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# Dsgn & Analy ISYE 6413

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This 0 page Class Notes was uploaded by Maryse Thiel on Monday November 2, 2015. The Class Notes belongs to ISYE 6413 at Georgia Institute of Technology - Main Campus taught by Staff in Fall. Since its upload, it has received 41 views. For similar materials see /class/234219/isye-6413-georgia-institute-of-technology-main-campus in Industrial Engineering at Georgia Institute of Technology - Main Campus.

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Date Created: 11/02/15

Unit 2 Experiments with a Single Factor One Way AN OVA Sources Sections 16 to 19 additional materials in these notes on random effects 0 Oneway layout with xed effects 0 Multiple comparisons 0 Quantitative factors and orthogonal polynomials 0 Residual analysis 0 Oneway layout with random effects Oneway layout and AN OVA An Example Re ectance data in pulp experiment each of four operators made ve pulp sheets re ectance was read for each sheet using a brightness tester Randomization assignment of 20 containers of pulp to operators and order of reading Table l Re ectance Data Pulp Experiment Operator A B C D 598 598 607 610 600 602 607 608 608 604 605 606 608 599 609 605 598 600 603 605 Objective determine if there are differences among operators in making sheets and reading brightness Model and ANOVA Model yij ncejz 1kj1ni where yij jth observation with treatment 139 T ith treatment effect 81 error independent N O 62 Model t yij 13 17 7 y sz 7i7 where means average over the particular subscript ANOVA Decomposition k 111 k k n 2 Z Vij 72 2175071 72 Z Z Vij yzt2 iljl i1f1 FTest ANOVA Table Degrees of Sum of Mean Source Freedom d f Squares Squares treatment k l SST 2L1 72 MST SSTdf residual N k SSE 25 2 yij 2 MSE SSE d f total N 1 25 231 yij y 2 The F statistic for the null hypothesis that there is no difference between the treatments ie H039C139cC7 is Zuni y 2k 1 A 2123 yij 72N k MSE which has an F distribution with parameters k 1 and N k 4 AN OVA for Pulp Experiment Degrees of Sum of Mean Source Freedom d f Squares Squares F operator 3 134 0447 420 residual 16 170 0106 total 19 3 04 O PF0bF3716 gt Z p value thus declaring a signi cant operatortooperator difference at level 002 0 Further question among the 6 pairs of operators what pairs show signi cant difference Answer Need to use multiple comparisons Multiple Comparisons For one pair of treatments it is common to use the t test and the 1 statistic 71 7i q 7 where 11 number of observations for treatment 139 2 RS Sdf in ANOVA declare treatments 1 and j different at level 0c if itiji gt tN koc2 Suppose k tests are performed to test H0 t1 Ck Experimentwise error rate EER Probability of declaring at least one pair of treatments signi cantly different under H0 Need to use multiple comparisons to control EER AVSB Avs C Avs D Bvs C Bvs D Cvs D O87 185 214 272 301 029 Bonferroni Method 0 Declare t different from I at level 06 if ltijl gt tNk72 where k no of tests Note error on page 27 one line above 148 k 0 For oneway layout with k treatments k kk l as k 2 increases k increases and the critical value tNk7 gets bigger ie method less powerful in detecting differences 0 Advantage It works without requiring independence assumption 0 For pulp experiment take oc 2005 k 4 k 6 t16700512 23008 Among the 6 ti values see p6 only the 1 value for BvsD 301 is larger Declare B and D different at level 005 Tukey Method Declare t different from Cj at level 06 if itiji gt qu koa where qk7Nk70L is the upper 06 point of the Studentized range distribution with parameter k and N k degrees of freedom See distribution table on p9 Tony Hayter Georgia Tech proved that its EER is at most 06 Proof in 156 not required For pulp experiment 1 l 405 qkw koos Equaoos 286 Again only BVsD has larger ti Value than 286 See p6 Tukey method is more powerful than Bonferroni method because 286 is smaller than 301 why Selected values of qkmoC for 06 005 k V 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 1797 2698 3282 3708 4041 4312 4540 4736 4907 5059 5196 5320 5433 5536 2 608 833 980 1088 1174 1244 1303 1354 1399 1439 1475 1508 1538 1565 3 450 591 682 750 804 848 885 918 946 972 995 1015 1035 1052 4 393 504 576 629 671 705 735 760 783 803 821 837 852 866 5 364 460 522 567 603 633 658 680 699 717 732 747 760 772 6 346 434 490 530 563 590 612 632 649 665 679 692 703 714 7 334 416 468 506 536 561 582 600 616 630 643 655 666 676 8 326 404 453 489 517 540 560 577 592 605 618 629 639 648 9 320 395 441 476 502 524 543 559 574 587 598 609 619 628 10 315 388 433 465 491 512 530 546 560 572 583 593 603 611 11 311 382 426 457 482 503 520 535 549 561 571 581 590 598 12 308 377 420 451 475 495 512 527 539 551 561 571 580 588 13 306 373 415 445 469 488 505 519 532 543 553 563 571 579 14 303 370 411 441 464 483 499 513 525 536 546 555 564 571 15 301 367 408 437 459 478 494 508 520 531 540 549 557 565 16 300 365 405 433 456 474 490 503 515 526 535 544 552 559 0Fupper tail probability Vdegrees of freedom IFnumber of treatments For complete tables corresponding to various values of OC refer to Appendix E OneWay AN OVA with a Quantitative Factor 0 Data 2 bonding strength of composite material a l l laser power at 40 50 60 watt Table 2 Strength Data Composite Experiment Laser Power watts 4O 50 60 2566 2915 3573 2800 3509 3956 2065 2979 3566 OneWay AN OVA Contd Table 3 ANOVA Table Composite Experiment Degrees of Sum of Mean Source Freedom Squares Squares F laser 2 224184 112092 1132 residual 6 59422 9904 total 8 283606 0 Conclusion from ANOVA Laser power has a signi cant effect on strength 0 To further understand the effect use of multiple comparisons is not useful here Why o The effects of a quantitative factor like laser power can be decomposed into linear quadratic etc Linear and Quadratic Effects 0 Suppose there are three levels of x low medium high and the corresponding y values are y1y2y3 yr Linear contrast y3 y1 401 yz y3 y1 Quadratic contrast yl 2y2 y3 12 l y2 y3 l0l and l2l are the linear and quadratic contrast vectors they are orthogonal to each other Linear and Quadratic Effects Contd 0 Using l0l and l2l we can write a more detailed regression model y X 3 8 where the model matrix X is given as in 158 o Normalization 39 Length of l0 l length of 1 2 l V5 divide each vector by its length in the regression model Why It provides a consistent comparison of the regression coef cients But the tstatistics in the next table are independent of such and any scaling Tests for Linear and Quadratic Effects Table 4 Tests for Polynomial Effects Composite Experiment Standard Effect Estimate Error 1 pValue linear 8636 1817 475 0003 quadratic 0381 1817 021 0841 0 Further conclusion Laser power has a signi cant linear but not quadratic effect on strength 0 Another question How to predict yvalue strength at a setting not in the experiment ie other than 40 50 60 Need to extend the concept of linear and quadratic contrast vectors to cover a whole interval for x This requires building a model using polynomials l4 Orthogonal Polynomials For three evenly spaced levels m A m and m A de ne the rst and second degree polynomials P1x xm l0andlforxm AmmA x m 2 2 P2x 3 A l2andlforxm AmmA Therefore P1 x and P2 x are extensions of the linear and quadratic contrast vectors Why Polynomial regression model y 30 31P1x BszX5 8 obtain regression ie least squares estimates 30 3103 31 8636 32 O381 Note B1 and 32 values are same as in Table 4 15 Prediction based on Polynomial Regression Model 0 Fitted model 9 310322 8636P1x 0381P2x6 0 To predict 9 at any x plug in the x on the right side of the regression equation Forx 55 2 50 10m 50A 10 P 55 55 50 1 1 10 2 55 50 2 2 5 P 55 3 2 10 3 4 9 3103228636O3536 O381 05103 342803 Estimation of Linear and Quadratic Effects 2566 2915 3573 2800 3509 3956 2065 2979 3566 l l l l l l l l 1 WE 0 WE WE 0 WE WE 0 WE 1 2 1 1 2 1 1 2 1 50 I31 54 B0 B1 361 denote respectively the intercept the linear effect and the quadratic effect Also let x and xq denote respectively the second and third columns of the X matrix By the least square theory 8 XI X1X y Running a multiple linear regression with response y and predictors x1 and xq we get BO 2 310322 8 8636 361 O381 Estimation of mean response Fitted model is y 310322 8636x1 038lxq where x and xq are the values of the rst and second degree polynomials P1x and P2 To predict the mean strength at laser power 55 W we have 4P1 55 03536 and xq 413255 125 Substituting these in the W2 W6 tted equation we get the predicted value ypred 342803 X Note thatypred x38 where X 1 03536 125 342803 A 95 con dence interval for the mean response at x x0 is Xo ilN p ipozs vX0TXX1X0 From the regression output not in the book 6 xMSE 3147 Thus the 95 con dence interval for the mean response is 282418 408825 Prediction interval for individual response A 95 prediction interval for an individual observation yo corresponding to X X0 is X03 i lN p 100256 1 XoT XX1X0 where 1 under the square root represents 62 variance of the new observation yo The 95 prediction interval for a new ie future observation on strength for laser power of 55 is 245998 445244 Note that this is much wider than the interval for the mean response since it incorporates the variability 6 of individual observations around the mean Residual Analysis Theory 0 Theory de ne the residual for the 1m observation x as 1 i yi 7i 9139 ij v 39 contains information given by the model ri is the difference between y observed and 39 tted and contains information on possible model inadequacy Vector of residuals r 13 1 y X3 0 Under the model assumption E y XB it can be shown that a M o b r and y are independent c variances of 13 are nearly constant for nearly balanced designs Residual Plots Plot ri vs 39 see Figure 1 It should appear as a parallel band around 0 Otherwise it would suggest model violation If spread of 13 increases as 9 increases error variance of y increases with mean of y Need a transformation of y Will be explained in Unit 3 Plot ri vs x see Figure 2 If not a parallel band around 0 relationship between yi and x not fully captured revise the X 3 part of the model Plot ri vs time sequence to see if there is a time trend or autocorrelation over time Plot ri from replicates per treatment to see if error variance depends on treatment residual Plot Of 739139 VS 9139 601 602 603 604 605 606 607 Figure 1 13 vs 39 Pulp Experiment 19 residual 06 04 02 00 02 04 Plot Of 739 VS Xi operator Figure 2 13 vs xi Pulp Experiment 20 BoxWhisker Plot 0 A powerful graphical display due to Tukey to capture the location dispersion skewness and extremity of a distribution See Figure 3 5th 5th 0 Q1 lower quartile 2 percentile Q3 upper quartile 7 percentile Q2 median location is the white line in the box Q1 and Q3 are boundaries of the black box I QR interquartile range length of box Q3 Q1 is a measure of dispersion Minimum and maximum of observed values within Q1 151 QR Q3 151QR are denoted by two whiskers Any values outside the whiskers are outliers and are displayed 0 If Q1 and Q3 are not symmetric around the median it indicates skewness 21 20 20 BOXWhisker Plot r 39 Figure 3 BOXWhisker Plot 22 Normal Probability Plot Original purpose To test if a distribution is normal eg if the residuals follow a normal distribution see Figure 5 More powerful use in factorial experiments will be discussed in Units 4 and 5 Let rm 3 g rm be the ordered residuals The cumulative probability for r is p i O5N Thus the plot ofpl vs rm should be Sshaped as in Figure 4a if the errors are normal By transforming the scale of the horizontal axis the Sshaped curve is straightened to be a line see Figure 4b Normal probability plot of residuals 1Piaquotltz gt a If the errors are normal it should plot roughly as a straight line See ilN CDnormalcdf Figure 5 23 Regular and Normal Probability Plots of Normal CDF a b 08 08 06 06 Figure 4 Normal Plot of ri Pulp Experiment 24 Normal Probability Plot Pulp Experiment 00 04 l l l l l 2 1 0 1 2 normal quantiles Figure 5 Normal Plot of ri Pulp Experiment 25 Pulp Experiment Revisited In the pulp experiment the effects I are called xed effects because the interest was in comparing the four speci c operators in the study If these four operators were chosen randomly from the population of operators in the plant the interest would usually be in the variation among all operators in the population Because the observed data are from operators randomly selected from the population the variation among operators in the population is referred to as random effects Oneway random effects model yij T Ci 817 eij s are independent error terms with N O 62 t are independent N O 6 and t and 81 are independent Why Give an example 62 and 6 are the two variance components of the model The variance among operators in the population is measured by 6 26 Oneway Random Effects Model AN OVA and Variance Components The null hypothesis for the xed effects model t1 Ck should be replaced by H0 6 0 Under H0 the F test and the ANOVA table in Section 16 still holds Reason under H0 SST N oleil and SSE N ozszvk Therefore the F test has the distribution Fk17Nk under H0 We can apply the same ANOVA and F test in the xed effects case for analyzing data For example using the results in Section 16 the F test has value 42 and thus H0 is rejected at level 005 However we need to compute the expected mean squares under the alternative of 6 gt O i for sample size determination and ii to estimate the variance components 27 Expected Mean Squares for Treatments 0 Equation 1 holds independent of 6 SSE 2 E MSE N k 6 0 Under the alternative 6 gt O and for 10 n SST EMST E k1 62n6 For unequal ni s n in 2 is replaced by 1 k 2 1 112 i n k1 n k 11 211 139 28 1 2 Proof of 2 The cross product term has mean 0 because I and 8 are independent It can be shownthat Ei cl C2 k 1c and E gig 2 k n152 MST quotk 1gtcltk 1oz T EWST EfS 1ZGZHG 29 AN OVA Tables nl n Source df SS MS EMS treatment k 1 SST MST EE Tl 62 716 residual N k SSE MSE 2 52 total N 1 Pulp Experiment Source df SS MS EMS treatment 3 134 0447 52 55 residual 16 170 0106 52 total 19 3 04 3O Estimation of 62 and 6 From equations 1 and 2 we obtain the following unbiased estimates of the variance components MST MSE n A2 61 62 MSE and Note that 6 2 0 if and only if MST 2 MSE which is equivalent to F 2 1 Therefore a negative variance estimate 6 occurs only if the value of the F statistic is less than 1 Obviously the null hypothesis H0 is not rejected when F g 1 Since variance cannot be negative a negative variance estimate is replaced by O This does not mean that 6 is zero It simply means that there is not enough information in the data to get a good estimate of 6 For the pulp experiment n 5 62 0106 6 0447 0106 5 0068 ie sheettosheet variance within same operator is 0106 which is about 50 higher than operatortooperator variance 0068 Implications on process improvement discuss in class try to reduce the two sources of variation also considering costs 31 Estimation of Overall Mean 1 o In random effects model T the population mean is often of interest From E yij T we use the estimate y A 9 62 k 0 Varm Var39c 8 whereN Zi1n A 2 2 For 11 n Varm 6 l g k n 1k2 1169 Using 2 Mn is an unbiased estimate of Var Con dence interval for T A MST ilk 1g nk o In the pulp experiment 6040 MST 0447 and the 95 con dence interval for T is O447 6040i3182 59926088 5 X 4 32

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