91影视

Confidence level is 90%.

Margin of error ( E) is 0.02.

The basic requirement is that the sample should be independent and randomly selected. In this case, the requirement has been satisfied.

The sample size can be determined with 2 different conditions. The formulae and the conditions are given below:

The critical value ${\mathbf{z}}_{\frac{\mathbf{伪}}{\mathbf{2}}}$is obtained from standard normal table at 90% level of confidence, which implies 0.10 level of significance. That is,

$$\begin{gathered} z_{\frac{伪}{2}}=z_{\frac{0.1}{2}} \\ =z_{0.05} \\ =1.645 \end{gathered}$$

a.

Sample size when the sample proportion is unknown,

$$\begin{gathered} n=\frac{{\left(z_{\frac{伪}{2}}\right)}^2脳0.25}{{\left(E\right)}^2} \\ =\frac{{\left(1.645\right)}^2脳0.25}{{\left(0.02\right)}^2} \\ =1691.266 \\ 鈮�1692. \end{gathered}$$

Therefore when there is no other prior information, the sample size is 1692.

b.

Sample size when the sample proportion is known to be 95%,

$$\begin{gathered} n=\frac{{\left(z_{\frac{伪}{2}}\right)}^2脳\hat{p}\hat{q}}{{\left(E\right)}^2} \\ =\frac{{\left(1.645\right)}^2脳0.95脳0.05}{{\left(0.02\right)}^2} \\ =321.3405 \\ 鈮�332. \end{gathered}$$

Therefore, the sample size when 95% adults have chickenpox is 332.

c.

In part b, with a known value of sample proportion, the sample size reduces significantly, leading to precise estimate.