數(shù)理統(tǒng)計課后題答案完整版

1、第一章3.解：因為Xi所以xacyi27=i0所以Xi2symicy一nacyniia-Yiniicyxacy成立因為s2i03521.5322291.54121.5341.5440.252Sxi002sy4.4025cycyicyyicy2i一又因為Syyiynii所以S2c2S2成立6.解：變換yii0x27*xyi-35-91234mi2341y-miyiniii35293i2434i01.5身154158162166170174178高：15：16：16：17：17：17：188260482組156160164168172176180中值學(xué)101426281282生數(shù)7解:*miiX1

2、li00i66i56i0i60i4i6426i72i2i6828i768i802s2-101001217233.44miixi15616621668141601662222616416628168166176216621801668解：將子樣值重新排列（由小到大）-4,0,0,Men12X70Me9解:Xi3.217.21X81.282221n1n1xini11n2n2xj%j1nixin2X2XiXi服從分布n1n2n1n2n1n2n2n1n2所以15.解:因為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同理審于1x1n2x2可知12.解:xi:PEXiDxii1,2,nX1X2X3,3N0,1EXe1nEx1DXd1n12nDxi13.解:a,bExDxii1,2,n在此題中ExDx121,2,n

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

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

6、nXii1上2由分布的性質(zhì)3可知3.解：因為總體X服從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從而有2.0,1).E(x)k(11b)212XS2、/3S、2nxdx令»p所以有2).其似然函數(shù)為L(P)n(1P)“1'1,2net27dtxe,2n4.解:L()lnL(1)dlnLxdx解之得：(2)母體E(x)k1p)pk(1p)k1x1n/pp(1nXin設(shè)x1，x2,Lxn為樣本觀察值則似然函數(shù)為:n(lnX的期望而樣本均值為:X1nXni1令E(x

7、)Xi)1,0-1xi1,i1,2,L,nlnxinln1xin1nlnxii1n-nlnxii1xf(x)dxxi1X5.。解：其似然函數(shù)為:xdxp)i1i 112L()n13ei2InL()nln(2)得:xi(2)由于i1201(2jxiXxedx2Xxe一01n,.1n,1E()'J)njxi)nnxini1的無偏估計量。6.解：其似然函數(shù)為:L()k(k)!x(k1)exexi()n(k1)!x(k1222.22.20.40338.解：取子樣值為則似然函數(shù)為: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：由題意知：均勻分布的母體平均數(shù)方差2321212用極大似然估計法求得極大似然估計量似然函數(shù)：L()選取使L達(dá)到最大nlnL()(xii1要使似然函數(shù)最大，則需9.解：取了樣值(x1x2,則其似然函數(shù)L(nxii1取min(x1,x2,=min(x1,x2,xn)(xi0)xixn)nxii1nXi1lnL()lnL()dnlnxi由題中數(shù)據(jù)可知-1x(3655100024515150250.0510.解:極差xi100xii135704545552565)200minximaxxii1(i)1in(2)maxx1in由以

9、上結(jié)論當(dāng)抽得容量為6的子樣數(shù)值,時20(1)由題中子樣值及題意知:6.21.54.7查表2-10.42994.72.0205平均極差R0.115,查表知0.32490.1150.045510.4299ds110.3249d1011/解：設(shè)U為其母體平均數(shù)的無偏估計，則應(yīng)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_則E(1)EX1一EX23313E(2)EX1EX24411E(3)2EX12EX2所以三個估計量1,q均為I,2,37,7的無偏估計CD(Xii11)D(Xi)0C22C(n1)i1要使E(

10、2只帝C12(n1)所以當(dāng)C15.證明:小時2.2為的無偏估計。參數(shù)的無偏估計量為0,由切比雪夫不等式d依賴于子樣容量n21D()叱X13X2)41DX1DX299limDn0故有l(wèi)impn151同理可得D(2)5，D(2)182即證為的相合估計量?？芍?的方差最小也亦2最有效。16證明：設(shè)X服從B(N,p),13解：XP()E(X),D(X)則分布律為P(Xk)cNPk(1P)k(k1,2,N)2E(S)1n,I。X)2這時E(X)NPD(X)NP(1P)1n22Ki1E(Xi)nE(X)EX2DX(EX)2NP(1P)N2P21n12)n(n2)所以E(P)(n2即S是的無偏估計dpD4N

11、P(12P)NNn羅一克拉美下界滿足EXNPNNP(1P)NnP(無偏)又因為1nE(X)E(1Xi)ni11E(nXi)1nKK-n-LnCNPK(1IRk0pP)NP2CNPK(1P)NP即X也是的無偏估計。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也是的無偏估計2、14.解：由題意：xN(,)因為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為優(yōu)效估計nN1()-17.解：設(shè)總體x的密度函數(shù)f(x)一一e22似然函數(shù)為得置信區(qū)間為(xus)n已知10.95s=40X=1000查表知u-21.96代入計算得2(Xi)2n(xi)nii21222-o22L()e2(2)2e2ii.2所求置信區(qū)間為(19.解：(1)已知)0.01cm(Xi)2LnL(2)2Ln22Ln2-則由UXN(0,1)PU.nu)-2解之得置信區(qū)間(又dLnLd2n(Xi1)2(X

13、i)2將n=16X=u7u0.051.6450.01因為(Lnf2X)2f(x)dx)2(X)22edx=Y【E(X)4E(X)2224=74824故2的羅一克拉美下界,241n2E(-(Xi)ni1Rn2又因E1n22-E(Xi)ni1口21n224且口()D(-(Xi)ni1n22所以是的無偏估計量且IR2D()22故是的優(yōu)效估計18.解：由題意：n=100,可以認(rèn)為此為大子樣,X所以U近似服從N(0,1)S.nPUu1代人計算得置信區(qū)間(2)未知解得置信區(qū)間為(又S220.1)PTt)2sht_).n2n=16t(15)1t0.05(15)1.7530.00029代入計算得置信區(qū)間為(解

14、：用T估計法Xt(n1)S/-：?n解之得置信區(qū)間(X將X6720S代入得置信區(qū)間為()。PTSn；t(n1)1"2*一SXt)n2220n=10查表t0.025(9)2.262221.解：因n=60屬于大樣本且是來自故由中心極限定理知nXinpi1np(1p)p山上vnp(1p)u1"2(01)分布的總體,nXnp近似服從N(0,1)np(1p)解得置信區(qū)間為(Xp(1p)uXp(1p)u)nwn工本題中將UA代替上式中的X由題設(shè)條件知UL0.25P(1P)nUn(nUn)0.05525.解：因Xn1與X1,X2,Xn相互獨立,查表知UnU0.0251.96所以Xn1與X

15、相互獨立，故代入計算的所求置信區(qū)間為（）Xn22.解:2,未知故XUN(0,1)、n12XN(0,(1-)n由PUu)2解得置信區(qū)間為q（X區(qū)間長度為2u,n2于是2un2L計算得nu,n2i42T2U,723.解:未知，用22估計法2(n1)S2P12_(n1)22(n1)2(n21)1解得的置信區(qū)間為（_2(n1)S2(1)當(dāng)n=10,S=時.2查表0.005(9)=20.995(9)=代人計算得的置信區(qū)間為（）(2)當(dāng)n=46,S=14時2查表0.005(45)=20.995(45)(n1)S2即為所求2(n1)又因nS22"2（n1）且與Xn1X相互獨立,有T分布的定義知26

16、.解：因Xi代入計算可得24.解：（1）先求的置信區(qū)間為（的置信區(qū)間由于）未知Xn1Xn1(nN(YjN(X“Tt(nS.n1)KT得置信區(qū)間為（XStn萬經(jīng)計算X5.12S0.2203查表t0.025(19)2.093n=20代人計算得置信區(qū)間為（2）未知用統(tǒng)計量(n1)S2-22(n1)P2)2得的置信區(qū)間為_2(n1)S_2(n1)S2)12查表(2025(19)=2，一、0.975(19)=代入計算得的置信區(qū)間為（nnS21)2Xn1XSn1n1t(n1)2)1,2,2)1,2,所以(X1)N(0,)，m由于X與Y相互獨立，則(X1)即(X又因2msx(Y(Y2)N0,(一m(Y2)2

17、2(m1)2nsy2emsx則2構(gòu)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證明：因抽取n>45為大子樣*2(n1)s22(n1)、2(n1)2（n1）由2分布的性質(zhì)3知近似服從正態(tài)分布N（0,1）所以PU2PUu12計算得置信區(qū)間為2(n1)3(n1)2(n1)u22(n1)u-2S(X1X2UJVS22mX2u22、，的置信區(qū)間為12n1u22n1u2，一._2_2把X11.71X21.67S10.035S220.0382nm100uu0.0251.96代入計算得置信

18、區(qū)間(0.0299,0.0501)22228.斛：因12未知，故用T統(tǒng)計重TX丫(12"m2)11sw-,nm其中sW(n謫1)S2nm231.解：由題意，u1,u2未知，則*2/S2/2FF(n21,n11)S1120.05n81.625145695,丫L(n2查表to.o25(4)2.144則PF(n21,n11)1.2經(jīng)計算得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的置信區(qū)間為故得置彳t區(qū)間(6.4

19、237,17.4237)1,n12S11)2,F2,S22(n21,n12132S229解：因122故用T統(tǒng)計量_*2_S10.245XaYb1swn-2(n1)S計算得置信區(qū)間為(XaXbSwt_(n2把Sw22)t(n-2(m1)S22)n(7)=2代入可得所求置信區(qū)間為(30.解：由題意用U統(tǒng)計量m2)Xb2S20.3570.05F0.025(5,8)4.82t_2F0.975(8,5)F0.025(5,8)14.820.207SWt_(n2m2).-)nm2帶入計算得1-的置信區(qū)間為:2232.解：2未知，則(0.142,4.639)。即:1)X1X2(12)U122123N(0,1)

20、S12S22；nmPTt(n1)1有：P(n*S1)1nS/則單側(cè)置信下限為：Xt(n1)上一將X6720S220nyr2xyn10to.o5(9)1.833帶入計算得6592.471即鋼索所能承受平其中XyniXYin2Xii1n2Yii1均張力在概率為95%的置信度下的置信下限為6592471。33.解：總體服從(0,1)分布且樣本容量n=100為大子樣。令X2.解:2690.14XY2736.511為樣本均值，由中心極限定理nXnPn2N(0,1)又因為2S22mx451.112my342.665代入得所以PnXnpunS2xyxy-2mx2736.5112690.140.8706451

21、.11則相應(yīng)的單側(cè)置信區(qū)間為90.140.87062667.508822my22mx342.6650.87062451.110.7487將X=S2m(1m)0.60.94uuo.051.645nn代人計算得所求置信上限為即為這批貨物次品率在置信概率為95%青況下置信上限為。3證明:dod1d0uvuvd1u2-2udod134.解:由題意:2(n1)S22(n1)doi1xcd1XQd1YiCod0ycdo(n1)1解得的單側(cè)置信上限為2(n1)S12(n1)d1Xcd1xqd1其中n=10,2,查表(n1)2-0.95(9)代人計算得的單側(cè)置信上限為。第五章1.解：對一元回歸的線性模型為Xi

22、1,2,nd0d0d1d1xYin離差平方和為QQyii1Xi的偏導(dǎo)數(shù),并令其為0,YiXid0變換得一XYini1因為所以解此方程得xy2XYiXiYi2YiXYi22Xixy2xy-2TxyxXIi11d2Xidovd°vYiCodoCodod1xixdod1Codod1C1c1d12一22ndoi1ViUidoMd0VC0d。d0d0d0d0d1Ui*2xyxy2mx15586210.4210.436.95636.9511.32mx*2*2xiCd1510929.843236.952812.373.517d0d14.解:假設(shè)H0:38H1:38用T檢驗法拒絕域為yi260025

23、002400230022002100200019001800支數(shù)y品質(zhì)指標(biāo)2_my34527.46代入得2211.2xy76061.67635.353132.130xyxy2-mx5.解:查表彳導(dǎo)t0.025xi將上面的數(shù)據(jù)代入得2mx3.1824所以接受H0即認(rèn)為6.解：(1)由散點圖看,1.89為38t0.0253X的回歸函數(shù)具有線性函數(shù)形式,76061.67635.3532211.2“”15.98認(rèn)為長度對于質(zhì)量的回歸是線性的O121198710度長9132.1302211.215.9835.3532776.1422_mx34527.462的無偏估計量(2)將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時y00.18217.50.182xa16b06.305由T分布定義Y0xOxOn1 1n一 2x為i 12X111Y0x0正規(guī)方程為X1X2k.025n20.95最小二乘估計為所以Y0的預(yù)測區(qū)間為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將（2）的數(shù)據(jù)代入得11解:*2nV42.450.182272.920.00750.0866計算得Y）的預(yù)測區(qū)間為8.9521,9.47219.解:利用第八題得到的公式(1)21y141.2Xy3138m.290xyxy代入得2mX313821141.21.929010.1Xi12xjXj2j1,215采用線性回歸模型Y1Xx2X2X15yi248.25y16.55i1151515yi24148.3125X1920x256734i1i1i115X161

26、.33Xi27257X2483.8i11515ii115Xi1X21Xi22352448944536615141.21.92性21回100.88X1V15170xi2yi12063925離差平方和為1,yiyi15可變換為2Xi2i,i1,2,nL112X11515Xi115673456426.66307.34L22152X215yii1iXi12Xi2L1215X235244893510936.613552.41515Xi1Xi215i1Xi115Xi2i14453664450962702的偏導(dǎo)數(shù)并令其為1Xi11Xi1Xi1X15L1yXM115152Xi22Xi2Xi1Xi2于是15為2

27、乎12Xi1yiX2Xi1Xi2Xi1Xi21n22 Xi2i1X15i1115Xi215i116.55li115yii1151701522656yi120639.25307.3427027013552.4,56,L156所以y10.504120103.25L1yL2y536565360可得10.21僅0.04X253621752.93 1085.62120012.解p18采用線性回歸模型于是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假設(shè)：H°：4321.022568.051752.97由于3507631920.223155.782xi31818x33078942723223557210139.5181x2i11818x1i118x210139.59053.881085.629659818xi1xi3

29、i19659818xi2xi3i118xyii1GV0.08x20.16x326,H1:265.2已知，故用統(tǒng)計量uJN(0,1)因顯著水平這時，就接受H0u的拒絕域u27.56265.240.05，則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的拒絕域u5.32532因顯著水平1.100.01，則u0.0052.576故此時拒絕H0：u5檢驗u4.8時犯第二類錯誤的概率22ndxi 1i 1令tu21%x近似u0N(0,1)拒絕域為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,認(rèn)為新工藝與舊工藝無顯著差異。0.99999790.719900.71806.解：由題意知，

31、母體X的分布為二點分布B(1,p),3.解:假設(shè)H0：3.25,f：3.25用t檢驗法拒絕域t(n1)20.01,作假設(shè)H0:pp0(p00.17)此時xm(m為n個產(chǎn)品中廢品數(shù))n因n400很大，故由中心極限定理知x近似服從正態(tài)分布。x3.252查表t0.0112(14)4.6041*2s0.00017,s0.0130代入計算T0.344t0.0112(14)mponPo(1Po)nN(0,1)故接受H0,認(rèn)為礦砂的銀含量為3.254解：改變加工工藝后電器元件的電阻構(gòu)成一個母體計算得拒絕域為則在此母體上作假設(shè)H0:2.64,用大子樣檢驗拒絕域為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故新加工工藝對元件電阻有顯著影響5.解：用大子樣作檢驗Ho:即接受Ho,認(rèn)為新工藝不顯著影響產(chǎn)品質(zhì)量。7解：金屬棒長度服從正態(tài)分布原假設(shè)H0:010.5,備擇假設(shè)Hi：0t二t(ni)拒絕域為|tts-n1x(10.410.610.7)10.4815樣本均方差212s(10.410.48)14_2(10,710.48)0.237x甲x乙22S1S2近似N(0.1)解得拒絕域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代入計算100x1x2故接受H0,認(rèn)為該機工作正常。8.解：原假設(shè)H0012100備擇假設(shè)H1:x0t(n1)，拒絕域為代入計算1252S2n1n2120.41210521101008.03查表uu00251.96因8.031.96U.-2故拒絕原假設(shè)即兩種槍彈速度有顯著差異。11.解：因兩種作物產(chǎn)量分別服從正態(tài)分布且t22.14211958,s2.153故拒絕原假設(shè)即認(rèn)為期望。9.假設(shè)H0:20.8H:使用新安眠藥睡眠平均時間1-(

34、26.722.0723.4)323,0.05假設(shè)H0t0.025(13)2.068故統(tǒng)計量G1Sw.一_t(n1n212)20.8其中Sw代入計算回1)s2(n21)s2拒絕域為Tn22t"224.063t_(n1n22)t0,005(18)萬287824.212-(26.724.2)2s22.296(23.424.2)2t30.9721.791124.063.10109.180.8510.756因為t0.852.878t0.005(18)3.42.296,.74046所以拒絕域為所以接受H0,認(rèn)為兩個品種作物產(chǎn)量沒有顯著差異。12.解：因兩臺機床加工產(chǎn)品直徑服從正態(tài)分布且母體方差相

35、等,t0.05(n1)由題意假設(shè)H0:12,H1:1查表t°.05(6)1.9434.046t故否定H0統(tǒng)計量T又因為x24.220.83故認(rèn)為新安眠藥已達(dá)到新療效。X1X2t(n1,11n22)10.原假設(shè)H0：甲乙，H1：甲乙Sw'n1n2S(n1SwV221)Si(n21)S2n1n22拒絕域為Tt_2數(shù)值代入計算sw0.5473XiX21-(20.5817(19.70.2164,s219.9)19.92519.2)200.3966拒絕域為Tt22.故接受H0:12019.9250.2650.54370.5175因t0.2652.160t0.025(13)故接受假設(shè)H0

36、,認(rèn)為直徑無顯著差異。13.解：由題意設(shè)施肥，未施肥植物中長勢良好率分別為p1,p2(均未知)則總體XB(1,p1),YB(1,p2)且兩樣本獨立假設(shè)H0:p1P2,H1P1P2既H0：E(x)E(y).f:E(x)E(y)而D(x),D(y)均未知，則xy_N(0,1)22ss2nn2由題意易得x1140.1,x2140.102.1009t0.025(18)2,即認(rèn)為兩種電池性能無顯著差異(2)檢驗要先假設(shè)其服從正態(tài)分布且15.解：由題意假設(shè)H0:由于未知。故2拒絕域為22得的觀測值查表彳導(dǎo)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)故應(yīng)拒絕H0,接受H1即認(rèn)為施肥的效果是顯著的。2S20.346.64660.051114.(1)解：假設(shè)