*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 *n*_{prov} and *n'*_{prov},
which is equal to 124.