Chapter 5.2, Problem 57E

To determine

To find: The margin of error.

Answer to Problem 57E

The margin of error is 0.04.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Sample size}\left(n\right)=400\\ \text{Sample proportion}\left(\widehat{p}\right)=\frac{317}{400}=0.7925\\ \text{Confidence level}=95\%=0.95\end{array}$

Formula used:

The formula to compute the margin of error ( E ) is:

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

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 margin of error at 95% confidence level can be calculated as:

  $\begin{array}{c}E=z_{\alpha /2}\times \sqrt{\frac{\widehat{p}\times \left(1-\widehat{p}\right)}{n}}\\ =1.96\times \sqrt{\frac{\left(0.7925\times \left(1-0.7925\right)\right)}{400}}\\ =0.0397\\ \approx 0.04\end{array}$

Hence, the correct option is (d), 0.04.