数理统计课后题答案完整版

1、第一章3.解：因为Xi所以xacyi27=i0所以Xi2symicy一nacyniia-Yiniicyxacy成立因为s2i03521.5322291.54121.5341.5440.252Sxi002sy4.4025cycyicyyicy2i一又因为Syyiynii所以S2c2S2成立6.解：变换yii0x27*xyi-35-91234mi2341y-miyiniii35293i2434i01.5身154158162166170174178高：15：16：16：17：17：17：188260482组156160164168172176180中值学101426281282生数7解:*miiX1

2、li00i66i56i0i60i4i6426i72i2i6828i768i802s2-101001217233.44miixi15616621668141601662222616416628168166176216621801668解：将子样值重新排列（由小到大）-4,0,0,Men12X70Me9解:Xi3.217.21X81.282221n1n1xini11n2n2xj%j1nixin2X2XiXi服从分布n1n2n1n2n1n2n2n1n2所以15.解:因为Xi:N0,1i1,2,nX1X2X3:N0,3ixi1xjj1n.一2n2x2n1n2n1n2n12S1一2x1n2s；22x2n

3、1x1一2n2x2n1n2n1n2n12S12n2s2n1一2x1n22x2n1x11一2n?x?n1n2n1n2n1n2n12S12n2s2n1n1n一22x1n2n1:n2x2n1n2n12n22S12n2S2_212n1ngxn1n2x22mn2x1x2n1n22n1n22n1S12n2S2n1n2x1一2x2n1n2n1n221122xi_2xnnn2i1n2E_23一30D所以X1X2X3同理审于1x1n2x2可知12.解:xi:PEXiDxii1,2,nX1X2X3,3N0,1EXe1nEx1DXd1n12nDxi13.解:a,bExDxii1,2,n在此题中ExDx121,2,n

4、EXe1nEx1DXD1nnDxii13n14.解：因为Xi：EXXi所以Xi0,11,2,n,2.由分布定义可知X1X2X3-3X4X5X632,分布的可加性,X1X2X3316.解:X4X5,3X6(1)因为Xi:N0,二n0,11,2,n所以Fky因为所以(2)fY1因为XiY1-2dxXi:Xi一:N0,1nx2n2n1y20,1,2,n2 ny0所以nXinY22故fY,ynyFy2ypY2PnY2nydx17.解：因为fY2ny存在相互独立的fY2nn_22n2y2nn22ny(3)因为所以fY3yf21Xi:NXii-:Nn:NXi.n0,0,1yf21(4)因为所以Fy4y0,

5、11,2,nX2u2彳V由定义可知xdx18解：因为12U2:2:F1,nXi:N0,1,2,：2-xeXi:<2nyeN0,Xi一、.nXi、n2n0,1W:1,2,ny_pY4f21xdx0'F4yf21nXi:i1nnmXi所以(2)所以因为Y20,1XiXiXinXii1nmXi2n10,1Xii1.nnmXi21,2,nmXiXi21mX1n2:Fn,mnmXi19.解：用公式计算20.019090V29OUo.0i1X in ) ln( 1P )查表得U0.012.33lnL(P)nIn工,20.01909031.26121.26dlnLdp20.解:因为2n解之得n

6、nXii1上2由分布的性质3可知3.解：因为总体X服从uaba,Xn-=：N、2nE(X)2令E(X)=XX)b)所以a-b)2120,1S2na+b/2n2nn(Xii1XD(X)X)2=S2,limPn1.xe从而有2.0,1).E(x)k(11b)212XS2、/3S、2nxdx令»p所以有2).其似然函数为L(P)n(1P)“1'1,2net27dtxe,2n4.解:L()lnL(1)dlnLxdx解之得：(2)母体E(x)k1p)pk(1p)k1x1n/pp(1nXin设x1，x2,Lxn为样本观察值则似然函数为:n(lnX的期望而样本均值为:X1nXni1令E(x

7、)Xi)1,0-1xi1,i1,2,L,nlnxinln1xin1nlnxii1n-nlnxii1xf(x)dxxi1X5.。解：其似然函数为:xdxp)i1i 112L()n13ei2InL()nln(2)得:xi(2)由于i1201(2jxiXxedx2Xxe一01n,.1n,1E()'J)njxi)nnxini1的无偏估计量。6.解：其似然函数为:L()k(k)!x(k1)exexi()n(k1)!x(k1222.22.20.40338.解：取子样值为则似然函数为:12(xhx2,L(lnL()nkIn(k1)ln(inXi)1xn),(xi(xi)exiXe-dx01)exid

8、InL()dnknXii1解得nk-nXii1Xf(x)1-,0x7.解：由题意知：均匀分布的母体平均数方差2321212用极大似然估计法求得极大似然估计量似然函数：L()选取使L达到最大nlnL()(xii1要使似然函数最大，则需9.解：取了样值(x1x2,则其似然函数L(nxii1取min(x1,x2,=min(x1,x2,xn)(xi0)xixn)nxii1nXi1lnL()lnL()dnlnxi由题中数据可知-1x(3655100024515150250.0510.解:极差xi100xii135704545552565)200minximaxxii1(i)1in(2)maxx1in由以

9、上结论当抽得容量为6的子样数值,时20(1)由题中子样值及题意知:6.21.54.7查表2-10.42994.72.0205平均极差R0.115,查表知0.32490.1150.045510.4299ds110.3249d1011/解：设U为其母体平均数的无偏估计，则应有-1又因x,(81403106226)4601.1,2.2即即知4212.解：XN(,1)E(xJ,D(Xi)1,(i1,2)_1_2_则E(1)EX1一EX23313E(2)EX1EX24411E(3)2EX12EX2所以三个估计量1,q均为I,2,37,7的无偏估计CD(Xii11)D(Xi)0C22C(n1)i1要使E(

10、2只帝C12(n1)所以当C15.证明:小时2.2为的无偏估计。参数的无偏估计量为0,由切比雪夫不等式d依赖于子样容量n21D()叱X13X2)41DX1DX299limDn0故有limpn151同理可得D(2)5，D(2)182即证为的相合估计量。可知3的方差最小也亦2最有效。16证明：设X服从B(N,p),13解：XP()E(X),D(X)则分布律为P(Xk)cNPk(1P)k(k1,2,N)2E(S)1n,I。X)2这时E(X)NPD(X)NP(1P)1n22Ki1E(Xi)nE(X)EX2DX(EX)2NP(1P)N2P21n12)n(n2)所以E(P)(n2即S是的无偏估计dpD4N

11、P(12P)NNn罗一克拉美下界满足EXNPNNP(1P)NnP(无偏)又因为1nE(X)E(1Xi)ni11E(nXi)1nKK-n-LnCNPK(1IRk0pP)NP2CNPK(1P)NP即X也是的无偏估计。又0,1_*2_*2E(aX(1)S)E(X)(1)E(S)EXi1NK_2K_K_NKn(LnCNKLnP(NP)Ln(1P)2CnPK(1P)K0PNKnPK0PNP,2_KKNKTV1Cnp(1P)(1)2因止匕X(1)S也是的无偏估计2、14.解：由题意：xN(,)因为n1E()2CE(XXi)2CD(Xi1Xi)(E(Xi1Xi)2i1222_2EX22NEX2EX2N22N

12、EXEX2n2P2P(1P)(1P)2NF(1P)nPnNnNN2P2N2PNP1P)n2p2(1P)2N22N2PNR1P)N2P2(1P)2P(1P)P(1P)rr所以IRDP即p为优效估计nN1()-17.解：设总体x的密度函数f(x)一一e22似然函数为得置信区间为(xus)n已知10.95s=40X=1000查表知u-21.96代入计算得2(Xi)2n(xi)nii21222-o22L()e2(2)2e2ii.2所求置信区间为(19.解：(1)已知)0.01cm(Xi)2LnL(2)2Ln22Ln2-则由UXN(0,1)PU.nu)-2解之得置信区间(又dLnLd2n(Xi1)2(X

13、i)2将n=16X=u7u0.051.6450.01因为(Lnf2X)2f(x)dx)2(X)22edx=Y【E(X)4E(X)2224=74824故2的罗一克拉美下界,241n2E(-(Xi)ni1Rn2又因E1n22-E(Xi)ni1口21n224且口()D(-(Xi)ni1n22所以是的无偏估计量且IR2D()22故是的优效估计18.解：由题意：n=100,可以认为此为大子样,X所以U近似服从N(0,1)S.nPUu1代人计算得置信区间(2)未知解得置信区间为(又S220.1)PTt)2sht_).n2n=16t(15)1t0.05(15)1.7530.00029代入计算得置信区间为(解

14、：用T估计法Xt(n1)S/-：?n解之得置信区间(X将X6720S代入得置信区间为()。PTSn；t(n1)1"2*一SXt)n2220n=10查表t0.025(9)2.262221.解：因n=60属于大样本且是来自故由中心极限定理知nXinpi1np(1p)p山上vnp(1p)u1"2(01)分布的总体,nXnp近似服从N(0,1)np(1p)解得置信区间为(Xp(1p)uXp(1p)u)nwn工本题中将UA代替上式中的X由题设条件知UL0.25P(1P)nUn(nUn)0.05525.解：因Xn1与X1,X2,Xn相互独立,查表知UnU0.0251.96所以Xn1与X

15、相互独立，故代入计算的所求置信区间为（）Xn22.解:2,未知故XUN(0,1)、n12XN(0,(1-)n由PUu)2解得置信区间为q（X区间长度为2u,n2于是2un2L计算得nu,n2i42T2U,723.解:未知，用22估计法2(n1)S2P12_(n1)22(n1)2(n21)1解得的置信区间为（_2(n1)S2(1)当n=10,S=时.2查表0.005(9)=20.995(9)=代人计算得的置信区间为（）(2)当n=46,S=14时2查表0.005(45)=20.995(45)(n1)S2即为所求2(n1)又因nS22"2（n1）且与Xn1X相互独立,有T分布的定义知26

16、.解：因Xi代入计算可得24.解：（1）先求的置信区间为（的置信区间由于）未知Xn1Xn1(nN(YjN(X“Tt(nS.n1)KT得置信区间为（XStn万经计算X5.12S0.2203查表t0.025(19)2.093n=20代人计算得置信区间为（2）未知用统计量(n1)S2-22(n1)P2)2得的置信区间为_2(n1)S_2(n1)S2)12查表(2025(19)=2，一、0.975(19)=代入计算得的置信区间为（nnS21)2Xn1XSn1n1t(n1)2)1,2,2)1,2,所以(X1)N(0,)，m由于X与Y相互独立，则(X1)即(X又因2msx(Y(Y2)N0,(一m(Y2)2

17、2(m1)2nsy2emsx则2构造t分布(X2msx27.2nsy2(m(X1)2222)N(0,)n2-)nN(0,1)2)(Y22(n1)2)1)2nsy(Y2)t(mn2)2mn21mn证明：因抽取n>45为大子样*2(n1)s22(n1)、2(n1)2（n1）由2分布的性质3知近似服从正态分布N（0,1）所以PU2PUu12计算得置信区间为2(n1)3(n1)2(n1)u22(n1)u-2S(X1X2UJVS22mX2u22、，的置信区间为12n1u22n1u2，一._2_2把X11.71X21.67S10.035S220.0382nm100uu0.0251.96代入计算得置信

18、区间(0.0299,0.0501)22228.斛：因12未知，故用T统计重TX丫(12"m2)11sw-,nm其中sW(n谪1)S2nm231.解：由题意，u1,u2未知，则*2/S2/2FF(n21,n11)S1120.05n81.625145695,丫L(n2查表to.o25(4)2.144则PF(n21,n11)1.2经计算得76.1252101.554,Sw123625代入得PF(n21,51-2_*21)工F(n21,n121""22F_("21)21色1S22)Sw11一一5,511.9237nm2解得122的置信区间为故得置彳t区间(6.4

19、237,17.4237)1,n12S11)2,F2,S22(n21,n12132S229解：因122故用T统计量_*2_S10.245XaYb1swn-2(n1)S计算得置信区间为(XaXbSwt_(n2把Sw22)t(n-2(m1)S22)n(7)=2代入可得所求置信区间为(30.解：由题意用U统计量m2)Xb2S20.3570.05F0.025(5,8)4.82t_2F0.975(8,5)F0.025(5,8)14.820.207SWt_(n2m2).-)nm2带入计算得1-的置信区间为:2232.解：2未知，则(0.142,4.639)。即:1)X1X2(12)U122123N(0,1)

20、S12S22；nmPTt(n1)1有：P(n*S1)1nS/则单侧置信下限为：Xt(n1)上一将X6720S220nyr2xyn10to.o5(9)1.833带入计算得6592.471即钢索所能承受平其中XyniXYin2Xii1n2Yii1均张力在概率为95%的置信度下的置信下限为6592471。33.解：总体服从(0,1)分布且样本容量n=100为大子样。令X2.解:2690.14XY2736.511为样本均值，由中心极限定理nXnPn2N(0,1)又因为2S22mx451.112my342.665代入得所以PnXnpunS2xyxy-2mx2736.5112690.140.8706451

21、.11则相应的单侧置信区间为90.140.87062667.508822my22mx342.6650.87062451.110.7487将X=S2m(1m)0.60.94uuo.051.645nn代人计算得所求置信上限为即为这批货物次品率在置信概率为95%青况下置信上限为。3证明:dod1d0uvuvd1u2-2udod134.解:由题意:2(n1)S22(n1)doi1xcd1XQd1YiCod0ycdo(n1)1解得的单侧置信上限为2(n1)S12(n1)d1Xcd1xqd1其中n=10,2,查表(n1)2-0.95(9)代人计算得的单侧置信上限为。第五章1.解：对一元回归的线性模型为Xi

22、1,2,nd0d0d1d1xYin离差平方和为QQyii1Xi的偏导数,并令其为0,YiXid0变换得一XYini1因为所以解此方程得xy2XYiXiYi2YiXYi22Xixy2xy-2TxyxXIi11d2Xidovd°vYiCodoCodod1xixdod1Codod1C1c1d12一22ndoi1ViUidoMd0VC0d。d0d0d0d0d1Ui*2xyxy2mx15586210.4210.436.95636.9511.32mx*2*2xiCd1510929.843236.952812.373.517d0d14.解:假设H0:38H1:38用T检验法拒绝域为yi260025

23、002400230022002100200019001800支数y品质指标2_my34527.46代入得2211.2xy76061.67635.353132.130xyxy2-mx5.解:查表彳导t0.025xi将上面的数据代入得2mx3.1824所以接受H0即认为6.解：(1)由散点图看,1.89为38t0.0253X的回归函数具有线性函数形式,76061.67635.3532211.2“”15.98认为长度对于质量的回归是线性的O121198710度长9132.1302211.215.9835.3532776.1422_mx34527.462的无偏估计量(2)将x17.5y9.49xy17

24、9.37215.98132.130786.692mx72.92xyxy2mx2my2.45代入得179.3717.59.490.18272.922020786.6918y210.4874.10xy15582my10929.84代入得(3)9.49x6.30516时y00.18217.50.182xa16b06.305由T分布定义Y0xOxOn1 1n一 2x为i 12X111Y0x0正规方程为X1X2k.025n20.95最小二乘估计为所以Y0的预测区间为X1X22X1YX22x?yx2yxix22xyx22%X222X1%物也2*2丫%2X1X222X1X212n一*Xt0.025n21n2

25、X)X_2XX一*X)ta.025n211n2X)X_2XX其中x1yyiX2yXi2yii1查表得t0.02542.776X1X2Xi1Xi2将（2）的数据代入得11解:*2nV42.450.182272.920.00750.0866计算得Y）的预测区间为8.9521,9.47219.解:利用第八题得到的公式(1)21y141.2Xy3138m.290xyxy代入得2mX313821141.21.929010.1Xi12xjXj2j1,215采用线性回归模型Y1Xx2X2X15yi248.25y16.55i1151515yi24148.3125X1920x256734i1i1i115X161

26、.33Xi27257X2483.8i11515ii115Xi1X21Xi22352448944536615141.21.92性21回100.88X1V15170xi2yi12063925离差平方和为1,yiyi15可变换为2Xi2i,i1,2,nL112X11515Xi115673456426.66307.34L22152X215yii1iXi12Xi2L1215X235244893510936.613552.41515Xi1Xi215i1Xi115Xi2i14453664450962702的偏导数并令其为1Xi11Xi1Xi1X15L1yXM115152Xi22Xi2Xi1Xi2于是15为2

27、乎12Xi1yiX2Xi1Xi2Xi1Xi21n22 Xi2i1X15i1115Xi215i116.55li115yii1151701522656yi120639.25307.3427027013552.4,56,L156所以y10.504120103.25L1yL2y536565360可得10.21仅0.04X253621752.93 1085.62120012.解p18采用线性回归模型于是y4.582L1085.623155.78336412003364355723x318146381.27718x1i1183231.5X111.944xi2758X2215L2y2216.5可得3231.

28、5L12216.542.11L3y7593759318xi3i1182xi222143507618182x11818X3181232xi318x1i13078642X14321.02所以第三章1.解:L222xi218181818X1X2L2118xi2xi318X1VL1y18L2yL3y1843.651.78x1假设：H°：4321.022568.051752.97由于3507631920.223155.782xi31818x33078942723223557210139.5181x2i11818x1i118x210139.59053.881085.629659818xi1xi3

29、i19659818xi2xi3i118xyii1GV0.08x20.16x326,H1:265.2已知，故用统计量uJN(0,1)因显著水平这时，就接受H0u的拒绝域u27.56265.240.05，则u1.21.211811820706.2118-x118i16382518x2yii118xi3yii11818xi11181811818i1X218754211818xi3i118x327645264451200i1182.解：x已知，故u；-0,nu"2u_u0.0251.961N(0,1)18xi396598932343364i120706.217474.73231.5yi638

30、2561608.52216.518yi18754217994975933.2u_2u的拒绝域u5.32532因显著水平1.100.01，则u0.0052.576故此时拒绝H0：u5检验u4.8时犯第二类错误的概率22ndxi 1i 1令tu21%x近似u0N(0,1)拒绝域为uS-,nu由2.2t22dt4.580.58et22dtn200o0.973x0994s0.16?0.05此。251.96(4.58)(0.58)0.0210.162.2001.8331.96(4.58)(0.58)1故接收H0,认为新工艺与旧工艺无显著差异。0.99999790.719900.71806.解：由题意知，

31、母体X的分布为二点分布B(1,p),3.解:假设H0：3.25,f：3.25用t检验法拒绝域t(n1)20.01,作假设H0:pp0(p00.17)此时xm(m为n个产品中废品数)n因n400很大，故由中心极限定理知x近似服从正态分布。x3.252查表t0.0112(14)4.6041*2s0.00017,s0.0130代入计算T0.344t0.0112(14)mponPo(1Po)nN(0,1)故接受H0,认为矿砂的银含量为3.254解：改变加工工艺后电器元件的电阻构成一个母体计算得拒绝域为则在此母体上作假设H0:2.64,用大子样检验拒绝域为PoPo(1Po)U22.把m56,n400,u

32、uo.0251.96，Po0.17由n200,x2.62,s0.06,0.01代入mp00.140.17n0.031.960.01880.037查表得u2.5752xo0.0203.330.06102.575故新加工工艺对元件电阻有显著影响5.解：用大子样作检验Ho:即接受Ho,认为新工艺不显著影响产品质量。7解：金属棒长度服从正态分布原假设H0:010.5,备择假设Hi：0t二t(ni)拒绝域为|tts-n1x(10.410.610.7)10.4815样本均方差212s(10.410.48)14_2(10,710.48)0.237x甲x乙22S1S2近似N(0.1)解得拒绝域u于是t0.02

33、s0.237.n.150.327xiSi而t0.025(14)2.144因0.3272.144n12805,x2120.41,s2140,n22680105.00代入计算100x1x2故接受H0,认为该机工作正常。8.解：原假设H0012100备择假设H1:x0t(n1)，拒绝域为代入计算1252S2n1n2120.41210521101008.03查表uu00251.96因8.031.96U.-2故拒绝原假设即两种枪弹速度有显著差异。11.解：因两种作物产量分别服从正态分布且t22.14211958,s2.153故拒绝原假设即认为期望。9.假设H0:20.8H:使用新安眠药睡眠平均时间1-(

34、26.722.0723.4)323,0.05假设H0t0.025(13)2.068故统计量G1Sw.一_t(n1n212)20.8其中Sw代入计算回1)s2(n21)s2拒绝域为Tn22t"224.063t_(n1n22)t0,005(18)万287824.212-(26.724.2)2s22.296(23.424.2)2t30.9721.791124.063.10109.180.8510.756因为t0.852.878t0.005(18)3.42.296,.74046所以拒绝域为所以接受H0,认为两个品种作物产量没有显著差异。12.解：因两台机床加工产品直径服从正态分布且母体方差相

35、等,t0.05(n1)由题意假设H0:12,H1:1查表t°.05(6)1.9434.046t故否定H0统计量T又因为x24.220.83故认为新安眠药已达到新疗效。X1X2t(n1,11n22)10.原假设H0：甲乙，H1：甲乙Sw'n1n2S(n1SwV221)Si(n21)S2n1n22拒绝域为Tt_2数值代入计算sw0.5473XiX21-(20.5817(19.70.2164,s219.9)19.92519.2)200.3966拒绝域为Tt22.故接受H0:12019.9250.2650.54370.5175因t0.2652.160t0.025(13)故接受假设H0

36、,认为直径无显著差异。13.解：由题意设施肥，未施肥植物中长势良好率分别为p1,p2(均未知)则总体XB(1,p1),YB(1,p2)且两样本独立假设H0:p1P2,H1P1P2既H0：E(x)E(y).f:E(x)E(y)而D(x),D(y)均未知，则xy_N(0,1)22ss2nn2由题意易得x1140.1,x2140.102.1009t0.025(18)2,即认为两种电池性能无显著差异(2)检验要先假设其服从正态分布且15.解：由题意假设H0:由于未知。故2拒绝域为22得的观测值查表彳导2(n22.2(n1)s20.00778900,xx(1x)100,y1)0.048,H1:2(n1)0.048,n0.007780.0487830.879000.113753八八0.531000.2491于是xy0.870.532酬2s20.11370.2491Jn1n2.900100查表U0.012.336.6466y(1y)故应拒绝H0,接受H1即认为施肥的效果是显著的。2S20.346.64660.051114.(1)解：假设