[1]

MA4104 Business StatisticsSpring 2008, Lecture 06

Process Monitoring Using Statistical Control Charts

[ Examples Class ]

[2]

x Chart Structure

UCLUCL

LCLLCL

Process mean when in controlProcess mean when in controlCenter LineCenter Line

Time Time

[3]

Control Limits for an x Chart

• Process Mean and Standard Deviation Known

n 3 = UCL

n 3 = LCL

[4]

The Standard Error [ of the sample mean ]

n

xSE

RAn 2 3

ASIDE: It is approximately true that

[5]

Control Limits for an x Chart

• Process Mean and Standard Deviation Unknown

where

x = overall sample mean

R = average range

A2 = a constant that depends on n; taken from

“Factors for Control Charts” table

RAx 2 = UCL

RAx 2 = LCL

==

__

[6]

Factors for x and R Control Charts

n d2 A2 d3 D3 D4

2 1.128 1.880 0.853 0 3.267

3 1.693 1.023 0.888 0 2.574

4 2.059 0.729 0.880 0 2.282

5 2.326 0.577 0.864 0 2.114

6 2.534 0.483 0.848 0 2.004

7 2.704 0.419 0.833 0.076 1.924

8 2.847 0.373 0.820 0.136 1.864

9 2.970 0.337 0.808 0.184 1.816

10 3.078 0.308 0.797 0.223 1.777

: : : : : :

[7]

Interpretation of Control Charts

• The location and pattern of points in a control chart enable us to determine, with a small probability of error, whether a process is in statistical control.

• A primary indication that a process may be out of control is a data point outside the control limits.

• Certain patterns of points within the control limits can be warning signals of quality problems.

– A large number of points on one side of the center line.

– Six or seven points in a row that indicate either an increasing or decreasing trend.

– . . . and other patterns.

[8]

Clip gap measurements in twenty five samples offive measurements each

Sample 1 2 3 4 5 6 7 8 9 10 11 12

65 75 75 60 70 60 75 60 65 60 80 85 Clip 70 85 80 70 75 75 80 70 80 70 75 75 gaps 65 75 80 70 65 75 65 80 85 60 90 85 65 85 70 75 85 85 75 75 85 80 50 65 85 65 75 65 80 70 70 75 75 65 80 70

Range 20 20 10 15 20 25 15 20 20 20 40 20

Sample 13 14 15 16 17 18 19 20 21 22 23 24 25

70 65 90 75 75 75 65 60 50 60 80 65 65 Clip 70 70 80 80 85 70 65 60 55 80 65 60 70 gaps 75 85 80 75 70 60 85 65 65 65 75 65 70 75 75 75 80 80 70 65 60 80 65 65 60 60 70 60 85 65 70 60 70 65 80 75 65 70 65

Range 5 25 15 15 15 15 20 5 30 20 15 10 10

[9]

Estimating

• First, calculate the average range from stable data,

– use the deletion principle

• then, convert to ̂

[10]

Applying the deletion principle

• identify the largest sample range value;

• calculate the average range from the remaining points;

• calculate a trial upper control limit and display the centre line and UCL on the line plot;

• check whether the largest and any other range values lie outside the limits:

– if none are outside, recalculate average range including the most extreme point,

– otherwise, delete the points outside the limits and repeat the whole process with the remaining points.

[11]

Clip gap measurements in twenty five samples offive measurements each

Sample 1 2 3 4 5 6 7 8 9 10 11 12

65 75 75 60 70 60 75 60 65 60 80 85 Clip 70 85 80 70 75 75 80 70 80 70 75 75 gaps 65 75 80 70 65 75 65 80 85 60 90 85 65 85 70 75 85 85 75 75 85 80 50 65 85 65 75 65 80 70 70 75 75 65 80 70

Range 20 20 10 15 20 25 15 20 20 20 40 20

Sample 13 14 15 16 17 18 19 20 21 22 23 24 25

70 65 90 75 75 75 65 60 50 60 80 65 65 Clip 70 70 80 80 85 70 65 60 55 80 65 60 70 gaps 75 85 80 75 70 60 85 65 65 65 75 65 70 75 75 75 80 80 70 65 60 80 65 65 60 60 70 60 85 65 70 60 70 65 80 75 65 70 65

Range 5 25 15 15 15 15 20 5 30 20 15 10 10

Sum of 25 Range values = 445. Average Range = 445/25 = 17.8

Exercise: Calculate deleted average range, , and UCL = 2.11×

Sample 1 2 3 4 5 6 7 8 9 10 11 12

65 75 75 60 70 60 75 60 65 60 80 85 Clip 70 85 80 70 75 75 80 70 80 70 75 75 gaps 65 75 80 70 65 75 65 80 85 60 90 85 65 85 70 75 85 85 75 75 85 80 50 65 85 65 75 65 80 70 70 75 75 65 80 70

Range 20 20 10 15 20 25 15 20 20 20 40 20

Sample 13 14 15 16 17 18 19 20 21 22 23 24 25

70 65 90 75 75 75 65 60 50 60 80 65 65 Clip 70 70 80 80 85 70 65 60 55 80 65 60 70 gaps 75 85 80 75 70 60 85 65 65 65 75 65 70 75 75 75 80 80 70 65 60 80 65 65 60 60 70 60 85 65 70 60 70 65 80 75 65 70 65

Range 5 25 15 15 15 15 20 5 30 20 15 10 10

R R

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Clip gaps Range chart;trial limits with Sample 11 deleted

5 10 15 20 25

Sample Number

0

10

20

30

40

50

Range

R = 18

UCL = 38

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Exclude Sample 11, delete Sample 21

Sum of 24 Range values = 405. R21 = 30

Exercise: Calculate deleted average range, ,

and UCL = 2.11 ×R

R

[14]

Range chart with Sample 11 excluded;trial limits with Sample 21 deleted

5 10 15 20 25

Sample Number

0

10

20

30

40

50

Range

R = 16

UCL = 34

[15]

Convert to ̂

[16]

Convert to ̂

Exercise: A2 = 0.58. Calculate . ̂

Did you get 7.26 ?

[17]

Exercise

Assuming a value of 7.3 mm for , use the Normal table to predict the proportion of clips whose gaps fail to meet the specification limits of 50 mm to 90 mm

(i) when the process mean is 70 mm,

Did you get 0.6 % ?

[18]

Estimating

• Identify stable data

– a form of deletion principle

• Calculate X

[19]

X-bar chart for clip gaps with historical centre line, sample 11 excluded

5 10 15 20 25

Sample Number

55

60

65

70

75

80

85

90

X barX = 71.5

LCL = 61.5

UCL = 81.5

[20]

Clip gaps X-bar chart with with redrawn control limits

5 10 15 20 25

Sample Number

55

60

65

70

75

80

85

90

X bar

8.73XBefore 75.66XAfter

[21]

Exercise, continued

Assuming a value of 7.3 mm for , use the Normal table to predict the proportion of clips whose gaps fail to meet the specification limits of 50 mm to 90 mm

(ii) when the process mean is 74 mm,

(iii) when the process mean is 67 mm.

(ii) Did you get 1.47 % ?

(iii) Did you get 1.07 % ?

[22]

Control Charts for COUNTS

np Chart

Used to monitor COUNTS, i.e. , the number of defective items in a subgroup [ sample of size n ] , with an overall proportion p defective.

[23]

)1(3 = UCL pnpnp

)1(3 = LCL pnpnp

assuming np > 5 and n (1-p) > 5

Note: If computed LCL is negative, set LCL = 0

Control Limits for an np Chart

[24]

The Standard Error [ of the sample total defective ]

)1( pnpnpSE

[25]

Clerical error counts in weekly samples of 100 forms for a 30 week period.

Week Error Week Error Week Error Week Error Week Error No. Count No. Count No. Count No. Count No. Count

1 5 7 12 13 5 19 3 25 6 2 5 8 10 14 6 20 4 26 5 3 4 9 6 15 8 21 6 27 5 4 6 10 3 16 8 22 7 28 8 5 3 11 8 17 6 23 5 29 4 6 9 12 3 18 5 24 7 30 7

For this example, assume that p = 0.05, i.e., 5% .

As n = 100 (forms), we have that np = 5

[26]

18.2

)05.01(05.0100

)1(

pnpnpSE

[27]

NP chart for Error Count in 100 formsfor a 30 week period

5 10 15 20 25 30

Sample Number

0

2

4

6

8

10

12

14

SampleCount

UCL=11.5

NP=5

LCL=0

[28]

Clerical error counts in weekly samples of 100 forms for a further 20 week period.

Week Error Week Error Week Error Week Error No. Count No. Count No. Count No. Count

31 1 36 4 41 4 46 2 32 1 37 4 42 1 47 1 33 4 38 4 43 6 48 5 34 4 39 1 44 5 49 2 35 1 40 6 45 1 50 3

[29]

NP chart for Error Count in 100 forms,extended to a 50 week period

5 10 15 20 25 30 35 40 45 50

Sample Number

0

2

4

6

8

10

12

14

SampleCount

UCL=11.5

NP=5

LCL=0

[30]

Post improvement :np chart with new limits

35 40 45 50

Sample Number

0

2

4

6

8

10

12

14

SampleCount

UCL=8.5

NP=3

LCL=0