91Ó°ÊÓ

The sample size of caffeine drinkers is 27.

The number of success be 11.

The confidence level is 99%.

The sampling error is 0.05.

The point estimate of the population proportion is obtained below:$$\begin{gathered} \hat{p}=\frac{Number of success in the sample \left(x\right)}{Sample\backslash \mathrm{kern}1pt size \left(n\right)} \\ =\frac{11}{27} \end{gathered}$$

Let the confidence level be 0.99.

$$\begin{gathered} For \left(1-\mathrm{Î\pm }\right)=0.99 \\ \mathrm{Î\pm }=0.01 \\ \frac{\mathrm{Î\pm }}{2}=0.005 \end{gathered}$$

The$Z_{\frac{\mathrm{Î\pm }}{2}}$obtained from the standard normal table is,

$$\begin{gathered} Z_{\frac{\mathrm{Î\pm }}{2}}=Z_{0.005} \\ =2.575 \end{gathered}$$

The formula for sample size is given below:

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

Where SE is the sampling error.

The value of the pq is unknown; it can be estimated by using the sample fraction of success,from a prior sample.

Let the sample proportion$\hat{p}$is 0.407.

Here, the value pq is unknown. Which can be obtained by using the sample fraction of success$\hat{p}$,$\hat{p}$.

The product of pq is computed as

$$\begin{gathered} pq=p\left(1-p\right) \\ =\left(0.407\right)\left(1-0.407\right) \\ =\left(0.407\right)\left(0.593\right) \\ =0.241351 \end{gathered}$$

The sample size is computed as

$$\begin{gathered} n=\frac{{\left(2.575\right)}^2\left(0.407\right)\left(0.593\right)}{{\left(0.005\right)}^2} \\ =\frac{1.6003}{0.0025} \\ =640.123 \\ ≈640 \end{gathered}$$

Hence, the required sample size is 640. So, the sample size is not large enough to estimate the true proportion of caffeine drinkers who are caffeine dependent to be within 0.05 with 99% confidence. Because the pq values are not closed to 0.5.