ISyE 5 Review for Midterm Istructor: Prof. Kaibo Liu Departmet of Idustrial ad Systems Egieerig UW-Madiso Email: kliu8@wisc.edu Office: Room 307 (Mechaical Egieerig Buildig) ISyE 5 Istructor: Kaibo Liu
Outlie About the eam Checklist Problems review ISyE 5 Istructor: Kaibo Liu
Whe Whe & Where /3 (Thursday) :30PM 3:45PM Where 63 MECH ENGR Buildig (last ame s iitial letter from A to J ) 540 Egieerig Hall Buildig (last ame s iitial letter from K to Z ) Please take a seat where the paper is distributed o ISyE 5 Istructor: Kaibo Liu 3
Eam Eam covers all the materials i the lecture util (icludig) Chapter 0 SPC with Autocorrelated Process Data. All homework problems ad eamples i lecture otes are importat. Thigs Allowed Oe-page (double-sided) cheat sheet Calculator Tables eeded i calculatio will be provided ISyE 5 Istructor: Kaibo Liu 4
5 problems totally Eam Problems st: Multiple choice questio d: Hypothesis testig, cofidece iterval, p value 3rd: Cotrol chart for variables 4th: Cotrol chart for attributes 5th: Type I ad type II calculatio uder differet decisio rules ISyE 5 Istructor: Kaibo Liu 5
Outlie About the eam Checklist Problems review ISyE 5 Istructor: Kaibo Liu 6
Distributios. Discrete Probability Distributio Hypergeometric distributio Biomial distributio Poisso Distributio. Cotiuous Probability Distributio Normal distributio Chi-Square distributio Studet t distributio ISyE 5 Istructor: Kaibo Liu 7
Useful Results o Mea ad Variace If is a radom variable ad a is a costat, the E(a+)=a+E() E(a*)=aE() V(a+)=V() V(a*)=a V() If,,, are radom variables, E( + + )=E( )+ +E( ) If they are mutually idepedet, ad a,,a are costats V(a + + a )=a V( )+ +a V( ) ISyE 5 Istructor: Kaibo Liu 8
INTERRELATIONSHIPS BETWEEN DISTRIBUTIONS Hypergeometric, Biomial, Poisso, Normal Samplig without replacemet i fiite populatio Hypergeometric fiite populatio if /N0. N: populatio size :sample size The sum of a sequece of Beroulli trials i ifiite populatio with probability of success p Number of defects per uit Poisso if 5 p=d/n, Biomial if large, small p <0., or large, large p > 0.9, p =-p Normal =, = If p>0 ad 0. p 0.9 =p, =p(-p) a 0.5 p a 0.5 p Pr( a) p( p) p( p) b 0.5 p a 0.5 p Pr( a b) p( p) p( p) Pr( pˆ ) p p( p) / p p( p) / ISyE 5 Istructor: Kaibo Liu 9
Iferece About Process Quality Estimatio poit estimatio iterval estimatio Hypothesis Testig Defiitio Testig o mea kow ad Ukow variace Testig o Variace ISyE 5 Istructor: Kaibo Liu 0
The Use of P-Values i Hypothesis Testig P-Value: the smallest level of sigificace that would lead to rejectio of the ull hypothesis if the predefied >P= mi, reject the ull hypothesis ISyE 5 Istructor: Kaibo Liu
Iferece o the MEAN of a Normal Populatio Variace Kow Assumig Kow Populatio Variace / ~ N(0,) Z Z / / pvalue H : 0 0 H Reject H 0 : 0 if Reject H 0 : < 0 if 0 > Z(/) / 0 < -Z() / 0 ( ) / 0 ( ) / Reject H 0 : > 0 if 0 / > Z() ( ) / 0 ISyE 5 Istructor: Kaibo Liu
Iferece o the MEAN of a Normal Populatio Variace Ukow Assumig Ukow Populatio Variace s/ H : 0 ~ t (-) H 0 Reject H 0 : 0 if Reject H 0 : < 0 if s t/ t/ 0 s/ 0 s/ > t(/, -) <- t(-) s pvalue CDF 0 t( ) ( ) s/ t ( 0 s / ) Reject H 0 : > 0 if 0 s/ > t(-) ( ) s / 0 t ISyE 5 Istructor: Kaibo Liu 3
Iterrelatioships betwee statistical ifereces Z Assumig Kow Populatio / Variace Z / Assumig Kow Populatio Variace ISyE 5 Istructor: Kaibo Liu / ~ N(0,) Reject H 0 : 0 if 0 Reject H 0 : < 0 Reject H 0 : > 0 / ~ N(0,) > Z(/) Reject / H 0 : 0 if if 0 / 0 < -Z() if > Z() Reject / H 0 : < 0 if P=[-( Z 0 0 )] with 0 > Z(/) / 0 < -Z() / two-sided H, i.e., H: 4
ISyE 5 Istructor: Kaibo Liu Iferece o the Differece i Meas of Two Populatios Variace Kow / / - - Z Z Assume Kow Populatio Variaces (0,) ~ - N Reject 0 : H if / - Z Reject 0 : H if Z - Reject 0 : H if Z - 0 : H H pvalue ) - ( ) - ( - [ ( )] Observatios i TWO samples are all i.i.d. 5
ISyE 5 Istructor: Kaibo Liu a) Assume Homogeeity ( ) ) ( ~ - t S p where ) ( ) ( s s S p Reject 0 : H if ), / ( - t S p Reject 0 : H if ), ( - t S p Reject 0 : H if ), ( - t S p, /, / - - p p S t S t ; 0 : H H Iferece o the Differece i Meas of Two Populatios Variace Ukow ( ) [ ( )] p t S pvalue ) ( ) ( S t p ) ( ) ( S t p i.i.d. Equal variace 6
ISyE 5 Istructor: Kaibo Liu b) Assume Heterogeeity ( ) ) ( ~ - v t s s where ) / ) / ) / / ( ( ( v s s s s Reject 0 : H if ), / ( - v t s s Reject 0 : H if ), ( - v t s s Reject 0 : H if ), ( - v t s s Iferece o the Differece i Meas of Two Populatios Variace UNKow 0 : H H 7
Iferece o the Variace of a Normal Distributio ( )s ~ ( ) ( ) s /, /, H0 : 0 H Reject H : if 0 o ( - ) s o > (/, - ) or ( - ) s o < ( - /, - ) Reject H 0 : < o if ( - ) s o < (-, - ) Reject H 0 : > o if ( - ) s o > (, - ) C.I. ( ) S ( ) S, Pr{ /, /, /, } / ISyE 5 Istructor: Kaibo Liu 8
S S / / ~ F, Iferece o the Variaces of Two Normal Distributios With H 0 : = for H : Reject H 0 for H : < Reject H 0 for H : > Reject H 0 C.I. S S ISyE 5 Istructor: Kaibo Liu if if if s s s s s s > F (/,, ) or > F (,, ) > F (,, ) S s s F /,, /,, /,, /,, /, F F F S < F ( /,, ) The two d.f. are echaged 9
Type I error ( producer s risk): Two Types of Hypothesis Test Errors = P{type I error} = P{reject H 0 H 0 is true} =P{coclude bad although actually good}= P{coclude statistically out of cotrol although the process is truly i cotrol} Type II error (cosumer s risk): = P{type II error} = P{fail to reject H 0 H 0 is false} =P{coclude good although actually bad} = P{coclude statistically i cotrol although the process is truly out of cotrol} Power of the test: Power = - = P{reject H 0 H 0 is false} Reality : You Coclude " H " 0 is True " H " is True " H 0 is True " C o f i d e c e P r o d u c e r E r r o r, " H is True " C o s u m e r E r r o r, P o w e r / LCL f() 0 UCL / ISyE 5 Istructor: Kaibo Liu 0
Geeral Model for a Cotrol Chart Let w be a sample statistic that measures some quality characteristics of iterest, ad suppose that the mea of w is w ad the stadard deviatio of w is w. The the ceter lie, the upper cotrol limit, ad the lower cotrol limit become UCL = w + L w Ceter lie = w LCL = w - L w 3 sigma cotrol limits: Actio limits: L=3 (p=0.007) Warig limits: L= (p=0.0455) where L is the "distace" of the cotrol limits from the ceter lie, epressed i stadard deviatio uits Probability limits (Wester Europe): Actio limits: 0.00 limits (p=0.00) Warig limits: 0.05 limits (p=0.050) ISyE 5 Istructor: Kaibo Liu
Average Ru Legth (ARL)-i cotrol ARL: The average umber of poits that must be plotted util a poit idicates a out-of-cotrol coditio. If the process is i-cotrol, The followig table illustrates the possible sequeces leadig to a "out of cotrol" sigal: Ru legth Probability ( ) ARL0 ARLicotrol 3 ( ) : : : k ( ) k Eample: ARL i-cotrol = /= /0.007 = 370. Eve the process is i cotrol, a out-of-cotrol sigal will be geerated every 370 samples o the average. If the process i actually i-cotrol, we wish to see as may icotrol samples as possible => less false alarm ISyE 5 Istructor: Kaibo Liu
Average Ru Legth (ARL) Out of Cotrol If the process is actually out-of-cotrol, ad the probability that the shift will be detected o the first sample is - the secod sample is (-) the rth sample is r- (-) The epected umber of samples take util the shift is detected is ARL ( ) ARLout of cotrol r r r Power: p(correctly detect o.o.c) power Remark: we wat to be large. Thus, the "out of cotrol" coditio ca be quickly detected. If the process i actually out-of-cotrol, we wish to see out-of-cotrol samples ASAP=>fewer false detectio ISyE 5 Istructor: Kaibo Liu 3
Sample Size ad Samplig Frequecy Larger sample size easier to detect small mea shift Average ru legth ad average time to sigal (ATS=ARL*samplig iterval h) are cosidered i desig ad the check the detectio power. ISyE 5 Istructor: Kaibo Liu 4
Cotrol Chart for X ad R Kow, Statistical Basis of the Charts suppose { ij, i=,,m, j=,,} are ormally distributed with ij,~n(, ), thus, ~ N(,( / ) ) X i X bar chart moitors betwee-sample variability (variability over time) ad R chart measures withi-sample variability (istataeous variability at a give time) If ad are kow, X bar chart is (if k=3) 3 3 A LCL A CL ULC A A 3 ISyE 5 Istructor: Kaibo Liu 5
Cotrol Chart for X ad R Kow, (Cot s) Rage R i =ma( ij )-mi( ij ) for j=,.. If ad are kow, the statistical basis of R charts is as follows: Defie the relative rage W=R/. The parameters of the distributio of W are a fuctio of the sample size. Deote W =E(W)=d, W =d 3, (d ad d 3, are give i Appedi Table VI of Tetbook P75) R =d, R =d 3, which are obtaied based o R=W R chart cotrol limits d 3d (d 3d ) R 3 R 3 3 Chart Statistic LCL D CL d ULC D D D d d 3d 3d 3 3 ISyE 5 Istructor: Kaibo Liu 6
ISyE 5 Istructor: Kaibo Liu Need to estimate ad X bar chart Cotrol Chart for ad R Ukow ad 7 X m R R ; d R ˆ ; m m X ˆ m i i m i j ij m i i R A d R / 3 ˆ 3 3ˆ ˆ R A ULC CL R A LCL d 3 A A is determied by, Appedi Table VI
ISyE 5 Istructor: Kaibo Liu Cotrol Chart for ad R Ukow ad (cot s) X 3 3 R m i i R d R d ˆ d ˆ ; m R R ˆ Need to estimate based o R=W, W =d 3, R chart (if k=3) R, R d R ˆ )R d d 3 ( d R d 3 R 3ˆ ˆ 3 3 R R R D ULC R CL R D LCL 4 3 3 4 3 3 d 3d D d 3d D D 3 ad D 4 are determied by, Appedi Table VI 8
Summary of Cotrol Charts (if k=3) Process Parameters X bar chart R chart S chart kow LCL A CL ULC A LCL D CL d ULC D LCL B CL c 4 5 ULC B 6 X bar & R chart ˆ X R ˆ d LCL A CL ULC A R R LCL D R CL R 3 ULC D R 4 X bar & S chart ˆ X ˆ S c 4 ; LCL A CL ULC A 3 3 S S LCL B CL S 3 ULC B 4 S S ISyE 5 Istructor: Kaibo Liu 9
OC Curve for bar ad R Chart X bar chart 0 UCL k / ; LCL k / ; 0 0 ( k) i cotrol d 0, P{ LCL UCL d} 0 Pr{ UCL } Pr{ LCL } k d k d out of cotrol The epected umber of samples take before the shift is detected (the process is o.o.c) ARL outofcotrol r r r ( ) If process is i cotrol: ARL is the epected umber of samples util a false alarm occurs ARL icotrol i k k ISyE 5 Istructor: Kaibo Liu
Wester Electric Rules (Zoe Rules for Cotrol Charts) Ehace the sesitivity of cotrol charts for detectig a small shift or other oradom patters. Oe or more poits outside 3-sigma limits.. Two of three cosecutive poits outside -sigma limits 3. Four of five cosecutive poits beyod the -sigma limits 4. A ru of eight cosecutive poits o oe side of the ceter lie. The rules above apply to oe side of the ceter lie at a time. Other sesitizig rules for Shewhart cotrol chart; Table 5-, P05 The overall type I error: the process is declared out of cotrol if ay oe of the rules is applied: k ( ) i is the type I error of usig oe rule i aloe i if all k rules are idepedet. i ISyE 5 Istructor: Kaibo Liu 3
Procedures for Establishmet of Cotrol Limits Ukow ad If ad are ukow, we eed to estimate ad based o the prelimiary i-cotrol data (ormally m=0~5). The cotrol limits established usig the prelimiary data are called trial cotrol limits, which are used to check whether the prelimiary data are i cotrol. First check R or S chart to esure all data i-cotrol, ad the check X-bar chart Collect Prelimiary Data X Estimate R or S Establish Trial Cotrol Limits Check Prelimiary Data I-cotrol Future Moitorig Update Estimatio Elimiate the Outliers due to Assigable Causes Out-of-cotrol ISyE 5 Istructor: Kaibo Liu 3
Estimatio of the Process Capability Get process specificatio limits (USL, LSL) Estimate based o ˆ R / (R chart) or ˆ S / c (S chart) d 4 Estimate the fractio of ocoformig products p (or p0 6 PPM) LSL USL pˆ Pr{ LSL} Pr{ USL} ( ) ( ) ˆ ˆ Process-Capability Ratio C p USL - LSL ; ˆ USL - LSL = C = ; p 6s 6ˆ s Estimated process stdev ISyE 5 Istructor: Kaibo Liu 33
Differeces amog NTL, CL ad SL ad Impact o Process Capability There is o relatioship betwee cotrol limits ad specificatio limits. C p is a ide relatig atural tolerace limits to specificatio limits. Eterally determied NTL: atural tolerace limits 3 Ceter lie o bar Distributio of bar values Distributio of idividual process measuremet LSL LNTL 3 LNTL 3 LSL UNTL 3 UNTL 3 USL USL C p >, P<00% C p =, P=00% ISyE 5 Istructor: Kaibo Liu LNTL LSL 3 3 USL Eterally determied Width defied by NTL s is larger tha that defied by CL, why? C p <, P>00% UNTL 34
How to Establish a p-chart? m=0-5 samples for costructig trial cotrol limits If p ukow, coduct a test ad trial cotrol limits with pˆ p m D i p m E(p) p i m i m pˆ i UCL p 3 Ceterlie LCL p 3 p( p) p p( p) Trial Cotrol Limits Is there a assigable cause for out-of-cotrol poits or a oradom patter? If so, fid the root causes ad delete these poits, ad the update cotrol limits. Whe a poit is ON a cotrol limit, it is cosidered as either out-ofcotrol or i-cotrol depedig o how the problem asks ISyE 5 Istructor: Kaibo Liu
p Cotrol Chart (The umber of ocoformig items) Rather tha plottig the fractio ocoformig, we plot the umber of ocoformig items with a p Chart : UCL X = p + 3 p( p) Ceter lie = p LCL X = p 3 p( p) If LCL X <0, set LCL X =0 Np ad p cotrol charts ca be trasferred to each other. From p to p, multiple to the UCL, CL, ad LCL; From p to p, divide to the UCL, CL, ad LCL. ISyE 5 Istructor: Kaibo Liu 36
Cotrol Charts for Nocoformities - c Chart Cotrol limits for the c chart with a kow c (kow mea ad variace) UCL c 3 c CL c If LCL<0, set LCL=0 LCL c 3 c If ukow c, c is estimated from prelimiary samples of ispectio uits for costructig trial cotrol limits m Nocoformities UCL c 3 c ci i total # of defects i all samples CL c ĉ c m umber of samples LCL c 3 c The prelimiary samples are eamied by the cotrol chart usig the trial cotrol limits for checkig out-of-cotrol poits Whe a poit is ON a cotrol limit, it is cosidered as O.O.C ISyE 5 Istructor: Kaibo Liu 37
Cotrol Charts for Nocoformities Per Uit - u Chart c: total ocoformities i a sample of ispectio uits ( is ot ecessary be iteger) u: average # of ocoformities per ispectio uit i a sample m u u LCL u 3 c i i i u ; u CL u i i m u UCL u 3 If ukow u, is estimated from prelimiary samples of ispectio uits for costructig trial cotrol limits u ISyE 5 Istructor: Kaibo Liu 38
Outlie About the eam Checklist Problems review ISyE 5 Istructor: Kaibo Liu 39
Hypothesis testig Problem 4 i HW The iside diameters of bearigs used i a aircraft ladig gear assembly are kow to have a stadard deviatio of 0.00 cm. A radom sample of 5 bearigs has a average iside diameter of 8.535 cm. (a) Test the hypothesis that the mea iside bearig diameter is 8.5 cm. Use a two-sided alterative ad α=0.05. (b) Calculate the p-value of the test i (a). (c) Costruct a 95% two-sided cofidece iterval o mea bearig diameter. ISyE 5 Istructor: Kaibo Liu
Hypothesis testig Solutio: (a) H 0 : u = 8.5 vs. H : u 8.5. Reject H 0 if Z 0 > Z α/ (b) Z value 6.78 is out of the rage i the table, therefore, P-value is close to 0, <0.00006 (c) ISyE 5 Istructor: Kaibo Liu
Problem 5 i HW Cotrol chart for variables ISyE 5 Istructor: Kaibo Liu 4
Cotrol chart for variables ISyE 5 Istructor: Kaibo Liu 43
Cotrol chart for attributes Problem i HW 3 Samples o the cotrol limits are regarded out-of-cotrol. ISyE 5 Istructor: Kaibo Liu
Cotrol chart for attributes 00, p 0.08, UCL 0.6, LCL 0 (a) p 00(0.08) 8 UCL p 3 LCL p 3 p( p( p) 8 3 8( 0.08) 6.4 p) 8 3 8( 0.08) 8 8.388 0 (b) p = 0.08 < 0. ad = 00 is large, so use Poisso approimatio to the biomial. Pr (type I error) Pr{ pˆ LCL p} Pr{ pˆ UCL p} Pr{ D LCL } Pr{ D UCL } Pr{ D 0 8} Pr{ D 6.4 8} POISSON(0,8, true) POISSON(6,8, true) 0.000 0.996 0.004 (c) p ew = 00(0.) = 0 > 0 ad 0. p 0. 9, so use the ormal approimatio to the biomial. ew If the studets use p-chart cotrol limits: [Pr{ pˆ UCL p} Pr{ pˆ LCL p}] 0.6 0. 0 0. 0.( 0.) 00 0.( 0.) 00 ( 0.975) ( 5) 0.648 0 0.648 UCL p p( p) LCL p p( p) (d) Pr(detect shift by at most 4 th sample) = Pr(ot detect by first four samples) = (0.648) 4 ISyE 5 Istructor: Kaibo Liu = 0.9996
Problem 4 i HW Type I ad Type II error calculatio uder differet decisio rules What is the Type-I error probability for each of these rules. If the mea of the quality characteristic shifts oe process stadard deviatio ( ), ad remais there durig the collectio of the et seve samples, what is the Type-II error probability associated with each decisio rule? ISyE 5 Istructor: Kaibo Liu
Type I ad Type II error calculatio uder differet decisio rules (a) Rule : Pr{out of cotroli cotrol} Pr{out of 7 beyod} Pr{ out of 7 beyod} Pr{7 out of 7 beyod} 7 0 7 - Pr{0 out of 7 beyod} (0.007) ( 0.007) 0 0.0875 0.985 Rule : Pr{out of cotroli cotrol} Pr{all 7 o oe side} (0.5) 0.0565 7 ISyE 5 Istructor: Kaibo Liu
Type I ad Type II error calculatio uder differet decisio rules (b) Rule : Sice ormal distributio is symmetric, the error should be the same o matter the mea shifts up or dow. Discussio o is eough. 3 / 3 / 0 Pr{sample withi out of cotrol} ( ) ( ) / / (3 5) ( 3 5) (0.76) 0.7764 (If, we have the same 0.) Pr{i cotrol out of cotrol} Pr{7 withi} ( ) 0.7764 7 7 0 Rule : Agai we oly discuss. Pr{fall i upper side} 0 ( ) ( 5) / (.3) 0.087 0.9873 0 Pr{fall i lower side} ( ) ( 5) / (.3) 0.087 0.7 Pr{i cotrol out of cotrol} Pr{all 7 o oe side} (0.9873 7 0.087 7 ) 0.0867 ISyE 5 Istructor: Kaibo Liu
ISyE 5 Istructor: Kaibo Liu 49