rather than the denominator as is usual in the standard deviation formula for a proportion.

I'm sure I'm being dense and missing something obvious, but I have looked and can't seem to find an answer. Can anyone help me? Thanks so much in advance.

Best Answer

The denominator $D$ isn't being ignored. It is calculated as sum of the numerators. The solution in the document is $n_1 = 363$ and $n_2 = 137$. Here's how to reproduce these numbers.

From the Solutions Document:

The values of $N_hs_h$ are 8000√(300 × 0.3 × 0.7) = 63498.03 for urban areas and 4000√(150 × 0.6 × 0.4) = 24000.00 for rural areas. "

  1. Sum the two expressions to get the denominator. $$ D = N_1s_1 + N_2s_2 = 63498.03 + 24000 = 87,498.03 $$
  2. Calculate the proportion allocated to the first stratum: $$ \frac{N_1s_1}{D} = \frac{63498.03}{87498.03}= 0.725708 $$
  3. The sample size allocated to the first stratum $n_1 = 0.725708\times 500 = 363$, rounded to the nearest integer. Then $n_2 = 500-363$.
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