天津大学版管理统计学答案.xls

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1、管管理理统统计计学学课课后后习习题题答答案案第一章题目1a调查规模为2500b定性c百分比d700题目2a定性b42.25%题目3a定量b定性 c定性 d定量题目4a定量b定量c定性 d定量e定性 题目5a40.00%b定性c略第二章题目1产品数量频数累计频数180220合计20题目2a38.00%b33.00%c29.00%题目3a时间频数0-2.953-5.9286-8.989-11.9612-14.93合计50b时间频数频率0-2.950.13-5.9280.566-8.980.169-11.960.1212-14.930.06合计501c上述数据显示了大多数人每周使用3-5.9小时计算

2、机第三章题目1a平均数=38.75bQ1=29.5Q3=47.5题目2a国内平均数115.13中位数111.56国外平均数36.62083中位数36.695b国内全距86.24样本标准差26.81995变异系数0.232954国外全距42.96样本标准差11.39883变异系数0.311266题目3AB全距28标准差0.6749485582.584139659A公司更具有一致性与可靠性题目4数据182168184190a全距22b方差75.2c标准差8.671793355d变异系数0.04871794题目5标准差0.0126192 0.005生产线要关闭题目6通话时间组中值频数组内和平方和47

3、5.5422184.968119.5547.539.2121513.5794.510.08161917.523554.08频率0-2.93-5.96-8.99-11.912-14.900.10.20.30.40.50.6202321.5121.584.64242725.5125.5174.24合计20246547.2平均数12.3方差28.8标准差5.366563146第四章题目1abc10%读报时间查表得78.12题目2abc题目3a22.8b7.757433711题目4a200b5cN(200,25)d中心极限定理5.0)0()1560-601560-XP(60)P(X02275.0)2(

4、1)-2()1560-301560-XP(30)P(X1.0)1560-x()1560-x1560-XP(x)P(X78.1x2085.11560-x6826.01)1(2)1()1(1)-X1P()X-P(9545.01)2(22)()2(2)-X2P()2X2-P(9973.01)3(23)()3(3)-X3P()3X3-P(题目5aN(250,50)bc题目5aN(320,187.5)b13.69306394cd第五章题目1=80S=10n=20总体方差未知，n=2030,因此用t分布估计。在90%置信度下，总体均值的置信区间为=1.7291置信区间为76.13361486 83.866

5、39 在95%置信度下，总体均值的置信区间为=2.0933置信区间为75.3192389 84.68076 题目2=6.525S=0.5437443总体方差未知，n=2030总体方差未知，n=6130,因此用正态分布估计。a在95%置信度下，总体均值的置信区间为=1.96（正态分布表中查0.975所对应u值）置信区间为21.14523858 23.65476 b在95%置信度下，总体均值的置信区间为=2置信区间为21.1196312 23.68037 题目4=2000=0.05=an=61.4656 取62bn=384.16 取385cn=1536.64 取1537题目5=6.25=0.05=

6、an=6.0025 取7bn=150.0625 取151题目6a=0.885714286b0.023569472c=1.645X XnSX,nSX025.0025.0nStX,nStX025.0025.0)6 0(t0 2 5.0)19(t025.0025.0025.0025.0p)p1(pn1)p1(pn1220 5.0)p1(pn1)p1(pn122nStX,nStX025.0025.0置信区间为0.8659327650.905495807 题目7a0.02=0.05n=2016.84 取2017b=0.237977194c单个总体比率的区间估计95%的置信区间为带入数据得0.219392

7、527 0.256562 第六章题目1 n=200S=1.5=1.28：3：34.7140452081.28因此拒绝，收看电视的时间显著增加题目2a拒绝规则为U=b n=40S=1.2=1.645：3:3-5.2704627671.28因此拒绝，该工序未达到标准题目4 n=500S=0.2=-1.28：0.15:0.151.628183 因此接受，可实行新包装题目5a n=30=224.24779435 时拒绝原假设b=3.417225575=0.99968370.000316c=1H2/0 1.00H0H1HppnXU/|00 1.0nppppU/)1(0000HnXU/020X)30/22

8、324.248(-1)30/22324.24830/223X(P)248.24X(P)30/22424.248(-1)30/22424.24830/224X(P)248.24X(P0.678612788=0.75130840.248692d=-0.690693606=0.24487910.755121题目6 n=30=0.8a意味着冲入量未达标准但错认为达标从而犯取伪错误的概率b=得到-5.383266-1.463266)30/22424.248(-1)30/22424.24830/224X(P)248.24X(P02/nXU/|0X)46.1()84.5()300.8/5.16-286.16

9、300.8/320X300.8/5.16714.15(P)286.16X714.15(P)30/22424.248(-1)30/22424.24830/224X(P)248.24X(P0.071697题目7=59令=得到查表得=1.285得到n=214.6225 取215或由公式计算题目8=30385=0.02=0.12.0551.285或由公式44.6224题目9=324=0.02=1-0.8=1nXU/005.0nX5645.1101.0)n/5n5645.11(-1)n/59n5645.110n/59X(P)n5645.110X(Pn/5n5645.1121022)()(n1021022

10、)()(n102.0550.845或由公式75.69题目10设：看前、看后总体均值分别为样本均值分别为方差分别为6.039337902=5.375因总体方差未知且是否相同未知，因此首先假设两总体方差是否相同：当显著性水平为0.01时0.1F=1.48419688F(7,7)=2.78当显著性水平为0.05时F=1.48419688F(7,7)=3.79所以认为两总体方差相等=1.616187：=0.8221053因此接受原假设，认为看该广告后并未提高潜在购买力题目112121022)()(n0H1H2121,21021n1n1SXXt21S,S21X,X2X1X1S2S0H1H21212nn1

11、)S(n1)S(nS212222110方差分别为0.201739801n1=0.086756364n2=：当显著性水平为0.05时F=5.407300359F(24,20)=2.08所以机器1有改进空间题目12设湿路干路方差分别为32n1=16n2=a：当显著性水平为0.05时F=4F(29,29)=1.86b湿路不易刹车，应减速慢行题目13设两个生产部方差为2.35.4：当显著性水平为0.1时21S,S1S2S0H1H212121SS21S,S1S2S0H1H21212212S,S12S22S0H1H2121F=0.425925926 所以拒绝原假设，认为两个生产部门产品有差异题目14国内航

12、班国际航班 合计一等舱292251商业/行政舱95121216全价经济舱/二等舱518135653合计642278920国内、国际航班各自机票类型比国内航班国际航班 合计一等舱0.04517134 0.079137 0.0554348商业/行政舱0.147975078 0.435252 0.2347826全价经济舱/二等舱0.806853583 0.485612 0.7097826合计111国内航班一等舱比例pa1=0.045171国际航班一等舱比例pa2=0.079137一等舱总比例pa=0.055435H0:pa1=pa22.0674021拒绝H0，一等舱旅行比例与航班类型有关国内航班商务

13、舱比例pb1=0.147975国际航班商务舱比例pb2=0.435252商务舱总比例pb=0.234783H0:pa1=pa29.4399796拒绝H0，商务舱旅行比例与航班类型有关国内航班经济舱比例pc1=0.806854国际航班经济舱比例pc2=0.485612经济舱总比例pc=0.709783)n,n(F1)n,n(F122/1212/1)n2/1n1/1)(pa1(papa2pa1U0 2 5.02/)n2/1n1/1)(pb1(pbpb2pb1U0 2 5.02/0 2 5.02/H0:pc1=pc29.8583452拒绝H0，经济舱旅行比例与航班类型有关综上，机票类型与航班类型有关

14、。第七章题目1a215b2.740326027c209.628961220.371,或用2代替1.96得题目2S=70B=15a=n=73.11917098取74题目3N1=300N2=600N3=S1=150S2=72S3=aB=20n=56.28820459 取58n1=18.32828659 取19n2=17.59515513 取18n3=20.36476288 取21bB=10n=22.23731539 取24n1=7.240804578 取8n2=6.951172395 取7n3=8.04533842 取9题目4N1=50N2=38N3=S1=80S2=150S3=an=107.39

15、61829109 取206n1=38.1006413739 取73n2=54.2934139655 取104n3=15.0021275415 取59bs1=s2=s3=100n=101.2480937 取195n1=41.15776166 取79n2=31.27989886 取60n3=28.81043316 取56题目4群MiXiai(xi-xc*Mi)21795110021832563025315190612254101402100合计50750154450)n2/1n1/1)(pc1(pcpc2pc1UniiniicMxx11a150.3M*xc=300*15=4500b121.4707

16、802总体均值95%的置信区间为12.0584395717.94156043 0.0484051总体总量的标准差为441.2340649c总体总量95%的置信区间为3617.531875382.46813 总体比率95%的置信区间为0.2031897620.396810238 第八章题目1工厂123加工时间202820262619niiniicMap1125300NMM1)(22nMxxMNnnNSicixc1)(22nMpaMNnnNSicipc)90000,300(300),(22cxcxSxNSxNccniiniicMxx11243123222722 和Ti9211284288Ti平方8

17、46412544705628064Yij平方和2136315017747060SE=44SA=104fA=2fE=9方差来源平方和自由度均方F比组间SA10425210.636364组内SE449 4.888889总合ST14811F0.05（2，9）=4.26F=10.6364.26，所以各工厂加工时间有差异题目2上级同事下属866565477654633745577565Ti464442132Ti平方2116193617645816Yij平方和276256234766SE=39SA=1fA=2fE=21方差来源平方和自由度均方F比组间SA120.5 0.2692308组内SE3921 1.

18、857143总合ST4023F0.05（2，21）=3.47F=0.2693.47，所以各信息渠道无差异差异题目3房地产代理商 建筑师股票经纪人814365486348686057加工时间695291545470627767766883565775616153658071645054693772837365858458755858方差分析：单因素方差分析SUMMARY组观测数求和平均房地产代理商151016 67.733333建筑师15917 61.133333股票经纪人1598765.8方差分析差异源SSdfMS组间345.37777782 172.68889组内6087.06666742

19、144.93016总计6432.44444444F0.05（2，42）=1.191532 b=11.74045 cANOVA表变差来源平方和自由度均方差回归249881.13781 249881.14误差85135.13784 21283.784总和3350005题目4a9bcs(b1)=1.3626题目5a地区编号综合空闲率平均租金率121.918.542633.7322.819.67418.121.01512.735.09614.519.4172025.28819.217.0291624.04106.631.421115.918.74129.226.761319.727.72142018.

20、2)(/11bsbt 21)()(XXMSEbsi)2(2/ntfeSSEfrSSRF/)2,1(nFxy21.70.202r158.3251617.129.781710.837.031811.128.64ab线性负相关cdF=12.276=4.49e=0.434 拟合程度不好f17.595g28.2673第十章题目1a230025bb1表示存货投资每增加1元，其他条件不变时销售收入增加25元b2表示存货投资广告费用每增加1元，其他条件不变时销售收入增加8元题目2每周总收入电视广告费报纸广告费9651.590229541.5922.52.59533.5943.52.3942.54.29432.

21、5ab系数不相同ca中系数b1表示电视广告费用每提高1千元每周总收入提高1.6千元b中系数b1表示其他条件不变时电视广告费用每提高1千元总收入提高2.28千元b中系数b2表示其他条件不变时报纸广告费用每提高1千元总收入提高1.27千元25201510500510152025303540综合空闲率平均租金率xy779.007.37)16,1(05.0F2r16.164.88xy2127.128.228.83xxy10349.29de32.81163 fT1=8.101393999 T2=4.420530872 回归统计Multiple R0.963955986R Square0.92921114

22、2Adjusted R Square0.900895599标准误差0.600852041观测值8方差分析dfSSMS回归分析2 23.69488 11.847442残差5 1.805116 0.3610232总计725.5Coefficients 标准误差t StatIntercept83.28284245 1.44206657.75245X Variable 12.283844253 0.2819088.101394X Variable 21.274961598 0.288418 4.4205309总计Coefficients 标准误差t StatIntercept88.63768116 1

23、.582367 56.015876X Variable 11.603864734 0.4777813.356905题目3竞争厂家的价格 该公司的价格 销售数量256240102280260100380352120260300773103204635030093250300262903006936040065)1/(/knSSEkSSRF)5,2(05.0F1(2/knt30035085ab系数b1表示其他条件不变时竞争厂商价格每提高1元销售数量提高0.5584千辆系数b1表示其他条件不变时该公司价格每提高1元销售数量减少0.4775千辆c85.277回归统计Multiple R0.673043

24、744R Square0.452987882Adjusted R Square0.296698705标准误差23.5150786观测值10方差分析dfSSMS回归分析2 3205.388 1602.6938残差7 3870.712 552.95892总计97076.1Coefficients 标准误差t StatIntercept57.84769518 57.27676 1.0099681X Variable 10.558385442 0.237968 2.3464689X Variable 2-0.477493643 0.236056-2.022798214775.05584.085.57x

25、xy170174上述数据显示了大多数人每周使用3-5.9小时计算机频率0-2.93-5.96-8.99-11.912-14.900.10.20.30.40.50.66826.01)1(2)1()1(1)-X1P()X-P(1.72912.093n=203099.52%1-99767.021)2.828(2)502502025050250X5025020250(P)20250X20250(P2%.841-921.021)1.414(2)502501025050250X5025010250(P)10250X10250(P76%.3564245.01)365.0(1)13.6932032513.69

26、320X(P)325X(P79%.92)92785.01(1)46.1()84.5()13.6932040013.69320X13.69320300(P)400X300(P2.0932.00031.961.96=1.96=0.019782025.0（正态分布表中查0.975所对应u值）p=0.3=1.963.5=0.0110%拒绝域1.28-1.6452=0.055%拒绝域-5.27046-1.962.39=0.02拒绝域拒绝域025.0XX)p1(pn1p,)p1(pn1-p)p1(pn1p,)p1(pn1-p2222近似为1.28-1.281.2888/500=0.176=0.0110%拒

27、绝域-1.28-5.27046-1.2825=0.02=-2.06时，即：=0.022524.2477923p拒绝原假设0)30/22324.248(-1)30/22324.24830/223X(P)248.24X(P)30/22424.248(-1)30/22424.24830/224X(P)248.24X(P=0.022524.2524.516=0.05=0.025=0.025=1.96161.9616.28628 或 15.7137216.5=3.66E-08=0.0716972/2/0 2 5.0X)30/22424.248(-1)30/22424.24830/224X(P)248.2

28、4X(P意味着冲入量未达标准但错认为达标从而犯取伪错误的概率)30/22424.248(-1)30/22424.24830/224X(P)248.24X(P)46.1()84.5()300.8/5.16-286.16300.8/320X300.8/5.16714.15(P)286.16X714.15(P=0.05=0.1=0.05-1.645109400取4525=0.020.21.005.00025取76241.946793n=n1=n2=81.59799接受原假设接受原假设1.96n1=642n2=2781.961.96n1=642n2=278所以拒绝原假设，认为两个生产部门产品有差异)n

29、,n(F1)n,n(F122/1212/10 2 5.02/0 2 5.02/1.961.96 209.5193 220.4807 0.05N=450500N=140010035N=12345B=45(ai-pc*Mi)2N=25n=41.21M=3000.362.2514.820 2 5.02/,或用2代替1.96得k=3m=4k=3m=8方差117.6381179.981137.1714FP-valueF crit1.191532 0.313814 3.2199423.2199423铁粉掺量抗压强度2137653.5268105397582268573.539891513765239907

30、3.5138675268602239732石膏掺量3.525276525569279933.5842.333365852953残差-156.443-191.77863.2657853.03606145.66182.72331-71.859536.47145-92.1114131.0339395.567X2.3061706.401 121549.81225000.3613511.7425000.0454037.651225000.160.000422500 1.97E-31)210(t2/2)Y-(Y2)(XXi矿渣掺量对强度影响最大，其次是石膏，再次是铁粉矿渣掺量取20%时强度大、石膏掺量取3

31、.5%时强度大，铁粉掺量取9%时强度大14001200100080060040020000500100015002000y=0.8958x+395.567R2=0.898205.02)Y-Y(30397.92625000.0930383.9825000.09249881.13350000.74=2.7764 回归系数显著=7.71 回归方程显著F比11.74045)2(2/nt)2,1(nF线性关系显著b1表示存货投资每增加1元，其他条件不变时销售收入增加25元b2表示存货投资广告费用每增加1元，其他条件不变时销售收入增加8元a中系数b1表示电视广告费用每提高1千元每周总收入提高1.6千元b中

32、系数b1表示其他条件不变时电视广告费用每提高1千元总收入提高2.28千元b中系数b2表示其他条件不变时报纸广告费用每提高1千元总收入提高1.27千元=5.79=2.5706b1显著b2显著F Significance F32.81629 0.001333P-value Lower 95%Upper 95%2.94E-08 79.57589 86.989790.000465 1.559178 3.0085110.006889 0.533559 2.016364P-value Lower 95%Upper 95%2.17E-09 84.76577 92.509590.015288 0.434777

33、 2.772952)5,2(05.0FF Significance F2.898396 0.121057P-value Lower 95%Upper 95%0.34614-77.5903 193.28570.051354-0.00432 1.1210910.082788-1.035680.08069系数b1表示其他条件不变时竞争厂商价格每提高1元销售数量提高0.5584千辆系数b1表示其他条件不变时该公司价格每提高1元销售数量减少0.4775千辆原始数据3.1从小到大排列 位次22221 Q1i=5 为整数，所以Q1=58242 Q2i=10 为整数，所以Q2=24293 Q3i=15 为整数

34、，所以Q3=5029429295523065731731318303294137104440114041124644132944143146153749163250174452184957192958203.2 国内从小到大排列 位次108.776.21112.2579.36294.0194.013144.0395.164162.44102.115161.61108.7676.2110.877 Me因为N为偶数，所以中位数Me=102.11112.258110.87114.16979.36121.8810129.04129.041195.16144.0312114.16161.6113121.

35、88162.4414 MEAN=115.13R=国外从小到大排列 位次29.8921.61141.1324.86240.6727.34340.4129.89443.0731.61524.8632.986 Me因为N为偶数，所以中位数Me=31.6140.41721.6140.67827.3441.13964.5741.311032.9843.071141.3164.5712 MEAN=36.621R=3.3AB118101091310711101111101011710151012全距28标准差0.67492.584139663.53.433.453.483.523.53.093.483.41

36、3.383.493.453.513.5方差0.01264.32610.2434125.44220.641560.841646.24230.0436174.241823.04254.841396.04541.6 60.17778平均值22.8标准差7.7574347.7574337115.266.65.876.36.27.25.76.476.56.266.57.27.37.66.866.2平均值6.525样本标准差 0.543744第6章-10之前之后6577433564987566平均值6.0393385.375样本标准差 1.9467931.597989809第6章-11机器1机器22.953.223.453.33.53.343.753.283.483.293.263.253.333.33.23.273.163.383.23.343.223.353.383.193.93.353.363.053.253.363.283.283.23.33.223.282.983.33.453.23.73.163.343.183.353.12平均值3.3043.2725样本标准差 0.201740.086756364252129.538.547.5111.5686.24 样本标准差26.8236.69542.96 样本标准差 11.399