**Exhibit 34.2 **90% of the observations fall within a range of ±1.65 standard
deviation from the mean.

When considering research programs like retail
measurement, consumer panels, or TV audience measurement, where continuous variables such as
sales, purchases, or viewership are tracked in a population, the determination of sample
size is guided by the Central Limit Theorem (CLT).

As previously discussed, the sample size required to estimate a variable
depends on the population variance, sampling design, and the desired level of precision,
which is defined by two key statistics — *confidence interval* or *margin of error*
and *confidence level*.

If the confidence level is set at 90% and the margin of error is denoted as
*e*, then based on the normal distribution (refer to Exhibit 34.2):
$$e = 1.65 σ$$

In general:

$$e = Z σ$$

*Where:*

*e: margin of error*
*Z: standardized value associated with the desired confidence level.
For different confidence levels: *
*90% confidence level corresponds to Z = 1.65*
*95% confidence level corresponds to Z = 1.96*
*99% confidence level corresponds to Z = 2.58*

Furthermore, according to the CLT the variance of the sampling distribution
(*σ*^{2}) is a function of the sample size (*n*), and the universe variance
(*S*^{2}):
$$ σ^2 = \frac {S^2}{n} $$

Substituting for *s *^{2} :

$$ e^2= \frac {Z^2 S^2}{n}$$

$$\mathbf {n=\frac{Z^2S^2}{e^2}=\left(\frac{ZS}{e}\right)^2, \;\; e = \frac {ZS}{\sqrt n}}$$

If we express the margin of error as a proportion of the parameter value,
denoted as e' = e/µ, and the relative sampling error as S' = RSE (relative sampling error) = S/µ,
the above formulas can be rewritten as follows:
$$\mathbf {n=\frac{Z^2S'^2}{e'^2}=\left(\frac{ZS'}{e'}\right)^2, \;\; e' = \frac {ZS'}{\sqrt n}}$$

*Example 1*: Consider a scenario where the standard
deviation of sales for a category in provision shops is *80*, and the average sales per
store, *μ*, is approximately *200*. We want to determine the required sample size to
ensure that the sales estimates fall within *±6%* of the true value with a confidence level
of *90%*. The calculation is as follows:
$$n_{prov}=\left(\frac{ZS}{e}\right)^2=\left(\frac{1.65×80}{0.06 × 200}\right)^2=121$$

Take note that in order to half a sampling error we would need to quadruple
the sampling size. For instance, if the sampling error was reduced to 3% (0.03), the required
sample size will be 484.