（完整版）假设检验习题及答案

relationship, established equivalent relationship 14, and subject: application problem (4)--scores and percentage application problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units \hen in-line solution. Category fraction multiplication word problem score Division applications engineering problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their significance in rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly used 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 measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)--line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angle of classification (slightly) 17, and subject: geometry preliminary knowledge (2)--plane graphics review content triangle, and edges shaped, and round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboid, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly 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：H0:??1000, H1:??1000构造统计量：此问题情形属于u检验，故用统计量： u=X??0?0n此题中：x?950 ?0?100 n=25 ?0?1000代入上式得：950-1000 u=??2.510025拒绝域： V=?u?u1???本题中：??0.05 u0.95?1.64即，u?u0.95拒绝原假设H0?认为在置信水平0.05下这批元件不合格。3.4某批矿砂的五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:?1??0?3.25 H1:?1??0构造统计量：本题属于?2未知的情形，可用t检验，即取检验统计量为： t=X??0Sn?1

本题中，x?3.252, S=0.0117, n=5代入上式得： t=3.252-3.25? 0.34190.01175?1否定域为： ?? V=?t>t?(n?1)??1-2?本题中，??0.01,t0.995(4)?4.6041Qt?t1?

?2?接受H0,认为这批矿砂的镍含量为3.25。

relationship, established equivalent relationship 14, and subject: application problem (4)--scores and percentage application problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units \hen in-line solution. Category fraction multiplication word problem score Division applications engineering problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their significance in rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly used 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 measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)--line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angle of classification (slightly) 17, and subject: geometry preliminary knowledge (2)--plane graphics review content triangle, and edges shaped, and round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboid, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surface area and volume 1, size 2, table ...和3.5确定某种溶液中的水分，它的10个测定值X?0.452%,S?0.035%,

设总体为正态分布N(?,?2),试在水平5%检验假设：

(i) H0:??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=X??0Sn?1本题中，X?0.452% S=0.035%代入上式得： t=拒绝域为： V=?t>t1-?(n?1)?本题中，??0.05 n=10 t0.95(9)?1.8331?t?4.1143?拒绝H00.452%-0.5%?-4.11430.035-1

(ii)构造统计量：?未知，可选择统计量 ??2nS2?02本题中，S?0.035% n=10 ?0?0.04%代入上式得： 210?（0.035%） ???7.65632（0.04%）否定域为：2 V=??2??12??(n?1)?本题中， ?21??(n?1)??20.95

(9)?16.919Q?2??12??(n?1)?接受H03.9设总体X:N(?,4),X1,K,X16为样本，考虑如下检验问题：

relationship, established equivalent relationship 14, and subject: application problem (4)--scores and percentage application problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units \hen in-line solution. Category fraction multiplication word problem score Division applications engineering problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their significance in rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly used 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 measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)--line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angle of classification (slightly) 17, and subject: geometry preliminary knowledge (2)--plane graphics review content triangle, and edges shaped, and round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboid, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surface area and volume 1, size 2, table ...和H0:??0 H1:???1(i) 试证下述三个检验（否定域）犯第一类错误的概率同为?=0.05 V1={2X?-1.645} ?? V=?2X??1.96或2X?1.96? V2= 1.50?2X?2.125 3

(ii) 通过计算他们犯第二类错误的概率，说明哪个检验最好？ 解：

(i)??P?x?VH0??0.05??X????即，P?U??U??U0.975??0.051????2??这里H0:??0?PX?2*1.96?0.05V1?2X??1.645?????????X?0?P2X??1.645?P???1.645???(?1.645)?1??(1.645)??????n? =1-0.95=0.05??????X?0??V2?1.50?2X?2.125??1.50??2.120??????n??P?V2H0???(2.215)??(1.50)?0.98?0.93?0.05??

?????X?0?V3?2X??1.96或2X?1.96?2X?1.96???1.96?????n???????P(V3H0)=1-P2X?1.96?2(1??(1.96))?0.05（ii）犯第二类错误的概率 ?=P??-VH1?V1: ?1=P2X??1.645???1????relationship, established equivalent relationship 14, and subject: application problem (4)--scores and percentage application problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units \hen in-line solution. Category fraction multiplication word problem score Division applications engineering problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their significance in rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly used 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 measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)--line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angle of classification (slightly) 17, and subject: geometry preliminary knowledge (2)--plane graphics review content triangle, and edges shaped, and round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboid, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surface area and volume 1, size 2, table ...和?????X?1? =P??0.355??1??(0.355)?0.36????n???V2:?2?1?P1.50?2X?2.125???1????X?1?? =1-P?3.50??4.125??????n?? =1-?(4.125)+?(3.50) =1V3:?3?P2X?1.96???1????????X?1?? =P?0.04??3.96??????n?? =?(3.96)-?(0.04) =0.99996092-0.516=0.48396092?V1出现第二类错误的概率最小，即V1最好。

3.10 一骰子投掷了120次，得到下列结果： 点数 出现次数 1 23 2 26 3 21 4 20 5 15 6 15 问这个骰子是否均匀？(??0.05) 解：

本题原假设为： H0:Pi?这里n=120,nPi?20本题采用的统计量为Pearson ?2统计量(ni?npi)2即， ???npii?12k1 i=1,2,L,66

代入数据为：222(ni?npi)2（23-20）?（26-20）?L?（15-20） ????=4.8npi20i?12k