Statistical Process Control & Statistical Methods for Process Improvement.
The following are answers to 8 questions designed to test your knowledge of Statistical Process Control, Process Capability, Statistical Analysis.Review the questions, formulate your answers, then check with the answers provided below.
Statistical Process Control Training. QUESTIONS >>>
Answer #1:
Mean = 2 cm.
Specification = 1.998 – 2.002 cm
Upper Specification limit = μ + 1σ
Lower Specification Limit = μ – 1σ
i) Calculate σ.
USL = μ + 1σ = 2.002
LSL = μ – 1σ = 1.998 (Subtract the LSL from the USL)
2σ = 0.004cm
σ = 0.002cm
ii) Calculate the process capability
Cpk = {USL – Mean}/3σshort or {Mean – LSL}/3σshort
Mean = 2cm
As the USL and the LSL are the same distance from the mean we can use either.
Using the USL
Cpk = {USL – Mean}/3σshort
= (2.002 – 2.0)/(3 * 0.002)
= 0.002/0.006
= 0.33
iii) What value of σ will deliver a 6σ process.
For a 6 σ process, the ULS (or LSL) must equal the mean + 6σ therefore,
2.0 + (6 * σ ) = 2.002
6 * σ = 0.002
σ = 0.002/6 = 0.00033
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Statistical Process Control Training – Answer #2:
a) You can measure variance once you have a sample greater than one.
b) Mean = (6 + 6.1 + 5.8 + 5.2)/4 = 5.8
Variance = (6-5.8)2 + (6.1 – 5.8)2 + (5.8 – 5.8)2 + (5.2-5.8)2
= 0.04 + 0.09 + 0 + 0.36
= 0.49
Note: In this example the unit of variance is (feet)2
(Feet to the power of 2)
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Answer #3:
Sample #1 ……..1996
2 ……..2090
3 ……..2010
4 ……..2008
5 ……..1835
6 ……..1820
7 ……..2180
8 ……..2118
a) The standard weight is 2000, therefore the nominal is 2000 grams.
The specification is +/- 10% of Nominal, therefore
Lower Specification Limit = 1600 grams
Upper Specification Limit = 2400 grams
Yes, all samples are within specification.
b) Mean, Variance and standard deviation of the samples
1996
2090
2010
2008
1835
1820
2180
2118
Mean = 2007 For the next step, we will call this value xav
Variance = Σ (xi – xav)2 / n-1
Sample #
………………..xi……..xav……..xi -xav………..(xi – xav)2……Σ (xi – x)2
……..1……….1996……..2007……..-11…………….121
……..2……….2090……..2007……..83…………….6889
……..3……….2010……..2007……..3………………..9
……..4……….2008……..2007……..1………………..1
……..5……….1835……..2007……..-172………….29584
……..6……….1820……..2007……..-187………….34969
……..7……….2180……..2007……..173…………..29929
……..8……….2118……..2007……..111…………..12321
…………………………………………………………………………………..113823
Therefore the Variance = 16260.43
Standard Deviation =
= Square root of the Variance
= √16260
= 127.5 = Standard Deviation
c) How many sigma is the process.
Specification limits are: 1600 – 2400 grams.
Nominal is 2000 grams.
The Mean is 2007 grams.
Taking the distance between the mean and the Upper Specification Limit (worst case)
2400 – 2007 = 293
393/127.5 = 3.08 ~ 3 Sigma
Approximately a 3 Sigma process.
d) What is the Capability of the process?
Cpk = {USL – Mean}/3σshort or {Mean – LSL}/3σshort
As the mean is closer to the Upper Specification Limit, we will use:
Cpk = {USL – Mean}/3σshort
USL = 2400
Mean = 2007
σ = 127.5
{2400 – 2007}/{3 * 127.5}
393/382.5
= 1.027
Is the process capable? Suggest it needs some further controls implemented. It is just capable at present.
e) Is the machine acceptable?
We should be aiming for a process that is closer to the 6σ range, this new machine is not really acceptable, it will produce rejects from day one, which is unacceptable.
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Statistical Process Control Training – Answer #4:
Set (before) No. 1…………………………………….Set (after) No.2
……………………….10………………………………………………..10
……………………….12………………………………………………..13
……………………….13………………………………………………..16
……………………….15………………………………………………..17
……………………….12………………………………………………..14
……………………….11………………………………………………..15
……………………….14………………………………………………..15
……………………….10………………………………………………..18
……………………….18………………………………………………..19
……………………….12………………………………………………..20
a) Impact of the process change in terms of the centering of the process between the specification limits
We need to calculate the mean for the process before and after the change.
Before…………………..After
10……………………….10
12……………………….13
13……………………….16
15……………………….17
12……………………….14
11……………………….15
14……………………….15
10……………………….18
18……………………….19
12……………………….20
Process Mean =
12.7……………………..15.7
Specification limits are:
9 to 25 degrees C, therefore the Nominal is midway = 17 C
Therefore the process is now closer to the nominal center.
b) The variability inherent within the process, both before and after the process change?
To consider the variability we should compare the standard deviations before and after the change
Lets first consider the before the change process
Variance = Σ (xi – xav)2 / n-1
Sample #
……………………..xi…………xav……..(xi – x)……..(xi – x)2……..Σ (xi – x)2
…………….1……..10………….12.3……..0.3………..5.29
…………….2……..12………….12.3……..0.3………..0.09
…………….3……..13………….12.3……..0.7………..0.49
…………….4……..15………….12.3……..2.7………..7.29
…………….5……..12………….12.3……..0.3………..0.09
…………….6……..11………….12.3……..0.3………..1.69
…………….7……..14………….12.3……..1.7………..2.89
…………….8……..10………….12.3……..0.3………..5.29
…………….9……..18………….12.3……..5.7………..32.49
…………….10…….12………….12.3……..-0.3……….0.09
…………………………………………………………………………………….55.7
Variance = 6.2
Standard Deviation =
= Square root of the Variance
= √6.2
= 2.49 = Standard Deviation
Lets first consider the after the change process
Variance = Σ (xi – xav)2 / n-1
Sample #
……………xi………xav……..(xi – x)…….(xi – x)2………Σ (xi – x)2
1…………10……..15.7…………0.7…………32.49
2…………13……..15.7…………0.7…………7.29
3…………16……..15.7…………0.3…………0.09
4…………17……..15.7…………1.3…………1.69
5…………14……..15.7…………0.7…………2.89
6…………15……..15.7…………0.7…………0.49
7…………15……..15.7…………0.7…………0.49
8…………18……..15.7…………2.3…………5.29
9…………19……..15.7…………3.3…………10.89
10……….20……..15.7…………4.3…………18.49
…………………………………………………………………………..80.1
…………………………………………………………Variance = 8.9
Standard Deviation =
= Square root of the Variance
= √8.9
= 2.98 = Standard Deviation
Therefore the standard deviation has gone from 2.49 to 2.98, which means there is a greater variability in the process as a result of the change.
c) What is the capability of the process both before and after the process change?
Cpk = {USL – Mean}/3σshort or {Mean – LSL}/3σshort
Lets consider the “before change” process.
Process Mean = 12.7
Specification limits are:
9 to 25 degrees C, therefore the Nominal is midway = 17 C
The mean of 12.7 is closer to the Lower Specification Limits, therefore we’ll use
Cpk = {LSL – Mean – LSL}/3σ short
{12.7-9}/{3 * 2.49}
= 3.7/7.47
= 0.495
Therefore the process is Not Capable before the process change.
Now consider the “after change” process.
Process Mean = 15.7
Specification limits are :
9 to 25 degrees C, therefore the Nominal is midway = 17 C
The mean of 15.7 is closer to the Lower Specification Limits, therefore we’ll use
Cpk = {Mean – LSL}/3σ short
{15.7-9}/{3 * 4.66}
= 6.7/13.9
= 0.482
Again the process is still not capable even after the process change.
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Answer #5:
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Statistical Process Control Training – Answer #6:
A process must be stable.
Cpk is about making an estimate about how the process will perform at some time in the future. Effectively we’re making a prediction about the future, to do this the process must be stable.
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Answer #7 :
e) The process mean is outside specifications.
Cpk is the smaller of (USL – Mean) / 3σ or (Mean – LSL) / 3σ
σ is always positive, therefore for Cpk to be negative the numerator must be negative.
Therefore the Mean must be > USL or < LSL, i.e. outside specification.
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Statistical Process Control Training – Answer #8:
None of the above. We would expect the CPK to increase due to the decrease in variability and the improved centering of the mean.
SPC & Statistical Methods for Process Improvement.
- Process Capability. Variability Reduction. Statistical Process Control.
- Pre-Control. R&R Studies.
- Process capability indices Cp, Cpk, Cpm, Capability ratio.
- Etc. … Etc. …
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