1、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: 3-7SAS分析结果如下:金球:及置信度为0.9的置信区间分别为: 铂球:及置信度为0.9的置信区间分别为: 3-13取假设H0:u1-u20和假设H1:u1-u20用sas分析结果如下: Sample Statistics Group N Mean Std. Dev. Std. Error - x 8 0.231875 0.0146 0.0051 y 10 0.2097 0.0097 0.0031 Hypothesis Test Null hypothesis: Mean 1 - Mean 2 = 0 Alternat
2、ive: Mean 1 - Mean 2 = 0 If Variances Are t statistic Df Pr t - Equal 3.878 16 0.0013 Not 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用sas分析如下: Hypothesis Test Null hypothesis: Variance 1 / Variance 2 = 1 Alternative: Variance 1 / Variance 2 = 1 - Degrees of F
3、reedom - F Numer. Denom. Pr F - 2.27 7 9 0.2501由p值为0.25010.05(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:Dependent 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 Model 11 82.8333333 7.5303030 1.39 0.2895 Error 12 65.0
4、000000 5.4166667 Corrected Total 23 147.8333333 R-Square Coeff Var Root MSE R Mean 0.560316 22.34278 2.327373 10.41667 Source DF Type I SS Mean Square F Value Pr F m 2 44.33333333 22.16666667 4.09 0.0442 n 3 11.50000000 3.83333333 0.71 0.5657 m*n 6 27.00000000 4.50000000 0.83 0.5684 Source DF Type I
5、II SS Mean Square F Value Pr F m 2 44.33333333 22.16666667 4.09 0.0442 n 3 11.50000000 3.83333333 0.71 0.5657 m*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。5-3用L9(34)确定配比试验方案: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(
6、0.1份)3(0.1份)2313121(0.5份)2(0.3份)3(0.1份)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 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:Dependent Variable: y Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr F Model 2 38.93714 19.46857 9.54 0.0033 Error
7、 12 24.47619 2.03968 Corrected Total 14 63.41333 Root MSE 1.42817 R-Square 0.6140 Dependent Mean 30.03333 Adj R-Sq 0.5497 Coeff Var 4.75530 Parameter Estimates Parameter Standard Variable DF Estimate Error t Value Pr |t| Intercept 1 19.03333 3.27755 5.81 F Model 1 76.24389 76.24389 7.45 0.0163 Error
8、 14 143.28371 10.23455 Corrected Total 15 219.52760 Parameter Standard Variable Estimate Error Type II SS F Value Pr F Intercept 8.22980 1.38618 360.74949 35.25 F Model 2 147.46551 73.73276 13.30 0.0007 Error 13 72.06209 5.54324 Corrected Total 15 219.52760 Parameter Standard Variable Estimate Error
9、 Type II SS F Value Pr F Intercept 18.33483 2.99803 207.32264 37.40 F Model 3 167.42492 55.80831 12.85 0.0005 Error 12 52.10268 4.34189 Corrected Total 15 219.52760 Parameter Standard Variable Estimate Error Type II SS F Value Pr F Intercept 19.49941 2.70837 225.06452 51.84 F 1 t9 1 0.3473 0.3473 17
10、5.752 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+0.00163+0.00728-2.193056-10(1)散点图如下:可以采用Logistic拟合此数据。(2)用Logistic模拟结果为: Dependent Variable y Method: Gauss-Newton Sum of Iter b c a Squares 0 3.7180 2.0000 21.0000 1124.1 1 3.6408 1
11、.8493 14.8393 570.9 2 3.5475 1.6684 14.9977 534.7 3 3.4704 1.5208 15.2362 499.4 4 3.4046 1.3963 15.4814 464.3 5 3.3469 1.2887 15.7348 429.2 6 3.2955 1.1943 15.9985 394.1 7 3.2491 1.1107 16.2735 359.1 8 3.2069 1.0360 16.5601 324.4 9 3.1684 0.9691 16.8579 290.5 10 3.1331 0.9091 17.1660 257.6 11 3.1008
12、 0.8551 17.4829 226.3 12 3.0712 0.8067 17.8067 196.8 13 3.0442 0.7632 18.1349 169.7 14 3.0195 0.7242 18.4653 145.1 15 2.9971 0.6892 18.7951 123.1 16 2.9768 0.6580 19.1220 104.0 17 2.9584 0.6300 19.4438 87.4241 18 2.9420 0.6050 19.7586 73.4129 19 2.9272 0.5827 20.0648 61.7148 20 2.9140 0.5628 20.3610
13、 52.0895 21 2.9023 0.5450 20.6464 44.2826 22 2.8920 0.5291 20.9203 38.0422 23 2.8828 0.5149 21.1822 33.1304 24 2.8748 0.5021 21.4319 29.3307 25 2.8679 0.4907 21.6693 26.4509 26 2.8618 0.4804 21.8946 24.3243 27 2.8566 0.4711 22.1078 22.8087 28 2.8522 0.4628 22.3094 21.7843 29 2.8484 0.4553 22.4995 21
14、.1514 30 2.8453 0.4486 22.6787 20.8276 31 2.8428 0.4425 22.8472 20.7456WARNING: Step size shows no improvement. WARNING: PROC NLIN failed to converge. Estimation Summary (Not Converged) Method Gauss-Newton Iterations 31 Subiterations 29 Average Subiterations 0.935484 R 0.653053 PPC(c) 0.012486 The N
15、LIN Procedure Estimation Summary (Not Converged) RPC . Object 0.00394 Objective 20.74557 Observations Read 15 Observations Used 15 Observations Missing 0 NOTE: An intercept was not specified for this model. Sum of Mean Approx Source DF Squares Square F Value Pr F Regression 3 3245.5 1081.8 625.77 1(
16、ii)内收缩,0(iii)收缩,01(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上习题记,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;1.2814-3SAS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
17、,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。14 PPT补充15单指标评价结果归一化处理得:权向量a=(0.30 0.25 0.15 0.20 0.10)aR=(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=10.1611+20.2223+30.4833+40.1333=2.588位于一般与良好之间,且该食品一般偏良(2)1000.1611+800.2223+600.4833+400.1333=68.2245评分值为:P=68.2245分
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