Chapter 8.3, Problem 56E

To determine

To calculate: The sample size needed to estimate the mean SAT score when the confidence interval is 95%.

Answer to Problem 56E

The required sample size is 2401.

Explanation of Solution

Given Information:

Margin of error $E=2$

Confidence interval = 95%

Standard deviation $\sigma =50$

Formula Used Sample size $E\ge z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{\sigma }{\sqrt{n}}$

Calculation:

Confidence interval is 95%.

Convert 95% into decimal.

  $\frac{95}{100}=0.95$

For confidence interval 0.95, use table A.

  $z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=1.96$

Find the sample size. Use the formula $E\ge z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{\sigma }{\sqrt{n}}$ .

  $\begin{array}{c}E\ge z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{\sigma }{\sqrt{n}}\\ 2\ge 1.96\frac{50}{\sqrt{n}}\\ 2\times \sqrt{n}\ge 1.96\times 50\\ \sqrt{n}\ge 49\end{array}$

Take the square in both side.

  $\begin{array}{c}{\left(\sqrt{n}\right)}^2\ge {49}^2\\ n\ge 2401\end{array}$

Hence, the required sample size is 2401.