For stratified samples, which are the norm for retail measurement services, the formula for sample size is as follows:
$$n'_i=\left(\frac{ZS_i}{e}\right)^2 × \frac{N_i {S_i}^2}{\sum_{j=1 \, to \, k}N_j {S_j}^2} $$Where:
Example 2:The national universe for a retail audit comprises provision stores, minimarkets and supermarkets. The standard deviation of the sale of a category in provision shops is 80, and the average sales per store is 200. In the case of minimarkets, the standard deviation of the sale of the category is 100, and the average sales per store is about 400. The required minimarket sample size so that the sales estimates fall within ±6% of its true value with a confidence level of 90% is therefore equal to:
$$n_{mini}=\left(\frac{ZS}{e}\right)^2=\left(\frac{1.65×100}{0.06×400}\right)^2=47$$The national universe comprises 200 supermarkets, 500 minimarkets and 2,000 provision stores. The standard deviation of the sales in supermarkets is 400 and average sales per store is 800. If the relative standard error at the national market breakdown is ±3%, the required sample size for minimarkets is then equal to:
$$n'_i=\left(\frac{ZS_i}{e}\right)^2 × \frac{N_i {S_i}^2}{\sum_{j=1 \, to \, k}N_j {S_j}^2} $$ $$n'_i=\left(\frac{1.65×100}{0.03×400}\right)^2 × \frac{500×100^2}{500×100^2 + 2000×80^2 + 200×400^2}=19 $$To meet both criteria for relative standard error at the minimarket breakdown and national level, we need a minimum sample size of 47 minimarkets.
Similarly, for provision stores, n'prov = 124, and the required sample size is the maximum of nprov and n'prov, which is equal to 124.
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