**Exhibit 33.2 **90% of the observations fall within a range of ±1.65 standard deviation from the mean.

Consider research programmes such as retail measurement
that track item sales in store, consumer panels that track the item purchases in homes or TAM (TV
audience measurement) that tracks TV viewership. Each of these platforms track a continuous
variable (e.g. sales, purchases or viewership) in a population.

The CLT (central limit theorem) provides the foundation for setting the sample size
for the variables measured in these programmes.

As mentioned before, the sample size required to estimate a variable is dependent on
population variance, sampling design and the desired level of precision. And the desired level of precision
is defined by two statistics — *confidence interval* or *margin of error* and *confidence level*.

If the confidence level is set at 90%, and the margin of error is *e*, then based
on the normal distribution (see Exhibit 33.2):

$$e = 1.65 σ$$

In general:

$$e = Z σ$$

Where:

e: margin of error

Z: standardized value associated with the level
of confidence. If level of confidence is:

90% then Z=1.65

95% then Z=1.96

99% then 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 the margin of error is expressed as a proportion of parameter
value, e' = e/µ, and S' = RSE (relative sampling error) = S/µ, the above formulae
may be rewritten as:

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

**Example 1**

If the standard deviation
of the sale of a category in provision shops is *80,* and the average
sales per store, *μ*,* *is about *200*, then the required sample
size, so that the sales estimates fall within *±6%* of its true
value with a confidence level of 90% is equal to:

$$n_{prov}=\left(\frac{ZS}{e}\right)^2=\left(\frac{1.65×80}{0.06 × 200}\right)^2=121$$

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

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