Problems 573
equivalent to an increase inE[−Xi]. Hence, we can detect a decrease in the mean value
of an item by running a one-sided cumulative sum chart on the negatives of the subgroup
averages. That is, for specified valuesdandB, not only do we plot the quantitiesSjas
before, but, in addition, we let
Wj=−Xj−(−μ)−dσ/√
n=μ−Xj−dσ/√
nand then also plot the valuesTj, where
T 0 = 0
Tj+ 1 =max{Tj+Wj+ 1 ,0}, j≥ 0The first time that eitherSjorTjexceedsBσ/
√
n, the process is said to be out of control.
Summing up, the following steps result in a cumulative sum control chart for detecting
a change in the mean value of a produced item: Choose positive constantsdandB; use the
successive subgroup averages to determine the values ofSjandTj; declare the process out
of control the first time that either exceedsBσ/
√
n. Three common choices of the pair of
valuesdandBared=.25,B=8.00, ord=.50,B=4.77, ord=1,B=2.49. Any
of these choices results in a control rule that has approximately the same false alarm rate as
does theX-control chart that declares the process out of control the first time a subgroup
average differs fromμby more than 3σ/
√
n. As a general rule of thumb, the smaller the
change in mean that one wants to guard against, the smaller should be the chosen value
ofd.
Problems..........................................................
- Assume that items produced are supposed to be normally distributed with mean
35 and standard deviation 3. To monitor this process, subgroups of size 5 are
sampled. If the following represents the averages of the first 20 subgroups, does it
appear that the process was in control?
Subgroup No. X Subgroup No. X
1 34.0 6 32.2
2 31.6 7 33.0
3 30.8 8 32.6
4 33.0 9 33.8
5 35.0 10 35.8
(continued)