试验与数据处理作业.doc

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 S090801010习题3-6（1）已知=0.6， SAS处理结果：Sample Statistics for x N Mean Std. Dev. Std. Error - 9 6.00 0.57 0.19 Hypothesis Test Null hypothesis: Mean of x = 0 Alternative: Mean of x = 0 With a specified known standard deviation of 0.6 Z Statistic Prob Z - - 30.000 t - 31.334 8 t - 4228.344 5 .0001 90 % Confidence Interval for the Mean Lower Limit: 6.67 Upper Limit: 6.68因此，的置信度为0.9的置信区间 （6.67，6.68）。 Sample Statistics for x N Mean Std. Dev. Variance - 6 6.6782 0.0039 15E-6 Hypothesis Test Null hypothesis: Variance of x = 1 Alternative: Variance of x = 1 Chi-square Df Prob - 0.000 5 t - 4967.052 4 .0001 90 % Confidence Interval for the Mean Lower Limit: 6.66Upper Limit: 6.67因此，置信度为0.9的的置信区间 （6.66，6.67）。 Sample Statistics for x N Mean Std. Dev. Variance - 5 6.664 0.003 9E-6 Hypothesis Test Null hypothesis: Variance of x = 1 Alternative: Variance of x = 1 Chi-square Df Prob - 0.000 4 t - 0.343 4 0.7489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H0：1000，H1：1000这是左边检验问题， =950，0=1000U=（-0）/（/）=-2.5 t - Equal 3.878 16 0.0013 Not Equal 3.704 11.67 0.0032说明：因为在t 检验中p-value 值0.0013 F - 2.27 7 9 0.2501说明：因为在F检验中p-value 值0.25010.05 (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 The ANOVA Procedure Class Level Information Class Levels Values c 5 1 2 3 4 5 Number of observations 20 The ANOVA ProcedureDependent Variable: y Sum of Source DF Squares Mean Square F Value Pr F Model 4 1480.823000 370.205750 40.88 F c 4 1480.823000 370.205750 40.88 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 The GLM Procedure Class Level Information Class Levels Values 浓度C 3 1 2 3 温度T 4 1 2 3 4 Number of observations 24 The GLM ProcedureDependent Variable: R Sum of Source DF Squares Mean Square F Value Pr F Model 11 82.8333333 7.5303030 1.39 0.2895 Error 12 65.0000000 5.4166667 Corrected Total 23 147.8333333 R-Square Coeff Var Root MSE Mean 0.560316 22.34278 2.327373 10.41667 Source DF Factoid SS Mean Square F Value Pr F C 2 44.33333333 22.16666667 4.09 0.0442 T 3 11.50000000 3.83333333 0.71 0.5657 C* T 6 27.00000000 4.50000000 0.83 0.5684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ABCD1234567891（0.1份）112（0.3份）223（0.2份）331（0.3份）2（0.4份）3（0.5份）1231231（0.2份）2（0.1份）3（0.1份）2313121（0.5份）2（0.3份）3（0.1）312231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=0.1=0.091B=0.3=0.272C=0.2=0.182D=0.5=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+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+。习题6-6（1）采用逐步回归法，SAS 运行结果如下： The REG Procedure Model: MODEL1 Dependent Variable: y Stepwise Selection: Step 1 Variable x3 Entered: R-Square = 0.6216 and C(p) = 14.7345 Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr F Model 1 10.58000 10.58000 9.86 0.0201 Error 6 6.44000 1.07333 Corrected Total 7 17.02000 Parameter Standard Variable Estimate Error Type II SS F Value Pr F Intercept 9.90000 0.36629 784.08000 730.51 F Model 2 13.22500 6.61250 8.71 0.0235 Error 5 3.79500 0.75900 Corrected Total 7 17.02000 Parameter Standard Variable Estimate Error Type II SS F Value Pr F Intercept 9.90000 0.30802 784.08000 1033.04 F Model 3 15.64500 5.21500 15.17 0.0119 Error 4 1.37500 0.34375 Corrected Total 7 17.02000 Parameter Standard Variable Estimate Error Type II SS F Value Pr F Intercept 9.90000 0.20729 784.08000 2280.96 F 1 x3 1 0.6216 0.6216 14.7345 9.86 0.0201 2 x1 2 0.1554 0.7770 9.0400 3.48 0.1209 3 x2 3 0.1422 0.9192 4.0000 7.04 0.0568在=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： The REG Procedure Model: MODEL1 Dependent Variable: y Stepwise Selection: Step 1 Variable x3 Entered: R-Square = 0.6216 and C(p) = 14.7345 Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr F Model 1 10.58000 10.58000 9.86 0.0201 Error 6 6.44000 1.07333 Corrected Total 7 17.02000 Parameter Standard Variable Estimate Error Type II SS F Value Pr F Intercept 9.90000 0.36629 784.08000 730.51 F 1 x3 1 0.6216 0.6216 14.7345 9.86 0.0201在=0.05下，回归方程为：=9.90+1.150x3。习题6-9 The REG Procedure Model: MODEL1 Dependent Variable: y Stepwise Selection: Step 1 Variable x2 Entered: R-Square = 0.2872 and C(p) = 12.2386 Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr F Model 1 63.04712 63.04712 5.64 0.0324 Error 14 156.48048 11.17718 Corrected Total 15 219.52760 Parameter Standard Variable Estimate Error Type II SS F Value Pr F Intercept 7.96038 1.64304 262.36439 23.47 0.0003 x2 0.08614 0.03627 63.04712 5.64 0.0324 Bounds on condition number: 1, 1- Stepwise Selection: Step 2 Variable x3 Entered: R-Square = 0.5361 and C(p) = 5.7737 Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr F Model 2 117.69530 58.84765 7.51 0.0068 Error 13 101.83230 7.83325 Corrected Total 15 219.52760 Parameter Standard Variable Estimate Error Type II SS F Value Pr F Intercept 17.05188 3.70671 165.77106 21.16 0.0005 x2 0.08614 0.03036 63.04712 8.05 0.0140 x3 -1.65300 0.62583 54.64818 6.98 0.0203 Bounds on condition number: 1, 4- All variables left in the model are significant at the 0.0500 level. The REG Procedure Model: M