文章目录
前言一、10. 1二、10. 3三、10. 4四、10. 5▌总结前言
\qquad计算器敲起来
仅供参考
一、10. 1
代入公式：
xi.=∑j=1nixij,i=1,2,...,a,x..=∑i=1a∑j=1nixijST=∑i=1a∑j=1nixij−x..2nSA=∑i=1axi.2ni−x..2nSE=ST−SAx_i. = \sum_{j=1}^{n_i}x_{ij} \,,i=1,2,...,a,\qquad x.. = \sum_{i=1}^a\sum_{j=1}^{n_i}x_{ij} \\ S_T = \sum_{i=1}^a\sum_{j=1}^{n_i}x_{ij} -\frac{{x^2_{..}}}{n}\\ S_A = \sum_{i=1}^a\frac{{x^2_{i.}}}{n_i}-\frac{{x^2_{..}}}{n}\\ S_E = S_T-S_Axi.=j=1∑nixij,i=1,2,...,a,x..=i=1∑aj=1∑nixijST=i=1∑aj=1∑nixij−nx..2SA=i=1∑anixi.2−nx..2SE=ST−SA
解：
设各机器生产的薄板厚度为：xij=μi+εiji=1,2,3j=1,2,3,4,5原假设H0:μ1=μ2=μ3备择假设H1：μi≠μj,至少有一对i,ja=3ni=5(i=1,2,3)n=15x..=∑i=13∑j=15xij=38x1.=12.1x2.=12.8x3.=13.1ST=2.362+2.382+...+2.622−38215=0.13SA=15(12.12+12.82+13.12)−38215=0.11SE=ST−SA=0.02SE,ST,SA的自由度分别为12,14,2MSA=0.11/2=0.055MSE=0.02/12=0.00167F=MSAMSE=32.93查表得：Fα(a−1,n−a)=3.89∵32.93>3.89∴拒绝H0，各机器生产的薄板厚度有显著差异设各机器生产的薄板厚度为：\\ x_{ij} = \mu_i + \varepsilon_{ij}\qquad i = 1,2,3 \ \ j = 1,2,3,4,5 \\原假设H_0:\mu_1=\mu_2=\mu_3\qquad备择假设H_1：\mu_i\ne\mu_j,至少有一对i,j\\ \qquad a =3\qquad n_i= 5(i=1,2,3)\qquad n=15\qquad x.. =\sum_{i=1}^3\sum_{j=1}^{5}x_{ij}= 38\\ x_{1.}=12.1 \qquad x_{2.}=12.8 \qquad x_{3.}=13.1\\ S_T = 2.36^2+2.38^2+...+2.62^2-\frac{38^2}{15} = 0.13\\ S_A = \frac15 (12.1^2+12.8^2+13.1^2)-\frac{38^2}{15} =0.11\\ S_E = S_T-S_A= 0.02 \\ S_E,S_T,S_A的自由度分别为12,14,2\\ MS_A = 0.11/2 = 0.055 \qquad MS_E = 0.02/12 = 0.00167\\ F = \frac{MS_A}{MS_E} = 32.93\\ 查表得： F_\alpha(a-1,n-a) = 3.89\\ \because32.93>3.89\qquad\therefore拒绝H_0，各机器生产的薄板厚度有显著差异设各机器生产的薄板厚度为：xij=μi+εiji=1,2,3j=1,2,3,4,5原假设H0:μ1=μ2=μ3备择假设H1：μi=μj,至少有一对i,ja=3ni=5(i=1,2,3)n=15x..=i=1∑3j=1∑5xij=38x1.=12.1x2.=12.8x3.=13.1ST=2.362+2.382+...+2.622−15382=0.13SA=51(12.12+12.82+13.12)−15382=0.11SE=ST−SA=0.02SE,ST,SA的自由度分别为12,14,2MSA=0.11/2=0.055MSE=0.02/12=0.00167F=MSEMSA=32.93查表得：Fα(a−1,n−a)=3.89∵32.93>3.89∴拒绝H0，各机器生产的薄板厚度有显著差异
二、10. 3
\qquad 设各测量值总体服从同方差的正态分布，试分析各类型电路对啊应时间有无显者影响(α=0.05)?(\alpha = 0.05)?(α=0.05)?.
和第一题一样：
xi.=∑j=1nixij,i=1,2,...,a,x..=∑i=1a∑j=1nixijST=∑i=1a∑j=1nixij−x..2nSA=∑i=1axi.2ni−x..2nSE=ST−SAx_i. = \sum_{j=1}^{n_i}x_{ij} \,,i=1,2,...,a,\qquad x.. = \sum_{i=1}^a\sum_{j=1}^{n_i}x_{ij} \\ S_T = \sum_{i=1}^a\sum_{j=1}^{n_i}x_{ij} -\frac{{x^2_{..}}}{n}\\ S_A = \sum_{i=1}^a\frac{{x^2_{i.}}}{n_i}-\frac{{x^2_{..}}}{n}\\ S_E = S_T-S_Axi.=j=1∑nixij,i=1,2,...,a,x..=i=1∑aj=1∑nixijST=i=1∑aj=1∑nixij−nx..2SA=i=1∑anixi.2−nx..2SE=ST−SA
解：
设各类型电路的响应时间为：xij=μi+εiji=1,2,3,4j=1,2,3,4,5(i取4时，j不取4,5)原假设H0:μ1=μ2=μ3=μ4备择假设H1：μi≠μj,至少有一对i,ja=4ni=5(i=1,2,3)n4=3n=18x..=386x1.=94x2.=141x3.=92x4.=59ST=192+222+...+192−386218=714.44SA=15(942+1412+922)+5923−386218=318.98SE=ST−SA=395.46ST,SA,SE的自由度分别为17,3,14MSA=318.98/3=106.33MSE=395.46/14=28.25F=MSAMSE=3.76查表得：F0.05(3,14)=3.34∵3.76>3.34∴拒绝H0，接受H1各类型电路的响应时间有显著差异设各类型电路的响应时间为：\\ x_{ij} = \mu_i + \varepsilon_{ij}\qquad i = 1,2,3,4 \ \ j = 1,2,3,4,5 (i取4时，j不取4,5) \\原假设H_0:\mu_1=\mu_2=\mu_3=\mu_4\qquad备择假设H_1：\mu_i\ne\mu_j,至少有一对i,j\\ \qquad a =4\qquad n_i= 5(i=1,2,3)\ \ n_4=3\qquad n = 18 \qquad x.. =386\\ x_{1.}=94\qquad x_{2.}=141\qquad x_{3.}=92\qquad x_{4.}=59\\ S_T = 19^2+22^2+...+19^2-\frac{386^2}{18} = 714.44\\ S_A = \frac15 (94^2+141^2+92^2)+\frac{59^2}{3}-\frac{386^2}{18} =318.98\\ S_E = S_T-S_A= 395.46 \\ S_T,S_A,S_E的自由度分别为17,3,14\\ MS_A = 318.98/3 = 106.33\qquad MS_E = 395.46/14= 28.25\\ F = \frac{MS_A}{MS_E} = 3.76\\ 查表得： F_{0.05}(3,14) = 3.34\\ \because3.76>3.34\qquad\therefore拒绝H_0，接受H_1\\ 各类型电路的响应时间有显著差异设各类型电路的响应时间为：xij=μi+εiji=1,2,3,4j=1,2,3,4,5(i取4时，j不取4,5)原假设H0:μ1=μ2=μ3=μ4备择假设H1：μi=μj,至少有一对i,ja=4ni=5(i=1,2,3)n4=3n=18x..=386x1.=94x2.=141x3.=92x4.=59ST=192+222+...+192−183862=714.44SA=51(942+1412+922)+3592−183862=318.98SE=ST−SA=395.46ST,SA,SE的自由度分别为17,3,14MSA=318.98/3=106.33MSE=395.46/14=28.25F=MSEMSA=3.76查表得：F0.05(3,14)=3.34∵3.76>3.34∴拒绝H0，接受H1各类型电路的响应时间有显著差异
三、10. 4
代入公式：
ST=∑i=1a∑j=1bxij2−x..2abSA=∑i=1axi.2b−x..2abSB=∑j=1bx.j2a−x..2abSE=ST−SA−SBS_T = \sum_{i=1}^a\sum_{j=1}^{b}x_{ij}^2 -\frac{{x^2_{..}}}{ab}\\ S_A = \sum_{i=1}^a\frac{{x^2_{i.}}}{b}-\frac{{x^2_{..}}}{ab}\\ S_B = \sum_{j=1}^b\frac{{x^2_{.j}}}{a}-\frac{{x^2_{..}}}{ab}\\ S_E = S_T-S_A-S_BST=i=1∑aj=1∑bxij2−abx..2SA=i=1∑abxi.2−abx..2SB=j=1∑bax.j2−abx..2SE=ST−SA−SB
解：
设产量为：xij=μ+αi+βj+εiji=1,2,3,4j=1,2,3,4,5,6原假设HA0:α1=α2=α3=α4=0HB0:β1=β2=...=β6=0备择假设HA1：αi≠0,至少有一个iHA1：βj≠0,至少有一个ja=4b=6ab=24α=0.05x..=1010.6由表可得：x1.=247.8x2.=248x3.=255.4x4.=259.4x.1=163.5x.2=162.1x.3=164.9x.4=169.8x.5=176.2x.6=174.1ST=∑i=14∑j=16xij2−1010.6224=83.34SA=∑i=1axi.2b−x..2ab=16.38SB=∑j=1bx.j2a−x..2ab=42.81SE=ST−SA−SB=24.15ST,SA,SB,SE的自由度分别为23,3,5,15MSA=5.46MSB=8.562MSE=1.61F1=MSAMSE=3.39F2=MSBMSE=5.32查表得：F0.05(3,15)=3.29F0.05(5,15)=2.90∵F1>3.29F2>2.90∴拒绝HA0，HB0，不同的机器、不同的运转速度对产量都有显著影响.x_{ij} = \mu +\alpha_i+\beta_j+ \varepsilon_{ij}\qquad i = 1,2,3,4 \ \ j = 1,2,3,4,5,6 \\原假设H_{A0}:\alpha_1=\alpha_2=\alpha_3=\alpha_4=0\qquad H_{B0}:\beta_1=\beta_2=...=\beta_6=0\\ 备择假设H_{A1}：\alpha_i\ne0,至少有一个i\qquad H_{A1}：\beta_j\ne0,至少有一个j\\ \qquad a =4\ \ b=6\qquad ab=24\qquad \alpha = 0.05 \qquad x.. =1010.6\\ 由表可得：x_{1.}=247.8\qquad x_{2.}=248\qquad x_{3.}=255.4\qquad x_{4.}=259.4\\ x_{.1}=163.5\quad x_{.2}=162.1\quad x_{.3}=164.9\quad x_{.4}=169.8\quad x_{.5}=176.2\quad x_{.6}=174.1 \\ S_T = \sum_{i=1}^4\sum_{j=1}^{6}x_{ij}^2 -\frac{{1010.6^2}}{24} = 83.34\\ S_A = \sum_{i=1}^a\frac{{x^2_{i.}}}{b}-\frac{{x^2_{..}}}{ab}=16.38\\ S_B = \sum_{j=1}^b\frac{{x^2_{.j}}}{a}-\frac{{x^2_{..}}}{ab}=42.81\\ S_E = S_T-S_A-S_B=24.15 \\ S_T,S_A,S_B,S_E的自由度分别为23,3,5,15\\ MS_A = 5.46\qquad MS_B = 8.562\qquad MS_E = 1.61\\ F_1 = \frac{MS_A}{MS_E} = 3.39\qquad F_2 = \frac{MS_B}{MS_E} =5.32\\ 查表得： F_{0.05}(3,15) = 3.29\qquad F_{0.05}(5,15) =2.90 \\ \because F_1>3.29\qquad F_2 > 2.90\\ \therefore拒绝H_{A0}，H_{B0}，不同的机器、不同的运转速度对产量都有显著影响.xij=μ+αi+βj+εiji=1,2,3,4j=1,2,3,4,5,6原假设HA0:α1=α2=α3=α4=0HB0:β1=β2=...=β6=0备择假设HA1：αi=0,至少有一个iHA1：βj=0,至少有一个ja=4b=6ab=24α=0.05x..=1010.6由表可得：x1.=247.8x2.=248x3.=255.4x4.=259.4x.1=163.5x.2=162.1x.3=164.9x.4=169.8x.5=176.2x.6=174.1ST=i=1∑4j=1∑6xij2−241010.62=83.34SA=i=1∑abxi.2−abx..2=16.38SB=j=1∑bax.j2−abx..2=42.81SE=ST−SA−SB=24.15ST,SA,SB,SE的自由度分别为23,3,5,15MSA=5.46MSB=8.562MSE=1.61F1=MSEMSA=3.39F2=MSEMSB=5.32查表得：F0.05(3,15)=3.29F0.05(5,15)=2.90∵F1>3.29F2>2.90∴拒绝HA0，HB0，不同的机器、不同的运转速度对产量都有显著影响.
四、10. 5
代入公式：
ST=∑i=1a∑j=1b∑k=1nxijk2−x...2abnSA=1bn∑i=1axi..2−x...2abnSB=1an∑j=1bx.j.2−x...2abnSA×B=1n∑i=1a∑j=1bxij.2−x...2abn−SA−SBSE=ST−SA−SB−SA×BS_T = \sum_{i=1}^a\sum_{j=1}^{b}\sum_{k=1}^n x_{ijk}^2 -\frac{{x^2_{...}}}{abn}\\ S_A =\frac1{bn} \sum_{i=1}^ax^2_{i..}-\frac{{x^2_{...}}}{abn}\\ S_B =\frac1{an} \sum_{j=1}^{b}x^2_{.j.}-\frac{{x^2_{...}}}{abn}\\ S_{A\times B} = \frac1n\sum_{i=1}^a\sum_{j=1}^{b}x^2_{ij.}-\frac{{x^2_{...}}}{abn}-S_A-S_B\\ S_E = S_T-S_A-S_B-S_{A\times B}ST=i=1∑aj=1∑bk=1∑nxijk2−abnx...2SA=bn1i=1∑axi..2−abnx...2SB=an1j=1∑bx.j.2−abnx...2SA×B=n1i=1∑aj=1∑bxij.2−abnx...2−SA−SBSE=ST−SA−SB−SA×B
解：
设燃烧速度为：xij=μ+αi+βj+γij+εijki=1,2,3,4j=1,2,3,4,5,6k=1,2原假设HA0:α1=α2=α3=0HB0:β1=...=β4=0HAB0:γij=0备择假设HA1：αi≠0,至少有一个iHA1：βj≠0,至少有一个jHAB0:γij≠0,至少有一对i,ja=3b=4n=2abn=24α=0.05x...=710.2由表可得：x1..=244x2..=237.4x3..=228.8x..1=189.6x..2=179.1x..3=170.3x..4=171.2ST=∑i=1a∑j=1b∑k=1nxijk2−x...2abn=21107.68−21016=91.68SA=1bn∑i=1axi..2−x...2abn=14.52SB=1an∑j=1bx.j.2−x...2abn=40.08SA×B=1n∑i=1a∑j=1bxij.2−x...2abn−SA−SB=22.17SE=ST−SA−SB−SA×B=14.91ST,SA,SB,SA×B,SE的自由度分别为23,2,3,6,12MSA=7.26MSB=13.36MSA×B=3.695MSE=1.24F1=MSAMSE=5.85F2=MSBMSE=10.77F3=MSA×BMSE=2.98查表得：F0.05(2,12)=3.89F0.05(3,12)=3.49F0.05(6,12)=3.00∵F1>3.89F2>3.49F3<3.00∴导弹系统推进器类型对燃烧速度有显著影响，它们的交互作用对燃烧速度无影响.x_{ij} = \mu +\alpha_i+\beta_j+\gamma_{ij} +\varepsilon_{ijk}\qquad i = 1,2,3,4 \ \ j = 1,2,3,4,5,6\ \ k=1,2 \\原假设H_{A0}:\alpha_1=\alpha_2=\alpha_3=0\quad H_{B0}:\beta_1=...=\beta_4=0\quad H_{AB0}:\gamma_{ij}=0\\ 备择假设H_{A1}：\alpha_i\ne0,至少有一个i\quad H_{A1}：\beta_j\ne0,至少有一个j \quad H_{AB0}:\gamma_{ij}\ne0,至少有一对i,j \\ \qquad a =3\ \ b=4\ \ n=2\qquad abn=24\qquad \alpha = 0.05 \qquad x... =710.2\\ 由表可得：x_{1..}=244\qquad x_{2..}=237.4\qquad x_{3..}=228.8\\ x_{..1}=189.6\quad x_{..2}=179.1\quad x_{..3}=170.3\quad x_{..4}=171.2\\ S_T = \sum_{i=1}^a\sum_{j=1}^{b}\sum_{k=1}^n x_{ijk}^2 -\frac{{x^2_{...}}}{abn}=21107.68-21016= 91.68\\ S_A =\frac1{bn} \sum_{i=1}^ax^2_{i..}-\frac{{x^2_{...}}}{abn}=14.52\\ S_B =\frac1{an} \sum_{j=1}^{b}x^2_{.j.}-\frac{{x^2_{...}}}{abn}=40.08\\ S_{A\times B} = \frac1n\sum_{i=1}^a\sum_{j=1}^{b}x^2_{ij.}-\frac{{x^2_{...}}}{abn}-S_A-S_B=22.17\\ S_E = S_T-S_A-S_B-S_{A\times B=14.91} \\ S_T,S_A,S_B,S_{A\times B} ,S_E的自由度分别为23,2,3,6,12\\ MS_A = 7.26\quad MS_B = 13.36\quad MS_{A\times B} = 3.695\qquad MS_E = 1.24\\ F_1 = \frac{MS_A}{MS_E} = 5.85\qquad F_2 = \frac{MS_B}{MS_E} =10.77\qquad F_3= \frac{MS_{A\times B}}{MS_E} =2.98\\ 查表得： F_{0.05}(2,12) = 3.89\quad F_{0.05}(3,12) =3.49\quad F_{0.05}(6,12) =3.00\\ \because F_1>3.89\qquad F_2 > 3.49 \qquad F_3 < 3.00\\ \therefore导弹系统推进器类型对燃烧速度有显著影响，它们的交互作用对燃烧速度无影响.xij=μ+αi+βj+γij+εijki=1,2,3,4j=1,2,3,4,5,6k=1,2原假设HA0:α1=α2=α3=0HB0:β1=...=β4=0HAB0:γij=0备择假设HA1：αi=0,至少有一个iHA1：βj=0,至少有一个jHAB0:γij=0,至少有一对i,ja=3b=4n=2abn=24α=0.05x...=710.2由表可得：x1..=244x2..=237.4x3..=228.8x..1=189.6x..2=179.1x..3=170.3x..4=171.2ST=i=1∑aj=1∑bk=1∑nxijk2−abnx...2=21107.68−21016=91.68SA=bn1i=1∑axi..2−abnx...2=14.52SB=an1j=1∑bx.j.2−abnx...2=40.08SA×B=n1i=1∑aj=1∑bxij.2−abnx...2−SA−SB=22.17SE=ST−SA−SB−SA×B=14.91ST,SA,SB,SA×B,SE的自由度分别为23,2,3,6,12MSA=7.26MSB=13.36MSA×B=3.695MSE=1.24F1=MSEMSA=5.85F2=MSEMSB=10.77F3=MSEMSA×B=2.98查表得：F0.05(2,12)=3.89F0.05(3,12)=3.49F0.05(6,12)=3.00∵F1>3.89F2>3.49F3<3.00∴导弹系统推进器类型对燃烧速度有显著影响，它们的交互作用对燃烧速度无影响.
▌总结
考神保佑！！！！！！！！！！！！考神保佑！！！！！！！！！！！！考神保佑！！！！！！！！！！！！附：
\qquad 第十一章传送门.
