The following MINITAB output presents a confidence interval for a proportion. $$\begin{array} { c r r c c }
\hline \text { Variable } & \mathrm { X } & \mathrm { N } & \text { Sample } \mathrm { p } & 95 \% \mathrm { CI } \
\mathrm { x } & 114 & 172 & 0.662791 & ( 0.592138,0.733444 ) \
\hline
\end{array}$$ Use the information in the display to construct a 98% confidence interval for p.
A) (0.592, 0.733)
B) (0.603, 0.722)
C) (0.579, 0.747)
D) (0.570, 0.756)

Question

The following MINITAB output presents a confidence interval for a proportion. $$\begin{array} { c r r c c } 
\hline \text { Variable } & \mathrm { X } & \mathrm { N } & \text { Sample } \mathrm { p } & 95 \% \mathrm { CI } \
\mathrm { x } & 114 & 172 & 0.662791 & ( 0.592138,0.733444 ) \
\hline
\end{array}$$ Use the information in the display to construct a 98% confidence interval for p.
A) (0.592, 0.733)
B) (0.603, 0.722)
C) (0.579, 0.747)
D) (0.570, 0.756)

Quizplus · Accepted Answer

To construct a 98% confidence interval, we need to find the z-value that corresponds to a 98% confidence level. Using a standard normal distribution table or calculator, we find that the z-value for a 98% confidence interval is 2.33. The formula for a confidence interval for a proportion is:

sample proportion +/- z-value * standard error

The standard error can be calculated as:

sqrt[(p * (1 - p)) / n]

Plugging in the values from the MINITAB output:

sample proportion = 0.662791 
n = 172

We can solve for p:

0.662791 +/- 2.33 * sqrt[(0.662791 * (1 - 0.662791)) / 172]

This gives us the interval:

(0.5787, 0.7469)

Rounding to three decimal places, the answer is (C) (0.579, 0.747).

The Following MINITAB Output Presents a Confidence Interval for a Proportion