The following MINITAB output presents a 95% confidence interval.
$\begin{array}{ccccc}
\hline {\text { The assumed sigma }=10.5211} & \
\text { Variable } & \mathrm{N} & \text { Mean } & \text { SE Mean } & 95 \% \mathrm{CI} \
\mathrm{x} & 58 & 39.501 & 1.3815 & (36.793,42.209) \
\hline
\end{array}$

Find the sample size needed so that the $95 \%$ confidence interval will have a margin of err 1.2.
A) 511
B) 296
C) 18
D) 416

Question

The following MINITAB output presents a 95% confidence interval. 
 $\begin{array}{ccccc}
\hline {\text { The assumed sigma }=10.5211} & \
\text { Variable } & \mathrm{N} & \text { Mean } & \text { SE Mean } & 95 \% \mathrm{CI} \
\mathrm{x} & 58 & 39.501 & 1.3815 & (36.793,42.209) \
\hline
\end{array}$

Find the sample size needed so that the $95 \%$ confidence interval will have a margin of err 1.2.
A) 511
B) 296
C) 18
D) 416

Quizplus · Accepted Answer

The margin of error (ME) is calculated as $ ME = z \times \frac{\sigma}{\sqrt{n}} $. For a 95% confidence interval, $ z \approx 1.96 $. Solving $ 1.2 = 1.96 \times \frac{10.5211}{\sqrt{n}} $ gives $ n \approx 296 $.

The Following MINITAB Output Presents a 95% Confidence Interval Find the Sample Size Needed So That The