数理统计教程课后重要答案习题

..第一章:统计量及其分布19.N2,和S,,2nn1~2,,n1.,n1N,n,12

nn1:n1n1n1N~N2而n1以2n2n~2n12.所n且S2n与n1以,,所n1n1.n1nSnn1n1~tnn1.n1t22,,的子样,即n1nn112设,,i,n,是取自二元正态分布Nii,121n,1nnniii1i1,S21n12S2,n和2nniii1i1ri12S22Sn1.S2rS

S2rSnini1ini2i1i12:E12.2DDD2cov,2n2212nn.1nn服从N布.所以122221212分nnn2nSSS2rS精品文档放心下载222

21212iiiii1i1i1

ni2i1i参考资料..iiS2S与2S.nSS~2n1,22S2rS2221212tn112SS22S2rSn1n1222222服.从1212nSS

S2rS)n1(2221212第二章：估计量设,,1np.:Eppˆn2ax,0xafx;aa2a计n,fx;a2ax,0xaa2a估其它.2xaaˆ3.:E

得a谢谢阅读aax令233a0fxa1x,0x1谢谢阅读aa??:(1)L111inx0xnnnxiiiii1nlnLnln1ln精品文档放心下载x.ii1令lnLnnlnx1i1i0，得ˆL1ni1nxi。参考资料..2Ln0故ˆ1ni1nxi是.ˆ21122L(2)由E111令1得221设.,,1n为取自参数为的普哇松分布的一个子样.试证子样平均和S21().,011S感谢阅读n2*2n1nin

i1.:ED,感谢阅读E1n2E.ESD.*2n1i1ni.又E1S1故与S感谢阅读*2

2nn故1*2Sn.设,,1n,为取自正态母体N,2的一个子样,试适当选择c,使S2c2.n12为i1:由EEjEiii1Dii,且,,i21nE2ijjnc2n1E2n1EcEE2EE感谢阅读22222i1i111

iii12cn1D取c1i2c2n1.,S为.2n1n2参考资料..设随机变量Pnx1nx,x,nxx试求.2:EnEDEn1nnnn1感谢阅读22

22故Enn1.N,感谢阅读222ˆ:i.i22Nnn1.:Ea02xaaˆ3.得aax令233a设1,,nN,2证明(i)Sn21n是22;ni11i.是,n证由in2ii12~n,n.22222nn24n,又N,2xfx112,故f2xe22222222f1对2224x22Ef211E2422424424或2E2L1124。2，故CRn222Sn2是.244.Ei12xixe222220y22参考资料..故Eˆ，即ˆ是.又21n2DˆDEE22n22n2n2n2222i111i2f1而EE22222222ˆ2n0.876。

故R，精品文档放心下载22n2n2.设,,1n.fx1xe,x0i是1nni1:Eix0ex2.是又E2x2e232x22，故Di20i2f211DE感谢阅读n.，而E2422,是.

故CR感谢阅读nD2P2n20n还是.设,,1n,fx;1,x,01，则Txnni1i是.:,,1nfx,,x1nn1xixi参考资料..设取k,,11,,是.12xi,kT1n2nn则inTni1于.:,,1nfx1xnienxx!ii取k12xien.,k2x!,:T1in.第三章：假设检验1设,12,,N(,H0:0,H:1:C():x1xx001x25c,c9~（，~（，）在H0,55（）PC21C感谢阅读00333055C32(1,53C,n,)N(,),2200H0:,H01:0的问题，取临界域C(x1xn):xc0.i.谢谢阅读ii00.5,200.04,0.05,n9求0.65.:在H0P0C2,~（，）0n0nC0P00000n,0参考资料..C00nu1C00nu100u是01）分布的C1在H12,~（，,nCn00n0PP1010un10C0=nnun感谢阅读001000当u1减少；反之当增ii1-。0.650.511un1u3谢谢阅读00.210.950=12.2510.6050.6050.7274.精品文档放心下载4精品文档放心下载~N,，2H0:1600在H0为真下，检验统计量观察值为ux0261.2578,0.05时uu1.96。uu1122，H0，52.640.06测的2.62?取显著性水平0.01。谢谢阅读ED0.06。感谢阅读22H0:2.64。2~N,感谢阅读参考资料..H0u2.622.64un1003.3300.060.01,uu。u3.33u。1122H0，3,

20.81.8,()为:23.4谢谢阅读谢谢阅读?(0.05)谢谢阅读ED1.8。谢谢阅读2H0:20.8，H1:1.827~N,谢谢阅读7H0u21.34un70.3701.80.05,uu1.645。u0.37u11。，H00X,X,XX~N12n0,2nX2i0U=2的1-i12U=ni1Xi022，2参考资料..22Xn2nn2i01i122故2的1-22XXnni0i0i1,。i1nn12216感谢阅读12和身高标准差的0.95置谢谢阅读（,，X170,S12,0.05。谢谢阅读2）2XU~t(n1),tS(nt(24)2.060.975故1n22U(n2~2(n2(n221

2(n22

222故的0.95(,)2(87.80,278.69)39.36412.401的0.9587.80,278.69(9.34,16.69).0.050.01感谢阅读XE(X)精品文档放心下载参考资料..X.当nUXXS0.05

nn布01X0.01n

PXP.即

0.050.05谢谢阅读n0.01nnn0.95,n296.04即0.055感谢阅读5n为或。89X=4.38m.谢谢阅读X==4.40m(=感谢阅读:4.40H(1)H01(2)UX4.40:4.4~N(0,1);nuZ0.9512(,1.64));,U04.384.400.33H0.204.4093666.5S为0.05精品文档放心下载nt352.0301t120.9751H0:设H1:2H0~tn1X0Sn参考资料..30.05t谢谢阅读t352.0301t120.975,精品文档放心下载4U1.42.0301,2.030166.5精品文档放心下载1536以H0,=1.6,谢谢阅读9X=S*2=1.19n=）H:1.6H:1.6222201n1S*2H成立时2U0n12n20~（3）查2分布表，找出临界值22n108n122.02512820拒.9752.180.感谢阅读911.194U3.73H感谢阅读01.620：

=.精品文档放心下载1H:80,:802H201精品文档放心下载1n1H在成立下2U~n2n023n1查精品文档放心下载221n120.959(4)X14265757871595768545562.4。参考资料..SXX1102*29nii1

9U013.7，即在0.051050505.35和6.11精品文档放心下载谢谢阅读感谢阅读0.05Y感谢阅读X~N,5.352,Y~N,6.11212谢谢阅读(1)H:01H1:122(2)选取统计量U~N0，1XY在H0成立下2212nm(3)0.05Z，174.341.67(4)U172.421.6405.356.1122ZH050507.1434563，谢谢阅读46854谢谢阅读：H0:p(p(p(16。H20(n)226i1iinpi2(5)2参考资料..组数i频数ninpnnpniiiii130.92/i298.1310.148-20.455-52.564-63.6215.6，2=11.07由于2。即等概率的假设不成立。2H07.15精品文档放心下载表如下：测试电压3.83.94.04.14.24.34.44.54.64.74.8感谢阅读击穿频数11127884641感谢阅读试对电缆耐压数据作分析检验（用概率图纸法和2解：用正态概率纸检验出数据基本上服从正态分布，下面2假设H0:N(ˆ,ˆ2)其中ˆ,ˆ2为和2的极大似然估计，其观察值精品文档放心下载参考资料..1nˆxˆs(xx)20.0484222nnii1感谢阅读所以要检验的假设H0:N(4.3744,0.04842)分组列表计算2统计量的观察值。频组距xixi1数ni标准化区间yi1yip(y)ii(y)npi1inii2/i4.150.10564.54080.04644.14.27-1.25-0.790.10874.24.38-0.79-0.340.15264.34.5-0.340.570.348814.99844.54.660.571.030.13284.650.312(npn)2ni1iinpi2.4852用0.1查表2(62222，所以不能否定正态0.9分布的假设。7.16用手枪对个靶各打发，只记录命中或不命中，射击结果列表如下谢谢阅读命中数xi：0123456789参考资料..频数fi：024420在显著水平0.05下用2拟合优度检验法检验射击结果所服从的分布。解对每一靶打一发，只记录命中或不命中可用二点分布描述，而对一个靶打十感谢阅读发，其射击结果可用二项分布b(K;10,p)来描述，其中p未知，可求其极大似然精品文档放心下载估计为pxf110100i0ixi0.5设是十发射击中射中靶的个数，建立假设H010:p(k)(0.5)K10K,K0,1,K,10用2拟合优度检验法列表如下：inpnpniiiii000.000970.0980.0982/i1270.9761.074240.009764.3950.0363511.7190.25240.0439420.5210.1075524.6090.07960.1171820.5210.3107811.7190.007840.205214.3950.0369220.9761.07400.246090.0980.098参考资料..40.2052120.1171880.0439450.0097650.000977取0.05，(npn)2103.1712iinpi0i2(111=2由于22H0。7.17在某细纱机上进行断头率测定，试验锭子总数为，测得断头总次数为感谢阅读次只锭子的断头次数纪律于下表。问每只锭子的纺纱条件是否相同？谢谢阅读每锭断头数012345679锭数（实测）31103谢谢阅读感谢阅读布，所以问题是要检验每只锭子的断头数p(K;)。其中未知，求其极大似感谢阅读参考资料..然估计为x列表H0.660:p(K;0.66)，由2拟合优度检验。i断头数Knipinpinii2/i1085.568219.668精品文档放心下载32249.532.684436.026谢谢阅读54-80.00472.06817.016感谢阅读82(npn)25i0iinpi40.962取0.05，2(511)=27.815，读感谢阅取0.01，2(51=20.95由于2。即认为每只锭子纺纱条件不相同。2H0第四、五章：线性回归