Statistical Quality Control Control Charts for Variables 4 (20) You are a new ma
ID: 3049091 • Letter: S
Question
Statistical Quality Control Control Charts for Variables 4 (20) You are a new manufacturing engineer in a Pharmaceutical industry. The operators give you some collected data over the last manufactured lots. The results are presented below. -2800 s2 = 180 -1 a. (10) Calculate the appropriate control limits b. (5) How would the control limits vary if the measure of dispersion for the data is split by half c. (5) The manager is challenging your selection of control chart based on what she remembers of the Central Limit Theorem. Justify and sustain your positionExplanation / Answer
Back-up Theory
Let there be n measurements on a quality characteristic taken at k equally-spaced intervals. Let xij represent the jth measurement at the ith interval, i = 1, 2, …., k and j = 1, 2, …., k.
The Xbar part of Xbar-s Chart has:
Central line: CL = Xdouble bar
Lower Control Limit: LCL = Xdouble bar – A3sbar
Upper Control Limit: UCL = Xdouble bar + A3sbar;
The s part of Xbar-s Chart has:
Central line: CL = sbar
Lower Control Limit: LCL = B3sbar
Upper Control Limit: UCL = B4sbar, where
Xdouble bar = (1/k)(i = 1, k)xibar, xibar = (1/n)(i = 1, n)xij
sbar = (1/k)(i = 1, k)si, si2 = {1/(n – 1)}(i = 1, n)(xij - xibar)2
A3, B3, B4 are constants which can be directly obtained from standard Control Chart Constants Table.
Preparatory Work
From the given data we find:
n = 5, k = 20, Xdouble bar = 2800/100 = 28 and Sbar = (180/20) = 3
Part (a)
All relevant computations are given below:
k
20
n
5
A3
1.427
B3
0
B4
2.089
Xdouble bar
28
sbar
3
Xbar Chart
CL
28
LCL
23.719
UCL
32.281
sChart
CL
3
LCL
0
UCL
6.267
DONE
Part (b)
When the measure of dispersion, i.e., sbar, is halved, the new sbar = 1.5, the revised control lomits would be:
Xbar Chart
CL
28
LCL
25.8595
UCL
30.1405
sChart
CL
1.5
LCL
0
UCL
3.1335
DONE
Part (c)
Since sample size is 5 and the number of samples is 20, the numbers are large enough for the Central Limit Theorem to be valid. Hence, by CLT, the sample averages can be taken to be Normally distributed and hence all calculations hold good. ANSWER
k
20
n
5
A3
1.427
B3
0
B4
2.089
Xdouble bar
28
sbar
3
Xbar Chart
CL
28
LCL
23.719
UCL
32.281
sChart
CL
3
LCL
0
UCL
6.267