非参数统计friedman检验.docx

非参数统计期末大作业一、Wilcoxon符号秩检验某个公司为了争夺竞争对手的市场，决定多公司重新定位进行宣传。在广告创意 中，预计广告投放后会产生效果。一组不看广告组和一组看广告，抽取16位被 调查者，让起给产品打分。现有数据如下分析广告效应是否显著。不看广告62839699716097100看广告87929086949582911、手算建立假设：H0：广告效应不显著111：广告效应显著不看广告组记为x,看广告组记为y。检验统计量计算表XYD=x-yIDI|D|的秩D的符号6287-25257-8392-992.5-9690661+998613134+7194-23236-6095-35358-978215155+10091992.5+由表可知：T+= 1+4+5+2. 5=12. 5=7+2. 5+6+8=23. 5根据n=8, T+和T-中较大者二23. 5,查表得，T+的右尾概率为0. 230到0. 273, 在显著性水平a = 0.05下，P值显然较大，故没有理由拒绝原假设，说明广告效 应不显著。四、k个独立样本的Kruskal-Wallis检验为检测四种防护服对人脉搏的影响，找来20人试穿，每种有5人试穿，测量试 穿者的脉搏，得到以下表格：试穿者防护服1防护服2防护服3防护服41130104123133211111611912831141061151304123981201125115104117110问：穿四种防护服测得的脉搏有无差异。1、手算建立假设：H0：测得的脉搏没有显著差异H1：测得的脉搏有显著差异 脉搏等级整理如下：防护服1防护服2防护服3防护服418.52.515.5206111317849.518.515.511479.52.5125秩和57.5216467.5计算检验统计量H:仁标为th"】） j=l 11257.52 + 212 4- 642 + 67.52=X3x21=70.854-63=7.854查表：自由度df=3,显著性水平a = 0.05,相对应的临界值卡方=7. 82。显然，11=7.854卡方:7.82,所以拒绝原假设，说明四种防护服对脉搏的影响 有显著差异。2、 spss输入20个观测值（数据4）在非参数检验中选择k个独立样本检验防护服分组定义为1到4操作如下列图：16：味傅133脑博防护服11301211113114141231611516104271162810629982101042111233121193131153141203匚151173164171284181304191124201104WormnetyzoQrephs UMitsa Add-Qns vymdow yelp日混口ReQorte2 -Descriptive Statistics防护TablesCotrjpore Meansvar Ivar |var空oncra Linear Model Gonerertzed Linear Mod* M 区ed ModelsCorreunte BegressionLoginearNeural NetworksClassttyQata Reduction由 2hl-Squ<*c四即om包画 Eurvs .囚 L-Sampie K-S.M. 21ndependent Samples KIndependent Samptes. 1M1 2 Related Samples . 班| K Related Samples .Sc*Norp肝ometrke Tc«t& Time SeriesSurvivalMissing Value An»tsis.Mutbpie Response Complex SamplesQuaiity Control|7"1 ROC Csye .切 Tests for Several Independent Samples1y?"r 团 Several Independent Samples:. 公jest vanatole UstRenge for Gromg Ver»aUeMntrum. |lMaxtrumj Continue Cancel 彳 t/pva假比4)Uenne Rer簧.Test T>e gKruskei-mhsH Q Medon O Jonckhecre-TerpstraOK £ftsteFe 沦 tOK £ftsteFe 沦 tCareer H&A输出结果如下（输出4）:Ranks防护服NMean Rank脉搏 1234Total55552011.504.2012.8013.50Test Statisticsa b脉搏Chi-SquaredfAsymp. Sig.7.8783.049a. Kruskal Wallis Testb. Grouping Variable:防护服由上表，卡方与手算十分接近，拒绝原假设，即说明四种防护服对脉搏的影 响有显著性差异。3、中位数检验20个数,中位数为115. 5,整理每个总体中大于或小于该中位数的观测值人数，如下表:1234>115.5214310=115.5341210555520计算Q检验量2232I242Q= 20-x 2 + x 2 + - x 2 + x 2 - 15 x 105 x 105 x 105 x 10=4Q统计量小于卡方=7.82,没有理由拒绝原假设，说明四种防护服对脉搏的影响 没有显著差异。Spss：唬test type中选择中位数，输出结果如下:Frequencies防护服1234脉搏> Median2143<=Median3412Test Statistics6脉搏N20Median115.50Chi-Square4.000adf3Asymp. Sig.261a. 8 cells (100.0%) have expected frequencies less than 5. The minimum expected cell frequency is 2.5.b. Grouping Variable:防护服卡方值为4,与手算结果一致，不拒绝原假设，即说明四种防护服对脉搏的影响没有显著性差异综上，两种算法的结果不一致！？ ？4、R语言输入语句(R语言2)：x=c(130, 111, 114, 123, 115, 104, 116, 106, 98, 104, 123, 119, 115, 120, 117, 133, 128, 130, 112, 110)y=c(l, 1,1,1,1,2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4,4,4)kruskal. test (x, y)输出结果：> kruskal. test (x, y)Kruskal-Wallis rank sum testdata: x and yKruskal-Wallis chi-squared = 7.878, df = 3, p-value = 0.0486 与以上的手算和KS检验法结果一致，拒绝原假设，说明四种防护服防护服对 脉搏的影响存在显著差异。五、列联表卡方检验一种原料来自三个不同的地区，原料质量被分成三个不同等级。从这批原料 中随机抽取500件进行检验，得样本数据如下表所示，要求检验地区与原料质量 之间有无依赖关系。一级二级三级合计地区1526424140地区2605952171地区3506574189合计1621881505001、手算：建立假设：H0：地区与原料质量无关H1：地区与原料质量相关地区等级f.(fij - eij)， Cij115245. 360.97126452. 642.451324427.71216055.40. 38225964.30. 44235251.30.01315061.242.06326571.060. 52337456.75. 28介计19. 82r c(2 = ££与肛=19.82i=l j=l "自由度 df= (r-1) (c-1) =4查表得，忘.05=949,由于319.82忌05=9.49,因此拒绝原假设，即说明地 区与原料质量是相关的。2、 SPSS：输入数据(数据5)后，对数量加权处理：DataWeight Cases在分析描述统计中选择列联表Crosstabs生成地区与等级列联表，卡方检验表，方向性测度和对称性测度。输出结果如F (输出5)：Bar Chart及 12 3等口123luno。4O-地区地区 * 等级 Crosstabulation等级Total123地区Count526424140Expected Count45.452.642.0140.0% within 地区37.1%45.7%17.1%100.0% within 等级32.1%34.0%16.0%28.0% of Total10.4%12.8%4.8%28.0%Count6059521712Expected Count55.464.351.3171.0% within 地区35.1%34.5%30.4%100.0% within 等级37.0%31.4%34.7%34.2% of Total12.0%11.8%10.4%34.2%Count5065741893Expected Count61.271.156.7189.0% within 地区26.5%34.4%39.2%100.0% within 等级30.9%34.6%49.3%37.8% of Total10.0%13.0%14.8%37.8%TotalCount162188150500Expected Count162.0188.0150.0500.0% within 地区32.4%37.6%30.0%100.0% within 等级100.0%100.0%100.0%100.0% of Total32.4%37.6%30.0%100.0%方向性测度表Directional MeasuresValueAsymp. Std.Error*Approx. TbApprox.Sig.Nominal byLambdaSymmetric.032.035.906.365Nominal地区Dependent.032.033.954.340等级Dependent.032.051.623.533Goodman and Kruskal 地区tauDependent.019,008.001c等级Dependent.019.008.001cUncertainty Coefficient Symmetric.019.0082.356.000d地区Dependent.019.0082.356.000。等级Dependent.019.0082.356.000dOrdinal by Ordinal Somers' dSymmetric.147.0383.880.000地区Dependent.147.0383.880.000等级Dependent.147.0383.880.000a. Not assuming the null hypothesis.Using the asymptotic standard error assuming the null hypothesis.b. Based on chi-square approximationLikelihood ratio chi-square probability.对称性测度表Symmetric MeasuresValueAsymp. Std.Error3Approx. TbApprox. Sig.Nominal by Nominal Phi.199.001Cramer's V.141.001Contingency Coefficient.195.001Ordinal by OrdinalKendall's tau-b.147.0383.880.000Kendall's tau-c.146.0383.880.000Gamma.220.0563.880.000N of Valid Cases500Not assuming the null hypothesis.a. Using the asymptotic standard error assuming the null hypothesis.卡方检验表Chi-Square TestsValuedfAsymp. Sig. (2-sided)Pearson Chi-SquareLikelihood RatioN of Valid Cases19.822a4.00120.7325004.000a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 42.00.由输出结果，检验统计量为19.822,精确双尾显著性概率P远远小于显著 性水平0.05,因此拒绝原假设，即说明地区与原料质量相关，与手算结果一致。六、Friedman检验（英文）来源： ：/ statisticslectures. com/topics/friednian/#videoThe Friedman Test is a version of the Repeated-Measures ANOVA that can be performed on ordinal(ranked) data.Friedman检验是一种通过顺序(排名)数据重复测量方差分析。Ordinal data is displayed in the table below. Is there a difference betweenWeeks 1, 2, and 3 using alpha = 0. 05原始数据显示在下面的表中。周1、2和3有什么区别，显著性水平。=0.05WeeklWeek 2Week 3Let's test to see if there are any differences with a hypothesis test. 让我们通过假设检验测试是否存在差异。Steps for Friedman TestFriedman检验的操作步骤1. Define Null and Alternative Hypotheses 定义零假设和备择假设State Alpha规定显著水平2. Calculate Degrees of Freedom 计算自由度State Decision Rule 陈述决策规那么3. Calculate Test Statistic 计算检验统计量State Results 声明结果4. State Conclusion 陈述结论Define Null and Alternative HypothesesHq there is no difference between the three conditions.Hi； there is a difference between the three conditions.HO：三周没有差异Hl：三周存在差异1. State Alphaalpha = 0. 05显著性水平为0. 052. Calculate Degrees of Freedomdf = k - 1, where k = number of groupsdf = 3 - 1 = 2自由度为2, k表示组的数量State Decision RuleWe look up our critical value in the Chi-Square Table and find a critical value of plus/minus 5.99.我们查询卡方表，找到临界值5. 99If x2 is greater than 5.99, reject the null hypothesis.如果卡方大于5. 99,那么拒绝零假设。3. Calculate Test StatisticFirst, we must rank the scores of every subject, as shown below in red:首先，我们必须对每个工程排序，如下方红色所示：2、 Spss在spss中输入八组数据（数据1）：不看广告看广告16287283923969049986571946609579782810091选择非参数检验中的两个相关样本检验FileEdit View Qata工r 所 sfortnAnalyzeGraphs Utilities Add-gns Window Help昌国Q3 ReportsDescriptive StatisticsTablesConjpare MeansGeneral Linear ModelGeneralized Linear Models Mixed ModelsCorrelateRegression应linearNeural NetworksClassifyData ReductionScale13:不看广告看广varvarvar162283396499571660797810091011Nonparametric TestsX2： Chi-Square.53 Binomial.丽 Runs.囚 1-Sample K-S.A. 2 Independent Samples. K Independent Samples.1213141516Time SeriesSurvival第 Missing Value Analysis.Multiple ResponseComplex Samples17Quality ControlM 2 Related Samples.18T ROC Curve.泗 K Related Samples.对话框中选择Wilcoxon,输出如下结果（输出1）：RanksNMean RankSum of Ranks1广告-不看广告Negative Ranks4a3.1212.501836237213092264143We then replace our original values with the rankings we've just found: 我们按排序重新排列我们的原始数值，如下：A Chi-Square value is then calculated using the sums of the ranks of each group:卡方值通过每组的秩和计算：Weekl Week 2 Week 312 v-i23，"丽包 W*3n(k + D12*=丽由啰 +132 + 142)76)(3 + 1)旌=2.336. State ResultsIf x2 is greater than 5.99, reject the null hypothesis.Xr 2.33Do not reject the null hypothesis.计算得卡方为2. 33,小于临界值5. 99,所以不拒绝零假设State ConclusionThere is no difference among the three groups, * = 2.33 (2, n = 6), p > .05.结论：三组之间没有差异。用Spss计算输入数据（数据6）在非参数统计中选择k个相关样本，在test type中选择Friedman, 选择精确计算，输出结果如下（输出6）：RanksTest Statistics3a. Friedman TestN6Chi-Square2.333df2Asymp. Sig.311Exact Sig.430Point Probability.177卡方2. 333,与上述英文手算结果一致，所以没有理由拒绝原假设，即三周没有 差异。Positive Ranks4D5.8823.50Ties0cTotal8a.看广告 < 不看广告 b.看广告 > 不看广告 c.看广告=不看广告由上表，负秩为4,正秩也为4,同分的情况为0,总共8。负秩和为12.5,正 秩和为23.5,与手算结果一致Test Statistics。看广告-不看广 告ZAsymp. Sig. (2-tailed)-.77V.441a. Based on negative ranks.b. Wilcoxon Signed Ranks Test由上表，Z为负，说明是以负秩为基础计算的结果，其相应的双侧渐进显著 性结果为0.441,明显大于0.05,因此在Q =0.05的显著性水平下，没有理由拒 绝原假设，即说明广告效应不显著，与手算的结论一致。3、R语言(尺语言1)输入语句：x=c(62,83,96,99,71,60,97,l 00)y=c(87,92,90,86,94,95,82,91)wilcox.tcst(x,y,cxact=F,cor=F)输出结果：Wilcoxon rank sum testdata: x and yW = 33, p-valuc = 0.9164alternative hypothesis: true location shift is not equal to 0由输出结果可知，P=0.9164,远大于。=0.05,因此没有理由拒绝原假设， 即广告效应并不显著，与以上结果一致。二 Wald-Wolfowitz 游程检验有低蛋白和高蛋白两种料喂养大白鼠，以比拟它们对大白鼠体重的增加是否 有显著不同的影响，为此对m=10,n=10只大白鼠分别喂养低蛋白和高蛋白两种饲 料，得增重量X, Y （单位：g）的表如下：饲料增重量低蛋白X64717275828384909196高蛋白Y42526165697578787881给定显著性水平a=0. 05,试用游程检验法检验两种饲料的影响有无显著差异。1、手算建立假设：H0：两种饲料对大白鼠无显著差异H1：两种饲料对大白鼠有显著差异将X,Y的数据按从小到大混合排列，得X,Y的混合样本序列：YYYXYYXXXYYYYYXXXXXX 故得游程总数U=6, m=10, n=10,查表得，U=6的概率为0.019,由于是双 侧检验，对于显著性水平a=0.05,对应的P值为2x 0.019 = 0.038 Vo.o5,因 此拒绝原假设，即说明两种饲料对大白鼠有显著差异。2、Spss在spss中输入数据（数据2）熠看量分组161 1271 1372 1475 1582 1683 1783 1890 1991 11096 11142 21252 21361 214652156921675 21778 21878 21978 22081 2在非参数检验中选择两个独立样本检验对话框:对话框:在Define Groups输入 1 和2。在Test Type选中Wald-Wolfbwitz runs。强 Two-Independent-Samples Tests屈 Two Independent Samples: Def.Group 1：Group 2:Options.Qefine Groups.Exact.Test TypeMann-Whitney UKoitnogorov-Smirnov ZMoses extreme reactions 回 Wald-Wolfowitz runsOK Easte Reset Cancel Help输出结果如下（输出2）：Frequencies分组N增重量 12Total101020Test Statistics110Number of RunsZExact Sig. (1-tailed)增重量 Minimum Possible6a-2.068.019Maximum Possible8a-1.149.128There are 2 inter-group ties involving 4 cases.a. Wald-Wolfowitz TestGrouping Variable:分组由上表，P值与手算结果一致，因此也拒绝原假设，即说明两种饲料对大白 鼠有显著差异。三、KolmogorovSmirnov 检验为了研究两家电信运营商套餐在目标市场的年龄维度上的分布是否相同，该电信 运营公司开展了一个社会调查活动。数据如下：两种通信套餐的用户年龄套餐1套餐218221848255122342442232626443138分析两种套餐的目标市场年龄的分布是否存在显著性差异。1、手算建立假设：H0：两种套餐的目标市场年龄分布不存在显著差异H1：两种套餐的目标市场年龄分布存在显著差异检验统计量D的计算表由上表，找出检验统计量 D=maxS(x)-S2(.y)| =7/9, m=7, n=9, mnD=7 x 9 x；= 49,查表得，相应的P值为0.008,在5%的显著性水平上，P值足够小，因 此拒绝原假设，说明两种套餐的目标市场年龄分布存在显著差异。年龄flf2ZflSI (x)S2(x)Sl(x)- S2(x)1810101/701/71810202/702/72211313/71/920/632310414/71/929/632410515/71/938/632510616/71/947/6326117212/97/931017313/96/934017414/95/938017515/94/942017616/93/944017717/92/948017818/91/95101791102、 spss输入数据(数据3)年龄运营商11812181325142215241623172618222948210512113421242213262144421531216382在非参数检验中选择两个独立样本检验：Edit yiew Qala 工ransform Analyze graphs Utilities Add-fins 凶 Mow Help年龄运营11821832542252462372682294810511134124213261444153116381厂18年龄运营11821832542252462372682294810511134124213261444153116381厂18ReportsDescriptive StatisticsTablesCompare MeansGeneral Linear ModelGenersized Linear Modelshixed ModelsCorrelateRegressionLoglinearNeural NetworksClassifyData ReductionScaleyonparametric TestsTine SeriesSurvival宵 hissing Value Analysis.Multiple ResponseComplex SagesQuality Control四ROC C”varPC ihi-SquareM Binomial.殛 Runs.囚 l-Sairvle K-S.曲 2 Independent Samples.K Independent Samples.2 Related Samples.K Related Samplesvar对话框：运营商1和2分类的变量输入到Grouping Variable,在Define Groups输入 1和2o在 Test Type 选中 Kolmogorov-Smirnov<>在点Exact时翻开的对话框中可以选择精确方法(Exact)o/ Two-lndependent-Samples TestsGroup 1：厘Group 2：gGroup 1：厘Group 2：gContinue C2 r图 p因 Two Independent Samples： Def. 岑Exe»ct.Options.Test TypeMann-V<iitney U< Kolmogorov-Smirnov ZPaste Reset Caned HdpMoses extreme reactions WaW-Wolfowitz runs输出结果如下（输出3）;Frequencies运营商N年龄 12Total7916Test Stati