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The percentage of adults who believe in astrology is to be estimated.

The sample size needs to be determined. The following values are given:

The margin of error is equal to 0.04.

The confidence level is equal to 99%.

a.

Let $\hat{p}$denote the sample proportion of adults who believe in astrology.

Let $\hat{q}$denote the sample proportion of adults who do not believe in astrology.

Here, nothing is known about the sample proportions.

The formula for finding the sample size is as follows:

$n=\frac{{\left[z_{\frac{伪}{2}}\right]}^{2}0.25}{E^2}$

The confidence level is equal to 99%. Thus, the level of significance is equal to 0.01.

The value of $z_{\frac{伪}{2}}$for $伪=0.01$from the standard normal table is equal to 2.5758.

Substituting the required values, the following value of the sample size is obtained:

$$\begin{gathered} n=\frac{{\left(2.5758\right)}^2脳0.25}{{\left(0.04\right)}^2} \\ =1036.68 \\ 鈮�1037 \end{gathered}$$

Hence, the required sample size is equal to 1037.

b.

The value of$\hat{p}$ is given to be equal to:

$$\begin{gathered} \hat{p}=26\% \\ =\frac{26}{100} \\ =0.26 \end{gathered}$$

Thus, the value of $\hat{q}$is computed below:

$$\begin{gathered} \hat{q}=1-\hat{p} \\ =1-0.26 \\ =0.74 \end{gathered}$$

The formula for finding the sample size is as follows:

$n=\frac{{\left[z_{\frac{伪}{2}}\right]}^2\hat{p}\hat{q}}{E^2}$

Substituting the required values, the following value of the sample size is obtained:

$$\begin{gathered} n=\frac{{\left(2.5758\right)}^2脳0.26脳0.74}{{\left(0.04\right)}^2} \\ =797.83 \\ 鈮�798 \end{gathered}$$

Hence, the required sample size is equal to 798.