社会统计学课后题答案(卢淑华)

3 - 100i 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、M =400 +dM =400 +050 -25402525 +18200 =525(元)200 =516.28(元)Q =200 +125 -1015200 =400.00(元)75 -65Q =600 + 200 =690.91(元)22Q =Q -Q =690.91 -400 =290.91(元) 3 1s2=n b 2i i( n b )i iNN2=52600 232760000 -5092400= =50924100 100s = 50924 =225.66(元)四、（)极差 R=5296=1464(百元）(2)将数据从小到大排序 :65 92 10 11 122 13 18 4 11529Q 的位置 = 110 +14=2.75Q 的位置 = 33 (10 +1) 4=8.25Q =92 +(106 -92) 0.75 =102.5(百元) 1Q =174 +(185 -174) 0.25 =176.75(百元) 3四分互差 Q =176.75 -102.5 =74.25(百元)(3)s=x2i( x )-NN2=2495204 -102674102= 178017.64 =421.92(百元)五、P ( A) =80100=0.820P ( B ) = =0.2 10012P ( B / A) = =0.15 8012P ( A / B ) = =0.62012P ( AB ) = =0.1210040P (C ) = =0.4 10032P (C / A) = =0.48012P ( A / B ) = =0.1580P ( AC ) =六、32100=0.32P ( A) =P ( B) =415715P ( AB) =110P ( AB) 1 15P ( A / B ) = = =0.214 P ( B ) 10 7P ( AB) 1 15P ( B / A) = = =0.375 P ( A) 10 4P ( A +B ) =P ( A) +P ( B ) -P ( AB) =七、4 7 1+ - =0.63315 15 10(1）10 口井皆产油的概率为： P (x=10) =C10100.3100.70=0.0000059(2) 0 口井皆不产油的概率为: P (x=0) =C0100.300.710=0.02825（3)该公司赢利的机会为: P (x2) =1 -C 0 0.3 0 0.7 10 -C 1 0.31 0.7 9 =0.8506910 10八、l=44 2P ( x =2) = e2!-4=8 2.71828-4=0.1465九、l=1.371.37 0 1.371P (0 x 1) =P ( x =0) +P ( x =1) = e -1.37 + e0! 1!十、m=0.1s =0.1-1.37=0.6022Z =0 -0.10.1=-1f(Z ) =f(-1)=1 -f(1) =1 -0.8413 =0.1587投资人投资于此种股票保证不亏的概率为：1-0.1587 =0.8413。 0.2 -0.1Z = =10.1f（Z）=f(1) =0.8413收益在20%以上的可能性为：1-0.8413 =0.1587。十一、(1)Z =15000 -100002000=2.51 -f(2.5) =1 -0.9938 =0.0062 =0.62%(2）Z =9500 -100002000100=-2.5f(-2.5) =1 -f(2.5) =0.0062 =0.62%（ )P (8000 x12000) =P ( -1 Z 1) =f(1) -f(-1)=0.8413 -(1 -0.8413) =68.26%(4)f(x -100002000) =0.90x -100002000=1.3x =2600 +10000 =12600十二、t =x -msnt0.025(24) =2.063930 -2.0639 3625,30 +2.0639 3625 =27.52,32.48十三、Z =x -msnZ=1.650.0521.5 -1.65 0.02259,21.5 +1.650.02259 =21.4175,21.5825十四、Z =p-PpqnZ0.005=2.580.9 -2.58 十五、0.9 0.1 0.9 0.1,0.9 +2.58 =0.8226,0.9774 100 100此人上班迟到的概率为： 0.4 30% +0.6 20% =24%此人某天迟到而乘地铁的概率为:0.6 20%24%=50%1 2 3 1 2 十六、np =100 20% =20n(1 -p ) =100 80% =80因此，可将此二项分布用正态分布来近似计算。 m=np =20s = npq = 100 20% 80% =4P ( x =15) P (14.5 x 15.5)14.5 -20 x -20 15.5 -20=P ( )4 4 4x -20=P ( -1.375 -1.125)4x -20=P (1.125 1.375)4=0.9154 -0.8696=0.0458=4.58%x 0.01061 =0.025 0.05x =10.025 0.01060.05=0.0053f(1.125) =0.0053 +f(1.10) =0.0053 +0.8643 =0.8696 x 0.00772 =0.025 0.05x =20.025 0.00770.05=0.0039f(1.375) =0.0039 +f(1.35) =0.0039 +0.9115 =0.9154“社会统计学”第六章习题答案一、设 X ，X ，X 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： 1 2 32 2 1X = X + X + X5 5 5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：1 1 1 X = X + X + X3 3 3作为总体均值的点估计值要差。 解答:根据方差的性质31 2 3 1 2 1 2 3 1 2 200 200D ( X2 2 1 ) =D ( X ) +D ( X ) +D ( X )5 5 52 2 1 =( ) 2 D ( X ) +( ) 2 D ( X ) +( )5 5 52D ( X )34 4 1= D ( X ) + D ( X ) + D ( X ) 25 25 259= s2251 1 1 D ( X ) =D ( X ) +D ( X ) +D ( X )3 3 31 1 1 =( ) 2 D ( X ) +( ) 2 D ( X ) +( )3 3 31 1 1= D ( X ) + D ( X ) + D ( X ) 9 9 91= s232D ( X )3因为 D ( X ) D ( 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 二、作为总体均值的点估计值更有效。解答： x =0.75 S =0.20a=1 -0.95 =0.05ta2=1.96代入式（6.2）置信区间为:0.20 0.20 0.75 -1.96 ,0.75 +1.96 = 0.7108,0.7892100 100 四、解答: x =4.5 S =5a=1 -0.95 =0.05ta2=1.96代入式(62)置信区间为:5 5 4.5 -1.96 ,4.5 +1.96 = 3.52,5.48100 100 五、解 答 : p=0.60q=1 -p=0.40a=1 -0.90 =0.10ta2=1.65代入式(。33）置信区间为： 0.60 0.40 0.60 0.40 0.60 -1.65 ,0.60 +1.65 =0.5428,0.6572i 第八章单总体假设检验一、解答：H ：u=2.5 H :u2。5 x -2.5 2.1 -2.5 -0.4z = = = =-3.641.1 0.11 0.11100-z =-1.65az=-3。41。65,故拒绝原假设。二、解答:H ：=0。 ：p0。6 1z =0.62 -0.60.6 0.4=0.020.0024=0.408100z =1.65az=0。08.6，故不能拒绝原假设三、解答：(1）H :u=75 H :u5% 1z =0.8 - 0.750.2100=2.5za/ 2=1.96z=2.96,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，即 0。四、解答： ：u=550 H :u50000 1n=6 x =60466.67x =362800ixi2=22043600000S =x2i( x )-nn -12=22043600000 -536280062=4610.710 A B 0 1 0 1 t =x -u 60466.67 -55000 =S 4610.71n 6=2.90t0.025(5) =2.57 0t =2.90 t (5) =2.57 6,故应拒绝原假设.a2第九章 二总体假设检验一、解答： : u -u =0 A BH ： u -u 0 z =69.2 -67.549.3 64.5+50 80=1.27z =1.65az =1.27 1。65,故不能拒绝原假设。二、解答:H : P -P =0 A BH ： P -P 0 A Bm +m 45 +63 P = A B =n +n 100 +100 A B=0.54z =0.45 -0.63 2 0.54 0.46 100 =-0.180.004968=-2.55-za2=-1.96z =-2.55 -196,故应拒绝原假设。四、解答：H : u -u =0A B(12 -1) 1.5 2 +(12 -1) 0.9 S 2 =12 +12 -22=1.53H ： u -u 0 A BS = 1.53 =1.237t0.05(22) =1.71710 A B d i 2 i n 6ijn nX -X 6.8 -5.3 t = A B =1 1 1 1S + 1.237 +n n 12 12A B=2.97t =2.97 1.71,故应拒绝原假设。五、解答:H : u -u =0A B5 +4 -1 +2 +6 +0d = =n62.67H : u -u 0 s 2 =dd2i( d ) 16 2- 82 -= =7.8667n -1 5s = 7.8667 =2.8048 dt =2.67 2.67= =2.33192.8048 1.1456t （）=。150。05t=2.3319t （)=2。0150,所以拒绝原假设，即可以认为新教学法优。05于原教学法.第十章列联表一、解答: x 2 =n c r n 2 -1 i=1 j =1 i * * j = 200 (452 47 2 39 2 26 2 212 22 2+ + + + + -1) 92 105 92 95 65 105 65 95 43 105 43 95= 200 (0.2096 +0.2527 +0.2229 +0.1095 +0.0977 +0.1185 -1) =1x 2 (2 -1)(3 -1) =5.9910.05x2=2.18 。991,故不能拒绝原假设。二、解答:（1） 2 ij n 2 ijn = 20+ 30+ 30= 27.5* j 26 26 26 24 24 26 24 24 86 66 86 28 86 80 88 66 88 28 88 80解 答:(1） 10 2 202 102 10 2 102 202 x 2 =80 + + + + + -140 20 40 30 40 30 40 20 40 30 40 30=6。664x 2 (2) =5.9910.05x 2 =6.664 599,故拒绝原假设（2） t=c r n 1 r-n n i =1 j =1 i * j =11 rn - n 2* jj =1n 2* jc r n 10 2 20 2 10 2 10 2 10 2 20 2= + + + + + =30 n 40 40 40 40 40 40i =1 j =1 i*1 r 2 1 (2 2 2 )n 80j =1t=30 -27.580 -27.5=0.048三、解答:（1) x2 19 2 7 2 7 2 17 2 =50 + + + -1 = 50 (0.5340 +2 0.0785 +0.5017 -1)=.6x2 (1) =3.8410.05x2=9.635 3。84,故拒绝原假设。（2) F =9.63550=0.44五、x2 46 2 10 2 30 2 20 2 18 2 50 2 =174 + + + + + -1 =174 （0.3728 +0.0415 +0.1308 +0.0689 +0.1315 +0.3551 -1） =174 （1.1006 -1）=17.5044x20.05(2) =5.991xy1 n 1 i i yyi i x 2 =17.5 5。，故拒绝原假设。(2) l=46 +50 -80 174 -80=0.17第十二章一、解答:（2)根据最小二乘法作回归线。28(3) x = =2.3312回归与相关y =19012=15.83Lb =Lxxx y - x yi i i=x 2 - ( x ) 2 ni=1485 - 28 190 12178 - 28 2 1241.666712.6667(4） y=3.289a = y -bx =15.83 -3.289 2.33 =8.156 =8.156 +3.289xx =1时，y x =2时，y x =3时，y x =4时，y=8.156 +3.289 =11.445 =8.156 +3.289 2 =14.734 =8.156 +3.289 3 =18.023 =8.156 3.289 4 =21.3121 1（5) TSS =L = y 2 - ( y ) 2 =3156 - 190 2 =147.67n 12RSSR =b Lxy=3.289 41.667 =137.04RSS =TSS -RSSR =147.67 -137.04 =10.63(6） r2=RSSR 137.04= =0.928TSS 147.67r=0。928 表明可以用 x 解释掉 y 的 9。28%的误差5 (）r=096r0.05(10) =0.576r=0.960.57,故 r 具有推论意义。 二、30解答:（1） x = =6533y = =6.65b =1222 - 30 33 51220 - 30 2524= =0.640a =y -bx =6.6 -0.6 6 =3 y =3 +0.6 x1(2) L =235 - 33 yy2=17.2r =LxyL Lxxyy=2440 17.2=0.915r2=L2xyL Lxxyy=0.8372判定系数 r=0832 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 8.72%，或可以用自变量 x 解释掉因变量 y83。72 的误差。(3) r0.05(3) =0.878r =0.915 0.88,故具有推论意义。 三、41解答：（1) x = =5.125838y = =4.758b =1255 - 41 38 81279 - 412860.25= =0.87568.875a =y -bx =4.75 -0.875 5.125 =0.2668 5 y=0.266 +0.875 x1(2) L =264 - 38 2 =83.5 yyr =LxyL Lxxyy=60.2568.875 83.5=0.794r 2 =0.6312判定系数 r2=。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 y.%的误差。（3） r0.05(6) =0.707r =0.794 0。7,故具有推论意义. 四、8250解答:(1) x = =16505855000y = =1710005b =11571500000 - 8250 8550005115062500 - 8250 25=1607500001450000=110.86a = y -bx =171000 -110.86 1650 =-11919y=-11919 +110.86 x1(2） L =164259000000 - 855000 yy2=18054000000 r =LxyL Lxxyy=1607500001450000 18054000000=0.9935r 2 =0.9871判定系数 r=0。81 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 9871%,或可以用自变量 x 解释掉因变量 y98。71%的误差.（3） r0.05(3) =0.878r =0.9935 0.88,故具有推论意义。第十三章方差分析一、m ni 1 ij. i . 8 10 91 () 解答： TSS =R -P = y 2 - T 2 =771 -507 =264ni =1 j =1BSS =Q -P =m 1 1 T 2 - T 2n ni =1 i482 452 24 2 = + + -507 =47.5 RSS =TSS -BSS =216.5F =F0.05BSS / m -1 47.5 / 3 -1= =2.63RSS / n -m 216.5 27 -3 (2,24) =3.4F =2.63 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 =45.51BSS = 44.372+33.62+26.42-518.02=541.20 -518.02 =23.18RSS =TSS -BSS =45.51 -23.18 =22.33F =F0.05BSS / m -1 23.18 / 3 -1= =9.34RSS / n -m 22.33 21 -3 (2,18) =3.55F =9.34 3。55,故三个城市中科研机构高级技术人才的拥有人数有显著 差异。BSS 23.18(2) eta 2 = = =0.5093TSS 45.51三、解答：TSS =a bi =1 j =1y 2 -ij1abT 2=(424621 +548634 +492398 +449695) - =1915348 -1671849=2434995172162 A i B j0. 505B 01 a T 2BSS = T 2 - b abi =11 5172 = (1173 2 +1416 2 +1314 2 +1269 2 ) -4 16=1679485.5 -1671849=7636.51 b T 2BSS = T 2 - a abj =121 5172 = (645 2 +1033 2 +1720 2 +1774 2 ) -4 16=1897147.5 -1671849 =225298.5RSS =TSS -BSS -BSSA B=243499 -7636.5 -225298.5=105642F =ABSSARSS=7636.5 / 310564 / 9=2.168F (3，9）=.6因为 F =2.8F （3,9）=3。86,所以不拒绝原假设，即认为各改革方案 。间收入无显著差异.F =BBSSBRSS=225298.5 / 310564 / 9=63.98因为 =63。9 (3,9)。86,所以拒绝原假设,即认为不同规模的工。厂间收入有显著差异。