.......学习帮手.王静龙《非参数统计分析》课后习题计算题参考答案习题一1.OneSamplet-testforaMeanSampleStatisticsforxNMeanStd.Dev.Std.Error-------------------------------------------------261.388.201.61HypothesisTestNullhypothesis:Meanofx=0Alternative:Meanofx^=0tStatisticDfProb>t---------------------------------0.861250.397695%ConfidenceIntervalfortheMean.......学习帮手.LowerLimit:-1.93UpperLimit:4.70则接受原假设认为一样习题二1.描述性统计乘机服务机上服务机场服务三者平均平均79.78平均54.46平均58.48平均64.24标准误差1.174661045标准误差2.085559681标准误差2.262605标准误差1.016376中位数82中位数55.5中位数58.5中位数64.5众数72众数60众数52众数65标准差8.306107908标准差14.74713393标准差15.99903标准差7.186861方差68.99142857方差217.4779592方差255.969方差51.65098峰度-1.059134152峰度0.083146927峰度0.41167峰度-0.04371偏度-0.164016852偏度0.264117712偏度-0.26232偏度-0.08186区域32区域65区域76区域34.66667最小值63最小值25最小值16最小值44最大值95最大值90最大值92最大值78.66667求和3989求和2723求和2924求和3212观测数50观测数50观测数50观测数50置信度(95.0%)2.360569749置信度(95.0%)4.191089091置信度(95.0%)4.546874置信度(95.0%)2.042483.......学习帮手.习题三1.1+01=1339:6500:650013=BINOMDIST(13,39,0.5,1)=0.026625957SnHmeHmePS另外：在excel2010中有公式BINOM.INV(n,p,a)返回一个数值，它使得累计二项式分布的函数值大于或等于临界值a的最小整数***0*0+1inf:2BINOM.INV(39,0.5,0.05)=141sup:1132S1313nmindinmminddmid=以上两种都拒绝原假设，即中位数低于65001.2****01426201inf:221inf:122BINOM.INV(40,0.5,1-0.025)=26d=n-c=40-26=14580064006200nnicnminccinmmixxmex.......学习帮手.2.+01=4070:6500:65002402*(1-BINOMDIST(39,70,0.5,1))=0.281978922SnHmeHmePS则接受原假设，即房价中位数是65003.1+01=15521552527207911::22n1039.5-1552+0.51552=5.33E-112519.75SnHpHpPS比较大，则用正态分布近似**+**0:=1552155252720791inf:221inf:122m=BINOM.INV(2079,0.5,0.975)=1084nnicnmiSnnccinmmi另外则拒绝原假设，即相信孩子会过得更好的人多3.2P为认为生活更好的成年人的比例，则1522=0.7465132079p的比估计是：4........学习帮手.00.90610.90618154157860:65:6510.9060.094~(,)181541BINOMDIST(18153,157860,0.094,1)=0SnHPHPpSbnpPS因为0〈0.05则拒绝原假设习题四1.车辆添加剂1添加剂2差值符号差的绝对值绝对值的秩122.3221.251.07+1.076225.7623.971.79+1.798324.2324.77-0.54-0.543421.3519.262.09+2.0910523.4323.120.31+0.311626.97260.97+0.974718.3619.4-1.04-1.045820.7517.183.57+3.5712924.0722.231.84+1.8491026.4323.353.08+3.08111125.4124.980.43+0.4321227.2225.91.32+1.327.......学习帮手.++0.025+W=6+8+10+1+4+12+9+11+2+7=70p2PW70n=12c=65p2PW65=0.05符号秩和检验统计量：值为，当得所以值小于即拒绝原假设2．被调查者x符号绝对值一个随机秩平均秩110+0004-1-112.561+122.5211+132.524-1-142.512+2575-2-26782+277152+287252+297143+31010.519-3-31110.524+4121474+41314124+41414174+41514.......学习帮手.224+4161495+51717.5105+51817.5188+819192611+1120203-13-13212113-14-1422221615+1523232016+16242423-23-232525++0.025+W=2.5+2.5+7+7+7+7+10.5+14+14+14+14+14+17.5+17.5+19+20+23+24=234.5p2PW234.5n=25c=236p2PW236=0.05符号秩和检验统计量：值为，当得所以值小于即接受原假设011826:0:02182*(1-BINOMDIST(17,25,0.5,1))=0.043285251SnHmeHmePS+符号检验：则拒绝原假设tt=0.861df=25p=0.3976检验：统计量接受原假设3．.......学习帮手.零售店豪华车普通车差值差值-100绝对值秩139027012020205239028011010102345035010000438030080-20205540030010000639034050-50508735029060-40407840032080-20205937028090-101021043032011010102（1）+0.0250.0250.025++=5+2+2=9833(1...
