The following MINITAB output presents a 95% confidence interval. $\begin{array} { c r r c c }
\hline { \text { The assumed sigma } = 8.7500 } & \\
\text { Variable } & \mathrm { N } & \text { Mean } & \text { SE Mean } & 95 \% \mathrm { CI } \
\mathrm { x } & 48 & 32.950 & 1.2630 & ( 30.475,35.425 ) \
\hline
\end{array}$

Find the sample size needed so that the 99% confidence interval will have a margin of error of 1.3.
A) 301
B) 14
C) 175
D) 246

Question

The following MINITAB output presents a 95% confidence interval. $\begin{array} { c r r c c } 
\hline { \text { The assumed sigma } = 8.7500 } & \\
\text { Variable } & \mathrm { N } & \text { Mean } & \text { SE Mean } & 95 \% \mathrm { CI } \
\mathrm { x } & 48 & 32.950 & 1.2630 & ( 30.475,35.425 ) \
\hline
\end{array}$

Find the sample size needed so that the 99% confidence interval will have a margin of error of 1.3.
A) 301
B) 14
C) 175
D) 246

Quizplus · Accepted Answer

The margin of error (E) for a confidence interval can be calculated using the formula $E = Z \times \frac{\sigma}{\sqrt{n}}$, where $Z$ is the Z-score corresponding to the confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size. To find the sample size needed for a 99% confidence interval with a margin of error of 1.3, we rearrange the formula to solve for $n$: $n = \left(\frac{Z \times \sigma}{E}\right)^2$. For a 99% confidence level, the Z-score is approximately 2.576. Given $\sigma = 8.75$ and $E = 1.3$, we calculate $n = \left(\frac{2.576 \times 8.75}{1.3}\right)^2$, which results in a sample size of approximately 301.

The Following MINITAB Output Presents a 95% Confidence Interval Find the Sample Size Needed So That the 99% Confidence