91Ó°ÊÓ

The confidence interval is 90%.

The sample standard deviation is 20%.

The sampling error is 5%.

The general formula for the sample size is given below:

$n={\left[\frac{Z_{\frac{\mathrm{Î\pm }}{2}}\left(\mathrm{σ}\right)}{SE}\right]}^2$

Where SE represents the sampling error.

The value of $\mathrm{σ}$is usually unknown. It can be estimated by the standard deviation, s from the prior sample.

For the confidence level of 90%, the level of significance is 0.90.

$$\begin{gathered} For\left(1-\mathrm{Î\pm }\right)=0.90 \\ \mathrm{Î\pm }=0.10 \\ \frac{\mathrm{Î\pm }}{2}=0.05 \end{gathered}$$

Thecorresponding to$Z_{\frac{\mathrm{Î\pm }}{2}}$ the standard normal table is,

$$\begin{gathered} Z_{\frac{\mathrm{Î\pm }}{2}}=Z_{0.05} \\ =1.645 \end{gathered}$$

The sample standard deviation is

$$\begin{gathered} s=\frac{20}{100} \\ =0.2 \end{gathered}$$

The sampling error is,

The sample size is computed as:

$$\begin{gathered} n={\left[\frac{\left(1.645\right)\left(0.2\right)}{0.05}\right]}^2 \\ ={\left[6.58\right]}^2 \end{gathered}$$

Hence, the number of swimming pools must be sampled to estimate the actual mean absolute deviation percentage to within 5% using a 90% confidence interval is 43.