91Ó°ÊÓ

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_{{\alpha \mathord{\left/

{\vphantom {\alpha 2}} \right.

\kern-\nulldelimiterspace} 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 \({z_{\frac{\alpha }{2}}}\) for \(\alpha = 0.01\) from the standard normal table is equal to 2.5758.

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

\(\begin{array}{c}n = \frac{{{{\left( {2.5758} \right)}^2} \times 0.25}}{{{{\left( {0.04} \right)}^2}}}\\ = 1036.68\\ \approx 1037\end{array}\)

Hence, the required sample size is equal to 1037.

b.

The value of \(\hat p\) is given to be equal to:

\(\begin{array}{c}\hat p = 26\% \\ = \frac{{26}}{{100}}\\ = 0.26\end{array}\)

Thus, the value of \(\hat q\) is computed below:

\(\begin{array}{c}\hat q = 1 - \hat p\\ = 1 - 0.26\\ = 0.74\end{array}\)

The formula for finding the sample size is as follows:

\(n = \frac{{{{\left( {{z_{{\alpha \mathord{\left/

{\vphantom {\alpha 2}} \right.

\kern-\nulldelimiterspace} 2}}}} \right)}^2}\hat p\hat q}}{{{E^2}}}\)

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

\(\begin{array}{c}n = \frac{{{{\left( {2.5758} \right)}^2} \times 0.26 \times 0.74}}{{{{\left( {0.04} \right)}^2}}}\\ = 797.83\\ \approx 798\end{array}\)

Hence, the required sample size is equal to 798.