Sample Control Chart Calculations. Here is a worked example of the x and R control chart calculations.

Transcription

1 Sample Control Chart Calculations Here is a worked example of the x and R control chart calculations. Step 1: The appropriate characteristic to measure was defined and the measurement methodology determined. A sample size of 8 was selected with measurements to be taken once a day. It is known that the nominal is 0 and the specifications are 13 and 7. Step : The data was collected and is shown in the table below. Day Sample Number Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Step 3: We then calculate the average of each sample. This is shown below.

2 Day Sample Number Average 01-Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Step 4: Calculate the range for each sample.

3 Day Sample Number Average Range 01-Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Step 5: Calculate the average of the averages and the average range.

4 Day Average Range 01-Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Jan Sum Average This then tells us that x 19.45and R We see that the process aim, at 19.45, is relatively close to the nominal of 0. Step 6: We now draw a histogram. We compare the observed shape to the expected shape. If it is not significantly different we continue. If there is a significant difference this means that there is most likely a special cause of variation present and we should stop and investigate. (This is the first of our four checks for a process characteristic being in a state of statistical control.)

5 # Observations Histogram to <= 15.0 to <= 16.0 to <= 17.0 to <= 18.0 to <= 19.0 to <= 0.0 to <= 1.0 to <=.0 to <= 3.0 to <= 4.0 to <= Class This could certainly be a developing bell shaped curve. At this point we can state we have passed check 1. It might be useful at this point to use a scorecard like the one shown below. Check # Description Results 1 Histogram Pass Points Within Xbar Limits 3 Points Within R Limits 4 Pattern of Variation Step 7: We then calculate the control limits for averages and ranges. We will use the Shewhart values from the table below. Each sample had 8 measurements.

6 n A A 3 d D 3 D 4 B 3 B UCLx LCLx x x - A A (.373)(5.857) R R (.373)(5.857) If only common cause variation is present in the process, that is, if the process is still doing like it always has we would expect to find all of the sample averages to fall between the control limits for averages. An examination of the data shows that all of the sample averages do, so we pass our second test for being in control.

7 We now calculate the control limits for Ranges. Check # Description Results 1 Histogram Pass Points Within Xbar Limits Pass 3 Points Within R Limits 4 Pattern of Variation UCLR LCLR D (1.864)(5. 857) D 3 4 R R (.136)(5.857) If only common cause variation is present in the process, that is, if the process is still doing like it always has we would expect to find all of the sample ranges to fall between the control limits for ranges. An examination of the data shows that all of the sample ranges do, so we pass our third test for being in control. Check # Description Results 1 Histogram Pass Points Within Xbar Limits Pass 3 Points Within R Limits Pass 4 Pattern of Variation Step 8: We draw the control chart and graph the points. The charts for Averages and Ranges are shown below. Xbar Chart 1.5 UCL= CEN= LCL= /1 1/ 1/3 1/4 1/5 1/6 1/7 1/8 1/9 1/10 1/11 1/1 1/13 1/14 1/15 1/16 1/17 1/18 1/19 1/0 1/1

8 UCL= R Chart CEN= LCL= /1 1/ 1/3 1/4 1/5 1/6 1/7 1/8 1/9 1/10 1/11 1/1 1/13 1/14 1/15 1/16 1/17 1/18 1/19 1/0 1/1 Step 9: We now evaluate the variation as time goes by. The pattern appears to be random, so we can state the process is in control. We have passed our 4 th check. Check # Description Results 1 Histogram Pass Points Within Xbar Limits Pass 3 Points Within R Limits Pass 4 Pattern of Variation Pass Since our process is in control, the six sigma mantra says we can now work to improve the process. We strive to improve processes that are not capable of meeting requirements before those that currently are capable. A sketch of the normal curve showing the spec limits looks like this. Normal Curve Sketch

9 We initially calculate the z values. In order to do this we calculate the Shewhart approximation to the standard deviation. R '.057 d.847 We then calculate our z values. Z u USL - x ' Z l LSL - x ' We then do our table look up. Corresponding to Z u we read Corresponding to Z l we read The total non-conforming is the sum of.0001 and or This may also be expressed as.097% non conforming. The sigma level is the smaller of the two Z values, or DPMO is the proportion times one million, or 970. We use the Cpk since the process is not centered at the midpoint of the specification limits. The Cpk is one third of the sigma level, or Cpk Sigma Level Our interpretation of all of these is that the process, by definition, is capable. The proportion nonconforming being less than.007, the DPMO being less than,700, the Sigma Level being greater than 3, and the Cpk being greater than 1 all show this. This table summarizes the criteria. Capability Measure Not Capable Capable More than Capable Proportion Non Conforming More than Less than.007 DPMO More than Less than 700 Sigma Level Less than More than 3.0 Cpk Less than More than 1.0

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11 Sample Control Chart Calculations Example Here is a worked example of the x and R control chart calculations. Step 1: The appropriate characteristic to measure was defined and the measurement methodology determined. A sample size of 8 was selected with measurements to be taken once a day. It is known that the nominal is 7 and the specifications are 0 and 14. Step : The data was collected and is shown in the table below. Sample May May May May May May May May May May May May May May May May May May Step 3: We then calculate the average of each sample. This is shown below.

12 Sample Average 7-May May May May May May May May May May May May May May May May May May Step 4: Calculate the range for each sample.

13 Sample Average Range 7-May May May May May May May May May May May May May May May May May May Step 5: Calculate the average of the averages and the average range.

14 Sample Average Range 7-May May May May May May May May May May May May May May May May May May Average: This then tells us that x 6.96and R.7. We see that the process aim, at 6.96, is relatively close to the nominal of 7. Step 6: We now draw a histogram. We compare the observed shape to the expected shape. If it is not significantly different we continue. If there is a significant difference this means that there is most likely a special cause of variation present and we should stop and investigate. (This is the first of our four checks for a process characteristic being in a state of statistical control.)

15 # of Observations Histogram Classes This could certainly be a developing bell shaped curve. At this point we can state we have passed check 1. It might be useful at this point to use a scorecard like the one shown below. Check # Description Results 1 Histogram Pass Points Within Xbar Limits 3 Points Within R Limits 4 Pattern of Variation Step 7: We then calculate the control limits for averages and ranges. We will use the Shewhart values from the table below. Each sample had 4 measurements.

16 n A A 3 d D 3 D 4 B 3 B UCLx 6.96 LCLx x x - A A (.79)(.7) R R (.79)(.7) If only common cause variation is present in the process, that is, if the process is still doing like it always has we would expect to find all of the sample averages to fall between the control limits for averages. An examination of the data shows that some of the averages are below the lower control limit and some are above the upper limit.

17 Check # Description Results 1 Histogram Pass Points Within Xbar Limits Fail 3 Points Within R Limits 4 Pattern of Variation At this point it is appropriate to stop the analysis and state that the process is not in control. We need to address the special cause(s) of variation before proceeding. (In addressing the special causes we either remove the special cause or develop a contingency plan to deal with it when it does occur.) For educational purposes we now calculate the control limits for Ranges. UCLR (.8)(. 7) LCLR D D 3 4 R R (0)(.7) If only common cause variation is present in the process, that is, if the process is still doing like it always has we would expect to find all of the sample ranges to fall between the control limits for ranges and no unusual patterns. An examination of the data shows that all of the sample ranges do fall within limits however, so we pass our third test for being in control. Check # Description Results 1 Histogram Pass Points Within Xbar Limits Fail 3 Points Within R Limits Pass 4 Pattern of Variation Step 8: We draw the control chart and graph the points. The charts for Averages and Ranges are shown below. Xbar Chart UCL=8.948 CEN= LCL= R Chart UCL=6.11 CEN=.7 LCL=

18 Step 9: We now evaluate the variation as time goes by. The pattern on the xbar chart appears to be an increasing trend and on the R chart we have 7 consecutive points below the centerline and 7 consecutive points above the centerline. These are both indications of special causes of variation, so we can state the process is not in control. We have failed our 4 th check. Check # Description Results 1 Histogram Pass Points Within Xbar Limits Fail 3 Points Within R Limits Pass 4 Pattern of Variation Fail Since our process is not in control, the six sigma mantra says we can now work to bring our process into control. Once we see any indication of special cause variation we must stop and address it. However, if we neglected to do the in control evaluation and proceeded directly to capability analysis it would be extremely misleading. A sketch of the normal curve showing the spec limits looks like this. We initially calculate the z values. In order to do this we calculate the Shewhart approximation to the standard deviation. R.7 ' 1.3 d.059

19 We then calculate our z values. Z u USL - x ' Z l LSL - x ' We then do our table look up. Corresponding to Z u we read Corresponding to Z l we read The total non-conforming is the sum of and or This may also be expressed as.00001% non conforming. The sigma level is the smaller of the two Z values, or 5.7. DPMO is the proportion times one million, or 1. We use the Cpk since the process is not centered at the midpoint of the specification limits. The Cpk is one third of the sigma level, or Cpk Sigma Level Our interpretation of all of these is that the process, by definition, is capable. The proportion nonconforming being less than.007, the DPMO being less than,700, the Sigma Level being greater than 3, and the Cpk being greater than 1 all show this. This table summarizes the criteria. Capability Measure Not Capable Capable More than Capable Proportion Non Conforming More than Less than.007 DPMO More than Less than 700 Sigma Level Less than More than 3.0 Cpk Less than More than 1.0 However, this is totally meaningless since the process is not in a state of statistical control. It appears that we have a highly capable process, but only because we neglected a critically important rule.