There is an average of 22.455 employees at 22 office furniture dealers in a major metropolitan area,with a standard deviation of 18.52.Construct a 99 percent confidence interval for the true mean number of full-time employees at office furniture dealers.
A) $[ 11.014 \quad 33.896 ]$
B) $[ 11.275 \quad 33.635 ]$
C) $[ 12.284 \quad 32.626 ]$
D) $[ 12.513 \quad 32.397 ]$
E) $\left[ \begin{array} { l l }
20.072 & 24.838
\end{array} \right]$

Question

There is an average of 22.455 employees at 22 office furniture dealers in a major metropolitan area,with a standard deviation of 18.52.Construct a 99 percent confidence interval for the true mean number of full-time employees at office furniture dealers.
A) $[ 11.014 \quad 33.896 ]$
B) $[ 11.275 \quad 33.635 ]$
C) $[ 12.284 \quad 32.626 ]$
D) $[ 12.513 \quad 32.397 ]$
E) $\left[ \begin{array} { l l } 
20.072 & 24.838
\end{array} \right]$

Quizplus · Accepted Answer

The 99% confidence interval is calculated using the formula: $\bar{x} \pm z \times \frac{s}{\sqrt{n}}$, where $\bar{x} = 22.455$, $s = 18.52$, $n = 22$, and $z$ for 99% confidence is approximately 2.576. This calculation results in the interval $[11.275, 33.635]$.

There Is an Average of 22 [11.01433.896][ 11.014 \quad 33.896 ][11.01433.896]

There Is an Average of 22 $[ 11.014 \quad 33.896 ]$