The following MINITAB output presents a $95 \%$ confidence interval.

$\begin{array}{ccccc}
\hline
{\text { The assumed sigma }=10.5215} & \
\text { Variable } & \mathrm{N} & \text { Mean } & \text { SE Mean } & 95 \% \mathrm{CI} \
\mathbf{x} & 41 & 31.580 & 1.6432 & (28.359,34.801) \
\hline
\end{array}$
Find the sample size needed so that the $95 \%$ confidence interval will have a margin of err $1.2$. rror of
A) 511
B) 18
C) 416
D) 296

Question

The following MINITAB output presents a $95 \%$ confidence interval.

$\begin{array}{ccccc}
\hline 
{\text { The assumed sigma }=10.5215} & \
\text { Variable } & \mathrm{N} & \text { Mean } & \text { SE Mean } & 95 \% \mathrm{CI} \
\mathbf{x} & 41 & 31.580 & 1.6432 & (28.359,34.801) \
\hline
\end{array}$
Find the sample size needed so that the $95 \%$ confidence interval will have a margin of err $1.2$. rror of
A) 511
B) 18
C) 416
D) 296

Quizplus · Accepted Answer

The Answer of The following MINITAB output presents a $95 \%$ confidence interval.

$\begin{array}{ccccc}
\hline 
{\text { The assumed sigma }=10.5215} & \
\text { Variable } & \mathrm{N} & \text { Mean } & \text { SE Mean } & 95 \% \mathrm{CI} \
\mathbf{x} & 41 & 31.580 & 1.6432 & (28.359,34.801) \
\hline
\end{array}$
Find the sample size needed so that the $95 \%$ confidence interval will have a margin of err $1.2$. rror of
A) 511
B) 18
C) 416
D) 296

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

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