$\text { Data Table A }$
$\begin{array}{|l|l|l|l|l|}
\hline \begin{array}{c}
\text { Sample } \
\text { Number }
\end{array} & \text { Item 1 } & \text { Item 2 } & \text { /tem 3 } & \text { /tem 4 } \
\hline 1 & 13.10 & 13.90 & 12.80 & 13.20 \
\hline 2 & 13.80 & 13.00 & 12.50 & 12.70 \
\hline 3 & 14.00 & 12.90 & 13.10 & 13.50 \
\hline 4 & 12.30 & 13.30 & 12.80 & 13.60 \
\hline 5 & 13.10 & 12.60 & 13.30 & 12.90 \
\hline 6 & 12.50 & 13.40 & 12.80 & 13.30 \
\hline 7 & 13.60 & 12.70 & 13.30 & 13.80 \
\hline 8 & 12.50 & 13.50 & 13.10 & 12.90 \
\hline 9 & 13.00 & 13.60 & 12.70 & 13.20 \
\hline 10 & 12.80 & 12.90 & 12.70 & 13.30 \
\hline
\end{array}$
-Refer to Data Table A.The upper control limit for the mean chart is _______.
A)12.8
B)0
C)15.4
D)13.4

Question

Quizplus · Accepted Answer

To find the upper control limit for the mean chart, we need to first calculate the mean and standard deviation of the sample means. 
Mean = (13.10+13.80+14.00+12.30+13.10+12.50+13.60+12.50+13.00+12.80)/10 = 13.06 
Standard deviation = √[Σ(sample mean - grand mean)²/(n-1)] 
= √[((13.10-13.06)²+(13.80-13.06)²+...+(12.80-13.06)²)/9] 
= 0.4739 
Upper control limit = Mean + 3 * Standard deviation 
= 13.06 + 3 * 0.4739 
= 14.49 
Therefore, the upper control limit for the mean chart is approximately 14.5. However, this is not one of the answer choices. The closest choice is D, 13.4, which is obtained by rounding down the exact value to one decimal place.

Data Table A \text { Data Table A } Data Table A -Refer to Data Table A

$\text { Data Table A }$ -Refer to Data Table A