2025年統(tǒng)計(jì)學(xué)期末考試：統(tǒng)計(jì)推斷與檢驗(yàn)非參數(shù)檢驗(yàn)試題試卷考試時(shí)間：______分鐘總分：______分姓名：______一、選擇題（每題2分，共20分）1.下列哪項(xiàng)不是非參數(shù)檢驗(yàn)的方法？A.Mann-WhitneyU檢驗(yàn)B.Kruskal-WallisH檢驗(yàn)C.獨(dú)立樣本t檢驗(yàn)D.Wilcoxon符號秩檢驗(yàn)2.以下哪個(gè)檢驗(yàn)適用于比較兩組數(shù)據(jù)的分布？A.獨(dú)立樣本t檢驗(yàn)B.配對樣本t檢驗(yàn)C.Mann-WhitneyU檢驗(yàn)D.Kruskal-WallisH檢驗(yàn)3.下列哪項(xiàng)描述了Mann-WhitneyU檢驗(yàn)的零假設(shè)？A.兩組數(shù)據(jù)的均值相等B.兩組數(shù)據(jù)的分布相等C.兩組數(shù)據(jù)的方差相等D.兩組數(shù)據(jù)的樣本量相等4.在Kruskal-WallisH檢驗(yàn)中，自由度df的計(jì)算公式為：A.df=n-1B.df=n1+n2-2C.df=n1+n2-1D.df=n1-1+n2-15.下列哪項(xiàng)描述了Wilcoxon符號秩檢驗(yàn)的適用條件？A.數(shù)據(jù)服從正態(tài)分布B.數(shù)據(jù)服從偏態(tài)分布C.數(shù)據(jù)服從均勻分布D.數(shù)據(jù)服從指數(shù)分布6.下列哪項(xiàng)描述了Mann-WhitneyU檢驗(yàn)的統(tǒng)計(jì)量U的計(jì)算公式？A.U=(n1*n2)/2B.U=(n1*n2)/2-(n1+n2)/2C.U=n1*n2D.U=n1*n2-(n1+n2)/27.下列哪項(xiàng)描述了Kruskal-WallisH檢驗(yàn)的統(tǒng)計(jì)量H的計(jì)算公式？A.H=(12/(n1*n2*n3))*Σ(Xi-X?)2B.H=(12/(n1*n2*n3))*Σ(Xi-X?)2/nC.H=(12/(n1*n2*n3))*Σ(Xi-X?)2/(n1+n2+n3)D.H=(12/(n1*n2*n3))*Σ(Xi-X?)2/(n1+n2)8.下列哪項(xiàng)描述了Wilcoxon符號秩檢驗(yàn)的統(tǒng)計(jì)量W的計(jì)算公式？A.W=Σ|Xi-X?|B.W=Σ|Xi-X?|/2C.W=Σ|Xi-X?|/nD.W=Σ|Xi-X?|/(n-1)9.在Mann-WhitneyU檢驗(yàn)中，若U值較大，則：A.接受零假設(shè)B.拒絕零假設(shè)C.無法判斷D.需要進(jìn)一步檢驗(yàn)10.在Kruskal-WallisH檢驗(yàn)中，若H值較大，則：A.接受零假設(shè)B.拒絕零假設(shè)C.無法判斷D.需要進(jìn)一步檢驗(yàn)二、簡答題（每題5分，共25分）1.簡述Mann-WhitneyU檢驗(yàn)的適用條件和原理。2.簡述Kruskal-WallisH檢驗(yàn)的適用條件和原理。3.簡述Wilcoxon符號秩檢驗(yàn)的適用條件和原理。4.簡述如何根據(jù)Mann-WhitneyU檢驗(yàn)的結(jié)果判斷兩組數(shù)據(jù)的分布是否存在顯著差異。5.簡述如何根據(jù)Kruskal-WallisH檢驗(yàn)的結(jié)果判斷多組數(shù)據(jù)的分布是否存在顯著差異。三、計(jì)算題（每題10分，共30分）1.已知兩組數(shù)據(jù)，數(shù)據(jù)組A：5,7,8,9,10；數(shù)據(jù)組B：3,4,6,7,8。請使用Mann-WhitneyU檢驗(yàn)判斷兩組數(shù)據(jù)的分布是否存在顯著差異。2.已知三組數(shù)據(jù)，數(shù)據(jù)組A：2,4,6,8；數(shù)據(jù)組B：3,5,7,9；數(shù)據(jù)組C：1,3,5,7。請使用Kruskal-WallisH檢驗(yàn)判斷三組數(shù)據(jù)的分布是否存在顯著差異。3.已知兩組數(shù)據(jù)，數(shù)據(jù)組A：5,7,8,9,10；數(shù)據(jù)組B：3,4,6,7,8。請使用Wilcoxon符號秩檢驗(yàn)判斷兩組數(shù)據(jù)的分布是否存在顯著差異。四、論述題（每題10分，共20分）1.論述非參數(shù)檢驗(yàn)在統(tǒng)計(jì)學(xué)中的優(yōu)勢和局限性。五、應(yīng)用題（每題10分，共20分）1.某公司為了比較兩種不同培訓(xùn)方法對員工工作效率的影響，隨機(jī)抽取了30名員工，分別采用兩種培訓(xùn)方法進(jìn)行培訓(xùn)。培訓(xùn)結(jié)束后，對員工的工作效率進(jìn)行了測試，得到了以下數(shù)據(jù)（單位：小時(shí)/天）：培訓(xùn)方法A：8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30培訓(xùn)方法B：6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29請使用合適的非參數(shù)檢驗(yàn)方法，判斷兩種培訓(xùn)方法對員工工作效率的影響是否存在顯著差異。六、分析題（每題10分，共20分）1.某研究人員對某地區(qū)不同年齡段人群的月收入進(jìn)行了調(diào)查，得到了以下數(shù)據(jù)（單位：元）：年齡段1：2000,2200,2300,2400,2500,2600,2700,2800,2900,3000年齡段2：2500,2600,2700,2800,2900,3000,3100,3200,3300,3400年齡段3：3000,3100,3200,3300,3400,3500,3600,3700,3800,3900請使用合適的非參數(shù)檢驗(yàn)方法，分析不同年齡段人群的月收入是否存在顯著差異。本次試卷答案如下：一、選擇題答案：1.C解析：獨(dú)立樣本t檢驗(yàn)屬于參數(shù)檢驗(yàn)，而題目要求的是非參數(shù)檢驗(yàn)的方法，因此排除A、B、C選項(xiàng)。2.C解析：Mann-WhitneyU檢驗(yàn)適用于比較兩組數(shù)據(jù)的分布，因此選擇C選項(xiàng)。3.B解析：Mann-WhitneyU檢驗(yàn)的零假設(shè)是兩組數(shù)據(jù)的分布相等。4.B解析：Kruskal-WallisH檢驗(yàn)的自由度df的計(jì)算公式為df=n1+n2-2。5.B解析：Wilcoxon符號秩檢驗(yàn)適用于比較兩組數(shù)據(jù)的分布，適用于偏態(tài)分布的數(shù)據(jù)。6.C解析：Mann-WhitneyU檢驗(yàn)的統(tǒng)計(jì)量U的計(jì)算公式為U=n1*n2。7.C解析：Kruskal-WallisH檢驗(yàn)的統(tǒng)計(jì)量H的計(jì)算公式為H=(12/(n1*n2*n3))*Σ(Xi-X?)2/(n1+n2+n3)。8.D解析：Wilcoxon符號秩檢驗(yàn)的統(tǒng)計(jì)量W的計(jì)算公式為W=Σ|Xi-X?|/(n1+n2)。9.B解析：在Mann-WhitneyU檢驗(yàn)中，若U值較大，則拒絕零假設(shè)，認(rèn)為兩組數(shù)據(jù)的分布存在顯著差異。10.B解析：在Kruskal-WallisH檢驗(yàn)中，若H值較大，則拒絕零假設(shè)，認(rèn)為多組數(shù)據(jù)的分布存在顯著差異。二、簡答題答案：1.解析：非參數(shù)檢驗(yàn)的優(yōu)勢在于對數(shù)據(jù)的分布沒有嚴(yán)格的要求，適用于偏態(tài)分布、小樣本量或未知總體分布的數(shù)據(jù)。其局限性在于檢驗(yàn)統(tǒng)計(jì)量的功效較低，無法提供關(guān)于數(shù)據(jù)分布的具體信息。2.解析：Kruskal-WallisH檢驗(yàn)適用于比較三個(gè)或更多獨(dú)立樣本的分布是否相等。其原理是將每個(gè)樣本排序，計(jì)算每個(gè)樣本的秩和，然后根據(jù)秩和計(jì)算H統(tǒng)計(jì)量，通過比較H統(tǒng)計(jì)量與卡方分布的臨界值來判斷分布是否相等。3.解析：Wilcoxon符號秩檢驗(yàn)適用于比較兩組配對樣本的分布是否相等。其原理是將配對數(shù)據(jù)相減得到差值，然后計(jì)算差值的秩和，根據(jù)秩和計(jì)算統(tǒng)計(jì)量W，通過比較W統(tǒng)計(jì)量與正態(tài)分布的臨界值來判斷分布是否相等。4.解析：根據(jù)Mann-WhitneyU檢驗(yàn)的結(jié)果，計(jì)算U統(tǒng)計(jì)量，然后查找對應(yīng)的臨界值表，若U統(tǒng)計(jì)量小于臨界值，則拒絕零假設(shè)，認(rèn)為兩組數(shù)據(jù)的分布存在顯著差異。5.解析：根據(jù)Kruskal-WallisH檢驗(yàn)的結(jié)果，計(jì)算H統(tǒng)計(jì)量，然后查找對應(yīng)的卡方分布表，若H統(tǒng)計(jì)量大于臨界值，則拒絕零假設(shè)，認(rèn)為多組數(shù)據(jù)的分布存在顯著差異。三、計(jì)算題答案：1.解析：使用Mann-WhitneyU檢驗(yàn)，首先將兩組數(shù)據(jù)合并并排序，得到合并后的數(shù)據(jù)：3,4,5,6,7,8,9,10,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30。然后計(jì)算U統(tǒng)計(jì)量：U=n1*n2-(n1+n2)/2=30*20-(30+20)/2=540。根據(jù)U統(tǒng)計(jì)量和樣本量，查找對應(yīng)的臨界值表，得到臨界值為450。由于U統(tǒng)計(jì)量大于臨界值，拒絕零假設(shè)，認(rèn)為兩組數(shù)據(jù)的分布存在顯著差異。2.解析：使用Kruskal-WallisH檢驗(yàn)，首先將三組數(shù)據(jù)合并并排序，得到合并后的數(shù)據(jù)：1,2,3,3,4,5,5,5,5,6,6,6,6,6,7,7,7,7,7,8,8,8,8,8,9,9,9,9,10,10,10,10,10,11,11,11,11,11,12,12,12,12,12,13,13,13,13,13,14,14,14,14,14,15,15,15,15,15,16,16,16,16,16,17,17,17,17,17,18,18,18,18,18,19,19,19,19,19,20,20,20,20,20,21,21,21,21,21,22,22,22,22,22,23,23,23,23,23,24,24,24,24,24,25,25,25,25,25,26,26,26,26,26,27,27,27,27,27,28,28,28,28,28,29,29,29,29,29,30,30,30,30,30,31,31,31,31,31,32,32,32,32,32,33,33,33,33,33,34,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