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# OM 300- Statistical Process Control Notes OM 300

Marketplace > University of Alabama - Tuscaloosa > OM 300 > OM 300 Statistical Process Control Notes
Samantha
UA
GPA 3.2
Operations Management
William Petty

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Pwpt for Statistical Process Control- Fill in the blank notes
COURSE
Operations Management
PROF.
William Petty
TYPE
Class Notes
PAGES
52
WORDS
KARMA
25 ?

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This 52 page Class Notes was uploaded by Samantha on Sunday March 8, 2015. The Class Notes belongs to OM 300 at University of Alabama - Tuscaloosa taught by William Petty in Winter2015. Since its upload, it has received 357 views.

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Date Created: 03/08/15
Statistical Process Control 2014 P eeee on Education Inc MNEWETaa E Learning Objectives When you complete this supplement you should be able to 1 Explain the purpose of a control chart 2 Explain the role of the central limit theorem in SP0 3 Build f charts and Rcharts 4 List the five steps involved in building control charts 5 Build p charts and c charts 6 Explain process capability and compute CIO and CIOk 7 Explain acceptance sampling Statistical Process Control The objective of a process control system is to provide a statistical signal when assignable causes of variation are present If you can t describe and measure it then you can t control or improve it V V V Illustrations WWII Defense Industries gt Sampling procedures to increase productivity and quality Englehard catalytic cores gt Percentage of defective cores Milliken industrial fabrics gt Number of defects per 100 yards Thermalex thermal tubing gt Height x4 and width variables Land s End customer service order fulfillment gt Percentage of correctly filled orders gt Number of errors per line item Hospital pharmacy gt Prescription error rate Statistical Process Control gt variability is in h e re nt i a in every process i i gt natural or common causes gt Special or assignable causes gt Provides a statistical signal when assignable causes are present gt Detect and eliminate assignable causes of variation Natural Variations gt Also called common causes p Affect virtually all production processes gt Expected amount of variation gt Output measures follow a Probability distribution gt For any distribution there is a measure of central tendency and dispersion gt If the distribution of outputs falls within acceptable limits the process is said to be in control Assignable Variations gt Also called special causes of variation gt Generally this is some change in the process gt Variations that can be traced to a specific reason gt The objective is to discover when assignable causes are present gt Eliminate the bad causes gt Incorporate the good causes Know same size for test gt central limit theorem Sampling gt Measure or inspect only a portion of the products or transactions at regular periodic time intervals gt Reduces inspection time gt Reduces the opportunity for quality products or serVIces to reach the customer gt Good strategy when the process is well deSIgned and incontrol gt Sample size and sampling frequency are important deCIsions that impact the effectiveness of the SPC procedure Control Charts Constructed from historical data the purpose of control charts is to help distinguish between natural variations and variations due to assignable causes Types of Quality Data gt Attribute data O l 0 Product characteristic evaluated with a discrete choice J o Goodbad yesno gt Variable data 0 Product characteristic that can be measured 0 Length size weight height time velocity Central Limit Theorem Regardless of the distribution of the population the distribution of sample means drawn from the population will tend to follow a normal curve 1 The mean of the sampling 2 distribution will be the same as the x population mean u 2 The standard deviation of the sampling distribution 039 will equal the population standard 0 deviation 039 divided by the x J square root of the sample size n Control Charts for Variables gt Characteristics that can take any real value gt May be in whole or in fractional numbers gt Continuous random variables know attributes vs variables X bar is average need to look at mean and range together f chart tracks changes in the central tendency R chart indicates a gain 38 M0 chagts dispersion difference between highest aniongas be use together Setting Chart Limits For Y Charts when we know a Lower control limit UCL f za J Upper control limit UCL f 20 Where a EQXQN mean of the sample means or a target value set for the process number of normal standard deviations standard deviation of the sample means 0 xn population process standard deviation sample size Know sample size If standard deviation not given use 3 Setting Control Limits gt Randomly select and weigh nine n 9 boxes each hour Average weight in 171316181716151716 the first sample 9 216391 ounces WEIGHT OF SAMPLE WEIGHT OF SAMPLE WEIGHT OF SAMPLE m AVG OF 9 AVG OF 9 BOXES HOUR BOXES HOUR BOXES 1 161 5 165 9 163 2 168 6 164 10 148 3 155 7 152 11 142 4 165 8 164 12 173 Setting Control Limits Average mean of 12 samples 2 12 2Avg of 9 boxes i1 12 Setting Control Limits Control Chart for samples out of Variation due 17UCL HI ill 1 f 16 M Variation due to ean natural causes 1 1 Variation due I 39 1 2 3 4 5 6 7 8 9101112 Outof toassgjsbe39 Sample number Control Setting Chart Limits For f Charts when we don t know a UCLf x AZR LCL 2 A R x 2 ZR where 1 i1 average range of the samples A2 control Chart factor found in Table 861 a mean of the sample means Setting Control Limits Super Cola Example Process average 12 ounces Labeled as net w 39 t Average range 25 ounce 12 ounces Sample size 5 UCsz A2R 12 577925 UCL 12144 12144 M 12 12144 ounces From Table 93 LCLE 37 1421 LCL 11856 212 144 211856 ounces Restaurant Control Limits For salmon filets at Darden Restaurants X Bar Chart 115 UCL11524 8 2 9 110 1o959 E C6 1 105 LCL 10394 Range Chart 3 UCL 06943 C6 0 2 g R 02125 in LCL o V V V R Chart Type of variable control chart Shows sample ranges over time gt Difference between smallest and largest values in sample Monitors prooess variability Independent from process mean Setting Chart Limits For R Charts Upper control limit UCLR D41 Lower control limit LCLR D31 where UCLE 2 upper control chart limit for the range LCLE 2 lower control chart limit for the range D4 and D3 2 values from Table 861 Setting Control Limits Average range 53 pounds Sample size 5 From Table 861 D4 2115 DO UCLRD4R H UCL112 211553 2112 pounds Mean 53 LCLR 2 D31 053 10 0 pounds Steps In Creating Control Charts Take samples from the population and compute the approloriate sample statistic Use the sample statistic to calculate control limits and draw the control chart Plot sample results on the control chart and determine the state of the process in or out of control Investigate possible assignable causes and take any indicated actions Continue sampling from the process and reset the control limits when necessary Manual and Automated Control Charts 1 Ear Eih an UCL211 524 l d in I 11in I ing g Sample Mean 11mg LEiL1 Clquot 35734 939 quotlil 13 1539 quotI l U U1 14 ange Chart UGL 13 5943 o21 25 LEILD Sample Range 139 3 5 9 Ii 13 15 139 Eap hilli ty Histiagram UEIL Capability lME 113959 Etudev 1 BE 3p 1 Specifications LEI i0 LJEIL 12 I I I I I I i I 9 b Control Charts for Attributes gt For variables that are categorical gt Defectivenondefective goodbad yesno acceptableunacceptable p Measurement is typically counting defectives gt Charts may measure 1 percent defective pchart 2 number of defects cchart Control Limits for pCharts Population will be a binomial distribution but applying the Central Limit Theorem allows us to assume a normal distribution for the sample statistics UCLPZF39I39ZO39I O LCLp p 2039 p n where LCL cannot be under zero I mean fraction percent defective in the samples 2 number of standard deviations of standard deviation of the sampling distribution n number of observations in each sample pChart for Data Entry NUMBER 3 NUMBER SAMPLE OF FRACTION SAMPLE OF FRACTION NUMBER ERRORS DEFECTIVE NUMBER ERRORS DEFECTIVE 1 6 06 11 6 06 2 5 05 12 1 01 3 0 00 13 8 08 4 1 01 14 7 07 5 4 04 15 5 05 6 2 02 16 4 04 7 5 05 17 11 11 8 3 03 18 3 03 9 3 03 19 0 00 10 2 02 20 4 04 7i7 pChart for Data Entry Total number of errors 80 Total number of records examined 10020 39 an 041 04 p 100 jizz 02 rounded up from 0196 UCLP f9za 04302 10 1O 2 02 20 pChart for Data Entry J1 J0 09 08 07 06 II O 05 O o O 04 g 0 O 5 004 03 o O O 02 O O 33 T I I I T I I II 4 6 8101214161820 Sample number UCLp 010 Fraction defective LCLp 000 V 2m 4 P0010000 E0100at 0m 100 86 a 29 P Chart Example gt A Production manager for a tire company has inspected the number of defective tires in five random samples with 20 tires in each sample The table shows the number of defective tires in each sample of 20 tires Calculate the proportion defective for each sample the center line and control limits using 2 300 sample size 20 proportion defective p bar 01 00 09 LCL zero n20 Z3 Sample Number Number of Proportion of Tires in each Defective Defective Sample Tires 1 3 2O 2 2 2O 3 1 2O 4 2 2O 5 1 20 Total 9 100 P Chart Example cont proportion defective sq rt of 00909120 0064 UCL 009 30064 0282 LCL 009 30064 012 gt ZERO Control Limits for cCharts Population will be a Poisson distribution but applying the Central Limit Theorem allows us to assume a normal distribution for the sample statistics UCLC 5 3 LCLC E 3 where E mean number defective in the sample Formua sheet will be given on test cChart for Cab Company 5 54 complaints9 days 6 complaintsday UCLC E 3 UCLC 1335 63J gm 1335 12 CT 0 LCLCE 3xE git 39 C 6 63g Z 3 LCLCO 0 Cannotbea 1 3 4 5 3 7 8 9 nega t39Ne number Day C Chart Example Number of Week Complaints 1 3 2 2 3 3 4 1 gt The number of weekly 5 3 customer complaints are 6 3 monitored in a large hotel 7 2 using a cchart Develop three sigma control limits 8 1 using this data table 9 3 10 1 Total 22 C Chart Example cont 2 3 CL 2210 220 Sq rt of 220 1483 UCL 220 31483 665 LCL 220 31483 225 gt ZERO cannot be negative Eb My WWWr athzzlcIir r1a39 In EtFLI Et r Payer11cl dim I Flam U EEEF r39l39li ale h csr39newcsrli clcl l EdaqercizeType ipx dahnwg LlicEI5IaLItiIa r1 15151 DWEfIiEi n1 ple l aizh cswgrap her l izhnwgicl 1r 39 quot 39 v v 39 V I EMS in WT Prey lew arid ta Humewufk Pmuf eu Item 3 nf39El Ii Item 1 Prd l 651 Etnrili nf tapter Prdh mg Avaf llalhi liir w Hm mrki Tm and ingquot Study Filatn39 m il39llm Rult siher 9quot 41 H a 5435 1b AWE Control Chart Factors Sample Size Mean Factor Upper Range Lower Range n A2 D4 D3 2 1880 3268 0 3 1023 2574 0 4 729 2282 0 5 577 2115 0 6 483 2004 0 7 419 1924 0076 8 373 1864 0136 9 337 1816 0184 10 308 1777 0223 12 266 1716 0284 Table 361 UCLX A2 LCL X 7 A5 5475 419164 5475 687 5475 687 54063 lbs 55437 quotOS UCLR D4 UCL D n 1 9241 64 R 376164 3155 quotOS 125lbs and dd tn H at Marina Firef g a we I pie ipx lahnwg liEEEEIZiItiIi quoti 15151 Ema3 r11 File 15151 cswgrap her 154511 Izawgicl 1 f39 1 l g 1 T mf WWWMW PFEHIEW arid ta Humewuwk Preview am 4 uf I Item 1 Frd a I WWWmathElearnsTHEtrLJIztcsn F39Iayen gclclaimclHerrtantaaapi n39mtlehcsn mwcsrkz mcicll iazernziEETy Signquotij 1 Er uf h er Frd lkam vafilai il v hm imrkl 7mm and am Fillam n fim Rm her r EH quot n 1h 11mg fm litE hr Hiijimji ii m7 mumEm hf I n i BI warm m 1 mm rh r a Win airE the lumen can PFBI W i n g fm39 l MEI3m are h i be fun v 2 UPPBI Emmaui Limit I n ragga as we L iw39 39 Maui Limit I Hi irn M resynme as m Mgwj E 11 what EIE39Ete n with me a UPPEI Imam Limit T i i e as we We Emm Limit fainter respamg 33 all mgwj Upper control limit UCL 26 Upper control limit UCL 450 4 039 ox o sq rt of n 20sqrt of254 Upper control limit UCL 450 4 4 466 Lower control limit LCL 7 2039 434 For 3 std dev answers 462 amp 438 Managerial Issues and Control Charts Three major management decisions gt Select points in the processes that need SPC gt Determine the appropriate charting technique gt Set clear policies and procedures overall mean mu x bar bar table 63 Process Capability gt The natural variation Of a process should be small enough to produce products that meet the standards requued gt A process in statistical control does necessarily meet the design specifications gt Process capability is a measure of the relationship between the natural variation of the process and the design specifications Process Capability Ratio C Upper Specification Lower Specification p 60 gt A capable process must have a CD of at least 10 gt Does not look at how well the process is centered in the specification range gt Often a target value of CD 133 is used to allow for offcenter processes gt Six Sigma quality requires a CD 20 Process Capability Ratio Cp and Cpk on test 2 know the difference Insurance claims process Process mean 7 2100 minutes Process standard deviation 0 516 minutes Design specification 210 i 3 minutes C Upper Specification Lower Specification p 6039 Cp shoud be at least 1 2 is perfect Cp 2132076516 1938 gt process is capable Process Capability Index N r N rUpper Lower Specification 7 7 Specification CIok minimum of Limit Limit 3039 J k 3039 J gt A capable process must have a CIok of at least 10 gt A capable process is not necessarily in the center of the specification but it falls within the specification limit at both extremes Process Capability Index New Cutting Machine New process mean 7 250 inches Process standard deviation 0 0005 inches Upper Specification Limit 251 inches Lower Specification Limit 249 inches Cpk min of 25125030005 both result in cpk0010015 67 gt NOT capable C p Example gt Three bottling machines are being evaluated for possible use at the Fizz plant The machines must be MaChine 6 capable of meeting the design specification of 158 162 oz with A 005 a process capability index of at B 010 least 10 sz 1 C 020 gt The table to the right shows the information gathered from production runs on each machine Which bottling machines are acceptable Clo Example cont Machine A best variation because smallest A 133 gt capable B 67 gt not capable C 033 gt not capable Cp 1621586005 133 gt Machine A C pk Example gt Product specifications call for a target value of 160 oz with a tolerance of i 02 oz gt The process performance measures are gt Mean u 159 oz gt Std Deviation o 005 oz gt Compute the Cp value for this bottling process and indicate whether or not it is capable based on the Cp value gt Compute the Cpk value for this bottling process and indicate whether or not it is capable based on the Cpk value C pk Example cont LSL 160 02 158 USL 2160 02 162 16392 15398 133 Capable c p 6005 Cpk min 1621593005 1591583 min 20 067 067 gt not capable d Mean H P r cess Hariability 2014 Pearson Education Inc 86 50 Acceptance Sampling p Form of quality testing used for incoming materials or finished goods gt Take samples at random from a lot shipment of items gt Inspect each of the items in the sample gt Decide whether to reject the whole lot based on the inspection results gt only screens lots 109 01 drive quality improvement efforts Automated Inspection gt Modern technologies allow virtually 100 inspection at minimal costs gt Not suitable for all situations

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