应用统计学试题2006.6(B).doc

华东理工大学20052006学年第二学期 应用统计学 课程期末考试试卷 B 2006.6开课学院： 商学院 ，专业： 考试形式：开卷，所需时间： 120 分钟考生姓名： 学号： 班级： 任课教师： 题序一二 三四五总 分得分评卷人一、农药收率提高问题（20分）某农药厂生产某种农药，为了提高农药收率，根据生产经验，发现影响农药收率的因子有4个，每个因子取2种水平，因子水平表1如下：表1 农药收率提高问题的因子水平因子水平A反应温度/B反应时间/hC原料配比D真空度/mmHg1802.51.1:16002603.51.2:1500根据经验，A与B两因子之间的交互作用应予以考虑。将因子A，B，C，D分别置于第1，2，4，7列上，测得8次试验结果（收率）分别为0.86，0.95，0.91，0.94，0.91，0.96，0.83，0.88。经SPSS软件的计算，列表如下，请根据计算结果进行分析。表2 ANOVADependent Variable: 收率/% SourceType III Sum of SquaresdfMean SquareFSig.A0.0008010.000803.2000.216B10.001807.2000.115C0.0060510.006050.039D0.0004510.000451.8000.312A * B0.00500120.0000.047Error0.0005020.00025 Corrected Total0.014607 a R Squared = 0.966 (Adjusted R Squared = 0.880)表3 Estimated Marginal Means反应温度 ADependent Variable: 收率/% 反应温度MeanStd. Error95% Confidence IntervalLower BoundUpper Bound800.9150.0080.8810.949600.8950.0080.8610.929反应时间/h BDependent Variable: 收率/% 反应时间/hMeanStd. Error95% Confidence IntervalLower BoundUpper Bound2.50.9200.0080.8860.9543.50.8900.0080.8560.924原料配比 CDependent Variable: 收率/% 原料配比MeanStd. Error95% Confidence IntervalLower BoundUpper Bound1.1:10.8780.0080.8430.9121.2:10.9330.0080.8980.967真空度/mmHg DDependent Variable: 收率/% 真空度/mmHgMeanStd. Error95% Confidence IntervalLower BoundUpper Bound6000.8980.0080.8630.9325000.9130.0080.8780.947反应温度 * 反应时间/h ABDependent Variable: 收率/% 反应温度反应时间/hMeanStd. Error95% Confidence IntervalLower BoundUpper Bound802.50.9050.0110.8570.9533.50.9250.0110.8770.973602.50.9350.0110.8870.9833.50.8550.0110.8070.903（1） 请完成表头设计。（2）表2中有些数据没给出，请根据方差分析表的原理将其计算出来，给出计算过程。（3）由表2分析哪些因子在什么显著性水平下是显著的，给出理由。（4）由表2和表3找出最优生产条件，给出理由。二、学习程度分析（20分）某班有15名学生，为了对该班学生的学习程度进行分类，今对每个学生考试了矩阵代数与数理统计两门课程，其成绩如下表：表4 学习程度数据学生编号课程123456789101112131415矩阵代数1254406-1-4-11-52-50数理统计30654142-311-51-32（表中的数据已作了适当变换）。用系统聚类法对此问题进行了分类，聚类结果见下面图表。表5 Case Processing Summary (a,b)CasesValidMissingTotalNPercentNPercentNPercent15100.00.015100.0a Euclidean Distance usedb Average Linkage (Between Groups)图1 学习程度聚类图（1）由表5，请写出此聚类问题分析中样品间采取何种距离以及采取何种系统聚类的方法。（2）由图1，若将这15名学生按学习程度分成三类，应该如何进行分类，并给出每类的特性。三、某地区春旱预测问题（20分）某气象站预报某地区有无春旱的观测资料中，与是与气象有关的综合预报因子。数据包括发生春旱的6个年份的，观测值和无春旱的8个年份的相应观测值（见表6）。表6 某地区有无春旱的观测数据：春旱：无春旱序号序号124.8-2.0722.1-0.7224.7-2.4821.6-1.4326.6-3.0922.0-0.8423.5-1.91022.8-1.6525.5-2.11122.7-1.5627.4-3.11221.5-1.01322.1-1.21421.4-1.3利用SPSS软件对此问题进行分析的结果见下表：表7 EigenvaluesFunctionEigenvalue% of VarianceCumulative %Canonical Correlation13.433(a)100.0100.00.880a First 1 canonical discriminant functions were used in the analysis.表8 Group Statistics分类Valid N (listwise)UnweightedWeighted春旱66.00066.000无春旱88.00088.000Total1414.0001414.000表9 Canonical Discriminant Function CoefficientsFunction10.803-0.605(Constant)-19.885Unstandardized coefficients表10 Classification Function Coefficients分类春旱无春旱48.55845.77572.32774.424(Constant)-530.389-460.606Fishers linear discriminant functions表11 Classification Results(b,c)分类Predicted Group MembershipTotal春旱无春旱OriginalCount春旱516无春旱088%春旱83.316.7100.0无春旱.0100.0100.0a Cross validation is done only for those cases in the analysis. In cross validation, each case is classified by the functions derived from all cases other than that case.b 92.9% of original grouped cases correctly classified.c 92.9% of cross-validated grouped cases correctly classified.（1）请解释“Unstandardized Standardized canonical discriminant function”与“Fishers linear discriminant functions”的差异所在。（2）请建立此问题的Canonical Discriminant Function(判别函数)。（3）若经观测，此地区今年的，请运用表10预测今年有无春旱，并说明理由。（4）对此判别问题的误判率进行分析。四、企业经济效益的分析（20分）某公司有20个工厂，现要对每个工厂作经济效益分析。从所取得的生产成果同所消耗的人力、物力、财力的比率，选取5个指标（变量）作分析：表示固定资产的产值率；表示净产值的劳动生产率；表示百元产值的流动资金占用率；表示百元产值的利润率；表示百元资金的利润率。今对这20个工厂同时按这5项指标收集数据，见表12。表12 企业经济效益数据表分量样品序号1243.87165216.4634.57149.852240.3182108.8916.9255.893211.151534910.0929.7780.134413.18167607.6724.14105.355349.6077216.4716.2799.416205.47812312.3318.4846.187298.11133085.0527.35138.768414.94137814.1016.6598.209287.25140434.2917.6758.3510303.93111267.6318.3974.2311608.40223922.9424.56223.3712433.92125080.6920.06118.7013572.63121022.7612.08110.4314533.78119903.8011.5975.5515545.7096783.559.4661.1916284.6165136.4112.8348.1517572.07186642.3117.76162.1118409.8673295.8912.2376.6819564.02143114.9328.50233.5820221.20644314.0830.2580.48经SPSS软件计算的结果如下：表13 CommunalitiesInitialExtraction固定资产的产值率1.0000.883净产值的劳动生产率1.0000.805百元产值的流动资金占用率1.0000.855百元产值的利润率1.0000.968百元资金的利润率1.0000.883Extraction Method: Principal Component Analysis.表14 Total Variance ExplainedComponentInitial EigenvaluesExtraction Sums of Squared LoadingsRotation Sums of Squared LoadingsTotal% of VarianceCumulative %Total% of VarianceCumulative %Total% of VarianceCumulative %12.67853.55953.5592.67853.55953.5592.23844.75244.75221.71734.33487.8931.71734.33487.8932.15743.14187.89330.3476.94094.83340.2124.23599.06850.0470.932100.000Extraction Method: Principal Component Analysis.表15 Component Matrix(a)Component12固定资产的产值率0.824-0.452净产值的劳动生产率0.8160.374百元产值的流动资金占用率-0.7760.502百元产值的利润率0.1150.977百元资金的利润率0.8470.407Extraction Method: Principal Component Analysis.a 2 components extracted.表16 Rotated Component Matrix(a)Component12固定资产的产值率0.9120.225净产值的劳动生产率0.3480.827百元产值的流动资金占用率-0.911-0.156百元产值的利润率-0.5770.797百元资金的利润率0.3480.873Extraction Method: Principal Component Analysis. Rotation Method: Varimax with Kaiser Normalization.a Rotation converged in 3 iterations.表17 Component Transformation MatrixComponent1210.7360.6772-0.6770.736Extraction Method: Principal Component Analysis. Rotation Method: Varimax with Kaiser Normalization.表18 Component Score Coefficient MatrixComponent12固定资产的产值率0.4050.014净产值的劳动生产率0.0770.366百元产值的流动资金占用率-0.4110.019百元产值的利润率-0.3540.448百元资金的利润率0.0720.389Extraction Method: Principal Component Analysis. Rotation Method: Varimax with Kaiser