Chapter 8.3, Problem 77E

To determine

To find: The sample size to have the margin of error to within ±1 at $99\%$ confidence level.

Answer to Problem 77E

The sample size is 373.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Standard}\text{deviation}\left(\sigma \right)=7.5\\ \text{Confidence level }=99\%\\ \text{Margin of error}\left(E\right)=1\end{array}$

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 $99\%$ confidence level is 2.575, which is obtained from the standard normal table.

The sample size can be calculated as:

  $\begin{array}{c}n={\left(\frac{z_{\frac{\alpha }{2}}\times \sigma }{m}\right)}^2\\ ={\left(\frac{2.575\times 7.5}{1}\right)}^2\\ =372.97\\ \approx 373\end{array}$

Hence, the sample size is 373.