The following MINITAB output presents a confidence interval for a proportion.
$\begin{array}{crrcc}
\hline \text { Variable } & \mathrm{X} & \mathrm{N} & \text { Sample } \mathrm{p} & 95 \% \mathrm{CI} \
\mathrm{x} & 42 & 106 & 0.396226 & (0.303113,0.489339) \
\hline
\end{array}$
Use the information in the display to construct a 98% confidence interval for p.
A) $( 0.286,0.507 )$
B) $( 0.318,0.474 )$
C) $( 0.303,0.489 )$
D) $( 0.274,0.519 )$

Question

The following MINITAB output presents a confidence interval for a proportion.
 $\begin{array}{crrcc}
\hline \text { Variable } & \mathrm{X} & \mathrm{N} & \text { Sample } \mathrm{p} & 95 \% \mathrm{CI} \
\mathrm{x} & 42 & 106 & 0.396226 & (0.303113,0.489339) \
\hline
\end{array}$
Use the information in the display to construct a 98% confidence interval for p.
A) $( 0.286,0.507 )$
B) $( 0.318,0.474 )$
C) $( 0.303,0.489 )$
D) $( 0.274,0.519 )$

Quizplus · Accepted Answer

To construct a 98% confidence interval for p, we need to adjust the margin of error for the increased confidence level. The original 95% CI is narrower than what a 98% CI would be because a higher confidence level requires a wider interval. Without the exact calculations (which depend on the z-score for 98% confidence and the standard error of the proportion), we can infer that option A provides a wider interval than the original 95% CI, which aligns with the expectation for a 98% CI.

The Following MINITAB Output Presents a Confidence Interval for a Proportion