The following MINITAB output presents a confidence interval for a population mean. $$\begin{array} { c r r r c c }
\hline \text { Variable } & \mathrm { N } & \text { Mean } & \text { StDev } & \text { SE Mean } & 99 \% \text { CI } \
\mathrm { x } & 30 & 89.4928 & 33.3397 & 6.0870 & ( 72.693,106.293 ) \
\hline
\end{array}$$ Find the critical value $t _ { \alpha / 2 }$ for a 98% confidence interval.
A) 2.462
B) 2.457
C) 2.750
D) 2.760

Question

The following MINITAB output presents a confidence interval for a population mean. $$\begin{array} { c r r r c c } 
\hline \text { Variable } & \mathrm { N } & \text { Mean } & \text { StDev } & \text { SE Mean } & 99 \% \text { CI } \
\mathrm { x } & 30 & 89.4928 & 33.3397 & 6.0870 & ( 72.693,106.293 ) \
\hline
\end{array}$$ Find the critical value $t _ { \alpha / 2 }$ for a 98% confidence interval.
A) 2.462
B) 2.457
C) 2.750
D) 2.760

Quizplus · Accepted Answer

The critical value for a 98% confidence interval with 29 degrees of freedom can be found using a t-distribution table or the TINV function in Excel. Using either method, we get that t_{0.01/2} = t_{0.005} = 2.462. Therefore, the best choice is A).

The Following MINITAB Output Presents a Confidence Interval for a Population