市场调查与数据分析.doc

市场调查与数据分析SPSS实验作业电子商务2009级2班王祎2009211049市场调查与数据分析电子商务2009级2班王祎2009211049一、对P62习题1中的数据进行描述性分析，同时进行再编码对成绩进行分组，分为60分以下、60-70分、70-80分、80-90分、90-100分，进行频数分析。描述性分析：操作过程：analyzedescriptive statisticsdescriptiveDescriptive Statistics NMinimumMaximumMeanStd. Deviation成绩5036.0099.0073.960010.63870Valid N (listwise)50 表Descriptive Statistics分析：由输出的图表可知，该数学成绩的样本总量是50，最低成绩为36，最高成绩为99，平均值为73.96，标准差为：10.6387。频数分析：分组过程：transformrecodein different variable.analyzedescriptive statisticsfrequentces. 成绩 FrequencyPercentValid PercentCumulative PercentValid9024.04.012.060-701530.030.042.070-801632.032.074.080-901326.026.0100.0Total50100.0100.0 分析：由其频数分析可知，成绩主要集中在60-70 70-80 80-90 三个阶段，所占比例高达88%，成绩优秀。二、对P106习题，首先进行相关性分析，然后进行回归分析。相关性分析：操作过程：生成比例，transformcompute 简单线形相关分析，grasspscattersimpledefineGraph分析：可以初步确定新药的浓度与杀死蟑螂的数目比例有近似线形趋势的相关关系。即使用不同浓度的药物能杀死不同数量的蟑螂。Correlations Descriptive Statistics MeanStd. DeviationN剂量6.57143.101467杀死数目7.57146.023767 Correlations Descriptive Statistics MeanStd. DeviationN杀死数目7.57146.023767蟑螂数17.142913.825447 Correlations 杀死数目蟑螂数杀死数目Pearson Correlation1.851(*)Sig. (2-tailed).015N77蟑螂数Pearson Correlation.851(*)1Sig. (2-tailed).015.N77* Correlation is significant at the 0.05 level (2-tailed). 剂量杀死数目剂量Pearson Correlation1.113Sig. (2-tailed).809N77杀死数目Pearson Correlation.1131Sig. (2-tailed).809.N77分析：计算结果表明：蟑螂数目与杀死蟑螂数目的相关关系较为密切，相关系数达到0.851，且在显著性水平为0.05时，不相关的概率为0.15%，呈现出明显的相关关系，而药物剂量与杀死蟑螂数目的相关关系较弱，相关系数只有0.113。回归分析：由第一步的线形分析可知，剂量浓度与杀死数目之间有一定的线形关系，需要进一步探索。操作过程：analyzeregressionlinearRegressionVariables Entered/Removed(b)ModelVariables EnteredVariables RemovedMethod1剂量(a).Entera All requested variables entered.b Dependent Variable: 比例 Model SummaryModelRR SquareAdjusted R SquareStd. Error of the Estimate1.977(a).955.946.08369a Predictors: (Constant), 剂量ANOVA(b)Model Sum of SquaresdfMean SquareFSig.1Regression.7431.743106.062.000(a)Residual.0355.007 Total.7786 a Predictors: (Constant), 剂量b Dependent Variable: 比例Coefficients(a)Model Unstandardized CoefficientsStandardized CoefficientstSig.BStd. ErrorBeta1(Constant)-.257.079 -3.255.023剂量.113.011.97710.299.000a Dependent Variable: 比例分析：表Model Summary 说明：该回归方程中变量的相关系数达0.977，判定系数达0.955。表Coefficients(a) 说明：根据资料所建立的回归方程为：=-0.257+0.113x，回归系数为零的概率为0.000（回归系数显著地不为零）。表ANOVA(b)说明：方差检验结果F值达106.062，其相应的概率为0.0000.10）。由于方差相等时t检验的样本数据的实际概率P（0.525）a(0.10)，所以检验结论为：没有足够的理由拒绝原假设，即认为两个程序完成某个动作的时间是相等的。四、对P139习题6进行独立性检验。操作过程：加权处理数据：dataweight cases主处理过程：analyzedescriptive statisticscrosstable.Case Processing Summary CasesValidMissingTotalNPercentNPercentNPercent性别 * 酒类150100.0%0.0%150100.0%性别 * 酒类 Crosstabulation 酒类Total淡啤酒普通啤酒黑啤酒性别男Count20402080Expected Count26.737.316.080.0% within 性别25.0%50.0%25.0%100.0% within 酒类40.0%57.1%66.7%53.3% of Total13.3%26.7%13.3%53.3%女Count30301070Expected Count23.332.714.070.0% within 性别42.9%42.9%14.3%100.0% within 酒类60.0%42.9%33.3%46.7% of Total20.0%20.0%6.7%46.7%TotalCount507030150Expected Count50.070.030.0150.0% within 性别33.3%46.7%20.0%100.0% within 酒类100.0%100.0%100.0%100.0% of Total33.3%46.7%20.0%100.0%Chi-Square Tests ValuedfAsymp. Sig. (2-sided)Pearson Chi-Square6.122(a)2.047Likelihood Ratio6.1782.046N of Valid Cases150 a 0 cells (.0%) have expected count less than 5. The minimum expected count is 14.00.Directional Measures ValueAsymp. Std. Error(a)Approx. T(b)Approx. Sig.Ordinal by OrdinalSomers dSymmetric-.189.075-2.526.012性别 Dependent-.169.067-2.526.012酒类 Dependent-.214.085-2.526.012a Not assuming the null hypothesis.b Using the asymptotic standard error assuming the null hypothesis.Symmetric Measures(c) ValueAsymp. Std. Error(a)Approx. T(b)Approx. Sig.Nominal by NominalPhi.202 .047Cramers V.202 .047Contingency Coefficient.198 .047Ordinal by OrdinalGamma-.333.127-2.526.012N of Valid Cases150 a Not assuming the null hypothesis.b Using the asymptotic standard error assuming the null hypothesis.c Correlation statistics are available for numeric data only.对以上5个表的分析表明：180名调查者的性别及啤酒偏好的分析属于两个变量的独立性分析，交叉分析表显示，男性与女性占总人数的比重分别为53.3%，46.7%；拥有淡啤酒偏好、普通啤酒偏好和黑啤酒偏好的人占总人数的比重分别是33.3%，46.7%和20%。卡方分析表计算结果发现， =6.122， P（0.047）(0.05),所以检验结论为：拒绝原假设，即认为这两个变量相互关联。性别和啤酒偏好之间存在显著关系。五、在P170习题1和习题2两题中任选一题，进行方差分析。习题1：操作过程：Analyzecompare meansone-way ANOVADescriptives产量 NMeanStd. DeviationStd. Error95% Confidence Interval for MeanMinimumMaximum Lower BoundUpper Bound 1530.00002.12132.9486827.366032.634028.0033.002526.00001.87083.8366623.677128.322924.0029.003530.60003.361551.5033326.426134.773927.0035.00Total1528.86673.15926.8157227.117130.616224.0035.00上表表明：早班和晚班的均值较大，说明两个组的满意程度较高。Test of Homogeneity of Variances产量 ：Levene Statisticdf1df2Sig.1.843212.200上表表明：Levene 统计量未超过临界值，可以接受原假设即方差齐性。ANOVA产量 Sum of SquaresdfMean SquareFSig.Between Groups62.533231.2674.860.028Within Groups77.200126.433 Total139.73314 上表表明：因为P=0.028=0.05,所以拒绝原假设：认为产量差异显著。Multiple ComparisonsDependent Variable: 产量 (I) 班次(J) 班次Mean Difference (I-J)Std. ErrorSig.95% Confidence IntervalLower BoundUpper BoundScheffe124.00001.60416.082-.47178.47173-.60001.60416.933-5.07173.871721-4.00001.60416.082-8.4717.4717 3-4.6000(*)1.60416.044-9.0717-.128331.60001.60416.933-3.87175.0717 24.6000(*)1.60416.044.12839.0717* The mean difference is significant at the .05 level.上表表明：在0.05的显著性水平上各组均值差异显著。Homogeneous Subsets产量 班次NSubset for alpha = .0512Tukey B(a)2526.0000 15 30.000035 30.6000Scheffe(a)2526.0000 1530.000030.000035 30.6000Sig. .082.933Means for groups in homogeneous