A sample of 100 items has a population standard deviation of 5.1 and a mean of 21.6.Construct a 95 percent confidence interval for ?.
A) $\left[ \begin{array} { l l }
11.60 & 31.60
\end{array} \right]$
B) $\left[ \begin{array} { l l }
21.16 & 22.04
\end{array} \right]$
C) $\left[ \begin{array} { l l }
20.60 & 22.60
\end{array} \right]$
D) $\left[ \begin{array} { l l }
20.76 & 22.43
\end{array} \right]$

Question

A sample of 100 items has a population standard deviation of 5.1 and a mean of 21.6.Construct a 95 percent confidence interval for ?.
A) $\left[ \begin{array} { l l } 
11.60 & 31.60
\end{array} \right]$
B) $\left[ \begin{array} { l l } 
21.16 & 22.04
\end{array} \right]$
C) $\left[ \begin{array} { l l } 
20.60 & 22.60
\end{array} \right]$
D) $\left[ \begin{array} { l l } 
20.76 & 22.43
\end{array} \right]$

Quizplus · Accepted Answer

The confidence interval is calculated using the formula: mean ± (Z * (standard deviation / sqrt(n))). For a 95% confidence level, Z is approximately 1.96. Thus, the interval is 21.6 ± (1.96 * (5.1 / sqrt(100))) = 21.6 ± 1.00, which gives the interval [20.60, 22.60].

A Sample of 100 Items Has a Population Standard Deviation [11.6031.60]\left[ \begin{array} { l l } 11.60 & 31.60\end{array} \right][11.60​31.60​]

A Sample of 100 Items Has a Population Standard Deviation $\left[ \begin{array} { l l } 11.60 & 31.60\end{array} \right]$