Control Charts for Process Variation

16.3 Control Charts for Process Variation

  The control charts discussed in the previous section were designed to control the location (equivalently, central tendency) of a process, with particular attention to the mean as a measure of location. It is equally important to ensure that a process is under control with respect to variation. In fact, most practitioners recommend that control

  be established on variation prior to constructing an X chart or any other chart for con- trolling location. In this section, we consider charts for variation based on the sample

  16.3 Control Charts for process Variation 691

  standard deviation S and also charts based on the sample range R. The former are generally preferred because the standard deviation gives a more efficient assessment of variation than does the range, but R charts were used first and tradition dies hard.

  the S chart

  We again suppose that k independently selected samples are available, each one consisting of n observations on a normally distributed variable. Denote the sample

  standard deviations by s 1 ,s 2 ,…, s k , with s 5 os i yk. The values s 1 ,s 2 ,s 3 ,… are plot-

  ted in sequence on an S chart. The center line of the chart will be at height s, and the 3-sigma limits necessitate determining 3s S (just as 3-sigma limits of an X chart

  required 3s X 5 3s yÏn, with s then estimated from the data).

  Recall that for any rv Y, V sYd 5 EsY 2 d 2 [EsYd] 2 , and that a sample variance S 2 is an unbiased estimator of s 2 , that is, E sS 2 d5s 2 . Thus

  V sSd 5 EsS 2 d 2 [EsSd] 2 5s 2 sa n s d 2 5s 2 s1 2 a 2 n d

  where values of a n for n 5 3,…, 8 are tabulated in the previous section. The stand- ard deviation of S is then

  s S 5 ÏVsSd 5 sÏ1 2 a n 2

  It is natural to estimate s using s 1 ,…, s k , as was done in the previous section namely,

  sˆ5 s ya n . Substituting sˆ for s in the expression for s S gives the quantity used to calculate 3-sigma limits.

  The 3-sigma control limits for an S control chart are

  LCL 5 s 2 3s Ï 12a 2 n ya n

  UCL 5 s 2 3s Ï 12a 2 n ya n

  The expression for LCL will be negative if n 5, in which case it is customary to use LCL 5 0.

  ExamplE 16.4 Table 16.2 displays observations on stress resistance of plastic sheets (the force, in psi, necessary to crack a sheet). There are k 5 22 samples, obtained at equally spaced time points, and n 5 4 observations in each sample. It is easily verified that

  os i 5 51.10 and s 5 2.32, so the center of the S chart will be at 2.32 (though because

  n5 4, LCL 5 0 and the center line will not be midway between the control limits).

  From the previous section, a 4 5 .921, from which the UCL is

  UCL 5 2.32 1 3 s2.32dsÏ1 2 s.921d 2 dy.921 5 5.26

  Table 16.2 Stress-Resistance Data for Example 16.4

  Sample No.

  Observations

  SD Range

  (continued )

  692 Chapter 16 Quality Control Methods

  Table 16.2 Stress-Resistance Data for Example 16.4 (continued)

  Sample No.

  Observations

  SD Range

  The resulting control chart is shown in Figure 16.3. All plotted points are well within the control limits, suggesting stable process behavior with respect to variation.

  Sample 0 5 10 15 20 number

  Figure 16.3 S chart for stress-resistance data for Example 16.4

  n

  the r chart

  Let r 1 ,r 2 ,…r k denote the k sample ranges and r 5 or i yk. The center line of an R chart will be at height r. Determination of the control limits requires s R , where R denotes the range (prior to making observations—as a random variable) of a random sample of size n from a normal distribution with mean value m and standard devia- tion s. Because

  R 5 max(X 1 ,…, X n ) 2 min(X 1 ,…, X n )

  5 s{max sZ 1 ,…, Z n d 2 minsZ 1 ,…, Z n d}

  16.3 Control Charts for process Variation 693

  where Z i 5 (X i 2m ) ys, and the Z i ’s are standard normal rv’s, it follows that

  standard deviation of the range of random sample

  1 of size n from a standard normal distribution 2

  s R 5s?

  5s? c n

  The values of c n for n 5 3,…, 8 appear in the accompanying table.

  n

  3 45678 c n .888 .880 .864 .848 .833 .820

  It is customary to estimate s by sˆ 5 r yb n as discussed in the previous section. This

  gives sˆ R 5 c n r yb n as the estimated standard deviation of R. The 3-sigma limits for an R chart are

  LCL 5 r 2 3c n r yb n UCL 5 r 1 3c n r yb n

  The expression for LCL will be negative if n 6, in which case LCL 5 0 should be used.

  ExamplE 16.5 In tissue engineering, cells are seeded onto a scaffold that then guides the growth

  of new cells. The article “On the Process Capability of the Solid Free-Form Fabrication: A Case Study of Scaffold Moulds for Tissue Engineering” (J.

  of Engr. in Med., 2008: 377–392) used various quality control methods to study

  a method of producing such scaffolds. An unusual feature is that instead of sub- groups being observed over time, each subgroup resulted from a different design dimension (mm). Table 16.3 contains data from Table 2 of the cited article on the deviation from target in the perpendicular orientation (these deviations are indeed all positive—the printed beams exhibit larger dimensions than those designed).

  Table 16.3 Deviation-from-Target Data for Example 16.5

  des dim

  mean

  range st dev

  694 Chapter 16 Quality Control Methods

  Table 16.3 yields or i 5 124, from which r 5 7.29. Since n 5 3, LCL 5 0. With

  b 3 5 1.693 and c 3 5 .888, UCL 5 7.29 1 3 ? s.888ds7.29dy1.693 5 18.76

  Figure 16.4 shows both an R chart and an X chart from the Minitab software pack- age (the cited article also included these charts). All points are within the appropri- ate control limits, indicating an in-control process for both location and variation.

  Sample Mean 5

  – R = 7.29

  Sample Range 5

  Figure 16.4 Control charts for the deviation-from-target data of Example 16.5

  n

  charts Based on Probability Limits

  Consider an X chart based on the in-control (target) value m 0 and known s. When

  the variable of interest is normally distributed and the process is in control,

  P sX i .m 0 1 3s yÏnd 5 .0013 5 PsX i ,m 0 2 3s yÏnd

  That is, the probability that a point on the chart falls above the UCL is .0013, as is the probability that the point falls below the LCL (using 3.09 in place of 3 gives .001 for each probability). When control limits are based on estimates of m and s, these probabilities will be approximately correct provided that n is not too small and k is at least 20.

  By contrast, it is not the case for a 3-sigma S chart that P sS i . UCL d5 P sS i , LCL d 5 .0013, nor is it true for a 3-sigma R chart that PsR i . UCL d5 P sR i , LCL d 5 .0013. This is because neither S nor R has a normal distribution even when the population distribution is normal. Instead, both S and R have skewed dis- tributions. The best that can be said for 3-sigma S and R charts is that an in-control process is quite unlikely to yield a point at any particular time that is outside the control limits. Some authors have advocated the use of control limits for which the “exceedance probability” for each limit is approximately .001. The book Statistical Methods for Quality Improvement (see the chapter bibliography) contains more information on this topic.

  16.4 Control Charts for attributes 695

  EXERCISES Section 16.3 (16–20)

  16. A manufacturer of dustless chalk instituted a quality control

  des dim

  observations

  program to monitor chalk density. The sample standard

  deviations of densities for 24 different subgroups, each

  consisting of n 5 8 chalk specimens, were as follows:

  Calculate limits for an S chart, construct the chart, and

  check for out-of-control points. If there is an out-of-

  9 15 control point, delete it and repeat the process. 22

  17. Subgroups of power supply units are selected once each

  hour from an assembly line, and the high-voltage output

  19. Calculate control limits for an S chart from the refractive

  of each unit is determined.

  index data of Exercise 11. Does the process appear to be

  a. Suppose the sum of the resulting sample ranges for

  in control with respect to variability? Why or why not?

  30 subgroups, each consisting of four units, is 85.2. Calculate control limits for an R chart. 2 20. When S is the sample variance of a normal random

  b. Repeat part (a) if each subgroup consists of eight

  sample, sn 2 1dS 2 ys 2 has a chi-squared distribution with

  units and the sum is 106.2.

  n2 1 df, so

  18. The following data on the deviation from target in the paral-

  sn 2 1dS 2

  1 .999,n21 s 2 .001,n21

  lel orientation is taken from Table 1 of the article cited in

  P x 2 ,

  ,x 2 5 .998

  Example 16.5. Sometimes a transformation of the data is appropriate, either because of nonnormality or because

  from which

  subgroup variation changes systematically with the sub-

  1 n2 1 n2 square root transformation for this data (the family of Box- 1 2

  s 2 x 2 .999,n21

  s 2 x .001,n21 2

  group mean. The authors of the cited article suggested a

  Cox transformations is y 5 x l , so l 5 .5 here; Minitab will

  This suggests that an alternative chart for controlling

  identify the best value of l). Transform the data as sug-

  process variation involves plotting the sample variances

  gested, calculate control limits for X, R, and S charts, and

  and using the control limits

  check for the presence of any out-of-control signals.

  LCL 5 s 2 x 2 .999,n21 y(n 2 1)

  des dim

  observations

  UCL 5 s 2 x 2 .001,n21 y(n 2 1)

  Construct the corresponding chart for the data of

  Exercise 11. [Hint: The lower- and upper-tailed chi-squared

  critical values for 5 df are .210 and 20.515, respectively.]