Theoretical Control Limits for XBAR Charts
Although theoretically possible, since we do not know either the population process mean or standard deviation, these formulas cannot be used directly and both must be estimated from the process itself. First the s chart is constructed. If the s chart validates that the process variation is in statistical control, the XBAR chart is constructed.
Steps in Constructing an s Chart
- Select k successive subgroups where k is at least 20, in which there are n measurements in each subgroup. Typically n is between 1 and 9. 3, 4, or 5 measurements per subgroup is quite common.
- Find the sample standard deviation of each subgroup s(i).
- Find the centerline for the s chart, denoted by
- Find the UCL and LCL with the following formulas: UCL= B(4)SBAR and LCL=B(3)SBAR with B(3) and B(4) can be found in the following table:
- Plot the subgroup data and determine if the process is in statistical control. If not, determine the reason for the assignable cause, eliminate it, and the subgroup(s) and repeat the previous 3 steps. Do NOT eliminate subgroups with points out of range for which assignable causes cannot be found.
- Once the s chart is in a state of statistical control and the centerline SBAR can be considered a reliable estimate of the range, the process standard deviation can be estimated using:
Table of B(3) and B(4)
n B(3) B(4) n B(3) B(4)
2 0 3.267 6 .03 1.970
3 0 2.568 7 .118 1.882
4 0 2.266 8 .185 1.815
5 0 2.089 9 .239 1.761
c(4) can be found in the following table:
n c(4) n c(4)
2 .7979 6 .9515
3 .8862 7 .9594
4 .9213 8 .9650
5 .9400 9 .9693
Steps in Constructing the XBAR Chart
- Find the mean of each subgroup XBAR(1), XBAR(2), XBAR(3)… XBAR(k) and the grand mean of all subgroups using:
- Find the UCL and LCL using the following equations:
- Plot the LCL, UCL, centerline, and subgroup means
- Interpret the data using the following guidelines to determine if the process is in control:
A(3) can be found in the following table:
n A(3) n A(3)
2 2.659 6 1.287
3 1.954 7 1.182
4 1.628 8 1.099
5 1.427 9 1.032
a. one point outside the 3 sigma control limits
b. eight successive points on the same side of the centerline
c. six successive points that increase or decrease
d. two out of three points that are on the same side of the centerline, both at a distance exceeding 2 sigmas from the centerline
e. four out of five points that are on the same side of the centerline, four at a distance exceeding 1 sigma from the centerline
f. using an average run length (ARL) for determining process anomalies
Example:
The following data consists of 20 sets of three measurements of the diameter of an engine shaft.
n meas#1 meas#2 meas#3 StdDev XBAR 1 2.0000 1.9998 2.0002 0.0002 2.0000 2 1.9998 2.0003 2.0002 0.0003 2.0001 3 1.9998 2.0001 2.0005 0.0004 2.0001 4 1.9997 2.0000 2.0004 0.0004 2.0000 5 2.0003 2.0003 2.0002 0.0001 2.0003 6 2.0004 2.0003 2.0000 0.0002 2.0002 7 1.9998 1.9998 1.9998 0.0000 1.9998 8 2.0000 2.0001 2.0001 0.0001 2.0001 9 2.0005 2.0000 1.9999 0.0003 2.0001 10 1.9995 1.9998 2.0001 0.0003 1.9998 11 2.0002 1.9999 2.0001 0.0002 2.0001 12 2.0002 1.9998 2.0005 0.0004 2.0002 13 2.0000 2.0001 1.9998 0.0002 2.0000 14 2.0000 2.0002 2.0004 0.0002 2.0002 15 1.9994 2.0001 1.9996 0.0004 1.9997 16 1.9999 2.0003 1.9993 0.0005 1.9998 17 2.0002 1.9998 2.0004 0.0003 2.0001 18 2.0000 2.0001 2.0001 0.0001 2.0001 19 1.9997 1.9994 1.9998 0.0002 1.9996 20 2.0003 2.0007 1.9999 0.0004 2.0003 SBAR CHART LIMITS: SBAR = 0.0002 UCL = B(4)*SBAR = 2.568*.0002 = 0.0005136 LCL = B(3)*SBAR = 0 * .0002 = 0.00 XBAR CHART LIMITS: XDBLBAR = 2.0000 UCL = XDBLBAR + A(3)*SBAR = 2.000+1.954*.0002 = 2.0003908 LCL = XDBLBAR - A(3)*SBAR = 2.000-1.954*.0002 = 1.9996092
s – Chart:
XBAR – Chart: