社会统计学课后题答案(卢淑华)

《社会统计学》课程练习题（ 1）答案一、略二、（1）对立事件 （2）互不相容事件 （3）互不相容事件 （1）对立事件三、Md4005025200525(元)40M040025200516.28(元)2518Q12002510200400.00(元)15Q36007565200690.91(元)22QQ3Q1690.91400290.91(元)ni2(nibi)2526002biN3276000050924002100N1005092410050924225.66(元)四、（1）极差R=1529-65=1464（百元）2）将数据从小到大排序：65921061181221351481741851529Q1的位置1012.754Q3的位置3(101)48.25Q192(10692)0.75102.5(百元)Q3174(185174)0.25176.75(百元)四分互差Q176.75-102.574.25(百元)（3）2(xi)226742xi2495204N10178017.64421.92(百元)N10五、80P(A)0.810020P(B)0.2100P(B/A)120.1580P(A/B)120.620P(AB)120.1210040P(C)0.4100P(C/A)320.480P(A/B)120.1580P(AC)320.32100六、P(A)415P(B)715P(AB)110P(A/B)P(AB)115P(B)100.2147P(B/A)P(AB)115P(A)100.3754P(AB)P(A)P(B)471P(AB)150.6331510七、（1）10口井皆产油的概率为：P(10)C10100.3100.700.0000059(2)10口井皆不产油的概率为：P(0)C1000.300.7100.02825（3）该公司赢利的机会为：P(2)1C1000.300.710C1010.310.790.85069八、4P(x2)42e482.7182840.14652!九、1.37P(0x1)P(x0)P(x1)1.3701.371.3711.370.6022ee0!1!十、0.10.100.1Z10.1(Z)(1)1(1)10.84130.1587投资人投资于此种股票 保证不亏的概率为：1 0.1587 0.8413。Z 0.2 0.1 10.1（Z） (1) 0.8413收益在以上的可能性为：0.8413。20%10.1587十一、15000-10000（1）Z20002.51(2.5)10.99380.00620.62%9500-100002.5Z2000（2）100(2.5)1(2.5)0.00620.62%(3)P(8000 12000) P(1 Z 1) (1) (1) 0.8413 (1 0.8413) 68.26%(4)(x 10000)0.902000x 100002000

1.3x 2600 10000 12600十二、xtsnt0.025(24)2.0639[302.063936,302.063936][27.52,32.48]2525十三、xZnZ0.051.65[21.51.650.02250.0225,21.51.65][21.4175,21.5825]99十四、?ZpP??pqnZ0.0052.58[0.92.580.90.1,0.92.580.90.1][0.8226,0.9774]100100十五、此人上班迟到的概率为：0.430%0.620%24%此人某天迟到而乘地铁的概率为：0.620%50%24%十六、np10020%20n(1p)10080%80因此，可将此二项分布用正态分布来近似计算。np20npq10020%80%4P(x15)P(14.5x15.5)P(14.520x2015.520)444P(1.375x201.125)4P(1.125x2041.375)0.91540.86960.04584.58%x10.01060.0250.050.0250.01060.0053x10.05(1.125)0.0053(1.10)0.00530.86430.8696x2 0.00770.025 0.050.025 0.0077x2 0.00390.05(1.375) 0.0039 (1.35) 0.0039 0.9115 0.9154“社会统计学”第六章习题答案一、设X1，X2，X3为简单随机抽样的3个观测值。如果采用如下不等权的平均值：X'2X12X21X3555作为总体均值的点估计值，试说明它将比采用等权的平均值：X1X11X21X3333作为总体均值的点估计值要差。解答：根据方差的性质D(X')D(2X1)D(2X2)D(1X3)555(2)2D(X1)(2)2D(X2)(1)2D(X3)5554D(X)4D(X)125D(X)252592251113)D(X)D(X1)D(X2)D(X333(1)2D(X1)(1)2D(X2)(1)2D(X3)333111D(X)D(X)D(X)999123因为D(X)pD(X')，所以采用等权的平均值 X作为总体均值的点估计值比采用不等权的平均值 X'作为总体均值的点估计值更有效。二、解答：x0.75S0.2010.950.05t1.962代入式（6.22）置信区间为：0.751.960.20,0.751.960.200.7108,0.7892100100四、解答：x4.5S510.950.05t1.962代入式（6.22）置信区间为：4.55,4.51.9651.963.52,5.48100100五、解答：?0.60??10.900.10t1.65pq1p0.402代入式（6.33）置信区间为：0.601.650.600.40,0.601.650.600.400.5428,0.6572200200第八章 单总体假设检验一、解答：H0：u=2.5H1：u<2.5x2.52.12.50.43.64z1.10.110.11100z1.65z=-3.64<-1.65 ，故拒绝原假设。二、解答：H0：p=0.6H1：p>0.60.620.60.020.408z0.40.00240.6100z 1.65z=0.408<1.65，故不能拒绝原假设。三、解答：（1）H0：u=75% H 1：u≠75%0.8-0.75z 2.50.2100z/2 1.96z=2.5>1.96，故应拒绝原假设。(2) 如果拒绝了食品费用占总收入的比例为75%的说法，则可能犯错误的概率为α，即0.05。四、解答：H0：u=55000H1：u≠55000n=6x60466.67362800222043600000ixix2(xi)2220436000003628002xin6S4610.71n15xu60466.6755000t4610.712.90Sn6t0.025(5)2.5706t2.90>t(5)2.57，故应拒绝原假设。062第九章 二总体假设检验一、解答：H0：uAuB0H1：uAuB>069.267.5z1.2749.364.55080z1.651.27<1.65，故不能拒绝原假设。二、0PB0H1PB≠0解答：H：PA：PA?mAmB45630.54PnAnB100100z0.450.630.182.5520.0049680.540.46100z1.962z2.55<-1.96，故应拒绝原假设。四、解答：H0：uAuB0H1：uAuB>0S2(121)1.52(121)0.921.5312122S 1.53 1.237 t0.05(22)1.7171tXAXB6.85.32.97S111.23711nAnB1212t 2.97>1.7171，故应拒绝原假设。五、0uB0H1uB>0解答：H：uA：uAddi5412602.67n6di2(di)216282n267.8667sdn15sd7.86672.8048t2.672.672.33192.80481.1456t0.05（5）=2.0150t=2.3319＞t0.05（5）=2.0150,所以拒绝原假设，即可以认为新教学法优于原教学法。第十章 列联表一、cr2解答：x2nnij1i1j1ni*n*j=200452472392262212222(105929565105659543105431)92952000.20960.25270.22290.10950.09770.11851=2.18x02.05(2 1)(3 1) 5.991x2 2.18<5.991，故不能拒绝原假设。二、解 答 ： （ 1 ）x2801022021021021022021402040304030402040304030=6.664x02.05(2)5.991x26.664>5.991，故拒绝原假设。crnij21r2ni*njn*j（2）i1j111r2nnjn*j1crnij210220210210210220230ni*404040404040i1j11r2120230230227.5njn*j8013027.50.0488027.5三、解答：（1）x250192722472172126262624262424=500.534020.07850.50171=9.635x0.052(1)3.841x29.635>3.841，故拒绝原假设。（2）9.6350.4450五、解答：（1）24621023022021825021x174668628868088668828888086174（0.37280.04150.13080.06890.13150.35511）174（）1.1006117.5044x20.05(2) 5.991x2 17.5>5.991，故拒绝原假设。（2）4650800.1717480第十二章 回归与相关一、解答：（2）根据最小二乘法作回归线。（3）282.33x1219015.83y12LxybLxxxiyi1yixinxi21(xi)2n1281901212821241.666712.66673.289aybx15.833.2892.338.156（4）y?8.1563.289x1时，y?2时，y?3时，y?4时，y?（5）TSS LyyRSSR b

8.1563.28911.4458.1563.289214.7348.1563.289318.0238.1563.289421.312yi21(yi)2315611902147.67n12Lxy3.28941.667137.04RSS TSS RSSR147.67 137.04 10.63(6)r2RSSR137.040.928TSS147.67r2=0.928表明可以用x解释掉y的9.28%的误差。(7)r=0.963r0.05(10) 0.576r=0.963＞0.576,故r具有推论意义。二、解答：（1）x306533y 6.652221303324b510.6220302405a y bx 6.6 0.6 6 3?0.6xy3（2）Lyy235133217.25rLxy24LxxLyy400.91517.2r2L2xy0.8372LxxLyy判定系数r2=0.8372说明用自变量父代受教育年限去预测因变量子代受教育年限，可以改善预测程度的83.72%,或可以用自变量x解释掉因变量y83.72%的误差。（3）r0.05(3)0.878r0.915>0.878，故具有推论意义。三、解答：（1）x415.1258y384.7582551413860.25b810.875279268.875418aybx4.750.8755.1250.266?0.2660.875xy（2）Lyy264138283.58rLxy60.25LxxLyy0.79468.87583.5r20.6312判定系数r2=0.6312说明用自变量生活期望值去预测因变量个人成就，可以改善预测程度的63.12%,或可以用自变量x解释掉因变量y63.12%的误差。（3）r0.05(6) 0.707r 0.794>0.707，故具有推论意义。四、解答：（1）x825016505y8550001710005157150000018250855000160750000b51110.86150625002145000082505aybx171000110.86165011919?11919110.86xy（2）Lyy16425900000018550002180540000005rLxy1607500000.9935LxxLyy145000018054000000r20.9871判定系数r2=0.9871说明用广告费用去预测因变量销售额，可以改善预测程度的98.71%,或可以用自变量x解释掉因变量y98.71%的误差。（3）r0.05(3) 0.8780.9935>0.878，故具有推论意义。第十三章方差分析一、mni212771507264解答：TSSRPyijnT..i1j1BSSQPm1212Ti.T..i1nin48245224250747.58109RSSTSSBSS216.5FBSS/m147.5/312.63RSS/nm216.5273F0.05(2,24) 3.42.63<3.4，故三个地区的平均家庭人口没有显著差异。二、解答：（1）TSS 563.53 518.02 45.511 2 2 2BSS