王静龙《非参数统计分析》课后计算题参考答案

1、王静龙非参数统计分析课后习题计算题参考答案习题1.One Sample t-test for a MeanSample Statistics for xMeanStd. Dev.Std. Error261.388.201.61Hypo thesis TestNull hypo thesis: Alternative:Mea n of x =0Mea n of x 人=0t StatisticDfProb > t0.861250.397695 % Co nfide nee In terval for the Mean-1.934.70Lower Limit:Upper Limit:则接受原

2、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

3、.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 =39H0: me = 6500 出：me c 6500P S + <13=BINOMDIST(13

4、,39,0.5,1)=0.026625957mJi=0J丿另外：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I * n m = i nf m : I -i 12n .*i =0BINO MN V(39,0.5,0.05)=14d=supd :"> =m-1 = 13S+ =13" =13以上两种都拒绝原假设,即中位数低于65001.2Y m*s丿i =0Zalli 丿2 jBINOM.INV(40,0.5,1-0.025)=26d=n-c=40-26=14Xg = 5800 X2

5、6 = 6400me 二 X20 = 62002.S+=40 n=70H0: me = 6500 H1 : m 65002P S+>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 = 20791 1Hq: p = 2 H1: p>- n比较大，则用正态分布近似ps+ >1552* II7519.75另外:S+=1552 n =1552 +527 =2079I *(1 )nJ 仆alI 12尼U丿2JI * m = inf m : I-i12丿+1Q39.

6、5-1552+Q.515.33E-112丿-n m*Zi=0:(n IJ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1522P的比估计是：需討7465134.s + = 18154 n = 157860H 0: P0.906 = 65 H1: P0.906 >65p =1 -0.906 =0.094s+ b(n,p)P s+>18154 =1 BINOMDIST(18153,157860,0.094,1)因为0 0.05则拒绝原假设=0习题四1.车辆添加剂1添加剂2差值符号差的绝对

7、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符号秩和检验统计量：W+=6+8+10+1+4+12+9+11+2+7=70P

8、值为2P(W+>70 )，当 n=12得Co.o25 =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+1523

9、232016+16242423-23-232525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6.025=236所以P值小于2P(W+ >236尸0.05即接受原假设符号检验：S* =18 n =26H 0:me =0 Hj :me 工02P S + >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

10、3.零售店豪华车普通车差值差值-100绝对值秩139027012020205239028011010102345035010000438030080-20205540030010000639034050-50508735029060-40407840032080-20205937028090-101021043032011010102(1)W+ =5+2+2=9 n =8 查表可得：0).025 = 33d _n(n+1) c _3U0.025 2一'0.025 Q2P (W+ <3) =0.052P(W+ <9)0.0550556065657070707575758080

11、80808080808585858585 909090909090959595959595100100100100100100100105105105 105 105110110110110110115115120AN=55则对称中心为日则接受原假设零售 店豪华 车普通 车差值Walsh 值 Walsh=(Xj+Xj )/21<i<j< ni=1i=2i=3i=4i=5i=6i=7i=8i=9i=1013902701201202390280110115110345035010011010510043803008010095908054003001001101051009010

12、06390340508580756575507350290609085807080556084003208010095908090657080937028090105100958595707585901043032011011511010595105808595100110Walsh平均由小到大排列:= W(N 十/2 厂 W28 = 90d =n(n+1 )/4-0.5-Uy/2 Jn(n +1 X 2 n+1 )/24 =27.5-0.5-1.96% JlOx 1仔 21/24 = 7.77101146 c =n(n +1 y4 +0.5+5£2 Jn(n +1 )(2 n+1

13、y 24 = 27.5+0.5+1.96咒山0咒 11% 21/24 =47.22898853A因为c不是整数，则0L介于w（k）与w（k+1）之间，其中k表示比d大的最小整数即为8A日L为70与75之间，即为72.5则H-L的点估计为9095%的区间估计为172.5,105习题五228002520026550265502690027350285002895029900301503045030450306503080031000313003135031350318003205032250323503275032900332503355033700339503410034800350503520

14、03550035600357003590036100363003670037250374003775038050382003820038800392003970040400410001.1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=5050个和在一起的中位数是(33250+33550) /2=33400P =2(i,24,25,50) =0.005060988p值很小，则拒绝原假设即认为女职工的收入比男职工的低。1.2Wilcoxon 秩和 W女=1+2+3.5+5+6+7+8+10+11.

15、5+11.5+13+15+16+17.5+17.5+20+22+24+26+29+31+32+34+35+36+44.5=478因为N=n+m=50，查不到表，则用其渐进正态分布求解 n=26,m=24,N=50, % = 478+(w女-n(N +1)/2p=2P(W <478 F© I 女 I IJmn(N +1)/12 丿则拒绝原假设，认为女职工的收入低=0.0003276432.Wilcoxo n 秩和 Wb=1+2+3+4+5+6+8+10+12=51 因为N=n+m=19，则用其渐进正态分布求解 n=9,m=10,N=19,W3 =51p=2P(W+ 51 )=2*

16、0 i忙n(N T)佟lJmn(N +1)/12 丿则拒绝原假设，认为A比B的作用好=0.001450862指数1116(11 月)1120(12 月)11251125113011471149114911511152秩12345678910指数1155116111661169117111761182118411841194秩111213141516171819207.Wilcoxon 秩和 W11月 =1+5+7+8+10+11+16+18+19+20=115 因为 n=n 1+ n2=20,n1=10,n 1 = 10,n=20, W|1 月=115所以，p=2P (W+>115)查表

17、可得： 5.025 = 158因为，158>115，所以p值一定大于0.05则接受原假设，认为11月和12月的波动相同位置参数检验:d =nm /2-O.5-U1£2 J nm(N +1)/12 =10*10 / 2-0.5-1.96710*10*101/12 =23.64 c =n m/2 +0.5 +642 Jnm(N +1)/12 =10*10/2 + 0.5 + 1.96J10*10*101/12 =76.36AA日L介于w（k）与w（k+1）之间，其中k表示比d大的最小整数即为24 6 L =-29AA日u介于与wc之间，其中c 表示比C小的最大整数即为76日U =1

18、7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myay5.053344135.1665.5342.255.55.198825685.21210.5182.2510.55.221514100145.251716.556.2516.55.27191925195.282121.56.2521.55.282221.56.2521.55.282321.56.2521.55.2924240245.32526.56.2521.55.32626.56.2521.55.32726.56.2521.55.32826.56.2521.55.333029.530.2518.55.34323264165.34333264165.353535121135.353635121135.363