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Ch. 17 Statistical Quality Control 1

Quality? Causes of variation Chance variation Assignable variation Diagnostic charts Outline To find quality problems which cause assignable variation Quality control charts To identify when assignable variation have entered the process Variable/Attribute control charts Acceptance sampling The quality of the incoming finished product: stock. 2

年魯理識立了理論年理念年什不年理 3

立利量不量不練領離說數練行 4

Six Sigma( 六 Sigma= =population standard deviation Under Normal assumption, P( X- >6 )=2 10-9 =0.0000002% http://www.sixsigma.org.tw/ 5

Six Sigma( 六 Six Sigma is a quality management program to achieve "six sigma" levels of quality. It was pioneered by Motorola in the mid-1980s and has spread to many other manufacturing companies, notably General Electric Corporation(GE). Six Sigma aims to have the total number of failures in quality, or customer satisfaction, occur beyond the sixth sigma of likelihood in a normal distribution of customers. Here sigma stands for a step of one standard deviation; designing processes with tolerances of at least six standard deviations will, on reasonable assumptions, yield fewer than 3.4 defects in one million 6

What is quality? Westinghouse : Total quality is performance leadership in meeting the customer requirements by doing the right things right the first time. AT&T : Quality is meeting customer expectations. 7

Causes of Variation Dr. Walter A. Shewhart(1924 異 Chance variation 異異異 Assignable variation 不異不不不不異異來復 8

兩來不異不異 9

異異良來理理異理行異行行 10

Diagnostic Charts To find the quality problems. Pareto Charts Fishbone Diagram, cause-and-effect diagram 11

Pareto charts Step by step : List the type of defects Rank the defects by their frequencies of occurrence X-axis = the type of defects (ordered), count, percentage, cumulative percentage. Y-axis = the frequency 12

Example. P591 The city manager of Grove City, Utah, is concerned with water usage, in single family homes. A sample of 100 homes are selected and investigated the purposes of typical daily water usage. What is the area of greatest usage? Where should be concentrated to reduce the water usage? 13

Reasons for water usage Gallons per day percentage cumu. Perc. Watering lawn Personal bathing Swimming pool Laundering Dishwashing Car washing 車 Drinking Cooking Conclusion : The activities of watering lawn, personal bathing, and pools account for 82.1% of the water usage. 14

Chart 17-1 15

Example. P592 Emily s Family Restaurant is investigating customer complaints. The five complaints heard most frequently are : 1. Discourteous service 2. Cold food 3. Long wait for seating 4. Few menu choices 5. Unruly young children. Suppose 1 was mentioned most frequently and 2 second. Two factors total more than 85%. 16

Fishbone Diagram The various causes and effects are organized on the diagram. The effect : a particular problem or a goal, put on the right hand side of the diagram. The major causes listed on the left-hand side of the diagram. Subcauses under each major causes may be producing the particular effect Investigate all subcauses, eliminate those not important, until the real cause is identified 17

Chart 17-2 Major causes 18

Example. P592 The Emily s restaurant receives complaints about cold food. All subcauses are listed as assumptions and investigated to find the real problem. See chart 17-3. 19

Quality control charts To identify when assignable causes of variation or changes have entered the process. Two types of control charts : For continuous measurements-- variable control chart Mean chart : change in central tendency Range chart : change in variation For nominal measurements attribute control chart Percent defect chart : based on p = sample proportion of defectives C-bar chart : based on c = number of defectives in a sample 20

Variable control chart Mean chart : a plot on sample means Data : A sample of n observations is selected in each period. There are k periods. The sample means and s.d. s are X1, X2, X3,... The overall/grand mean is found by X X The overall standard deviation is S. k i 1 = = k i 21

Recall : Empirical rule : There is 99.74% probability that µ 3 SE(X) X µ + 3 SE(X) The unknown can be estimated by X And the standard error is SE(X) = σ n s n 22

Mean chart : to monitor the central tendency Mean chart : a plot on sample means The control limits for the mean are The upper control limit, UCL, is given by s µ + 3 SE(X) X + 3 n The lower control limit, LCL, is given by These limits are called the 3-sigma limits. UCL s µ 3 SE(X) X 3 LCL n 23

Alternatively, The ranges can be used to replace the sample s.d. S. Another control limits for the mean: UCL = X + A2 R LCL = X A2 R Where R = (R is the mean range of the k 1 + L+ R k ) / k samples. A2 can be found in Appendix I. 24

Example. P595 Statistical Software, Inc., offers a toll-free number where customers call with problems from 7 AM until 11PM daily. To understand their process, the company decides to develop a control chart describing the total response time. Data : one day, for the 16 hours of operation, 5 calls were sampled each hour. See table on P596. 25

Raw data sample number Time AM7 PM12 1 2 3 4 5 6 7 8 9 10 26

Table 17-1 : in minutes sample number Time mean range AM7 PM12 1 2 3 4 5 6 7 8 9 10 total mean 27

The grand mean is X = 9.4125 The mean range is R = 6.375 The constant A2=0.577 control limits for the mean: UCL = X + A2 R = 9.4125 + 0.577(6.375) = 13.0909 LCL = X A2 R = 9.4125 0.577(6.375) = 5.7341 Based on 16 samples of five calls, 99.74% of the time the mean response time will be between 5.7341 and 13.09009 28

Chart 17-4 29

Range chart : to monitor the variation Range chart : a plot on k ranges, R1,,Rk The 99.7% control limits for the range are The upper/lower control limit, UCL/LCL, are given by UCL = D4 R, LCL = D3R D3, D4 Reflect 3 sigma limits for range Can be found in Appendix B. 30

Example. P598 (Continue) Statistical Software, Inc.. Develop a control chart for the range. Does it appear that there is any time when there is too much variation in the operation? Since n=5, D3=0, D4=2.115, R = 6.375 UCL=2.115(6.375)= 13.4831, CL=0(6.375)=0 See chart 17-5. All the ranges are within the control limits. The variation is in control. 31

Minitab : P 599 32

In-control situation 1. The sample means/ranges are clustered close to the centerlines. The process is in control. 33

Out-control situation 2. The sample means are in control, but the last two ranges are out of control. There is considerable variation from piece to piece. An adjustment in the process is probably necessary. 34

Out-control situation 3. There is an upward trend in the sample means. The variation is in control. An adjustment in the process is probably necessary. 35

Attribute Control Charts Data : k samples of n observations are selected. The sample proportions are p1, p2,p3,k 數不良例 The numbers of defectives are c1,c2,c3,k 度不良數 Percent defective chart A plot of proportions of defectives in samples C-bar chart A plot of numbers of defectives in samples 36

Recall : Empirical rule : for sample proportion p and population proportion, there is 99.74% probability that π 3 SE(p) p π + 3 SE(p) The unknown can be estimated by And the standard error is SE(p) = p k π(1 π) n p i 1 = = k i p(1 p) n s p 37

Percent defective chart : to monitor the central tendency Percent defective chart : a plot of sample proportions The 99.74% control limits for the proportion are The upper control limit, UCL, is given by p(1 p) π + 3 SE(p) p + 3 UCL n The lower control limit, LCL, is given by p(1 p) π 3 SE(p) p 3 LCL n These limits are called the 3-sigma limits. 38

Example. P603 The Credit Department at Global National Bank is responsible for entering each transaction charged to the customer s monthly statement. To guard against errors, each data entry clerk rekeys a sample of 1,500 of their batch of work a second time and computer programs checks that the numbers mismatched. 7 people were investigated. See the following result. 39

Inspector n number mismatched, ci pi M R G S Re W Rea sum mean P=41/10500=(0.0027+ +0.0273)/7=0.0039 p(1 p) 0.0039(1 0.0039) UCL = p + 3 = 0.0039 + 3 = 0.0039 + 0.0048 = 0.0087 n 1500 p(1 p) p 3 n LCL = 0 = 0.0039 3 0.0039(1 0.0039) 1500 = 0.0039 0.0048 < 0 40

Chart 17-6 41

C-bar chart : to monitor the central tendency C-bar chart : a plot of sample counts The 99.74% control limits for the count are The upper control limit, UCL, is given by c + 3 c UCL The lower control limit, LCL, is given by c 3 c LCL These limits are called the 3-sigma limits. 42

Recall : C~Poisson( ) C = no. of occurrence within a period 數 Then Population distribution is Poisson ( ), is a unknown parameter The population of C has mean The population of C has variance The population of C has s.d. λ Thus, the 3-sigma limits for C are λ ± 3 λ c ± 3 c 43

Example. P604 The publisher of the Oak Harbor Daily Telegraph is concerned about the number of misspelled words in the daily newspaper. To control the problem, a control chart is to be instituted. The number of misspelled words found in the final edition of the paper for the last 10 days is : 5, 6, 3, 0, 4, 5, 1,2,7,4. Determine the appropriate control limits and interpret the chart. 44

Since c = (5 + 6 + 3 + 0 + 4 + 5 + 1+ 2 + 7) /10 = Thus the control limits are UCL = c + 3 c = 3.7 + 3 Q c 3 See Chart 17-7 c = 3.7 3 0,LCL All numbers are less than the upper control limit, they are in control. 3.7 < 3.7 = 9.47 = 0 3.7 45

Acceptance sampling : example on P607 Sims Software purchased CDs from CDs International. The CDs are packaged in lots of 1000. Sims Software has agreed to accept lots with 10% defective CDs. A random sample of 20 CDs are selected and inspected. If there are 2 defectives in the sample, the lot will be accepted. 46

Acceptance sampling is a statistical sampling plan to screen the quality of incoming parts The population consists the lots of N units Eg. N=1000 A sample of n units is randomly selected without replacement and inspected. Eg. The n=20 Let X = the number of defective in the sample The critical number/acceptance no. is predetermined as c. Eg. The c=2 The lot is accepted, if X c; rejected, otherwise. 47

Acceptance sampling is a decision-making process. The states of nature could be : Good Lot, or Bad Lot. Let S=total defectives in the population =population proportion of defective = S/N Good Lot : 0.1 H1 Bad Lot : >0.1 H0 Consumer s risk : falsely accept a bad lot. Type I error Producer s risk : a good lot been falsely rejected. Type II error State of Nature Decision Good Lot Bad Lot Accept Lot Correct Consumer's risk Reject Lot Producer's risk Correct 48

To evaluate a sampling plan and determine that it is fair to both the producer and the consumer, the operating characteristic curve (OC curve) is constructed. Find P(Accept the lot ) = OC( ) at every possible OC curve = a plot of the function 49

Example. Let = true proportion of defectives in the lot of 1000 CDs X = no. of defectives in a sample of n=20 observations, sampling without replacement. Then, as n/n 0 X ~ Hypergeometric(N, S, n=20) Binomial(20, ) The lot is accepted if X 2, then OC( ) =P(Accept the lot ) = P(X 2 ) 2 20 x 20 x = π (1 π) x= 0 x For example, at = 0.1, OC(0.1)=0.677 See Chart 17-8 50

Chart 17-8 =1-producer s risk =Consumer s risk 51

Exercise Pareto chart : 17 Quality Control Charts Variable control chart : 19, 21 Attribute control chart : 27, 33 Acceptance sampling : 29 52