西南交通大学：《管理统计学》（双语版） 第10章 方差分析

第10章方差分析( ANOVA Analysis of Variance

第10章 方差分析(ANOVA) Analysis of Variance

本章概要 The completely randomized model: One-Factor Analysis of Variance o F-Test for Difference in c Means n The Tukey-Kramer Procedure 口 ANOVA AsSumptions The factorial Design Model: Two-Way Analysis of Variance a Examine effect of factors and Interaction Kruksa-Wallis Rank Test for Differences in c Medians

• The Completely Randomized Model: One-Factor Analysis of Variance  F-Test for Difference in c Means  The Tukey-Kramer Procedure  ANOVA Assumptions • The Factorial Design Model: Two-Way Analysis of Variance  Examine Effect of Factors and Interaction • Kruksal-Wallis Rank Test for Differences in c Medians 本章概要

One-Factor Analysis of Variance 单因子方差分析 Evaluate the Difference Among the Means of 2 or More(c) Populations n e.g., Several Types of Tires, Oven Temperature Settings Assumptions: a Samples are Randomly and Independently Drawn (This condition must be met. a Populations are Normally distributed (F test is Robust to moderate departures from normality. n Populations have Equal Variances

One-Factor Analysis of Variance 单因子方差分析 • Evaluate the Difference Among the Means of 2 or More (c) Populations  e.g., Several Types of Tires, Oven Temperature Settings • Assumptions:  Samples are Randomly and Independently Drawn (This condition must be met.)  Populations are Normally Distributed (F test is Robust to moderate departures from normality.)  Populations have Equal Variances

多个独立样本均值的比较-单因素方差分析 1、资料类型 样本1样本2 样本k 11 12 lk XC 21 22 2k n22 Xu.k 注意,k个样本含量不必相等!!!

多个独立样本均值的比较--单因素方差分析 1、资料类型 样本1 样本2 …… 样本k 注意，k 个样本含量不必相等！！！               n n n k k k k x x x x x x x x x ... ... ... ... ... ... ... 1 2 2 1 2 2 2 1 1 1 2 1 1 2

2、总变异的分解…方差分析的关键!!! Xm=D∑∑x1/n,n=n1+n2+…+nk ∑ 1.2.k ∑∑(x1-Xm)21 ∑1n1(x1-Xa)2+21∑1(x1-X SS between + SS internal MS,= SS between /(k-1),MS=SS Interna (n-k) F=MSh/MS F(k-l,n-k)

2、总变异的分解……方差分析的关键！！！ / ~ ( 1, ) /( 1), /( ) ( ) ( ) [ ( ) ] [ ]/ , 1,2,..., ; [ ]/ , ... ; int int 1 1 2 1 2 1 2 1 1 1 2 1 1 F MS MS F k n k MS SS k MS SS n k SS SS n X X x X SS x X X x n j k X x n n n n n b i b between i ernal between ernal k j n i i j j k j j j total k j n i total i j total j n i j i j k k j n i total i j j j j j = − − = − = −  + = − + − = − = = =  + + +       = = = = = = = = 

One-Factor ANOVA Test Hypothesis Ho: ui=H2=p ●●● μc .All population means are equal No treatment effect (NO variation in means among groups HI: not all the k are equal .At least oNE population mean is different (Others may be the same!) There is treatment effect Does not mean that all the means are different 1≠μ2≠…≠μ

One-Factor ANOVA Test Hypothesis H0 : 1 = 2 = 3 = ... = c •All population means are equal •No treatment effect (NO variation in means among groups) H1 : not all the k are equal •At least ONE population mean is different (Others may be the same!) •There is treatment effect Does NOT mean that all the means are different: 1  2  ...  c

3、[实例分析 三组销售不同包装饮料的商品的日均销售量 瓶装组罐装组袋装(老包装) 75 74 60 70 78 64 66 72 65 60 68 55 71 58 问题:三种包装的日平均销售量是否有显著差异?

3、[实例分析] 三组销售不同包装饮料的商品的日均销售量 瓶装组 罐装组 袋装（老包装） 75 74 60 70 78 64 66 72 65 69 68 55 71 63 58 问题：三种包装的日平均销售量是否有显著差异？

方差分析结果0:A1=/2=/3 anova for SALE SS df Mean Square F Sig Between 420.367 2 210.183 14.661.001 Within172.0331214.336 Total 592.40014 结论:P=0.001-0.05,可以认为不同包装下的饮 料平均销售量整体上表现出统计学意义上的 差。注意,要想知道是哪两种包装之间有差 异,尚须做两量两比较

方差分析结果 ANOVA for SALE SS df Mean Square F Sig. Between 420.367 2 210.183 14.661 .001 Within 172.033 12 14.336 Total 592.400 14 结论：P=0.001<0.05，可以认为不同包装下的饮 料平均销售量整体上表现出统计学意义上的 差。注意，要想知道是哪两种包装之间有差 异，尚须做两量两比较。 0 1 2 3 H :  =  = 

One-Factor ANOVA No Treatment effect H0:p1=2=p3 请注意其含义 Hi: not all the pk are equal The null Hypothesis is True 1=12=13

One-Factor ANOVA: No Treatment Effect  =  =  H0 : 1 = 2 = 3 = ... = c H1 : not all the k are equal The Null Hypothesis is True 请注意其含义

One factor ANOVa Treatment Effect Present H0:p1=2=3 Hr: not all the uk are equal The null Hypothesis is NOT True μ1=≠3

One Factor ANOVA: Treatment Effect Present  =   H0 : 1 = 2 = 3 = ... = c H1 : not all the k are equal The Null Hypothesis is NOT True