2022年2022年假设检验习题及答案

1、精选学习资料 - - - 欢迎下载relationship、 establi she d equivalent relati onship 14、 and subje ct: a ppli cation problem 4-score s and percentage a ppli cation problem review conte nt overview a nswer s scores、 a nd perce ntage applicati on pr oblem of key is: accordi ng to meani ng、 1 determine sta ndar d vol

2、 ume units "1" 2 find associate "volume rate corresponds to" relationship、 the n in -line sol ution. category fracti on multi plicati on w ord pr oblem score division applicati ons engi neeri ng problem problem xv、 a subje ct: review of the measur ement of the am ount of capa cit

3、y、 measurement a nd units of measurement of com mon units of measureme nt and their sig nifica nce i n rate 1、 currency、 le ngth、 area、 v olume、 unit si ze、 vol ume、 weight a nd rate. omitted 2、 commonly use d time unit s and t heir relationships. slightly with a measurement units zhijia n of of pol

4、y 1、 and of method 2、 a nd poly method 3、 a nd of method and poly meth od of relationshi p measureme nt dista nce of method 1、 and tool mea sureme nt 2、 and estimates 16、 and subject: ge ometry preliminary k nowle dge 1-li ne and a ngle review conte nt line、 a nd segment、 and ray、 a nd vertical 、 an

5、d parallel、 a nd a ngle angl e of classification slightly 17、 and subje ct: geometry pr eliminary knowle dge 2-plane gra phi cs review conte nt triangle、 a nd e dges shape d、 a nd round、 and fan axisymmetric graphi cs perimeter and are a com binati on graphics of area subje ct : preliminary knowledg

6、e 3-review of soli d content categ ory 1-d shape s are divided int o: cyli nder a nd cone 2、 colum n is divi ded i nto: cuboi d、 square 3、 cone cone of the features of cuboids a nd cubes relati onship betwee n characteristi cs of cir cular cone is slig htly soli d surfa ce area and vol ume 1、 size 2

7、、 table . 和第三章假设检验3.2 一种元件 、要求其使用寿命不低于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）；已知这种元件寿命听从标准差10（0 小时） 的正态分布,试在显著水平0.05 下确定这批元件为否合格；精品学习资料精选学习资料 - - - 欢迎下载提出假设：h 0 :1000、h1 :1000精品学习资料精选学习资料 - - - 欢迎下载构造统计量：此问题情形属于u检验,故用统计量：u= x00n精品学习资料精选学习资料 - - - 欢迎下载此题中：x9500100n=2501000精品学习资料精选学习资料 - - - 欢迎

8、下载精品学习资料精选学习资料 - - - 欢迎下载代入上式得：u=950-1000 100252.5精品学习资料精选学习资料 - - - 欢迎下载拒绝域：v=uu1精品学习资料精选学习资料 - - - 欢迎下载精品学习资料精选学习资料 - - - 欢迎下载此题中：0.05u0.951.64精品学习资料精选学习资料 - - - 欢迎下载即,uu0.95拒绝原假设 h 0精品学习资料精选学习资料 - - - 欢迎下载认为在置信水平0.05下这批元件不合格；精品学习资料精选学习资料 - - - 欢迎下载3.4 某批矿砂的五个样品中镍含量经测定为（% ）： 3.253.273.243.263.24设测

9、定值听从正态分布,问在0.01下能否接受假设,这批矿砂的镍含量为精品学习资料精选学习资料 - - - 欢迎下载提出假设：h 0 :103.25h 1 :10精品学习资料精选学习资料 - - - 欢迎下载构造统计量：此题属于2未知的情形,可用t检验,即取检验统计量为：精品学习资料精选学习资料 - - - 欢迎下载t=x0sn1精品学习资料精选学习资料 - - - 欢迎下载此题中,x3.252、s=0.0117、n=5精品学习资料精选学习资料 - - - 欢迎下载代入上式得：精品学习资料精选学习资料 - - - 欢迎下载t=3.252-3.250.0117510.3419精品学习资料精选学习资料

10、- - - 欢迎下载否定域为：v=t>tn11-精品学习资料精选学习资料 - - - 欢迎下载此题中,q tt20.01、t0.995 44.6041精品学习资料精选学习资料 - - - 欢迎下载12精品学习资料精选学习资料 - - - 欢迎下载接受 h 0 、认为这批矿砂的镍含量为3.25；精品学习资料精选学习资料 - - - 欢迎下载relationship、 establi she d equivalent relati onship 14、 and subje ct: a ppli cation problem 4-score s and percentage a ppli ca

11、tion problem review conte nt overview a nswer s scores、 a nd perce ntage applicati on pr oblem of key is: accordi ng to meani ng、 1 determine sta ndar d vol ume units "1" 2 find associate "volume rate corresponds to" relationship、 the n in -line sol ution. category fracti on mult

12、i plicati on w ord pr oblem score division applicati ons engi neeri ng problem problem xv、 a subje ct: review of the measur ement of the am ount of capa city、 measurement a nd units of measurement of com mon units of measureme nt and their sig nifica nce i n rate 1、 currency、 le ngth、 area、 v olume、

13、 unit si ze、 vol ume、 weight a nd rate. omitted 2、 commonly use d time unit s and t heir relationships. slightly with a measurement units zhijia n of of poly 1、 and of method 2、 a nd poly method 3、 a nd of method and poly meth od of relationshi p measureme nt dista nce of method 1、 and tool mea sure

14、me nt 2、 and estimates 16、 and subject: ge ometry preliminary k nowle dge 1-li ne and a ngle review conte nt line、 a nd segment、 and ray、 a nd vertical 、 and parallel、 a nd a ngle angl e of classification slightly 17、 and subje ct: geometry pr eliminary knowle dge 2-plane gra phi cs review conte nt

15、triangle、 a nd e dges shape d、 a nd round、 and fan axisymmetric graphi cs perimeter and are a com binati on g raphics of area subje ct : preliminary knowledge 3-review of soli d content categ ory 1-d shape s are divided int o: cyli nder a nd cone 2、 colum n is divi ded i nto: cuboi d、 square 3、 cone

16、 cone of the features of cuboids a nd cubes relati onship betwee n characteristi cs of cir cular cone is slig htly soli d surfa ce area and vol ume 1、 size 2、 table . 和精品学习资料精选学习资料 - - - 欢迎下载3.5 确定某种溶液中的水分,它的10 个测定值 x0.452%、 s0.035%、精品学习资料精选学习资料 - - - 欢迎下载精品学习资料精选学习资料 - - - 欢迎下载设总体为正态分布n、2 、 试在水平 5%

17、 检验假设：精品学习资料精选学习资料 - - - 欢迎下载i h 0 :0.5%h1 :0.5%ii h0 :0.04%h1 :0.0.4%i 构造统计量：本文中未知,可用 t检验；取检验统计量为t=x0sn1此题中,x0.452%s=0.035%精品学习资料精选学习资料 - - - 欢迎下载代入上式得：t=拒绝域为：0.452%-0.5%0.035%10-1-4.1143精品学习资料精选学习资料 - - - 欢迎下载v=t >t 1-n1精品学习资料精选学习资料 - - - 欢迎下载此题中,0.05n=10精品学习资料精选学习资料 - - - 欢迎下载t0.95 91.8331t拒绝h

18、 04.1143精品学习资料精选学习资料 - - - 欢迎下载ii 构造统计量：未知,可挑选统计量22 ns 2 0精品学习资料精选学习资料 - - - 欢迎下载此题中,s0.035%n=1000.04%精品学习资料精选学习资料 - - - 欢迎下载代入上式得：21否定域为：10 （0.035%）2（0.04%）27.6563精品学习资料精选学习资料 - - - 欢迎下载精品学习资料精选学习资料 - - - 欢迎下载v=此题中,22n1精品学习资料精选学习资料 - - - 欢迎下载221n10.95 916.919q221n1接受h 0精品学习资料精选学习资料 - - - 欢迎下载3.9 设总

19、体 x :n 、 4、 x 1 、k、 x 16为样本,考虑如下检验问题：精品学习资料精选学习资料 - - - 欢迎下载relationship、 establi she d equivalent relati onship 14、 and subje ct: a ppli cation problem 4-score s and percentage a ppli cation problem review conte nt overview a nswer s scores、 a nd perce ntage applicati on pr oblem of key is: accordi

20、 ng to meani ng、 1 determine sta ndar d vol ume units "1" 2 find associate "volume rate corresponds to" relationship、 the n in -line sol ution. category fracti on multi plicati on w ord pr oblem score division applicati ons engi neeri ng problem problem xv、 a subje ct: review of

21、the measur ement of the am ount of capa city、 measurement a nd units of measurement of com mon units of measureme nt and their sig nifica nce i n rate 1、 currency、 le ngth、 area、 v olume、 unit si ze、 vol ume、 weight a nd rate. omitted 2、 commonly use d time unit s and t heir relationships. slightly

22、with a measurement units zhijia n of of poly 1、 and of method 2、 a nd poly method 3、 a ndof method and poly method of relationshi p measureme nt dista nce of m ethod 1、 and tool mea sureme nt 2、 and estimates 16、 and subject: ge ometry preliminary k nowle dge 1-li ne and a ngle review conte nt line、

23、 a nd segment、 and ray、 a nd vertical 、 and parallel、 a nd a ngle angl e of classification slightly 17、 and subje ct: geometry pr eliminary knowle dge 2-plane gra phi cs review conte nt triangle、 a nd e dges shape d、 a nd round、 and fan axisymmetric graphi cs perimeter and are a com binati on graphi

24、cs of area subje ct : preliminary knowledge 3-review of soli d content categ ory 1-d shape s are divided int o: cyli nder a nd cone 2、 colum n is divi ded i nto: cuboi d、 square 3、 cone cone of the features of cuboids a nd cubes relati onship betwee n characteristi cs of cir cular cone is slig htly

25、soli d surfa ce area and vol ume 1、 size 2、 table . 和h 0 :0h 1 :1精品学习资料精选学习资料 - - - 欢迎下载i ii试证下述三个检验（否定域）犯第一类错误的概率同为v1=2x-1.645v 2 =1.502x2.125v3 =2x1.96或2x1.96通过运算他们犯其次类错误的概率,说明哪个检验最好？=0.05精品学习资料精选学习资料 - - - 欢迎下载解：精品学习资料精选学习资料 - - - 欢迎下载i pxv h 00.05精品学习资料精选学习资料 - - - 欢迎下载精品学习资料精选学习资料 - - - 欢迎下载即,

26、puxuu 0.975120.05精品学习资料精选学习资料 - - - 欢迎下载这里 h 0 :0px2*1.960.05v12 x1.645精品学习资料精选学习资料 - - - 欢迎下载p2 x1.645px01.6451.64511.645精品学习资料精选学习资料 - - - 欢迎下载n=1-0.95=0.05精品学习资料精选学习资料 - - - 欢迎下载v 21.502 x2.1251.50x02.120精品学习资料精选学习资料 - - - 欢迎下载精品学习资料精选学习资料 - - - 欢迎下载p v2 h 0n2.2151.500.980.930.05精品学习资料精选学习资料 - -

27、- 欢迎下载精品学习资料精选学习资料 - - - 欢迎下载v32 x1.96或2 x1.962 x1.96x01.96精品学习资料精选学习资料 - - - 欢迎下载精品学习资料精选学习资料 - - - 欢迎下载p v3（ ii ）h 0 =1-p2 xn1.96211.960.05精品学习资料精选学习资料 - - - 欢迎下载犯其次类错误的概率=p-v h1精品学习资料精选学习资料 - - - 欢迎下载v1 :1 =p 2 x1.6451精品学习资料精选学习资料 - - - 欢迎下载relationship、 establi she d equivalent relati onship 14、

28、 and subje ct: a ppli cation problem 4-score s and percentage a ppli cation problem review conte nt overview a nswer s scores、 a nd perce ntage applicati on pr oblem of key is: accordi ng to meani ng、 1 determine sta ndar d vol ume units "1" 2 find associate "volume rate corresponds t

29、o" relationship、 the n in -line sol ution. category fracti on multi plicati on w ord pr oblem score division applicati ons engi neeri ng problem problem xv、 a subje ct: review of the measur ement of the am ount of capa city、 measurement a nd units of measurement of com mon units of measureme nt

30、 and their sig nifica nce i n rate 1、 currency、 le ngth、 area、 v olume、 unit si ze、 vol ume、 weight a nd rate. omitted 2、 commonly use d time unit s and t heir relationships. slightly with a measurement units zhijia n of of poly 1、 and of method 2、 a nd poly method 3、 a nd of method and poly meth od

31、 of relationshi p measureme nt dista nce of method 1、 and tool mea sureme nt 2、 and estimates 16、 and subject: ge ometry preliminary k nowle dge 1-li ne and a ngle review conte nt line、 a nd segment、 and ray、 a nd vertical 、 and parallel、 a nd a ngle angl e of classification slightly 17、 and subje c

32、t: geometry pr eliminary knowle dge 2-plane gra phi cs review conte nt triangle、 a nd e dges shape d、 a nd round、 and fan axisymmetric graphi cs perimeter and are a com binati on g raphics of area subje ct : preliminary knowledge 3-review of soli d content categ ory 1-d shape s are divided int o: cy

33、li nder a nd cone 2、 colum n is divi ded i nto: cuboi d、 square 3、 cone cone of the features of cuboids a nd cubes relati onship betwee n characteristi cs of cir cular cone is slig htly soli d surfa ce area and vol ume 1、 size 2、 table . 和=px10.35510.3550.36精品学习资料精选学习资料 - - - 欢迎下载v2 :2n1p 1.502x2.12

34、51精品学习资料精选学习资料 - - - 欢迎下载精品学习资料精选学习资料 - - - 欢迎下载=1-p3.50x14.125精品学习资料精选学习资料 - - - 欢迎下载n=1-4.125+3.50=1精品学习资料精选学习资料 - - - 欢迎下载v3 :3p2x1.961精品学习资料精选学习资料 - - - 欢迎下载精品学习资料精选学习资料 - - - 欢迎下载=p0.04x13.96精品学习资料精选学习资料 - - - 欢迎下载n=3.96-0.04=0.99996092-0.516=0.48396092v1显现其次类错误的概率最小,即v1最好；3.10 一骰子投掷了 120 次,得到以

35、下结果：点数123456显现次数232621201515问这个骰子为否匀称？0.05精品学习资料精选学习资料 - - - 欢迎下载解：此题原假设为：h10 : pi6i=1、2、l 、6精品学习资料精选学习资料 - - - 欢迎下载这里 n=120、npi20精品学习资料精选学习资料 - - - 欢迎下载此题采纳的统计量为pearson2统计量精品学习资料精选学习资料 - - - 欢迎下载即,2kni i 12npi npi精品学习资料精选学习资料 - - - 欢迎下载代入数据为：k2222精品学习资料精选学习资料 - - - 欢迎下载2nii 1npi npi（ 23-20 ） （ 26-2

36、0 ）l20（ 15-20 ） =4.8精品学习资料精选学习资料 - - - 欢迎下载relationship、 establi she d equivalent relati onship 14、 and subje ct: a ppli cation problem 4-score s and percentage a ppli cation problem review conte nt overview a nswer s scores、 a nd perce ntage applicati on pr oblem of key is: accordi ng to meani ng、

37、1 determine sta ndar d vol ume units "1" 2 find associate "volume rate corresponds to" relationship、 the n in -line sol ution. category fracti on multi plicati on w ord pr oblem score division applicati ons engi neeri ng problem problem xv、 a subje ct: review of the measur ement

38、of the am ount of capa city、 measurement a nd units of measurement of com mon units of measureme nt and their sig nifica nce i n rate 1、 currency、 le ngth、 area、 v olume、 unit si ze、 vol ume、 weight a nd rate. omitted 2、 commonly use d time unit s and t heir relationships. slightly with a measurement units zhijia n of of poly 1、 and of method 2、 a nd poly method 3、 a ndof method and poly method of relationshi p measureme nt dista nce of m ethod 1、 and tool mea sureme nt 2、 and estimates 16、 and subject: ge ometry