SPSS方差分析作业.doc

统计作业（3）1、抽查某地区三所小学五年级男学生的身高，数据见文件：“男生身高”。设各小学五年级男学生的身高服从同方差的正态分布。问该地区三所小学五年级男学生的平均身高是否有显著差异（=0.05）？解：Test of Homogeneity of Variances身高Levene Statisticdf1df2Sig.5.243215.019ANOVA身高Sum of SquaresdfMean SquareFSig.Between Groups465.8812232.9414.372.032Within Groups799.2551553.284Total1265.13617因为sig=0.0190.05，所以所用样本的方差不相等。所以选择Tamhane又因为P=0.0320.05，所以三所小学五年级男学生的平均身高有显著差异。Multiple Comparisons身高Tamhane(I) VAR00001(J) VAR00001Mean Difference (I-J)Std. ErrorSig.95% Confidence IntervalLower BoundUpper Bound12-10.85005.0393.187-26.3654.6653-10.7333*2.4928.009-18.389-3.0782110.85005.0393.187-4.66526.3653.11674.65561.000-15.50815.7413110.7333*2.4928.0093.07818.3892-.11674.65561.000-15.74115.508*. The mean difference is significant at the 0.05 level.由上表可得，因为1与3：sig=0.0090.05，所以在=0.05，第一小学跟第三小学中男生的平均身高有显著差异同理可得：无法证明在=0.05，第一小学跟第二小学中，第二小学跟第三小学中男生的平均身高显著有差异2、某钢厂检查一月上旬内的五天中生产钢锭重量，数据见文件：“钢锭重量”，设各日所生产的钢锭重量服从同方差的正态分布，试检验不同日期生产的钢锭的平均重量有无显著差异（=0.05）？Test of Homogeneity of Variances重量Levene Statisticdf1df2Sig.3.478415.034ANOVA重量Sum of SquaresdfMean SquareFSig.Between Groups227680.000456920.0003.950.022Within Groups216175.0001514411.667Total443855.00019因为sig=0.0340.05，所以所用样本的方差不相等。所以选择Tamhane又因为P=0.0220.05，所以不同日期生产的钢锭的平均重量有显著差异。Multiple Comparisons重量Tamhane(I) 日期(J) 日期Mean Difference (I-J)Std. ErrorSig.95% Confidence IntervalLower BoundUpper Bound12197.5000117.0025.802-342.111737.1114277.500065.6220.208-194.719749.719912.500066.94211.000-430.253455.25310107.500091.1386.964-284.132499.13221-197.5000117.0025.802-737.111342.111480.000097.3824.998-631.155791.1559-185.000098.2768.808-873.725503.72510-90.0000116.1178.998-629.988449.98841-277.500065.6220.208-749.719194.7192-80.000097.3824.998-791.155631.1559-265.0000*16.5831.001-352.631-177.36910-170.000064.0312.540-630.154290.15491-12.500066.94211.000-455.253430.2532185.000098.2768.808-503.725873.7254265.0000*16.5831.001177.369352.6311095.000065.3835.930-335.288525.288101-107.500091.1386.964-499.132284.132290.0000116.1178.998-449.988629.9884170.000064.0312.540-290.154630.1549-95.000065.3835.930-525.288335.288*. The mean difference is significant at the 0.05 level.由上表可得，因为4与9：sig=0.0090.05，所以在=0.05下，日期4和日期9生产的钢锭的平均重量有显著差异同理可得：在=0.05下，无法说明日期1和日期2，日期1和日期4，日期1和日期9，日期1和日期10，日期2和日期4，日期2和日期9，日期2和日期10，日期4和日期10，日期9和日期10生产的钢锭的平均重量有显著差异3、在某种橡胶的配方中，考虑了3种不同的促进剂，4种不同分量的氧化剂。各种配方各实验一次，测得300%定强数据见文件：“橡胶配方定强”。假定各种配方的定强服从同方差的正态分布。试问不同促进剂、不同分量氧化锌分别对定强有无显著影响（=0.05）？Between-Subjects FactorsValue LabelN促进剂1促进剂A142促进剂A243促进剂A34氧化锌1氧化锌B132氧化锌B233氧化锌B334氧化锌B43Levenes Test of Equality of Error VariancesaDependent Variable:配方定强Fdf1df2Sig.110.Tests the null hypothesis that the error variance of the dependent variable is equal across groups.a. Design: Intercept + 促进剂 + 氧化锌 + 促进剂 * 氧化锌Tests of Between-Subjects EffectsDependent Variable:配方定强SourceType III Sum of SquaresdfMean SquareFSig.Corrected Model96.729a118.794.Intercept16465.021116465.021.促进剂28.292214.146.氧化锌66.063322.021.促进剂 * 氧化锌2.3756.396.Error.0000.Total16561.75012Corrected Total96.72911a. R Squared = 1.000 (Adjusted R Squared = .)由上表可得：由于sig=0.0000.05，所以在=0.05下，无法证明这四种灯丝生产的灯泡，其使用寿命有显著差异根据上图，可以知道配料方案1最佳5、用四种饲料喂猪，共19头猪分成四组，每组用一种饲料，一段时间后称重。猪体重增加数据见文件DATA12-01，试比较这四种饲料对猪体重增加的作用有无显著差异（=0.05）。通过均值多重比较，说明那种饲料最合适。解：Test of Homogeneity of VariancesweightLevene Statisticdf1df2Sig.024315.995由于sig=0.9950.05，所以选用LSDANOVAweightSum of SquaresdfMean SquareFSig.Between Groups20538.69836846.233157.467.000Within Groups652.1591543.477Total21190.85818P0.0000.05，则这四种饲料对猪体重增加的作用有显著差异（=0.05）Multiple ComparisonsweightLSD(I) fodder(J) fodderMean Difference (I-J)Std. ErrorSig.95% Confidence IntervalLower BoundUpper BoundAB-18.68000*4.17024.000-27.5687-9.7913C-56.36000*4.17024.000-65.2487-47.4713D-87.41500*4.42321.000-96.8428-77.9872BA18.68000*4.17024.0009.791327.5687C-37.68000*4.17024.000-46.5687-28.7913D-68.73500*4.42321.000-78.1628-59.3072CA56.36000*4.17024.00047.471365.2487B37.68000*4.17024.00028.791346.5687D-31.05500*4.42321.000-40.4828-21.6272DA87.41500*4.42321.00077.987296.8428B68.73500*4.42321.00059.307278.1628C31.05500*4.42321.00021.627240.4828*. The mean difference is significant at the 0.05 level.由上表可得：比较A与B，由于sig=0.000，所以在=0.05下，饲料A与饲料B对猪体重增加的作用有显著差异同理可得：饲料A与饲料C，饲料A与饲料D，饲料B与饲料C，饲料B与饲料D，饲料C与饲料D对猪体重增加的作用有显著差异ReportweightfodderMeanNStd. DeviationA1.3336E256.80794B1.5204E256.95723C1.8972E256.35035D2.2078E246.10594Total1.7151E21934.31137由上表可得：D对应的均值最大，所以饲料D最佳6、研究两种治疗缺铁性贫血药物的疗效，随机抽取12个病人分成四组，给以不同的疗效：第一组使用一般疗法；第二组使用一般疗法外加药物A；第三组使用一般疗法外加药物B；第四组在一般疗法外家用药物A药物B。一个月后观察红细胞增加数（百万/mm3），数据见文件：DATA12-04。请问药物A,药物B是否分别对红细胞增加数有显著疗效？同时使用药物A和药物B是否对红细胞增加数有显著疗效（=0.05）？绘制均值散布图，说明那种治疗方案最佳。解：Between-Subjects FactorsValue LabelN药A0不用61用6药B0不用61用6Tests of Between-Subjects EffectsDependent Variable:红细胞增数SourceType III Sum of SquaresdfMean SquareFSig.Corrected Model2.963a3.98898.750.000Intercept19.508119.5081.951E3.000druga1.68811.688168.750.000drugb.9081.90890.750.00