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1、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.One Sample t-test for a MeanSample Statistics for xN Mea n Std. Dev. Std. Error 261.388.201.61Hypothesis TestNull hypothesis: Mea n of x = 0Alternative:Mea n of x A= 0t Statistic Df Prob t0.861250.397695 % Con fide nee In terval for the MeanLower Limit:-1.93Upper Limit:4.70
2、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217.477959方差255.969方差51.650928峰度-1.059134152峰度0.083146927峰度0.41167峰度-0.0437
3、1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习题二1.1S+=13 n =39Ho: me 二 6500： me ： 6500PS 3l=BINOMDIST(13,39
4、,0.5,1)=0.026625957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r1 *广1、n m*n m :Eii=0J丿 CL jBINO MN V(39,0.5,0.05)=14v d* rd=sup d* :*2)ZS+ =13 = 13以上两种都拒绝原假设,即中位数低于65001.2r1 *5、n m (n xa |m = i nf $ m :S |1 1 1I2z U丿2j. * i -c1JIn) a nzBINOM.INV(40,0.5,1 -0.025)=26d=n-c=40-26
5、=14x14 二 5800 x26 二 6400 me = x20 = 62002.S+=40 n =70H0: me =6500： me= 65002PS 40； -2*(1-BIN0MDIST(39,70,0.5,1)=0.281978922则接受原假设，即房价中位数是 65003.1S+=1552 n =1552 527 二 207911Ho : p 出：p 22PS -15521039.5-1552+0.5519.75=5.33E-112n比较大，则用正态分布近似另外:S+=1552 n =1552 527 = 2079I *(1 5 c =inf c : In m*zi =012 丿
6、 i-1 -2I *(1 : m = inf m : 12丿m=BINOM.INV(2079,0.5,0.975)=1084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p的比估计是：1522=0.74651320794.S” =18154 n =157860H 0 : F0.906 = 65H 1 : P0.906- 65p =1 -0.906 =0.094S b(n, p)P fS _18154： -1 -BIN0MDIST(18153,157860,0.094,1)=0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
7、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符号秩和检验统计量：+W =6+8+10+1+4+12+9+11+2+7=70p值为2P(
8、W+70 )，当 n=12得5.025=65 所以p值小于2P W+ -65 =0.05 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+41514224+4161495+51717.5105+51817.5188+819192611+1120203-13-13212113-14-1422221615+1523232016+1624242
9、3-23-232525符号秩和检验统计量：W+ =2.5+2.5+7+7+7+7+10.5+14+14+14+14+14+17.5+17.5+19+20+23+24=234.5p值为2P(W+ 234.5，当 n=25 得c.25=236所以p值小于2P W+ -236 =0.05即接受原假设符号检验：+S =18 n =26H 0: me = 0 已：me = 02PS _18.；=2*(1-BIN0MDIST(17,25,0.5,1)=0.043285251则拒绝原假设t检验：t 统计量=0.861 df=25 p=0.3976接受原假设3.零售店豪华车普通车差值差值-100绝对值秩139
10、027012020205239028011010102345035010000438030080-20205540030010000639034050-50508735029060-40407840032080-20205937028090-101021043032011010102(1)W+=5+2+2=9 n =8 查表可得：00.025 =33 d -n(n 1) c _30.0250.025 22P(W+ 乞3) =0.052P(W+ 乞9)0.05Walsh平均由小到大排列：50 55 60 65 65 7085 9090 90 90 90则接受原假设(2)零售 店豪华 车普通 车差
11、值Walsh 值Walsh=(Xj+Xj )/212 兰 jW ni=1i=2i=3i=4i=5i=6i=7i=8i=9i=1013902701201202390280110115110345035010011010510043803008010095908054003001001101051009010063903405085807565755073502906090858070805560840032080100959080906570809370280901051009585957075859010430320110115110105951058085951001107070757575808080 808080808585 858590959595959595100
