王静龙《非参数统计分析》课后计算题参考答案解析

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 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.则接受原假设，即房价中位数是65003.1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4.因为00.05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+1.3272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：5055606565707070757575808080808080808585858585 909090909090959595959595100100100100100100100105105105105 105110110110110110115115120N=55 则对称中心为因为c不是整数，则为70与75之间，即为72.5习题五1.1228002520026550265502690027350285002895029900301503045030450306503080031000313003135031350318003205032250323503275032900332503355033700339503410034800350503520035500356003570035900361003630036700372503740037750380503820038200388003920039700404004100050个和在一起的中位数是（33250+33550）/2=33400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=501.227.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在区间内则认为11月和12月的波动相同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b1，则认为机器一更有机会改进质量*答案是自己做的，但是有一次发现有个地方错了啊，还没改过来，仅作参考！enjoythetrustof得到.的信任have/puttrustin信任intrust受托的，代为保管的take.ontrust对.不加考察信以为真truston