Question: Evaporation from swimming pools. Refer to the Heating/ Piping/Air Conditioning Engineering (April 2013) study of evaporation from occupied swimming pools, Exercise 6.34 (p. 349). The researchers desired an estimate of the mean absolute value of the deviation between the actual and estimated evaporation level (recorded as a percentage). Using a small sample, the researchers obtained the following summary statistics for absolute deviation percentage x = 18%, s = 20%. How many swimming pools must be sampled to estimate the true mean absolute deviation percentage to within 5% using a 90% confidence interval?

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

$n={\left[\frac{Z_{\frac{\alpha }{2}}\left(\sigma \right)}{SE}\right]}^2$

Where SE represents the sampling error.

The value of$\sigma$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

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

The$Z_{\frac{\alpha }{2}}$corresponding to the standard normal table is:

$$\begin{gathered} Z_{\frac{\alpha }{2}}=Z_{0.05} \\ =1.645 \end{gathered}$$

The sample standard deviation is,

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

The sampling error is,

$$\begin{gathered} SE=\frac{5}{100} \\ =0.05 \end{gathered}$$

The sample size is computed as:

$$\begin{gathered} n={\left[\frac{\left(1.645\right)\left(0.2\right)}{0.05}\right]}^2 \\ ={\left[6.58\right]}^2 \\ =43.2964 \\ \approx 43 \end{gathered}$$

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.