Answer to Question #152645 in Statistics and Probability for Shavetra

Question

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 appear below.
Weight
Day Package 1 Package 2 Package 3 Package 4
Monday 23 22 23 24
Tuesday 23 21 19 21
Wednesday 20 19 20 21
Thursday 18 19 20 19
Friday 18 20 22 20
(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. [3 Marks]
(d) What is upper and lower control limit for the R-Chart?
(c) Based on the x-bar and R charts, is this process in control?

Answer 1

(a) Sample means:

$\overline{x}_{Mon} = (23+22+23+24)/4=23$

$\overline{x}_{Tue} = (23+21+19+21)/4=21$

$\overline{x}_{Wed} = (20+19+20+21)/4=20$

$\overline{x}_{Thu} = (18+19+20+19)/4=19$

$\overline{x}_{Fri} = (18+20+22+20)/4=20$

Mean of all sample means:

$\overline{\overline{x}} = (23+21+20+19+20)/5=20.6$

(b) Subgroup ranges:

$R_{Mon} =24-22=2$

$R_{Tue} =23-19=4$

$R_{Wed} =21-19=2$

$R_{Thu} =20-18=2$

$R_{Fri} =22-18=4$

Average R value:

R-bar = (2+4+2+2+4)/5 = 2.8

A₂ = 0.729 (for subgroup size 4)

X-bar Chart Upper Control Limit:

UCL(X-bar) = $\overline{\overline{x}}$ + (A₂ * R-bar) = 20.6+(0.729*2.8) = 22.6

X-bar Chart Lower Control Limit

LCL(X-bar) = $\overline{\overline{x}}$ - (A₂ * R-bar) = 20.6-(0.729*2.8) = 18.6

(c) D₄ = 2.282 (for subgroup size 4)

UCL(R) = R-bar * D₄ =2.8*2.282 = 6.39

D₃ = 0 (for subgroup size 4)

LCL(R) = R-bar * D₃ =2.8*0 = 0

(d) $\overline{x}_{Mon}$ is above UCL(X-bar). The process is out of control.