91影视

The percentage of women who give birth is to be estimated.

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

The margin of error is equal to 0.02 (two percentage points).

The confidence level is equal to 99%.

a.

Let$\hat{p}$ denote the sample proportion of women who give birth.

Let $\hat{q}$denote the sample proportion of women who do not give birth.

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:

$$\begin{gathered} n=\frac{{\left(2.5758\right)}^2脳0.25}{{\left(0.02\right)}^2} \\ =4146.72 \\ 鈮�4147 \end{gathered}$$

Hence, the required sample size is equal to 4147.

b.

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

$$\begin{gathered} \hat{p}=82\% \\ =\frac{82}{100} \\ =0.82 \end{gathered}$$

Thus, the value of is computed below:

$$\begin{gathered} \hat{q}=1-\hat{p} \\ =1-0.82 \\ =0.18 \end{gathered}$$

The formula for finding the sample size is as follows:

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

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

$$\begin{gathered} n=\frac{{\left(2.5758\right)}^2脳0.82脳0.18}{{\left(0.02\right)}^2} \\ =2448.22 \\ 鈮�2448 \end{gathered}$$

Hence, the required sample size is equal to 2448.

c.

It is known that women bear children after a certain age and until their menopause.

A randomly selected sample of women will also include younger women who won鈥檛 go for a baby until several years and women who have passed the age of reproduction.

Therefore, to accurately estimate the proportion of women who give birth, the population should include only adult women.