1第7章参数估计习题参考答案7.1参数的点估计习题答案1解(1)总体X的期望()EXmp,从而得到方程11ˆniimpXn所以p的矩估计量为111ˆniipXXmnm.(2)总体X服从二项分布,则有()(1),0,1,...,xxmxmPXxCppxm从而似然函数为11121()(1)...(1)nniiiiiniinxmnxxxmxxxxmmmmiLpCppCCCpp取对数1211ln()ln(...)ln()ln(1)nnnxxxmmmiiiiLpCCCxpmnxp,令1111ln()()01nniiiidLpxmnxdppp,解得p的极大似然估计值为111ˆniipxxmnm,故极大似然估计量为111ˆniipXXmnm.2.解(1)110()1EXxxdx,从而得到方程1ˆ1ˆ1niixxn所以的矩估计值为ˆ1xx.(2)似然函数为1121()(,)(...)nniniLfxxxx取对数1ln()ln(1)lnniiLnx,令1ln()ln0niidnLxd,得的极大似然估计值为1ˆlnniinx27.2估计量的评选标准习题答案1.解(1)1123123111111ˆ()442442EEXXXEXEXEX2123123111111ˆ()623623EEXXXEXEXEX3123123111111ˆ()333333EEXXXEXEXEX,123ˆˆˆ,,均为的无偏估计量。(2)211231113ˆ161648DDXDXDX,同理227ˆ18D,231ˆ3D.312ˆˆˆDDD,3ˆ最有效.2.证明22ˆˆˆ()EED,又ˆ是的无偏估计量,即ˆE,且ˆ0D,222ˆˆED.故2ˆ不是2的无偏估计量.3.解总体服从正态分布2(,)N,从而21(0,2)iiXXN.11122221111111ˆ[()]()[()]()nnniiiiiiiiiiiEECXXCEXXCEXXDXX122122(1)niCCn故令12(1)Cn,则有22ˆE,即此时2ˆ是2的无偏估计量.7.3参数的区间估计习题1.解25,19,10.95,nx即0.05,查标准正态分布表可得20.0251.96uu.从而22191.9618.2165xun,22191.9619.7845xun,故的置信度为0.95的置信区间为(18.216,19.784).2.解10,19,20,10.95,nxs即0.05,(1)查(1)tn分布表可得20.025(1)(9)2.2622tnt.3从而220(1)15002.26221485.6910sxtnn220(1)15002.26221514.3110sxtnn,故的置信度为0.95的置信区间为(1485.69,1514.31).(2)查表得2220.025(1)(9)19.023n,2220.9751(1)(9)2.700n.从而2222(1)920189.24(1)19.023nsn,22221(1)9201333.33(1)2.700nsn,故2的置信度为0.95的置信区间为(189.24,1333.33).3.解1250,80,3.1,10.99,nnxy即0.01,20.0052.57uu从而2212122252050803.12.570.874nnxyu2212122252050803.12.575.326nnxyu,故12的置信度为0.99的置信区间为(0.874,5.326).4.解由样本数据计算得212215.0125,0.094514.99,0.0261xsys又128,9,10.95nn,即0.05,(1)查表得2120.025(1,1)(7,8)4.53FnnF2212210.0251(1,1)1/(1,1)1/(8,7)1/4.900.2041FnnFnnF从而22212120.799(1,1)ssFnn,2221212117.74(1,1)ssFnn故方差比的置信度为0.95的置信区间为(0.799,17.74).(2)0.0225xy,0.05,查表得2120.025(2)(15)2.1315tnnt,有222112212(1)(1)0.05802wnsnssnn.从而412211111289(2)0.02252.13150.05800.227wnnxytnns1221112(2)0.272wnnxytnns,故12的置信度为0.95的置信区间为(-0.227,0.272)。5.解(1)由于(1)XttnSn,于是(1)1XPtnSn即(1)1SPXtnn.此题中20.0510,10.95,576.4,75.16,(9)1.8331nxst,故75.16(1)576.41.8331571.3710sxtnn,即总体均值的置信度为0.95的单侧置信下限为571.37.(2)由于22222211(1)()(1)niinSXXn于是22222(1)(1)(1)1(1)nSnSPnPn此题中220.0510,10.95,576.4,75.16,(9)16.919nxs故标准差的置信度为0.95的单侧置信上限为22(1)676.46.32(1)16.919nSn.本章参数估计复习题1.解似然函数为12222221111()(,)2(2)niiixxnniniiLfxee,取对数221122ln()ln(2)ln2ln22nniiiixxLnnn5令2122ln()022niixdnLd,解得2的极大似然估计值为221ˆx.2.解记12min(,,...,)nnXXXX,此时的似然函数等价于1,()0,niixnnnexLx所以只有当nx时,才有可能使()L取到最大值.又()L对nx的是增函数,故当nx取到其最大值.即0()max()nLxL所以的极大似然估计值为12ˆmin(,,...,)nnxxxx.3.解由于[,1]XU,故总体的期望为212EX,从而得到方程ˆ21,2X解得1ˆ2X.所以的矩估计量为1ˆ2X.又111ˆ()()()222EEXEXEX,故1ˆ2X是的无偏估计量.4.证明2221122111ˆ[()]()1(2)nniiiiniiiEEXEXnnEXEXn2222211(2)nin故2ˆ是2的无偏估计量。65.解220.108,9,4.484,10.95,nx即0.05,查标准正态分布表可得20.0251.96uu.从而20.1084.4841.964.413xun20.1084.4841.964.553xun,故该厂铁水的平均含碳量的置信度为0.95的置信区间为(4.41,4.55).6.解1220,6,10.99,nnxy即0.01,查标准正态分布表可得20.0052.57uu.从而2222121220.050.05202062.576.04nnxyu2222121220.050.05202062.575.96nnxyu,故12的置信度为0.99的置信区间为(-6.04,-5.96).7.解22120.5419,0.6065ss,1210,10.95nn,即0.05,查表得2120.025(1,1)(9,9)4.03FnnF2212210.0251(1,1)1/(1,1)1/(9,9)1/4.03FnnFnnF从而22212120.222(1,1)ssFnn,222121213.601(1,1)ssFnn故方差比的置信度为0.95的置信区间为(0.222,3.601).