The following MINITAB output presents a 95% confidence interval.

$\begin{array}{ccccc}
\hline{\text { The assumed sigma }=10.2726} \
\text { Variable } & \mathrm{N} & \text { Mean } & \text { SE Mean } & 95 \% \mathrm{CI} \
\mathrm{x} & 46 & 48.611 & 1.5146 & (45.642,51.580) \
\hline
\end{array}$

Find the sample size needed so that the 95% confidence interval will have a margin of error of 1.1.
A) 472
B) 579
C) 19
D) 336

Question

The following MINITAB output presents a 95% confidence interval.

$\begin{array}{ccccc}
\hline{\text { The assumed sigma }=10.2726} \
\text { Variable } & \mathrm{N} & \text { Mean } & \text { SE Mean } & 95 \% \mathrm{CI} \
\mathrm{x} & 46 & 48.611 & 1.5146 & (45.642,51.580) \
\hline
\end{array}$

Find the sample size needed so that the 95% confidence interval will have a margin of error of 1.1.
A) 472
B) 579
C) 19
D) 336

Quizplus · Accepted Answer

The margin of error (E) in a confidence interval is given by $E = Z \times \frac{\sigma}{\sqrt{n}}$, where $Z$ is the Z-score corresponding to the confidence level (for 95%, $Z \approx 1.96$), $\sigma$ is the population standard deviation (assumed to be 10.2726 as given), and $n$ is the sample size. Rearranging to solve for $n$, we get $n = \left(\frac{Z \times \sigma}{E}\right)^2$. Substituting $E = 1.1$, $Z = 1.96$, and $\sigma = 10.2726$, we can solve for $n$. This calculation gives a value close to 336, making option D the correct choice.

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