spss实践题分析及答案.docx

1、spss实践题分析及答案SPSS实践题习题1分析此班级不同性别的学生的物理和数学成绩的均值、最高分和最低分。Case Processing SummaryCasesIncludedExcludedTotalNPercentNPercentNPercent数学 * 性别26100.0%0.0%26100.0%物理 * 性别26100.0%0.0%26100.0%Report性别数学物理男生Mean80.076974.5385N1313Std. Deviation5.751255.17390Minimum72.0069.00Maximum95.0087.00女生Mean80.769276.1538

2、N1313Std. Deviation8.917728.32512Minimum70.0065.00Maximum99.0091.00TotalMean80.423175.3462N2626Std. Deviation7.360296.84072Minimum70.0065.00Maximum99.0091.00结论：男生数学成绩 最高分: 95 最低分: 72 平均分: 80.08 物理成绩 最高分: 87 最低分: 69 平均分: 74.54 女生数学成绩 最高分: 99 最低分: 70 平均分: 80.77 物理成绩 最高分: 91 最低分: 65 平均分: 76.15习题2分析此班级的

3、数学成绩是否和全国平均成绩85存在显著差异。One-Sample StatisticsNMeanStd. DeviationStd. Error Mean数学2680.42317.360291.44347One-Sample TestTest Value = 85 tdfSig. (2-tailed)Mean Difference95% Confidence Interval of the DifferenceLowerUpper数学-3.17125.004-4.57692-7.5498-1.6040结论：由分析可知相伴概率为0.004，小于显著性水平0.05，因此拒绝零假设，即此班级数学成绩

4、和全国平均水平85分有显著性差异习题3分析兰州市2月份的平均气温在90年代前后有无明显变化。Group Statistics分组NMeanStd. DeviationStd. Error Mean二月份气温011-4.5272731.2034043.3628400118-3.2000001.3006786.3065729Independent Samples TestLevenes Test for Equality of Variancest-test for Equality of MeansFSig.tdfSig. (2-tailed)Mean DifferenceStd. Error

5、Difference95% Confidence Interval of the DifferenceLowerUpper二月份气温Equal variances assumed1.019.322-2.74027.011-1.3272727.4843246-2.3210246-.3335208Equal variances not assumed-2.79422.599.010-1.3272727.4750156-2.3108823-.3436632结论：由分析可知, 方差相同检验相伴概率为0.322,大于显著性水平0.05,因此接受零假设，90年代前后2月份温度方差相同。双侧检验相伴概率为0

6、.011, 小于显著性水平0.05，拒绝零假设，即2月份平均气温在90年代前后有显著性差异习题4分析15个居民进行体育锻炼3个月后的体质变化。Paired Samples StatisticsMeanNStd. DeviationStd. Error MeanPair 1锻炼前65.20157.5231.943锻炼后53.27155.8731.516Paired Samples CorrelationsNCorrelationSig.Pair 1锻炼前 & 锻炼后15-.300.277Paired Samples TestPaired DifferencestdfSig. (2-tailed)

7、MeanStd. DeviationStd. Error Mean95% Confidence Interval of the DifferenceLowerUpperPair 1锻炼前 - 锻炼后11.93310.8462.8005.92717.9404.26114.001结论：由分析可知，锻炼前后差值与零比较，相伴概率小于显著性水平，拒绝零假设，即锻炼前后有显著性差异习题5为了农民增收，某地区推广豌豆番茄青菜的套种生产方式。为了寻找该种方式下最优豌豆品种，进行如下试验：选取5种不同的豌豆品种，每一品种在4块条件完全相同的田地上试种，其它施肥等田间管理措施完全一样。根据表中数据分析不同豌豆品

8、种对平均亩产的影响是否显著。ANOVA产量Sum of SquaresdfMean SquareFSig.Between Groups(Combined)13195.70043298.9254.306.016Linear TermContrast3591.02513591.0254.687.047Deviation9604.67533201.5584.179.025Within Groups11491.50015766.100Total24687.20019Multiple ComparisonsDependent Variable:产量(I) 品种(J) 品种Mean Difference

9、(I-J)Std. ErrorSig.95% Confidence IntervalLower BoundUpper BoundLSD12-13.2500019.57166.509-54.966028.46603-5.7500019.57166.773-47.466035.96604-21.5000019.57166.289-63.216020.2160551.50000*19.57166.0199.784093.21602113.2500019.57166.509-28.466054.966037.5000019.57166.707-34.216049.21604-8.2500019.571

10、66.679-49.966033.4660564.75000*19.57166.00523.0340106.4660315.7500019.57166.773-35.966047.46602-7.5000019.57166.707-49.216034.21604-15.7500019.57166.434-57.466025.9660557.25000*19.57166.01015.534098.96604121.5000019.57166.289-20.216063.216028.2500019.57166.679-33.466049.9660315.7500019.57166.434-25.

11、966057.4660573.00000*19.57166.00231.2840114.716051-51.50000*19.57166.019-93.2160-9.78402-64.75000*19.57166.005-106.4660-23.03403-57.25000*19.57166.010-98.9660-15.53404-73.00000*19.57166.002-114.7160-31.2840*. The mean difference is significant at the 0.05 level.产量品种NSubset for alpha = 0.0512Student-

12、Newman-Keulsa54212.500014264.000034269.750024277.250044285.5000Sig.1.000.696Means for groups in homogeneous subsets are displayed.a. Uses Harmonic Mean Sample Size = 4.000.结论：由以上分析可知，F统计量F(4,15)=4.306，对应的相伴概率为0.016，小于显著性水平，拒绝零假设，即不同品种豌豆与亩产量之间存在显著性差异。1、2、3、4号品种与5号有明显差异, 5号品种产量最低, 因此购种选择前四种均可。习题6由于时间安排紧张，公司决定安排4名员工操作设备A、B、C各一天，得到日产量数据如表所示。试分析4名员工和3台设备是否有显著性差异，以便制定进一步的采购计划。Tests of Between-Subjects EffectsDependent Variable:日生产量SourceType III Sum of SquaresdfMean SquareFSig.Corrected Model433.167a586.63315.831.002Intercept31212.000131212.0005703.716.000equipment318.