# Sample Size — Stratified Sampling

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:

• n'i: Sample size for stratai (k in all)
• Ni: Stratai population
• Si2: Stratai population variance
• Z: Standardized z value associated with the level of confidence
• e: Margin of error

Example 2: Consider a situation where the national universe for a retail audit consists of provision stores, minimarkets, and supermarkets. We have the following information:

• For provision shops: standard deviation of category sales = 80, average sales per store = 200.
• For minimarkets: standard deviation of category sales = 100, average sales per store = 400.
• For supermarkets: standard deviation of category sales = 400, average sales per store = 800.

We want to determine the required sample size for minimarkets to ensure that the sales estimates fall within ±6% of the true value with a confidence level of 90%. The calculation is as follows: $$n_{mini}=\left(\frac{ZS}{e}\right)^2=\left(\frac{1.65×100}{0.06×400}\right)^2=47$$

Now, considering the national universe breakdown, which comprises 200 supermarkets, 500 minimarkets, and 2,000 provision stores, and aiming for a relative standard error of ±3% at the national market breakdown, we calculate the required sample size for minimarkets: $$n'_i=\left(\frac{ZS_i}{e}\right)^2 × \frac{N_i {S_i}^2}{\sum_{j=1 \, to \, k}N_j {S_j}^2}$$ $$n'_i=\left(\frac{1.65×100}{0.03×400}\right)^2 × \frac{500×100^2}{500×100^2 + 2000×80^2 + 200×400^2}=19$$

To satisfy both criteria for the relative standard error at the minimarket breakdown and the national level, we need a minimum sample size of 47 minimarkets.

Similarly, for provision stores, n'prov = 124. Therefore, the required sample size is the maximum of nprov and n'prov, which is equal to 124.

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