A sample set of weights in pounds are 1.01,.95,1.03,1.04,.97,.97,.99,1.01,and 1.03.Assume the population of weights is normally distributed.Find a 99 percent confidence interval for the mean population weight.
A) $\left[ \begin{array} { l l }
.965 & 1.035
\end{array} \right]$
B) $\left[ \begin{array} { l l }
.969 & 1.031
\end{array} \right]$
C) $\left[ \begin{array} { l l }
.973 & 1.027
\end{array} \right]$
D) $[ .941 \quad 1.059 ]$

Question

A sample set of weights in pounds are 1.01,.95,1.03,1.04,.97,.97,.99,1.01,and 1.03.Assume the population of weights is normally distributed.Find a 99 percent confidence interval for the mean population weight.
A) $\left[ \begin{array} { l l } 
.965 & 1.035
\end{array} \right]$
B) $\left[ \begin{array} { l l } 
.969 & 1.031
\end{array} \right]$
C) $\left[ \begin{array} { l l } 
.973 & 1.027
\end{array} \right]$
D) $[ .941 \quad 1.059 ]$

Quizplus · Accepted Answer

The correct answer is A because the confidence interval is calculated as follows:mean +/- z * (standard deviation / square root of sample size)where z is the z-score corresponding to the desired confidence level. For a 99% confidence level, z = 2.576. The standard deviation is 0.032 and the sample size is 9. Plugging these values into the formula, we get:1.00 +/- 2.576 * (0.032 / sqrt(9)) = [0.965, 1.035]

A Sample Set of Weights in Pounds Are 1 [.9651.035]\left[ \begin{array} { l l } .965 & 1.035\end{array} \right][.965​1.035​]

A Sample Set of Weights in Pounds Are 1 $\left[ \begin{array} { l l } .965 & 1.035\end{array} \right]$