Asked by Wendy Thurmond on Jun 20, 2024

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A quality analyst wants to construct a sample mean chart for controlling a packaging process. He knows from past experience that the process standard deviation is two ounces. Each day last week, he randomly selected four packages and weighed each. The data from that activity appears below.  Weight  Day  Package 1 Package 2 Package 3  Package 4  Monday 23222324 Tuesday 23211921 Wednesday 20192021 Thursday 18192019 Friday 18202220\begin{array} { | l | l | l | l | l | } \hline & { \text { Weight } } \\\hline \text { Day } & \text { Package } 1 & \text { Package } 2 & \text { Package 3 } & \text { Package 4 } \\\hline \text { Monday } & 23 & 22 & 23 & 24 \\\hline \text { Tuesday } & 23 & 21 & 19 & 21 \\\hline \text { Wednesday } & 20 & 19 & 20 & 21 \\\hline \text { Thursday } & 18 & 19 & 20 & 19 \\\hline \text { Friday } & 18 & 20 & 22 & 20 \\\hline\end{array} Day  Monday  Tuesday  Wednesday  Thursday  Friday  Weight  Package 12323201818 Package 22221191920 Package 3 2319202022 Package 4 2421211920 (a) Calculate all sample means and the mean of all sample means.
(b) Calculate upper and lower control limits that allow for natural variations.
(c) Is this process in control?

Process Standard Deviation

A measure of variability or dispersion in a process or system, showing how much variation exists from the average.

Control Limits

Statistical boundaries within a control chart that signal the range of variability expected in a process's stable and controlled state.

  • Proceed with the computation and clarification of control limits and process capabilities in multiple control chart types, such as X-bar, R-chart, p-chart, and c-chart.
  • Review sample data to conclude whether a process is in alignment with statistical control.
  • Assess the central tendency and dispersion for each sample and the process as a whole.
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LW
Lemar WhiteJun 21, 2024
Final Answer :
(a) The five sample means are 23, 21, 20, 19, and 20. The mean of all sample means is 20.6.
(b) UCL = 20.6 + 2.2/ 4\sqrt { 4 }4 = 22.6; LCL = 20.6 - 2.2/ 4\sqrt { 4 }4 = 18.6.
(c) Sample 1 is above the UCL; all others are within limits. The process is out of control.