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Case Study 5: Statistical Process Control

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CASE STUDY 5: Statistical Process Control

Ikrom Abdullaev M0158958
Shamil Tlukashaev
Chuan Lai

Exercise A: Control Charts for Variables
Step 1: Gather data Four samples of five observations (launches) each were arranged in tabular form. The mean and range for each sample determined and computed the mean of the means and the mean of the ranges.
Data Table SampleNumber | Observation 1 2 3 4 5 | SampleMean x | Sample Range R | 1 | 7.2 | 8.1 | 8 | 8.5 | 9.2 | 8.20 | 2 | 2 | 8.4 | 8.2 | 7.6 | 9.3 | 10.1 | 8.72 | 2.2 | 3 | 10 | 9.1 | 7.4 | 7.9 | 9.4 | 8.76 | 2.6 | 4 | 10.1 | 9.2 | 7.8 | 7.3 | 10.4 | 8.96 | 3.1 |



Step 2: Develop an R-chart
Using the data gathered and the appropriate and values, we computed the upper and lower three-sigma control limits for the range. Because the average range for Ris 2.475, to compute we used value for =2.115 according to the table given in the book.
=2.115(2.475) =5.234
The result we got is 5.234
To compute we used formula given below
=0(2.475) =0 According to table given in the book we already know that =0.
So for now we know that
=2.115(2.475) =5.234
=0(2.475) =0
We drew an R-chart, after we have tabulated all the numbers in the tables and equations. From this chart we can see that R is within the tolerance. We also can see from this chart that the pattern of values as rising above and then falling below 2.4, and staying in continuous samples.

Step 3: Develop an x-chart
=8.66+0.577(2.475) =10.088
=8.66-0.577(2.475) =7.231
An x-chart was graphed after computing all equations. This graph shows us that the pattern changes in sample 7 and 8. In sample 8 the value almost meets the UCLx and upper tolerance.

Step 4: Observe the process
Two more samples of five trials each were collected. We observed what affected the process. The chart indicated that the process is operating in the same pattern as with the first the way when collecting data.
Data table for additional observations SampleNumber | Observation 1 2 3 4 5 | SampleMean x | Sample Range R | 5 | 8.1 | 8.4 | 9.2 | 9.4 | 9.5 | 8.92 | 1.4 | 6 | 8.8 | 8.3 | 8 | 9.1 | 8.1 | 8.46 | 1.1 |

Step 5: Observe a changed process SampleNumber | Observation 1 2 3 4 5 | SampleMean x | Sample Range R | 7 | 9.2 | 9.6 | 9.8 | 10.6 | 10.5 | 9.94 | 1.4 | 8 | 9.7 | 9.2 | 10.5 | 10.1 | 10.8 | 10.06 | 1.6 |
After we made change we find out that the range values kept within a narrow zone, while performance increased. But it still stays under statistical control. Since the process changed, we feel that this change is real and not just due to the particular sample chosen. The pattern is consistent and reflects continued improvement in the production operation.
Data table for additional observations

Exercise B: Control Charts for Attributes
Step 1: Gather data SampleNumber | Observation 1 2 3 4 5 6 7 8 9 10 | Misses | p | 1 | M | M | H | H | M | H | M | M | M | H | 6 | 0,6 | 2 | H | M | H | M | M | M | H | M | M | M | 7 | 0,7 | 3 | H | H | H | M | M | H | M | M | M | H | 5 | 0,5 | 4 | M | H | M | H | H | H | M | M | H | M | 5 | 0,5 | To find p we used formula
Step 2: Develop a p-chart
First, to calculate the average fraction defective, we used formula. So the answer we got is =23/40=0,575
Next, we calculated the standard deviation of the distribution of p. Here n represents the sample size (in this case 10).
Then we determined the confidence level for the UCL and LCL. For that we used 3 sigmas (z = 3).
=0,575+3(0, 0244) =0, 6482
=0,575-3(0, 0244) =0, 5080

We think that the process statistically out of control. Because amount of defectives has been decreasing but falling outside the limits. Even though it shows an improvement in the process, defective percentage is decreasing.
Step 3: Observe the process SampleNumber | Observation 1 2 3 4 5 6 7 8 9 10 | Misses | p | 1 | H | M | H | H | M | H | M | M | M | H | 5 | 0,5 | 2 | H | M | H | H | H | M | H | H | M | M | 4 | 0,4 | 3 | H | H | H | M | M | H | M | H | H | H | 3 | 0,3 | 4 | M | H | M | H | H | H | H | M | H | M | 4 | 0,4 |
The process is still in control. We believe with confidence that it is not out of control. We determined the control limits for a 95 percent confidence level. With these limits, our revised process was still in control.
=16/40=0, 4
=0, 4(1-0, 4)10=0, 024
=0, 4+3(0, 024) =0, 472
=0, 4-3(0, 024) =0, 328
From the P-chart with control limits (95%), we can see that the process was still out of control. But the defective level will be lower because we can control as well as monitor the performance.

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