(整理)实验七双因素方差分析.

1、精品文档实验七双因素方差分析实验目的：掌握两因素方差分析的基本理论、统计意义与方法、掌握利用SAS进行模型的建立与显著性检验.实验要求：编写程序，结果分析.实验内容：1.写出两因素等重复实验下方差分析模型，检验步骤；模型：Y一因素A和B,水平 A, A2,，Aa , BB2,，Bb ,组合水平（A，Bj）观测值 yji, yj2,，yjcyijk 二t . ；ijk -. ： i . L . ij . ；ijk, i =1,2，a, j =1,2，b, k-1,2, ,c -2Sijk N(0,cr ), 且诸 却相互独立 abab、0,工匕=0, '、 ij =0, '、 i

2、j =0 i 1j 4i 4j 4SSt 二SSA SSb SSab SSe检验步骤：(1)统计模型Y一因素A和B,水平A1,A2,，Aa,B”B2,，Bb,组合水平(Ai,Bj)观测值yij 1, yij 2，yijcJiyjk=-ij +% =2+% +1+嘛,i =1,2，a, j =1,2，b, k = 1,2，c«qjk N(0,仃2), 且诸 时相互独立ababE j = 0, g B j = 0, X - j = 0, Z 彳 j = 0、id：jdi 苴j4SS = SSa SSb SSab SSe(2)显著检验假设：Ha。：% =。2 ="" =

3、% =0修 H A1 ：至少有某个% #0H B0邛1 = £=Pb =01 H B1 :至少有某个Pj #0H Abo ： Zj =0, i =1,2,，a, j =1,2,，bi H abo :至少% ¥0.、，一MQ hao真MQ hb0真统计重FAa F(a -1, ab(c -1)Fbb F(b-1, ab(c-1)MSeMSeMS 2 HA° 真Fab =VSA1 F(a-1)(b-1), ab(c-1) MSe检3P p值：如p <a ,拒绝相应假设Pa =Pha°(Fa - fA) =P(F(a-1, ab(c-1) 一 £

4、;a)Pb =Pha°(Fb - fB) = P(F(b -1, ab(c-1) fB)Pab =Phab°(Fab -fAB)=P(F(a-1)(b-1)，ab(c-1) - fAB)(3)无交互效应的各因素均值的估计与比较b ci1偏估计内y/bcj 1 kW色.1 « 置信区间：(y .土t .ab(c1)，MSE/bc)一2林一九 置信区间：(yi«-yi2«±t -ab(c-1)j2MSE/bc), 1万m =a(a -1)/2个 代一匕2 的Bonferroni同时置信区间U1 靖一y2_a(ab(c -1)J2MSe /

5、bc)-2m(4)有交互效应时因素各水平组合上的均值估计与比较(A,Bj)上均值均无偏估计：£ =yj. (i=1,2,|l,a, j=1,2川，b)匕的置信区间：(yj.+q .ab(c-1)jMSE /c), 一2由一匕上置信度不小于1 一口的Bonferroni同时置信区间(yi1j. yi2j. t ：.(ab(c -1) 2MSe/c)1 2m3.63.7 (选作)程序如下：(1)data examp3_6;input tielizi $ jldj $ jldjdz ;cards ;al bl 0.71al b1 1.66a1 b1 2.01a1 b1 2.16a1 b1

6、2.42a1 b1 2.42a1 b1 2.56a1 b1 2.60a1 b1 3.31a1 b1 3.64a1 b1 3.74a1 b1 3.74a1 b1 4.39a1 b1 4.50a1 b1 5.07a1 b1 5.26a1 b1 8.15a1 b1 8.24a1 b2 2.20a1 b2 2.93a1 b2 3.08a1 b2 3.49a1 b2 4.11a1 b2 4.95a1 b2 5.16a1 b2 5.54a1 b2 5.68a1 b2 6.25a1 b2 7.25a1 b2 7.90a1 b2 8.85a1 b2 11.96a1 b2 15.54a1 b2 15.89a1

7、b2 18.30a1 b2 18.59a1 b3 2.25a1 b3 3.93a1 b3 5.08al b3 5.82al b3 5.84al b3 6.89al b3 8.50al b3 8.56al b3 9.44al b3 10.52al b3 13.46al b3 13.57al b3 14.76al b3 16.41a1 b3 16.96a1 b3 17.56a1 b3 22.82a1 b3 29.13a2 b1 2.20a2 b1 2.69a2 b1 3.54a2 b1 3.75a2 b1 3.83a2 b1 4.08a2 b1 4.27a2 b1 4.53a2 b1 5.32a2

8、 b1 6.18a2 b1 6.22a2 b1 6.33a2 b1 6.97a2 b1 6.97a2 b1 7.52a2 b1 8.36a2 b1 11.65a2 b1 12.45a2 b2 4.04a2 b2 4.16a2 b2 4.42a2 b2 4.93a2 b2 5.49a2 b2 5.77a2 b2 5.86a2 b2 6.28a2 b2 6.97a2 b2 7.06a2 b2 7.78a2 b2 9.23a2 b2 9.34a2 b2 9.91a2 b2 13.46a2 b2 18.40a2 b2 23.89a2 b2 26.39a2 b3 2.71a2 b3 5.43a2 b3

9、6.38a2 b3 6.38a2 b3 8.32a2 b3 9.04a2 b3 9.56a2 b3 10.01a2 b3 10.08a2 b3 10.62a2 b3 13.80a2 b3 15.99a2 b3 17.90a2 b3 18.25a2 b3 19.32a2 b3 19.87a2 b3 21.60a2 b3 22.25；run ;procanova data =examp3_6;classtielizi jldj;modeljldjdz=tielizi jldj tielizi*jldj ;meanstielizi jldj;run ;精品文档样本均值与标准值如下:me 者掰 sys

10、temThe ANOVA ProcedureLevel of t ielIzlal嘘N5454j1dj dzMean7.884258268.40277770Std Dev6.143609546.03399690Levs 1 ofj i oj azJldjNMeanStd ttevbl364.817777B2,66785212b2368.91805566.05675435b33612.19472226.49327982The SAS SystemThe 直机ProcedureClass Levs I Informat ionClassLeveIsValuestielizi24 a2j Idj3b

11、l b2 b3Number of observations 108则 c=18n=abc=108 a=2,b=3注：SSr =SSa +SSb +SSab ,因素变量为自变量，个数为自由度p-1=ab-1=5f=i-、卜(心-1,岫9-1=先检验 H 0 ： ij =0,统计量F(5,102),_ '拎建=7 31J - 7$P = f H5r102 )> / )观测值 山”，检验p值电"ofSourceDFSquaresMeanSquareF VluePr > FMode 151054.1G7G52210.3335307.32<.0001Error1022

12、838.20881128,815918Carrected Total1078892.8712B3R-SquareCceffVarRootM8EjIdjdz Mean0.26404562,094095,3671146,643519SourceDFAnova SSMeanSquareF ValuePr > Ft is 1 2162.259259362.25325932.160.1446Jldj2963.817578649 L000788817.07<.0001t i 811 z i 1 dj28.29081304,14540650J40.8661The SAS System11:37

13、FirThe ANOVA Procedurejz由结果可知，产泅=0 8661 > °-05 ,接受h Abo ,认为铁离子与剂量等级的交互不显著，即 两种形式的铁离子在不同剂量下在动物体内的存留量认为相同.而检验HA0和HB0,检验p值分另I为pA=0.1446和Pb =0.001 ,接受Hao,拒绝Hbo.说明因素A寸剂量等级的值 的影响不显著。由输出结果可看出，高剂量水平的标准差与其他两个的标准差是有显著性差 异的，中剂量和低剂量水平的标准差无明显差异，所以可以认为各组和水平上的标准差有显著相差异，而认为误差的等方差性不合理。(2)程序如下：data change3_6;

14、set examp3_6;aaa = log (jldjdz);run ;proc anova data = change3_6;class jldj;model aaa = jldj;means jldj;run ;20:44 ThursdiThe SAS SystemThe ANOVA ProcedurediI 12 3 J b b baaaMeanStd Dev1.420526281.995634722.341685040.583637570.616140060.60676504由以上输出结果可以看出，各组合水平上的标准差趋于一致(3)程序如下：proc anova data = cha

15、nge3_6;class tielizi jldj;model aaa = tielizi jldj tielizi*jldj;means tielizi jldj tielizi*jldj;means tielizi jldj/bon cldiff alpha =0.05 ;run ;flnova SSMean SquareF ValuePrF2.0738394 I2.07383941S, 990.01EI15-588407977>79420398£2.52J MOI0.810269470.405134731.1?0.3143SourceDFSun of SquaresMea

16、n SquareF ValuePr > FMode)S18.472516843.6945033710.68< .«00tError10235.2958G0180.34603784Corrected Total10?53.76837702R-SquareCoeff UarRoot MSEaaa Mean0.34355730 M4 9470.5802501.9192Q2存留量的方差分析表、.、.广. 力左来源自由度离子12.07382.07385.990.0161剂量215.58847.794222.520.0001交互 效应20.81030.40511.170.314310

17、235.29590.3460总和10753.7648由此结果可看出离子与剂量之间的交互效应是不显著的，也就是说离子的不同对剂量差异无影响；同样，剂量之间的差异与铁离子的差异无关。两种离子在小白鼠体内存留量百分比是有显著差异的，不同剂量在小白鼠体内存留量百分比也是有显著差异的。(4)程序如下：means tielizi jldj/bon cldiff alpha=0.05;run;The SHS System14:54 ThursdayThe ANOVA ProcedureLeve 1of 一- Witie】i z iNMeanStd Deval541.780709950.77926M3日254

18、2.057054080.60672273Levs i TL口口口JldJNMeanStd Devbl361.420526280.58363757b?361.995634720.61614006b3362.341605040.60076504TheANOVA ProcedureBonferron i(Dunn) t Tests foraaa>ntrols the Type I exper imentw ise error rate, but i t generally has i error rate than Tukey's for al 1 pairwise compar i

19、sons -AlphaQ.0SError Degrees of Freedom102Error (lean Square0.346038Cr i t ical Value oft1.98350M in imun Signif icant D jfference0,2245Compar isons signi f icant at the 0,05 1 eve 1 are ind icated by *.t i e 1 i z i Comparison-al al - a2D i fferenceBetween Means0.2771-0.2771Simultaneous95M Conf ide

20、nceLimits0,0526 0.5017 *-0.5017 -0.0526 *%. =20578 7=0一055对于专业因素A,由上述结果求得其，,=1-7807,个差匕Z，故m=1,皿7)= &41吟=°O5 ,忆匕,的置信度).因此认为 .父也.，即三价铁离子对小白鼠为95%勺置信区间为(-0.5017 -0.0526的影响显著差于二价铁离子的影响。Bonferron i (Dunn) t Tests for aaa：the Type I exper imen ise error rate, but i t genera 1 Iy has error rate than Tulkey 1 s