# Sample Size — Continuous Parameters #### 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 (σ2) is a function of the sample size (n), and the universe variance (S2): $$σ^2 = \frac {S^2}{n}$$

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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