The following MINITAB output presents a confidence interval for a population mean. $$\begin{array} { c r r r c c }
\hline \text { Variable } & \mathrm { N } & \text { Mean } & \text { StDev } & \text { SE Mean } & 99 \% \text { CI } \
\mathrm { x } & 28 & 108.1268 & 30.4888 & 5.7618 & ( 92.161,124.093 ) \
\hline
\end{array}$$ Use the information in the output to construct a 98% confidence interval.
A) (92.161,124.093)
B) (105.546,110.707)
C) (93.878,122.376)
D) (105.235,111.018)

Question

The following MINITAB output presents a confidence interval for a population mean. $$\begin{array} { c r r r c c } 
\hline \text { Variable } & \mathrm { N } & \text { Mean } & \text { StDev } & \text { SE Mean } & 99 \% \text { CI } \
\mathrm { x } & 28 & 108.1268 & 30.4888 & 5.7618 & ( 92.161,124.093 ) \
\hline
\end{array}$$ Use the information in the output to construct a 98% confidence interval. 
A) (92.161,124.093)
B) (105.546,110.707)
C) (93.878,122.376)
D) (105.235,111.018)

Quizplus · Accepted Answer

To construct a 98% confidence interval, we need to look at the t-value for a 98% confidence level with 27 degrees of freedom. Using a t-distribution table or calculator, we find the t-value to be approximately 2.48. Then, we can use the formula:

$ 	ext{CI} = 	ext{sample mean} \pm t_{\alpha/2} 	imes 	ext{standard error} $

Substituting the given values, we get:

$ 	ext{CI} = 108.1268 \pm 2.48 	imes 5.7618 $

$ 	ext{CI} = (93.878, 122.376) $

Therefore, the best choice is C.

The Following MINITAB Output Presents a Confidence Interval for a Population