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实验设计与数据处理课后答案

《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H0:u1-u2≤0和假设H1:u1-u2>0用sas分析结果如下:Sample StatisticsGroup N Mean Std. Dev. Std. Error----------------------------------------------------x 8 0.231875 0.0146 0.0051y 10 0.2097 0.0097 0.0031Hypothesis TestNull hypothesis: Mean 1 - Mean 2 = 0Alternative: Mean 1 - Mean 2 ^= 0If Variances Are t statistic Df Pr > t----------------------------------------------------Equal 3.878 16 0.0013Not Equal 3.704 11.67 0.0032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 = 1Alternative: Variance 1 / Variance 2 ^= 1- Degrees of Freedom -F Numer. Denom. Pr > F----------------------------------------------2.27 7 9 0.2501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: ySum ofSource DF Squares Mean Square F Value Pr > FModel 4 1480.823000 370.205750 40.88 <.0001Error 15 135.822500 9.054833Corrected Total 19 1616.645500R-Square Coeff Var Root MSE y Mean0.915985 13.12023 3.009125 22.93500Source DF Anova SS Mean Square F Value Pr > Fc 4 1480.823000 370.205750 40.88 <.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 ofSource DF Squares Mean Square F Value Pr > FModel 11 82.8333333 7.5303030 1.39 0.2895Error 12 65.0000000 5.4166667Corrected Total 23 147.8333333R-Square Coeff Var Root MSE R Mean0.560316 22.34278 2.327373 10.41667Source DF Type I SS Mean Square F Value Pr > Fm 2 44.33333333 22.16666667 4.09 0.0442n 3 11.50000000 3.83333333 0.71 0.5657m*n 6 27.00000000 4.50000000 0.83 0.5684Source DF Type III SS Mean Square F Value Pr > Fm 2 44.33333333 22.16666667 4.09 0.0442n 3 11.50000000 3.83333333 0.71 0.5657m*n 6 27.00000000 4.50000000 0.83 0.5684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。

4-4 解:(1) 不用协变量做方差分析由分析结果可知,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。

(2) 引入协变量作方差分析由分析结果可见,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L 9(34)确定配比试验方案: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091.05.02.03.01.011.0=+++⨯=A 272.05.02.03.01.013.0=+++⨯=B182.05.02.03.01.012.0=+++⨯=C 455.05.02.03.01.015.0=+++⨯=D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:由分析结果得回归方程为:32115000.155000.057500.090000.9x x x y +++= 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:3115000.157500.090000.9x x y ++= 6-9 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 VarianceSum of MeanSource DF Squares Square F Value Pr > F Model 1 76.24389 76.24389 7.45 0.0163 Error 14 143.28371 10.23455 Corrected Total 15 219.52760Parameter StandardVariable Estimate Error Type II SS F Value Pr > F Intercept 8.22980 1.38618 360.74949 35.25 <.0001 t9 0.01056 0.00387 76.24389 7.45 0.0163Bounds on condition number: 1, 1------------------------------------------------------------------------------------------------------Stepwise Selection: Step 2Variable t13 Entered: R-Square = 0.6717 and C(p) = 84.4265Analysis of VarianceSum of MeanSource DF Squares Square F Value Pr > FModel 2 147.46551 73.73276 13.30 0.0007Error 13 72.06209 5.54324Corrected Total 15 219.52760Parameter StandardVariable Estimate Error Type II SS F Value Pr > FIntercept 18.33483 2.99803 207.32264 37.40 <.0001t9 0.01173 0.00287 92.81733 16.74 0.0013t13 -1.89938 0.52989 71.22162 12.85 0.0033------------------------------------------------------------------------------------------------------Stepwise Selection: Step 3Variable t5 Entered: R-Square = 0.7627 and C(p) = 60.2727Analysis of VarianceSum of MeanSource DF Squares Square F Value Pr > FModel 3 167.42492 55.80831 12.85 0.0005Error 12 52.10268 4.34189Corrected Total 15 219.52760Parameter StandardVariable Estimate Error Type II SS F Value Pr > FIntercept 19.49941 2.70837 225.06452 51.84 <.0001t5 0.00163 0.00076176 19.95941 4.60 0.0532t9 0.00728 0.00328 21.44626 4.94 0.0462t13 -2.19305 0.48856 87.48515 20.15 0.0007Bounds on condition number: 1.8185, 13.825------------------------------------------------------------------------------------------------------All 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 Variable Number Partial ModelStep Entered Removed Vars In R-Square R-Square C(p) F Value Pr > F1 t9 1 0.3473 0.3473 175.752 7.45 0.0163 2 t13 2 0.3244 0.6717 84.4265 12.85 0.0033 3 t5 3 0.0909 0.7627 60.2727 4.60 0.0532由结果可知,y=19.49941+0.0016321x x +0.0072842x x -2.193053x6-10 解:(1)散点图如下:可以采用Logistic 拟合此数据。

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