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 1.5 ounces. Each day last week, he randomly selected four packages and weighed each. The data from that activity appear below.
(a) Calculate all sample means and the mean of all sample means.
(b) Calculate upper and lower 2-sigma x-bar chart control limits that allow for natural variations.
(c) Based on the x-bar chart, is this process in control? Explain either natural variation causes or assignable variation causes.
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Expert's answer
2020-12-24T17:15:09-0500
(a) Let's calculate sample means:
xMon=(23+22+23+20)/4=22
xTue=(23+21+19+21)/4=21
xWed=(20+19+20+21)/4=20
xThu=(18+19+20+17)/4=18.5
xFri=(18+20+22+22)/4=20.5
Mean of all sample means:
x=(22+21+20+18.5+20.5)/5=20.4
(b) Upper and lower 2-sigma x-bar chart control limits:
UCL=x+2∗nσ=20.4+2∗41.5=21.9
LCL=x−2∗nσ=20.4−2∗41.5=18.9
(c) xMon is above UCL(X-bar) and xThu is below LCL(X-bar). The process is out of control. So, we have assignable variation causes.
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