试验设计与数据处理作业doc.docx

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试验设计与数据处理作业doc

试验设计与数据处理作业

姓名:

班级:

学号:

3-7

SAS分析结果如下:

金球:

置信度为0.9的置信区间分别为:

铂球:

置信度为0.9的置信区间分别为:

3-13

取假设H0:

u1-u2≤0和假设H1:

u1-u2>0用sas分析结果如下:

SampleStatistics

GroupNMeanStd.Dev.Std.Error

----------------------------------------------------

x80.2318750.01460.0051

y100.20970.00970.0031

    HypothesisTest

Nullhypothesis:

Mean1-Mean2=0

Alternative:

Mean1-Mean2^=0

IfVariancesAretstatisticDfPr>t

----------------------------------------------------

Equal3.878160.0013

NotEqual3.70411.670.0032

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

用sas分析如下:

HypothesisTest

Nullhypothesis:

Variance1/Variance2=1

Alternative:

Variance1/Variance2^=1

-DegreesofFreedom-

FNumer.Denom.Pr>F

----------------------------------------------

2.27790.2501

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

4-1

Sas分析结果如下:

DependentVariable:

y

Sumof

SourceDFSquaresMeanSquareFValuePr>F

Model41480.823000370.20575040.88<.0001

Error15135.8225009.054833

CorrectedTotal191616.645500

R-SquareCoeffVarRootMSEyMean

0.91598513.120233.00912522.93500

SourceDFAnovaSSMeanSquareFValuePr>F

c41480.823000370.20575040.88<.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

SAS分析如下:

TheGLMProcedure

DependentVariable:

R

Sumof

SourceDFSquaresMeanSquareFValuePr>F

Model1182.83333337.53030301.390.2895

Error1265.00000005.4166667

CorrectedTotal23147.8333333

R-SquareCoeffVarRootMSERMean

0.56031622.342782.32737310.41667

SourceDFTypeISSMeanSquareFValuePr>F

m244.3333333322.166666674.090.0442

n311.500000003.833333330.710.5657

m*n627.000000004.500000000.830.5684

SourceDFTypeIIISSMeanSquareFValuePr>F

m244.3333333322.166666674.090.0442

n311.500000003.833333330.710.5657

m*n627.000000004.500000000.830.5684

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。

5-3

用L9(34)确定配比试验方案:

试验方案

因素

试验号

A

B

C

D

1

2

3

4

5

6

7

8

9

1(0.1份)

1

1

2(0.3份)

2

2

3(0.2份)

3

3

1(0.3份)

2(0.4份)

3(0.5份)

1

2

3

1

2

3

1(0.2份)

2(0.1份)

3(0.1份)

2

3

1

3

1

2

1(0.5份)

2(0.3份)

3(0.1份)

3

1

2

2

3

1

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1。

在1号条件中,四种数值分别是

其余实验条件按相同方法可得。

6-5

(1)作出散点图如下:

(2)SAS分析结果如下:

DependentVariable:

y

AnalysisofVariance

SumofMean

SourceDFSquaresSquareFValuePr>F

Model238.9371419.468579.540.0033

Error1224.476192.03968

CorrectedTotal1463.41333

RootMSE1.42817R-Square0.6140

DependentMean30.03333AdjR-Sq0.5497

CoeffVar4.75530

ParameterEstimates

ParameterStandard

VariableDFEstimateErrortValuePr>|t|

Intercept119.033333.277555.81<.0001

t111.008570.356432.830.0152

t21-0.020380.00881-2.310.0393

故回归方程为:

6-6

(1)作线性回归分析结果如下:

由分析结果得回归方程为:

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)

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:

6-9

分析结果如下:

DependentVariable:

y

StepwiseSelection:

Step1

Variablet9Entered:

R-Square=0.3473andC(p)=175.7517

AnalysisofVariance

SumofMean

SourceDFSquaresSquareFValuePr>F

Model176.2438976.243897.450.0163

Error14143.2837110.23455

CorrectedTotal15219.52760

ParameterStandard

VariableEstimateErrorTypeIISSFValuePr>F

Intercept8.229801.38618360.7494935.25<.0001

t90.010560.0038776.243897.450.0163

Boundsonconditionnumber:

1,1

------------------------------------------------------------------------------------------------------

StepwiseSelection:

Step2

Variablet13Entered:

R-Square=0.6717andC(p)=84.4265

AnalysisofVariance

SumofMean

SourceDFSquaresSquareFValuePr>F

Model2147.4655173.7327613.300.0007

Error1372.062095.54324

CorrectedTotal15219.52760

ParameterStandard

VariableEstimateErrorTypeIISSFValuePr>F

Intercept18.334832.99803207.3226437.40<.0001

t90.011730.0028792.8173316.740.0013

t13-1.899380.5298971.2216212.850.0033

------------------------------------------------------------------------------------------------------

StepwiseSelection:

Step3

Variablet5Entered:

R-Square=0.7627andC(p)=60.2727

AnalysisofVariance

SumofMean

SourceDFSquaresSquareFValuePr>F

Model3167.4249255.8083112.850.0005

Error1252.102684.34189

CorrectedTotal15219.52760

ParameterStandard

VariableEstimateErrorTypeIISSFValuePr>F

Intercept19.499412.70837225.0645251.84<.0001

t50.001630.0007617619.959414.600.0532

t90.007280.0032821.446264.940.0462

t13-2.193050.4885687.4851520.150.0007

Boundsonconditionnumber:

1.8185,13.825

------------------------------------------------------------------------------------------------------

Allvariablesleftinthemodelaresignificantatthe0.1500level.

Noothervariablemetthe0.1500significancelevelforentryintothemodel.

SummaryofStepwiseSelection

VariableVariableNumberPartialModel

StepEnteredRemovedVarsInR-SquareR-SquareC(p)FValuePr>F

1t910.34730.3473175.7527.450.0163

2t1320.32440.671784.426512.850.0033

3t530.09090.762760.27274.600.0532

由结果可知,y=19.49941+0.00163

+0.00728

-2.19305

6-10

(1)散点图如下:

可以采用Logistic拟合此数据。

(2)用Logistic模拟结果为:

DependentVariabley

Method:

Gauss-Newton

Sumof

IterbcaSquares

03.71802.000021.00001124.1

13.64081.849314.8393570.9

23.54751.668414.9977534.7

33.47041.520815.2362499.4

43.40461.396315.4814464.3

53.34691.288715.7348429.2

63.29551.194315.9985394.1

73.24911.110716.2735359.1

83.20691.036016.5601324.4

93.16840.969116.8579290.5

103.13310.909117.1660257.6

113.10080.855117.4829226.3

123.07120.806717.8067196.8

133.04420.763218.1349169.7

143.01950.724218.4653145.1

152.99710.689218.7951123.1

162.97680.658019.1220104.0

172.95840.630019.443887.4241

182.94200.605019.758673.4129

192.92720.582720.064861.7148

202.91400.562820.361052.0895

212.90230.545020.646444.2826

222.89200.529120.920338.0422

232.88280.514921.182233.1304

242.87480.502121.431929.3307

252.86790.490721.669326.4509

262.86180.480421.894624.3243

272.85660.471122.107822.8087

282.85220.462822.309421.7843

292.84840.455322.499521.1514

302.84530.448622.678720.8276

312.84280.442522.847220.7456

WARNING:

Stepsizeshowsnoimprovement.

WARNING:

PROCNLINfailedtoconverge.

EstimationSummary(NotConverged)

MethodGauss-Newton

Iterations31

Subiterations29

AverageSubiterations0.935484

R0.653053

PPC(c)0.012486

TheNLINProcedure

EstimationSummary(NotConverged)

RPC.

Object0.00394

Objective20.74557

ObservationsRead15

ObservationsUsed15

ObservationsMissing0

NOTE:

Aninterceptwasnotspecifiedforthismodel.

SumofMeanApprox

SourceDFSquaresSquareFValuePr>F

Regression33245.51081.8625.77<.0001

Residual1220.74561.7288

UncorrectedTotal153266.3

CorrectedTotal14949.9

Approx

ParameterEstimateStdErrorApproximate95%ConfidenceLimits

b2.84280.004882.83212.8534

c0.44250.003210.43550.4494

a22.84720.574921.594724.0998

ApproximateCorrelationMatrix

bca

b1.00000000.7483922-0.2191675

c0.74839221.0000000-0.4085662

a-0.2191675-0.40856621.0000000

故得

7-1

n=9

作变换

,则x1=1,x2=2,x3=3,…,x9=9。

并可设

=b0+b1Ф1(x)+b2Ф2(x)+b2Ф2(x)+b4Ф4(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,

b3,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:

代入上式得所求多项式回归方程:

7-3

本题属于一次回归的正交设计

采用二水平,做变换:

x1=

,x2=

,x3=

,x4=

试验方案

试验号

x1

x2

x3

x4

y’

1

2

3

4

5

6

7

8

1

1

1

-1

-1

-1

-1

-1

1

1

-1

-1

1

1

-1

-1

1

-1

1

-1

1

-1

1

-1

1

-1

-1

1

-1

1

1

-1

5.4

3.4

0.8

0.3

-2.5

-1

-3.3

-3.6

程序如下:

dataE1;

inputx1-x4y;

cards;

11115.4

11-1-13.4

1-11-10.8

1-1-110.3

-111-1-2.5

-11-11-1

-1-111-3.3

-1-1-1-1-3.6

;

procregdata=E1;

modely=x1-x4;

run;

8-1

安排方案如下:

因素

试验号

Z1

Z2

Z3

1

2.0

600

24

2

3.4

800

22

3

4.8

500

20

4

6.2

700

18

9-1

(1)反射点E的坐标(5,

,0)

(2)(i)扩大,а>1(ii)内收缩,а<0(iii)收缩,0<а<1

(3)新单纯形各点坐标B(2,4,3),A’(1.5,3,3.5),C’(2.5,2.5,2.5),D”(3,3.5,2)

10-4

分析结果如下:

经分析可知,在a=0.05的情况下,一次项均不显著,二次项、交互项均是显著的。

 

12-1PPT上习题

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

,查标准正态表得到标准分依次是:

-1.64,-0.67,0.25;1.28

14-3

SAS分析结果如下:

分成9类:

3,5一类,其他各为一类;

分成8类:

1,3,5一类,其他各为一类;

分成7类:

1,3,5一类,8,10一类,其它各为一类;

分成6类:

1,3,5,6一类,8,10一类,其他各为一类;

分成5类:

1,3,5,6一类,8,9,10一类,其他各为一类;

分成4类:

1,2,3,5,6一类,8,9,10一类,其他各为一类;

分成3类:

1,2,3,5,6,8,9,10一类,4,7各为一类;

分成2类:

1,2,3,4,5,6,8,9,10一类,7为一类。

14PPT补充

15

单指标评价结果归一化处理得:

权向量a=(0.300.250.150.200.10)

a·R=(0.1333,0.4833,0.2223,0.1611)

(1)

根据最大隶属度原则0.48333所对应的评语为一般。

根据秩加权平均原则,用1、2、3、4分别代表差、一般、良好、优

A=1×0.1611+2×0.2223+3×0.4833+4×0.1333=2.588

位于一般与良好之间,且该食品一般偏良

(2)

100×0.1611+80×0.2223+60×0.4833+40×0.1333=68.2245

评分值为:

P=68.2245分

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