试验与数据处理作业.docx

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试验与数据处理作业.docx

试验与数据处理作业

实验设计与数据处理课后题作业

机械工程包装工程09硕任艳玲S0

习题3-6

(1)已知σ=0.6,

SAS处理结果:

SampleStatisticsforx

NMeanStd.Dev.Std.Error

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

96.000.570.19

HypothesisTest

Nullhypothesis:

Meanofx=0

Alternative:

Meanofx^=0

Withaspecifiedknownstandarddeviationof0.6

ZStatisticProb>Z

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

30.000<.0001

95%ConfidenceIntervalfortheMean

LowerLimitUpperLimit

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

5.616.39

所以,μ的置信度为0.95的置信区间为(5.61,6.39)。

3-6

(2)σ未知,

SampleStatisticsforx

NMeanStd.Dev.Std.Error

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

96.000.570.19

HypothesisTest

Nullhypothesis:

Meanofx=0

Alternative:

Meanofx^=0

tStatisticDfProb>t

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

31.3348<.0001

95%ConfidenceIntervalfortheMean

LowerLimit:

5.56

UpperLimit:

6.44

所以,μ的置信度为0.95的置信区间为(5.56,6.44)。

习题3-7

(1)

SampleStatisticsforx

NMeanStd.Dev.Std.Error

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

66.680.000.00

HypothesisTest

Nullhypothesis:

Meanofx=0

Alternative:

Meanofx^=0

tStatisticDfProb>t

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

4228.3445<.0001

90%ConfidenceIntervalfortheMean

LowerLimit:

6.67

UpperLimit:

6.68

因此,μ的置信度为0.9的置信区间(6.67,6.68)。

SampleStatisticsforx

NMeanStd.Dev.Variance

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

66.67820.003915E-6

HypothesisTest

Nullhypothesis:

Varianceofx=1

Alternative:

Varianceofx^=1

Chi-squareDfProb

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

0.0005<.0001

90%ConfidenceIntervalfortheVariance

LowerLimitUpperLimit

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

676E-80.0001

因此,σ的置信度为0.9的置信区间(676E-8,0.0001)。

(2)SampleStatisticsforx

NMeanStd.Dev.Std.Error

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

56.660.000.00

HypothesisTest

Nullhypothesis:

Meanofx=0

Alternative:

Meanofx^=0

tStatisticDfProb>t

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

4967.0524<.0001

90%ConfidenceIntervalfortheMean

LowerLimit:

6.66

UpperLimit:

6.67

因此,μ置信度为0.9的的置信区间(6.66,6.67)。

SampleStatisticsforx

NMeanStd.Dev.Variance

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

56.6640.0039E-6

HypothesisTest

Nullhypothesis:

Varianceofx=1

Alternative:

Varianceofx^=1

Chi-squareDfProb

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

0.0004<.0001

90%ConfidenceIntervalfortheVariance

LowerLimitUpperLimit

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

379E-8507E-7

因此,σ的置信度为0.9的置信区间(379E-8,507E-7)。

习题3-10SampleStatisticsforx

NMeanStd.Dev.Std.Error

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

53.250.010.01

HypothesisTest

Nullhypothesis:

Meanofx=3.25

Alternative:

Meanofx^=3.25

tStatisticDfProb>t

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

0.34340.7489

说明:

因为在t检验中p-value值0.7489>0.01(显著性水平),所以接受原假设,即这批矿砂的镍含量的均值为3.25。

习题3-11

解:

按题意需检验假设

H0:

μ≥1000,H1:

μ<1000

这是左边检验问题,

=950,μ0=1000

U=(

-μ0)/(σ/

)=-2.5<-μ0.95=-1.645,U的值落在拒绝域内,所以我们在显著性水平0.05下拒绝H0,即这批元件不合格。

习题3-13

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

说明:

因为在t检验中p-value值0.0013<0.05(显著性水平),所以拒绝原假设,即认为两个作家所写的小品文中包含由3个字母组成的词的比例有显著的差异。

习题3-14

SampleStatistics

GroupNMeanStd.Dev.Variance

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

x80.2318750.01460.000212

y100.20970.00970.000093

HypothesisTest

Nullhypothesis:

Variance1/Variance2=1

Alternative:

Variance1/Variance2^=1

-DegreesofFreedom-

FNumer.Denom.Pr>F

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

2.27790.2501

说明:

因为在F检验中p-value值0.2501>0.05(显著性水平),所以接受原假设,即认为两总体方差无显著性差异。

习题4-1

TheANOVAProcedure

ClassLevelInformation

ClassLevelsValues

c512345

Numberofobservations20

TheANOVAProcedure

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

由于Pr>F,且概率小于0.0001,即小于0.01,说明这些抗生素百分比的均值有高度显著差异。

习题4-2

TheGLMProcedure

ClassLevelInformation

ClassLevelsValues

浓度C3123

温度T41234

Numberofobservations24

TheGLMProcedure

DependentVariable:

R

Sumof

SourceDFSquaresMeanSquareFValuePr>F

Model1182.83333337.53030301.390.2895

Error1265.00000005.4166667

CorrectedTotal23147.8333333

R-SquareCoeffVarRootMSEMean

0.56031622.342782.32737310.41667

SourceDFFactoidSSMeanSquareFValuePr>F

C244.22.4.090.0442

T311.3.0.710.5657

C*T627.000000004.0.830.5684

由于浓度C的Pr>F,且概率为0.0442,小于0.05,说明浓度C的作用是显著的。

而温度T的Pr为0.5657,大于0.05,

C*T的Pr为0.5684,大于0.05,说明温度与温度和浓度的交互作用都是不显著的。

习题5-3解:

用L9(34)确定配比试验方案。

如表所示

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)

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

A=0.1ⅹ

=0.091

B=0.3ⅹ

=0.272

C=0.2ⅹ

=0.182

D=0.5ⅹ

=0.455

其余实验条件可按照相同方法得出。

习题6-5

(1)散点图如下:

(2)令t1=x;t2=x2,则转化为线性回归模型y=b0+b1t1+b2t2+ε

取因变量为三次抗压强度之和,即y=

SAS分析结果如下:

由结果可得模型方程为:

=19.0200+1.0149t1-0.0206t2,即回归方程为

=19.0200+1.0149x-0.0206x2+ε。

习题6-6

(1)采用逐步回归法,SAS运行结果如下:

TheREGProcedure

Model:

MODEL1

DependentVariable:

y

StepwiseSelection:

Step1

Variablex3Entered:

R-Square=0.6216andC(p)=14.7345

AnalysisofVariance

SumofMean

SourceDFSquaresSquareFValuePr>F

Model110.5800010.580009.860.0201

Error66.440001.07333

CorrectedTotal717.02000

ParameterStandard

VariableEstimateErrorTypeIISSFValuePr>F

Intercept9.900000.36629784.08000730.51<.0001

x31.150000.3662910.580009.860.0201

Boundsonconditionnumber:

1,1

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

StepwiseSelection:

Step2

Variablex1Entered:

R-Square=0.7770andC(p)=9.0400

AnalysisofVariance

SumofMean

SourceDFSquaresSquareFValuePr>F

Model213.225006.612508.710.0235

Error53.795000.75900

CorrectedTotal717.02000

ParameterStandard

VariableEstimateErrorTypeIISSFValuePr>F

Intercept9.900000.30802784.080001033.04<.0001

x10.575000.308022.645003.480.1209

x31.150000.3080210.5800013.940.0135

TheREGProcedure

Model:

MODEL1

DependentVariable:

y

StepwiseSelection:

Step2

Boundsonconditionnumber:

1,4

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

StepwiseSelection:

Step3

Variablex2Entered:

R-Square=0.9192andC(p)=4.0000

AnalysisofVariance

SumofMean

SourceDFSquaresSquareFValuePr>F

Model315.645005.2150015.170.0119

Error41.375000.34375

CorrectedTotal717.02000

ParameterStandard

VariableEstimateErrorTypeIISSFValuePr>F

Intercept9.900000.20729784.080002280.96<.0001

x10.575000.207292.645007.690.0501

x20.550000.207292.420007.040.0568

x31.150000.2072910.5800030.780.0052

Boundsonconditionnumber:

1,9

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

Allvariablesleftinthemodelaresignificantatthe0.1500level.

Allvariableshavebeenenteredintothemodel.

SummaryofStepwiseSelection

VariableVariableNumberPartialModel

StepEnteredRemovedVarsInR-SquareR-SquareC(p)FValuePr>F

1x310.62160.621614.73459.860.0201

2x120.15540.77709.04003.480.1209

3x230.14220.91924.00007.040.0568

在а=0.1下,回归方程为:

=9.90+0.575x1+0.550x2+1.150x3。

(2)采用逐步回归法,SAS运行结果如下:

TheREGProcedure

Model:

MODEL1

DependentVariable:

y

StepwiseSelection:

Step1

Variablex3Entered:

R-Square=0.6216andC(p)=14.7345

AnalysisofVariance

SumofMean

SourceDFSquaresSquareFValuePr>F

Model110.5800010.580009.860.0201

Error66.440001.07333

CorrectedTotal717.02000

ParameterStandard

VariableEstimateErrorTypeIISSFValuePr>F

Intercept9.900000.36629784.08000730.51<.0001

x31.150000.3662910.580009.860.0201

Boundsonconditionnumber:

1,1

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

Allvariablesleftinthemodelaresignificantatthe0.0500level.

Noothervariablemetthe0.0500significancelevelforentryintothemodel.

SummaryofStepwiseSelection

VariableVariableNumberPartialModel

StepEnteredRemovedVarsInR-SquareR-SquareC(p)FValuePr>F

1x310.62160.621614.73459.860.0201

在а=0.05下,回归方程为:

=9.90+1.150x3。

习题6-9TheREGProcedure

Model:

MODEL1

DependentVariable:

y

StepwiseSelection:

Step1

Variablex2Entered:

R-Square=0.2872andC(p)=12.2386

AnalysisofVariance

SumofMean

SourceDFSquaresSquareFValuePr>F

Model163.0471263.047125.640.0324

Error14156.4804811.17718

CorrectedTotal15219.52760

ParameterStandard

VariableEstimateErrorTypeIISSFValuePr>F

Intercept7.960381.64304262.3643923.470.0003

x20.086140.0362763.047125.640.0324

Boundsonconditionnumber:

1,1

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

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