Chapter 8, Problem T8.4SPT

To determine

To find: The smallest number of consumers to have the margin of error at most 0.05 at 99% confidence level.

Answer to Problem T8.4SPT

The smallest number of consumers is 600.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Confidence level }=99\%\\ \text{Margin of error}\left(E\right)=0.05\end{array}$

Formula used:

The formula to compute the sample size is:

  $n=\widehat{p}\left(1-\widehat{p}\right){\left(\frac{z_{\frac{\alpha }{2}}}{E}\right)}^2$

Calculation:

Here, no information about $\widehat{p}$ has been provided. Thus, assume is to be 0.50.

The value of $z_{\frac{\alpha }{2}}$ corresponding to 99% confidence level is 2.576, which is obtained from the standard normal table.

The sample size can be calculated as:

  $\begin{array}{c}n=\widehat{p}\left(1-\widehat{p}\right){\left(\frac{z_{\frac{\alpha }{2}}}{E}\right)}^2\\ =0.50\left(1-0.50\right){\left(\frac{2.576}{0.05}\right)}^2\\ =664\end{array}$

Here, the smallest number of consumers is 600. Thus, the correct option is (b).