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1、精选学习资料 - - - - - - - - - relationship, establishe d equivalent relatinowle onship 14, and subject: appli cation problem (4)-scores and percentage a a pplication problem review content overview answers scores, and perce jentage application pr problem of key is: accordi i ng to meaning, (1) determine
2、standard volume (units 1) (2) find associate volume rate corresponds to relationship, T he n in -line sol ution. Category fraction multiplication word problem score Division applications engi gi neeri ng problem problem XV, a subjeo: cyli nder a ct: review of the measurement of the amount of capacit
3、y, measurement a t a nd units of measurement of common units off measurement and their sig g nifica nce in rate 1, currency, length, area, v v olume, unit size, volume, weight and rate. (Omitted) 2, commonly use d time units and their relationships. (Slightly) with a measurement units Zhijian of of
4、poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary ke dge (1)-li i ne and angle review content line, and segment, and Ray, and vertical, and parallel,
5、and angle angle of classification (slightly) 17, and subject: geometry preliminary knowledge (2)-plane graphi cs review conte nt triangle, and edges shaped, a nd round, and fan axisymmetric graphics perimeter and area com binati ti on graphics of area subject : Preliminary knowledge (3)-review of so
6、lid content category 1-d shapes are divided int a nd cone 2, column is divided into: cuboi d, square 3, , cone cone of of the features of cuboids and cubes relationship between characteristics of circular cone is slig lig htly solid surface area and volume 1, size 2, table .和第三章假设检验3.2 一种元件 ,要求其使用寿命
7、不低于 1000（小时） ,现在从一批这种元件中随机 抽取 25 件,测得其寿命平均值为 950（小时）。已知这种元件寿命服从标准差100（小时）的正态分布，试在显著水平0.05 下确定这批元件是否合格。提出假设：H 0 : 1000, H : 1 1000构造统计量：此问题情形属于 u检验，故用统计量： u= X 00 n此题中：x 950 0 100 n=25 0 1000代入上式得： u=950-10002.510025拒绝域： V= u u 1本题中：0.05 u 0.95 1.64即，u u 0.95 拒绝原假设 H 0认为在置信水平 0.0 5下这批元件不合格。3.4 某批矿砂的五
8、个样品中镍含量经测定为（% ）：3.25 3.27 3.24 3.26 3.24 设测定值服从正态分布，问在0.01下能否接受假设，这批矿砂的镍含量为提出假设：H0:103.25 H1:1t0构造统计量：本题属于2未知的情形，可用检验，即取检验统计量为： t=Xn01S本题中，x3.252, S=0.0117, n=5代入上式得： t=3.252-3.251 0.34190.01175否定域为：名师归纳总结 V=tt1-2(n1)3.25。第 1 页，共 10 页本题中，0.01,t0.995(4)4.6041Qtt12认为这批矿砂的镍含量为接受H0,- - - - - - -精选学习资料 -
9、 - - - - - - - - relationship, establishe d equivalent relationship 14, and subje je ct: appli cation n problem (4)-scores and percentage a a pplication problem review content overview answers scores, and percentage application problem of key is: accordi i ng to meaning, (1) determine standard volum
10、e (units 1) (2) find associate volume rate corresponds to relationship, T he n in -line sol ution. Category fraction multiplication word problem score Division applications engi gi neeri ng problem problem XV, a subject: review of the measurement of the amount of capacity, measurement a t a nd units
11、 of measurement of common units of measurement and their significa nce in rate 1, currency, length, area, v v olume, unit size, volume, weight and rate. (Omitted) 2, commonly use d time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly
12、 method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)-li i ne and a ngle review content line, and segment, and Ray, and vertical al , and parallel, and angle angle of classi
13、fication (slightly) 17, and subje je ct: geometry pr pr eliminary knowledge (2)-plane graphi cs review conte nt triangle, and edges shaped, a nd round, and fan axisymmetric graphics perimeter and area com binati ti on graphics of area subject : Preliminary knowledge (3)-review of solid content categ
14、ory 1-d shapes are divided into: cyli nder and cone 2, column is divided into: cuboi d, square 3, , cone cone of of the features of cuboids and cubes relationship between characteristi i cs of circular cone is slig lig htly solid surface area and volume 1, size 2, table .和S0.035%,3.5 确定某种溶液中的水分，它的10
15、 个测定值X0.452%,设总体为正态分布 N( , 2), 试在水平 5%检验假设：( ) H : 0.5% H : 0.5%( ) H : 0 0.04% H : 1 0.0.4%( ) 构造统计量：本文中未知，可用 检验。取检验统计量为 t= 本题中，XX0Sn10.452% S=0.035%代入上式得： t=0.452%-0.5%-4.11430.035%10-1拒绝域为： V= t t 1-( n 1)本题中，0.05 n=10 t 0.95 (9) 1.833 1 t 4.1143拒绝 H 0( ) 构造统计量：未知，可选择统计量22 nS20本题中，S0.035% n=10 00
16、.04%代入上式得：210（0.035%）27.6563（0.04%）2否定域为： V=22(n1)1本题中，名师归纳总结 2(n1)1)2 0.95(9)116.919第 2 页，共 10 页1Q22( n1接受H0X:N( , 4),X,K,X16 为样本，考虑如下检验问题：3.9 设总体- - - - - - -精选学习资料 - - - - - - - - - relationship, establishe d equivalent relationship 14, and subject: appli cation problem (4)-scores and percentage
17、a a pplication problem review content overview answers scores, and percentage application problem of key is: accordi i ng to meaning, (1) determine standard volume (units 1) (2) find associate volume rate corresponds to relationship, Th cs perimeter and aree n in -line sol ution. Category fraction m
18、ultiplication word problem score Division applications engi gi neeri ng problem problem XV, a subjeo: cyli nder a ct: review of the measurement of the amount of capacity, measurement a t a nd units of measurement of common units of measurement and their significa nce in rate 1, currency, length, are
19、a, volume, unit size, volume, weight and rate. (Omitted) 2, commonly use d time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measur
20、ement 2, and estimates 16, and subject: geometry preliminary knowle dge (1)-li i ne and angle review content line, and segment, and Ray, and vertical al , and parallel, and angle angle of classification (slightly) 17, and subje je ct: geometry pr pr eliminary knowledge (2)-plane graphi cs review con
21、te nt triangle, and edges shaped, a nd round, and fan axisymmetric graphia com bination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided int a nd cone 2, column is divided into: cuboi d, square 3, cone cone of of the features of cuboids and
22、 cubes relationship between characteristi i cs of circular cone is slig lig htly solid surface area and volume 1, size 2, table .和H0:0 H :1( ) 试证下述三个检验（否定域）犯第一类错误的概率同为=0.05(ii) V =2X-1.645 V = 1.502X2.125 V = 2X1.96或2X1.96通过计算他们犯第二类错误的概率，说明哪个检验最好？解：( )P xV H00.0520U0.9750.05( 1.645)1(1.645)即， PUXU1这
23、里H0:00.051.645PX2*1.96V 12X1.645PXP2X1.645n =1-0.95=0.05V21.502X2. 1251.50X02.120nP V2H0(2.215)(1.50)0.980.930.0501.96V 32X1.96或2X1.962X1.96XnP( V3H )=1-P2X1.962(1(1.96)0.05（ ）ii犯第二类错误的概率名师归纳总结 V 1:=P1-VH11.6451第 3 页，共 10 页=P2X- - - - - - -精选学习资料 - - - - - - - - - relationship, establishe d equivale
24、nt relatinowle onship 14, and subje je ct: appli cation n problem (4)-scores and percentage a a pplication problem review content overview answers scores, and percentage application pr problem of key is: accordi i ng to meaning, (1) determine standard volume (units 1) (2) find associate volume rate
25、corresponds to relationship, Th cs perimeter and aree n in -line sol ution. Category fraction multiplication word problem score Division applications engi gi neeri ng problem problem XV, a subject: review of the measurement of the amount of capacity, measurement a t a nd units of measurement of comm
26、on units of measurement and their significa nce in rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly use d time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method a
27、nd poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary ke dge (1)-li i ne and a ngle review content line, and segment, and Ray, and vertical, and parallel, and angle angle of classification (slightly) 17, and subje
28、 je ct: geometry preliminary knowledge (2)-plane graphi cs review conte nt triangle, and edges shaped, a nd round, and fan axisymmetric graphia com binati ti on graphics of area subject : Preliminary knowledge (3)-review w of solid content category 1-d shapes are divided into: cyli nder and cone 2,
29、column is divided into: cuboi d, square 3, , cone cone of of the features of cuboids and cubes relationship between characteristi i cs of circular cone is slig lig htly solid surface area and volume 1, size 2, table .和X1 =P0.3551(0.355)0.36nV 2:21P1.502 X2.1251 =1-P3.50X14.125n =1-(4.125)+(3.50) =1V
30、 3:3P2 X1.961 =P0.04X13.96n =(3.96)-(0.04) =0.99996092-0.516=0.48396092 V出现第二类错误的概率最小，即 V最好。3.10 一骰子投掷了 120 次，得到下列结果：点数1 (2 3 4 5 6 出现次数23 26 21 20 15 15 问这个骰子是否均匀？0.05)解：本题原假设为： Hi0:P1 i=1,2,L,66这里 n=120,nPi202统计量Pearson 本题采用的统计量为即，2k(nnpi)2npii 1代入数据为：名师归纳总结 2k(n inp i)2（ 23-20 ） （ 26-20 ）2 2L2（15
31、-20 ） =4.8第 4 页，共 10 页i 120np i- - - - - - -精选学习资料 - - - - - - - - - relationship, establishe d equivalent relationship 14, and subject: appli cation n problem (4)-scores and percentage application problem review content overview answers scores, and percentage application problem of key is: accordi
32、i ng to meaning, (1) determine standard volume (units 1) (2) find associate volume rate corresponds to relationship, T he n in -line sol ution. Category fraction multiplication word problem score Division applications engi gi neeri ng problem problem XV, a subjeo: cyli nder a ct: review of the measu
33、rement of the amount of capacity, measurement a t a nd units of measurement of common units of measurement and their significa nce in rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly use d time units and their relationships. (Slightly) with a measurem
34、ent units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)-li i ne and angle review content line, and segment, and Ray, a
35、 a nd vertical, and parallel, and angle angl gl e of classification (slightly) 17, and subje je ct: geometry pr pr eliminary knowledge (2)-plane graphi cs review conte nt triangle, and edges shaped, a nd round, and fan axisymmetric graphics perimeter and area com bination graphics of area subject :
36、Preliminary knowledge (3)-review of solid content category 1-d shapes are divided int a nd cone 2, column is divided into: cuboi d, square 3, cone cone of of the features of cuboids and cubes relationship between characteristi i cs of circular cone is slig lig htly solid surface area and volume 1, s
37、ize 2, table .和2（k-1）=0.95 2（ ）=11.0711由于22（k-1） 所以接受H01即认为这个是均匀的。3.11 某电话站在一小时内接到电话用户的呼唤次数按每分钟记录的如下表：呼吸次数0 1 2 3 4 5 6 =7 频数8 16 17 10 6 2 1 0 试问这个分布能看作为泊松分布吗？ （=0.05）解：检验问题为：名师归纳总结 H :P xk)ke参数为2第 5 页，共 10 页k!已知 的最大似然估计Xnp081*16L6*17*0L260606060P 1P X020e2e20.13530!P 2P X11 2e22*e20.27071!P 3P X22
38、 2e22*e20.27072!P 4P X33 2e21.5*e20.2 0303!P 5P X44 2e22*e20.09024!3P 6P X55 2e24*e2 0.03615!15P 7P X66 2e24*e2 0.01206!45P 8P X71P X602ik( n inp i)2(860*0.1353)2(1660*0.2707)2L(1 60*0.0120)1np i60*0.135360*0.270760*0.0120 =0.6145由于2（k-1） =2 0.95（5）=11.0711Q22（k-1）1接受H0,即分布可以看作为泊松分布。3.13 从一批滚珠中随机抽取了
39、50 个，测得他们的直径为（单位：mm）: - - - - - - -精选学习资料 - - - - - - - - - relationship, establishe d equivalent relatinowle onship 14, and subject: appli cation problem (4)-scores and percentage a pplication problem review content overview answers scores, and perce jentage application pr problem of key is: accori
40、 ng to meaning, (1) determine standard volume (units 1) (2) find associate volume rate corresponds to relationship, Th cs perimeter and aren in -line sol ution. Category fraction multiplication word problem score Division applications engi neeri ng problem problem XV, a subjeo: cyli nder a ct: review of the measurement of the amount of capacity, measurement a nd units of measurement of common units of measurement and their significa nce in rate 1, curren