# 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

**Questions >>>**

## 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)

**Questions >>>**

## 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.

**Questions >>>**

## 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.

**Questions >>>**

## Answer #5:

**Questions >>>**

## 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.

**Questions >>>**

## 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.

**Questions >>>**

## 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. …
**Information, Training, current Best Practice >>>**