...wd...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 Mean Sample Statistics for x N Mean Std. Dev. Std. Error ------------------------------------------------- 26 1.38 8.20 1.61 Hypothesis Test Null hypothesis: Mean of x = 0 Alternative: Mean of x ^= 0 t Statistic Df Prob > t --------------------------------- 0.861 25 0.3976 95 % Confidence Interval for the Mean Lower Limit: -1.93 Upper Limit: 4.70 那么承受原假设 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习题三1.1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的最小整数以上两种都拒绝原假设，即中位数低于65001.22.那么承受原假设，即房价中位数是65003.1那么拒绝原假设，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，那么4.因为0〈0.05那么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+1.3272．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+16242423-23-2325253．零售店豪华车普通车差值差值-100绝对值秩139027012020205239028011010102345035010000438030080-20205540030010000639034050-50508735029060-40407840032080-20205937028090-101021043032011010102〔1〕〔2〕零售店豪华车普通车差值Walsh值i=1i=2i=3i=4i=5i=6i=7i=8i=9i=101390270120120239028011011511034503501001101051004380300801009590805400300100110105100901006390340508580756575507350290609085807080556084003208010095908090657080937028090105100958595707585901043032011011511010595105808595100110Walsh平均由小到大排列：50 55 60 65 65 70 70 70 75 75 75 80 80 80 80 80 80 80 85 85 85 85 8590 90 90 90 90 90 95 95 95 95 95 95 100 100 100 100 100 100 100 105 105 105 105105 110 110 110 110 110 115 115 120N=55 那么对称中心为因为c不是整数，那么为70与75之间，即为72.5习题五1.122800 25200 26550 26550 26900 27350 28500 28950 29900 30150 30450 30450 30650 30800 31000 31300 31350 31350 31800 32050 32250 32350 32750 32900 33250 33550 33700 33950 34100 34800 35050 35200 35500 35600 35700 35900 36100 36300 36700 37250 37400 37750 38050 38200 38200 38800 39200 39700 40400 4100050个和在一起的中位数是〔33250+33550〕/2=33400工资<33400元工资>33400元合计男职工N11=7N12=17N1+=24女职工N21=18N22=8N2+=26合计N+1=25N+2=25N=501.22．7.指数1116(11月)1120(12月)11251125113011471149114911511152秩12345678910指数1155116111661169117111761182118411841194秩11121314151617181920所以，区间为：[-29,17] 即0在区间内那么认为11月和12月的波动一样8机器〔y〕平均秩myay5.053344135.1665.5342.255.55.198825685.21210.5182.2510.55.221514100145.251716.556.2516.55.27191925195.282121.56.2521.55.282221.56.2521.55.282321.56.2521.55.2924240245.32526.56.2521.55.32626.56.2521.55.32726.56.2521.55.32826.56.2521.55.333029.530.2518.55.34323264165.34333264165.353535121135.353635121135.363837.5182.2510.55.384039.5240.258.5和511.52269.25346.5因为b<1，那么认为机器一更有时机改进质量*答案是自己做的，但是有一次发现有个地方错了啊，还没改正来，仅作参考！。
