Chapter 8, Problem R8.7RE

To determine

To find: The sample size to have the margin of error within 1000 pounds at 95% confidence level.

Answer to Problem R8.7RE

The sample size is 35.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Standard}\text{deviation}\left(\sigma \right)=3000\\ \text{Confidence level }=95\%\\ \text{Margin of error}\left(E\right)=1000\end{array}$

Formula used:

The formula to compute the sample size is:

  $n={\left(\frac{z_{\frac{\alpha }{2}}\times \sigma }{E}\right)}^2$

Calculation:

The value of $z_{\frac{\alpha }{2}}$ corresponding to 95% confidence level is 1.96, which is obtained from the standard normal table.

The sample size can be calculated as:

  $\begin{array}{c}n={\left(\frac{z_{\frac{\alpha }{2}}\times \sigma }{E}\right)}^2\\ ={\left(\frac{1.96\times 3000}{1000}\right)}^2\\ =34.5744\\ \approx 35\end{array}$

Hence, the sample size is 35