Chickenpox : You plan to conduct a survey to estimate the percentage of adults who have had chickenpox. Find the number of people who must be surveyed if you want to be 90% confident that the sample percentage is within two percentage points of the true percentage for the population of all adults.

a. Assume that nothing is known about the prevalence of chickenpox.

b. Assume that about 95% of adults have had chickenpox.

c. Does the added knowledge in part (b) have much of an effect on the sample size?

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

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

a.

Sample size when the sample proportion is unknown,

$$\begin{gathered} n=\frac{{\left(z_{\frac{\alpha }{2}}\right)}^2\times 0.25}{{\left(E\right)}^2} \\ =\frac{{\left(1.645\right)}^2\times 0.25}{{\left(0.02\right)}^2} \\ =1691.266 \\ \approx 1692. \end{gathered}$$

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%,

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

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.