机械工程专业的《试验设计与数据处理》课程

1、试验设计与数据处理专业：机械工程班级：机械 11 级专硕 学号： S110805035姓名：赵龙第三章：统计推断3- 13解：取假设HO: u1-u2w 0和假设H1: u1-u2 > 0用sas分析结果如下:Sample StatisticsGroupNMeanStd. Dev.Std. Errorx80.2318750.01460.0051y100.20970.00970.0031Hypothesis TestNull hypothesis:Mean 1 - Mean 2= 0Alternative:Mean 1 - Mean 2A= 0If Varianees Aret stati

2、stie DfPr > tEqual3.878160.0013Not Equal3.70411.670.0032由此可见 p 值远小于 0.05，可认为拒绝原假设， 即认为 2 个作家所写的小品文中 由 3 个字母组成的词的比例均值差异显著。3-14解：用sas分析如下：Hypothesis TestNull hypothesis:Variance 1 / Variance 2 = 1Alter native:Varia nee 1 / Varia nee 2 A= 1- Degrees of Freedom -FNumer. Denom.Pr > F2.27 7 9 0.2501

3、由 p 值为 0.2501>0.05（显著性水平），所以接受原假设， 两方差无显著差异第四章：方差分析和协方差分析4- 1 解：Sas分析结果如下：Dependent Variable: ySoureeDFSum ofSquares Mean Square F ValuePr > FSourceDFType I SSMean SquareF ValuePr > Fm244.3333333322.166666674.090.0442n311.500000003.833333330.710.5657m*n627.000000004.500000000.830.5684Source

4、DFType III SSMean SquareF ValuePr > Fm244.3333333322.166666674.090.0442n311.500000003.833333330.710.5657m*n627.000000004.500000000.830.5684Model1480.823000 370.20575040.88 <.0001Error15 135.822500 9.054833Corrected Total19 1616.645500Source cR-Square Coeff Var Root MSE y Mean0.915985 13.12023

5、3.009125 22.93500DF Anova SSMean SquareF ValuePr > F41480.823000370.20575040.88<.0001由结果可知，p 值小于 0.001，故可认为在水平 a=0.05 下，这些百分比的均值有显著差异。4-2 解：The GLM ProcedureDependent Variable: RSum ofSourceDFSquaresMean SquareF ValuePr > FModel1182.83333337.53030301.390.2895Error1265.00000005.4166667Correc

6、ted Total23147.8333333R-Square Coeff VarRoot MSER Mean0.56031622.34278 2.32737310.41667由结果可知， 在不同浓度下得率有显著差异， 在不同温度下得率差异不明显， 交 互作用的效应不显著。4-4 解： (1) 不用协变量做方差分析The GLM ProcedureDepsidait ariable: ySum ofSourceDFSquaresM«aa SquareF ValuePr>FModd3】Q4l 二91箕：347J638S?1.380"<5Eitot205018 166

7、66725O.9OB333C crrrected Total236059.953333R'SquareRoot MSEy皿逊'0117191422.8875415.3JOO?59.20S33S DirreeDFType I SSMsan. SquareF ValuePrFy1330.04166671.320.2650驭1693.37500001.160 112011S.3 7500000.070?S95SourceDFType IH 阴hleaKi SquareF ValuePrFy1330.041667裁9剧弱仇1.32D.26j0驭1693.j7jM>00絵左WQ2.

8、760.112011.S.j?500000 070.7 &9J由分析结果可知，花的品种、温度和两者的交互作用对鲜花产量的影响都是 不显著的。引入协变量作方差分析The GLM ProcedureD epeadent Vari able: yS&urceModdEircfCorrected TotalSuin ofF Value12： 92Pr>F0001DF斗1923SquaiH5832.19036127.1391476059,958333Mean Square145S.l_9n_2I1.9J996OR-Square以喊册RwtNfSEyhkaa0.96250243.45

9、S31869.20833SourceDFType】 SSilean SquareFPr = Fy1350.W16672.605XH1693.37500037 J7<000111S.3"5DOO1 540.23031*19092719400.5S<0001SciirceDFTypelUSSMean SquareF ValuPrFy123.94J2S12.000.1733毋:4798886540.07uDOT：3175.07199S6.1S0.0215由分析结果可见，引入协变量后，v、m、和x对鲜花产量的影响都是显著地第五章：正交试验设计5- 3解:用L9 （34）确定配比试

10、验方案： 试验方案因素试验号ABCD11（ 0.1 份）1（ 0.3 份）1（ 0.2 份）1（ 0.5 份）212（ 0.4 份）2（ 0.1 份）2（ 0.3 份）313（ 0.5 份）3（ 0.1 份）3（ 0.1 份）42（ 0.3 份）123522316231273（ 0.2 份）1328321393321以1号条件为例，表中四个数值的组成比为:A: B： C: D=0.1: 0.3： 0.2： 0.5配比方案中，要求各行四个比值之和为1A =0.10.0910.1 +0.3 十0.2 +0.51。在1号条件中，四种数值分别是1B=0.30.2720.1+0.3 + 0.2 + 0.

11、5= 0.210.10.30.20.5= 0.182=0.510.10.3 0.2 0.5=0.455其余实验条件可按照相同方法得出第六章：回归分析6- 6 解：（1）作线性回归分析结果如下:D epraident 7ari able: yAnalysis of VarianceSum cfSourceDFS quarssModelJ(5.64500Error二1.3300Corrected Total717 02000SquareF ValuePrF5.21500J5170.0110.54375RwtMSE0.58630R-SquareQ91929.90000jR-Sq0.3586Pirmi

12、rtef EjhmattsParameterStandardVarimblmDFEsttmaiErrort ValuePr=- ft|lXLltlCtpt1A900000 2077047.76（X）010.5750002.7?O.OfOl10.550002.650 056S11.15MG007255.55由分析结果得回归方程为：y =9.90000 0.57500xj 0.55000X2 1.15000x3由p值都小于0.1可知，每项都是显著的，方程也是显著的(2)Stepwiae Sei ecu del: Step 3VariableEntered:R-Square = O 9192d Gt

13、pO = 4 0000Analysis of VartanceSum ofMean*-SourceDFSqxLOfesSqiureF ValuePr 才 F315.M5OO5-2150015.170.0119Ertcr41 375000 34375Csrgwd Toxal717 02000ParamecerStaxidvdVnSableErrorTypeWSSF ValuePr =>FIntercept9.900000.20729784080002280.96=00010-575000.207292-645007.690.05010550000.207292-420007.040.05

14、6S序31150000.2072910-5800030730.0052Hoimds on conditBDn number: 1.9All v arsab les left in tiie model are signcficziit a.t tiie 0-1500 leel.All variables have been entered into tiie model-SiirnTi：iar'呂亡即低-乜栏 SelectionVariabledumberPartialModelStepEnteredFLemov ed£5驭.加R- SquireC£p>F V

15、sdueP171x310 15216枣.14.73459.860.0201220l 133-10.77709.04003.4S0.12Q9330.14Z2O 91924 OOOO7 04O.O5&S由分析结果可知，在a=0.05下，仅有x3和x1应当引入方程。故所求方程为：y =9.90000 0.57500X11.15000X36-9解：分析结果如下：Dependent Variable: yStepwise Selection: Step 1Variable t9 Entered: R-Square = 0.3473 and C(p) = 175.7517Analysis of V

16、arianceSum ofMeanSourceDFSquaresSquareF ValuePr > FModel176.2438976.243897.450.0163Error14143.2837110.23455Corrected Total15219.52760ParameterStandardVariableEstimateError Type II SS F ValuePr > FIntercept8.229801.38618360.7494935.25<.0001t90.010560.0038776.243897.450.0163Bounds on conditio

17、n number: 1, 1Stepwise Selection: Step 2Variable t13 Entered: R-Square = 0.6717 and C(p) = 84.4265Analysis of VarianceSum ofMeanSourceDFSquaresSquareF ValuePr > FModel2147.4655173.7327613.300.0007Error1372.062095.54324Corrected Total15219.52760ParameterStandardVariableEstimateError Type II SS F V

18、aluePr > FIntercept18.334832.99803207.32264 37.40<.0001t90.011730.0028792.81733 16.74 0.0013t13-1.899380.5298971.22162 12.85 0.0033Stepwise Selection: Step 3Variable t5 Entered: R-Square = 0.7627 and C(p) = 60.2727Analysis of VarianceSum ofMeanSourceDFSquaresSquare F ValuePr > FModel3167.42

19、49255.8083112.850.0005Error1252.102684.34189Corrected Total15219.52760ParameterStandardVariableEstimateError Type II SS F Value Pr > FIntercept19.499412.70837225.06452 51.84<.0001t50.001630.0007617619.95941 4.600.0532t90.007280.0032821.44626 4.940.0462t13-2.193050.4885687.48515 20.150.0007Boun

20、ds on condition number: 1.8185, 13.825All variables left in the model are significant at the 0.1500 level.No other variable met the 0.1500 significance level for entry into the model.Summary of Stepwise SelectionVariable VariableNumber PartialModelStepEnteredRemovedVars In R-Square R-Square C(p) F V

21、aluePr > F1t910.34730.3473175.7527.450.01632t1320.32440.671784.426512.850.00333t530.09090.762760.27274.600.0532由结果可知，y=19.49941+0.00163xiX2+0.00728x2X4 -2.19305x36-10解：(1)散点图如下：2015 ¥11 24681012-X1可以采用Logistic拟合此数据。(2) 用Logistic模拟结果为：Dependent Variable y Method: Gauss-NewtonSum ofIterbcaSqua

22、res03.71802.000021.00001124.113.64081.849314.8393570.923.54751.668414.9977534.733.47041.520815.2362499.443.40461.396315.4814464.353.34691.288715.7348429.263.29551.194315.9985394.173.24911.110716.2735359.183.20691.036016.5601324.493.16840.969116.8579290.5103.13310.909117.1660257.6113.10080.855117.482

23、9226.3123.07120.806717.8067196.8133.04420.763218.1349169.7143.01950.724218.4653145.1152.99710.689218.7951123.1162.97680.658019.1220104.0172.95840.630019.443887.4241182.94200.605019.758673.4129192.92720.582720.064861.7148202.91400.562820.361052.0895212.90230.545020.646444.2826222.89200.529120.920338.

24、0422232.88280.514921.182233.1304242.87480.502121.431929.3307252.86790.490721.669326.4509262.86180.480421.894624.3243272.85660.471122.107822.8087282.85220.462822.309421.7843292.84840.455322.499521.1514302.84530.448622.678720.8276312.84280.442522.847220.7456WARNING: Step size shows no improvement.WARN

25、ING: PROC NLIN failed to converge.Estimation Summary (Not Converged)MethodGauss-NewtonIterations31Subiterations29Average Subiterations0.935484R0.653053PPC(c)0.012486The NLIN ProcedureEstimation Summary (Not Converged)RPCObject0.00394Objective20.74557Observations Read15Observations Used15Observations

26、 Missing0NOTE: An intercept was not specified for this model.Sum ofMeanApproxSourceDFSquaresSquareF ValuePr > FRegression33245.51081.8625.77 <.0001Residual1220.74561.7288Uncorrected Total153266.3Corrected Total14949.9ApproxParameterEstimateStd ErrorApproximate 95% Confidence Limitsb2.84280.004

27、882.83212.8534c0.44250.003210.43550.4494a22.84720.574921.594724.0998Approximate Correlation Matrixbcab1.00000000.7483922-0.2191675c0.74839221.0000000-0.4085662a-0.2191675-0.40856621.0000000故得'VS ： 1 I： 、.；：； -.'I. I ： 1 第七章：回归正交设计7- 1 解：n=9z 39z 38 5作变换 X12z - 77，贝U x1=1, x2=2, x3=3,，x9=9。0.

28、50.5并可设 y=bO+b1 1(x)+ b22(x)+ b22(x)+ b44(x)对于n=9,查附表6,利用SAS软件进行回归多项式分析，结果如下：根据结果分析b0，b1，b2的P值小于0.05，是显著的，The SAS System 21:18 Vednesdeiy November 14, 20102The ORTHOREG ProcedureDependent Variable： ySourceDFSum ofSquaresMean SquareF ValuePr > FModel41.54309602720.385773756B45.670.0014Error40.0338

29、6052840.0084G51321Corrected Total01.5769555556Rgt USE 0,0920061525R-Square 0,9785279121VariableDFParameter Est imateStandard Errort ValuePr > ItlIntercept12*072222222222220.030666717567.57<.0001wl10 J1950.011877943210.080.0005i210.00174751240.740.0003础1-0.006292929292920,0028241476-2.150.0878w

30、410.00206628220.090,9363b3, b4的P值大于0.05是不显著的，所以只需要配要二次项就行了 当n=9时，有：1(x) = ；(x) =x -57582(x)二 2 2(x) =3 (x2 -10x) =3x2 -30x 3796y = b。6 1(x) b2 2(x)=2.072220.1195 (x - 5)0.01528(3 x2 - 30 x 379)将x = 2z-77代入上式得所求多项式回归方程:2y =304.5489 -14.79452z0.18336z7-3解：本题属于一次回归的正交设计z1 - 310 z2 - 25 z3 - 225 z4 - 90

31、采用一水平，做变换：x1=，x2=，x3=，x4=1052510试验方案试验号x1x2x3x4y'111115.4211-1-13.431-11-10.84-1-1-110.35-111-1-2.56-11-11-17-1-111-3.38-1-1-1-1-3.6程序如下:data E1; input x1-x4 y; cards;11115.411-1-13.41-11-10.81-1-110.3-111-1-2.5-11-11-1-1-111-3.3-1-1-1-1-3.6proc reg data=E1;model y=x1-x4;run;The SAS System22.03

32、Wednesday. November 14, 20101The REQ Procedure Model： MODEL I Dependent Variable： yflnalysis of VarianceSourceOFSin ofSquaresMeanSquareF YaluePr > FModel468*4850017J212512.730.0317Error34.033761.84468CorrectedTotal772.51976Root MSEL159GBR-Square0.8444DependentMean-0.0625CAdj R-Sq0.870JCoeff Var-1

33、 OGG.28872Parameter Est i matesVariableOFParameter Est imateStandardErrort ValuePr > ItlIntercept1-0,062500.40897-DJ50.B9B6xl12.537500*40997G.190*0085x21L387500.409570.0430x810J6250Q.4D9370.40Q.7188x410,412600.409971J10.38B5y" =-0.0625 2.5370& 1.38750X2 0.16250x3 0.41250x4即y -87 =-0.0625

34、 2.5370归310 1.38750空 25 0.16250z3 一225 0.41250z4 90 1052510y =3.822 0.2537z1 0.2775z2 0.0065z3 0.041250z4第八章：均匀设计8- 1解：安排方案如下:.因素试验号Z1Z2Z312.06002423.48002234.85002046.270018第九章：单纯形优化设计9- 1解：(1)反射点E的坐标(5，10，0)3(2) ( i)扩大，a >1 (ii )内收缩，a <0 (iii)收缩，0< a <1(3) 新单纯形各点坐标 B(2，4, 3)，A'1.5,

35、 3, 3.5)，C'(2.5, 2.5, 2.5)，D” (3,3.5， 2)。第十章：析因试验设计10- 4解：分析结果如下:ANOVA for Y1Source CFXIX2X3X1*X1X1*X2X1*X3X2*X2X?*X3X3«X3ModelErrorTotalMaster Model0.4060.4050.3959690.55881110.4050.4D&0.3959690.55681lC.791250.7612&0J63810.42212310.78125P.791250.763610.J221235.611255.611255.4859870

36、.06619715.611255.611255.4859870.06619734.441634.441633.672740.002143134.441634.441633.672740.00214332.4932.4931.764710.002438132.4932.4931.764710.01243826.0126.0125.429360.003958126.012G.0125.429360.00395831.6B0063L6B00630J72850.002579131.880063k6B00630,97235Q.OQ2573109.2025109*2025106.76470.0001461

37、109,2025109. ms10t,764?0,00014$3B.&D1G88.601B87.442110.i)016713801688.501637.B42110.00167J65.209Bf9.4877829 JC9930.0009749£65-205829.467PC2909930.000074&.1141671.02203355.1141671.022933270.32414270.324Predict Ive ModelF Pr > F DFSSMSF Pr > F9 5 4r it St at i st p cs i or Y1Master

38、Mode I Fred ii ct ii ve Mcicl曰 IRsqufareAdj. R-square RMSCCV32.5298. 1 I需94.70X 1.01135? 3.1099392XX2951053.1739CM - -193S41O9 .8 0 1Master ModslPredict ive MadeflINio cffects a 1 i ascd.No> ef f cct® a 1 i asedl.a U j as- St ructure f or ¥ IFred i ct i ve Mode) f 口厂 ¥1Coded Levs H

39、 s(-I>I)：Y1 = 37,43333 - O.225*X1 - 0,3125*X2 - 0,8375X3 - 3,054167n<X1*X1 - 2,85*X1*X? 一 2,55i«Xf»tX3 一 2.929 167X2X2 一 5*225wX2X3 一 3.229 167X3X3Uincoded Leve I s :Y1 = 87.48333 - 0-225n*K1 0.81?5*X2 - 0.8375*K8 - 8.054187*X1 *X1 - 2,05#X1*X2 -2_55X1X3 一 2 S291B7*X2X2 一 5225*X2>

40、t«X3 一 3 - 229167>t«X3*X3Effect Estintes for Y1Master ModelPredictive ModelTerm123233K*x绷*x嚨啪XIK2K3XIXIXIX1X2X3-d.2250.357567-0.829250+556811-C.225O+357567-0.G29Z5i*556811-0,31250.357567-0.0?W0.422128-1.31250.957G87-0.B79960.422123-0.8375O.3575S7-2.84222Q,Q6B187-J.83750.357567-2.342220,

41、0B8197-3.054U70.5263245.00md.002143-3.0541E70.G26924-E.802B20.002143-2*SBD.50567&-5.63IGO20.002480-2.950.506676-5.636020.00243B-2.560.506B7G-5.04:2750,003960-?,550.506676-5.042750.003858-2.91670.52682-G.58533(1.002576-2.3291670.56824-6.565330,002678-5.2250.50567&-10.33270.000140-5.2250.50G97

42、6-10.3327Q.00014G-3,2291670+526324-8.186820J01B7'8.2281B70.526924-6.135320.001S7Estivate Std ErrEat inate Std EnrPr > It I经分析可知，在a=0.05的情况下，一次项均不显者，二次项、交互项均是显 著的。第十二章：多指标综合评价概论12-4 解：记P(U : U：J八，其中优等品10件，一等品30件，二等品40件，20件三等品，所有标准分依次是-U 0.95；-Uo.75；Uo.6；U 0.9，查标准正态表得到标准分依次是：-1.64, -0.67, 0.25;

43、 1.28;第十三章：主成分分析法和因子分析法ppt第一题解：利用SAS软件进行主成分分析主要输出结果如下:EigenvaIues of thw Correlation MeitrixE i阿軀1 ueDifferenceProportion伽ulalive13.104912520.207470900.3061ouaai22.887441B21.987228080.8B220.750330J30215G60.288093290.11S3(M刪4Q.642122260.333038180.03030.94665Q.304084130.217486370.03300J84860刖昭?州Q. 064

44、413380.995770.032184330.02974261D.00400J99780.002441780.00031.000014:38 Monday, December 2010The PRINCOMP ProcedureEijervectorsPrlnlPririZPrin3Prin4PrinSPrinCPrin?卩血Xi0.4766500.2959910.1041900.045803-.184219*.0658540J578190.246000x20.4728080.2779940JB29B3-.174010.305448-.046451-.E1B4130.527105x3D.42

45、38450.3773510J662550.05867D0.017475o.oaao4«-J 74045-.730540x4-.21288!Q.45140G<.0086440.GW088-.5394070.287855-.2434270,220126x5-.3884600.9309450.321133-.1984160.4490990.5822990.2328690.030G23xG-.9524270.4027370.1451440.279257Q.31BS36-J135710.066436-J42366X?0.214S35-J7741B0 J 404590J681B90.418

46、2010J986370,0528420.041160述0.0550340.272738*.8511620.0710550.3222010.12216S0.067111-J03300要求累积方差贡献率85%，所以取前三个作为主成分即可,Prin 1=0.476650x1+0.472808x2+0.423845x3+ 0.055034x8Prin2=0.295991x1+0.277894x2+0.377951x3+0.272736x8Prin3=0.104190x1+0.162983x2+0.156255x3 - -0.891162x8第一主成分，各项数值彼此相差不大，表示各单项指标对综合的经济效

47、益起 着相当的作用，因此第一主成分可以理解为全面能力综合指标；第二主成分，指标x4的系数明显比其它的指示系数大，而且系数为正，表 明第二主成分主要由x4决定；第三主成分，指标x8的系数明显比其它的指示系数大，而且系数为负的, 表明第三主成分主要是反应与x8相反的指标。排序结果为如下:The SAS System 20：40 Friday, Itovener 23T 2010 iPrirtS prinl23 rankprinIZ!PrinZPrinlfls id od- At 1 Ab- cdw _rn- -uu 11 Qu Phu Ab- 1 9L ro 9L- Ou A/- <=* d" OD Ab- Qv no 4" Ru 1 4, s- Qu 1 Ou _FD rD 1 -4 _rQ Qu *u oo Jul JI _u- 4 fio2.78308LH3571.120482.119140.20704帕L101120,