Name ……………………………………………………….STAT 5031 test, Apr 9, 2003 Page 1 of 4
STAT 5031 Statistical Methods for Quality Improvement Hawkins
Midterm test, April 9, 2003
Time 50 minutes
Write your answers in the space provided. If this is not enough, continue on the back of the sheet or use
additional pages.
Question 1. PCB’s are toxic chemicals produced in some manufacturing operations. Every three months we
sample the waste water and measure the amount of PCB in it. While in control this has been found to follow a
normal distribution with mean 40 parts per million and standard deviation 10 ppm. For legal reasons we need
to know if the mean goes above 50 ppm.
(a) Find the reference value we should use in the cusum (2 points)
Midway between the in-control and target out-of-control mean, namely 45. Or, equivalently, K=5. Either
answer is fine.
(b) A decision interval cusum from the PCB sampling is on page 4. Describe and interpret as fully as you can
any signals you see in it. (5 points)
The cusum ran in control for a while, but then signaled an upward shift. The signal occurred at time 10. The
last in-control time seems to be time 5. The DI cusum at the time of the signal (reading off the graph) is about
50. [[The exact value turns out to be 48.9, so this is really close]] So we estimate that the mean PCB
concentration has increased to 45+50/(10-5) = 55 ppm.
(c ) The decision interval chosen is 40. What in-control ARL does this correspond to? (2 points)
Turning this into standard units, it corresponds to h=40/10 = 4. The reference value corresponds to k=5/10=0.5.
Going to the cusum ARL table for these two values, the one-sided ARL is 335.4. (As we are really only
bothered about increases in PCB, this answer is worth full points. Since the figure also shows a downward
cusum, it would also be valid to argue that the ARL as half this, or 167.7)
(d) What would be the ARL if the mean shifted to 45 ppm? (2 points)
A shift to a mean of 45 ppm corresponds to 0.5 standard units. Looking at the ARL table for k=(0.5-0.5) = 0
and h=4.0, the ARL would be 26.68. (This is worth full points. It would also be valid to argue that we seem to
be doing the cusum two-sided, and so the correct ARL would account for the possibility of a signal on the
wrong side. The ARL to this would be 14511.5, for an overall ARL just a shade under 26.68.)
Question 2. The variability of the PCB is also a concern. While in control, the standard deviation is 10 ppm.
To monitor variability, every three months a set of 8 samples are taken and their standard deviation found.
(a) Find the center line and three-sigma control limits for an S chart (4 points)
We have an n of 8. For this n, c4=0.9650, giving control limits 10*0.178 = 1.78, and 10*1.752 = 17.52
(b) Using the nearest number in some suitable table, find the approximate in-control ARL of the chart (3 points)
Name ……………………………………………………….STAT 5031 test, Apr 9, 2003 Page 2 of 4
We have 7 degrees of freedom. So the upper and lower limits correspond to chi-squareds of 7*0.1782 = 0.222
and 7*1.7522 = 21.479. The lower limit is well below even the smallest entry in the chi-squared table, so we’ll
ignore its contribution to the ARL. Nearby the upper limit of 21.479, fractiles on the chi-squared table are
20.278 with a tail area of 0.005 and 24.322 with a tail area of 0.001. The first of these is nearer, giving an ARL
of 200. This will do as a ball-park figure.
(c ) What would the ARL be if the standard deviation went from 10 to 13 ppm? (3 points)
If the SD went up by 30%, the upper limit of 21.479 would correspond to a chi-squared value of 21.479/1.32 =
12.7. This is just above the 90% fractile, 12.017, and corresponds to a right tail area of roughly 0.1 and an ARL
of around 10.
Question 3. The manufacturer develops a quick color test for PCB concentration. You add a test chemical to
the sample of waste waster. If the PCB concentration is above 50 ppm,the water turns red, and if it is below 50
ppm the water stays clear.
(a) If the process is in control (PCB concentration ~ N(40,102)), find the probability that a randomly selected
sample will turn red. (2 points)
This corresponds to a Z score of 1, and a probability of 0.159.
(b) A new QC scheme is proposed. They will take 50 samples of water every three months, and count the
number that turn red when used with this test. They will chart this number of red samples. Find the center line
and three-sigma control limits for this chart (4 points)
This is a NP chart. The center line will be np=50*0.159 = 7.95, and control limits of 7.95 + 3√(0.159*0.841) =
0.19 and 15.71
(c ) Another scheme being considered is a cusum of the number of reds. This should be ‘tuned’ for best
performance if the true mean PCB concentration goes to 50 ppm. Find the reference value for this cusum? (4
points)
 1  p1 
 1  0.5 
ln 
ln 


 1  p0   50
 1  0.159 
The reference value will be n
= 15.61
 p1 (1  p0 ) 
 0.5(1  0.159) 
ln 
ln 


p
(1

p
)
 0.159(1  0.5) 
1 
 0
Name ……………………………………………………….STAT 5031 test, Apr 9, 2003 Page 3 of 4
PCB Cusum
70
60
50
40
30
20
CUUP
10
CUDOWN
0
HPLUS
-10
HMINUS
-20
-30
-40
-50
0
5
Sample number
10