Single Factor Experiments

Transcription

1 Single Factor Experiments Bruce A Craig Department of Statistics Purdue University STAT 514 Topic 4 1

2 Analysis of Variance Suppose you are interested in comparing either a different treatments a levels of one treatment factor You randomly assign EUs to the a groups using CRD For analysis, could do ( a 2) two-sample t-tests but... Linear model/anova provides method of joint inference { i = 1,2...a y ij = µ+τ i +ǫ ij j = 1,2,...n i µ+τ i - ith treatment mean ǫ ij N(0,σ 2 ) - error or noise component H 0 : τ 1 = τ 2 =... = τ a = 0 H 1 : τ i 0 for at least one i STAT 514 Topic 4 2

3 Partitioning the Variability of y ij Notation y i. = n i j=1 y ij y i. = y i. /n i (trt sum/average) y.. = y ij y.. = y.. /N (grand sum/average) Decomposition of y ij : Rewrite y ij as y.. +(y i. y.. )+(y ij y i. ) Thus, we can think of y ij as ˆµ+ ˆτ i +ˆǫ ij where ˆµ = y.. ˆτ i = (y i. y.. ) ˆǫ ij = y ij y i. Note that there are some built-in estimate properties niˆτ i = 0 ˆǫij = 0 for each treatment i STAT 514 Topic 4 3

4 Computing the Sum of Squares Given y ij y.. = (y i. y.. )+(y ij y i. ) Can show Total SS = Treatment SS + Error SS (yij y.. ) 2 = n i (y i. y.. ) 2 + (y ij y i. ) 2 SS T = SS Treatments + SS E To test H 0, look at n iˆτ 2 i = n i (y i. y.. ) 2 Will be small if all ˆτ i s 0 Thus, if large we reject H 0, but how large is large? STAT 514 Topic 4 4

5 F statistic To determine large, need to standardize n iˆτ i 2 to account for inherent variability of response within a treatment ni (y F 0 = i. y.. ) 2 /(a 1) (yij y i. ) 2 /(N a) = SS Treatment/(a 1) SS E /(N a) = average squared deviation between trts average squared deviation within trts STAT 514 Topic 4 5

6 Why use F 0? From general sum of squares result, can show 1. SS E /(N a) is unbiased estimate of σ 2 2. Under H 0, SS Treatment /(a 1) is also unbiased estimate of σ 2 Call these estimates Mean Squares Can show the expected values are E(MS E )=σ 2 E(MS Treatment ) = σ 2 + n i τ 2 i /(a 1) The test statistic F 0 is a ratio of mean squares If statistic much larger than one, then reject H 0 What is the test statistic s distribution under H 0? STAT 514 Topic 4 6

7 Test Statistic Distribution I Assume independent y ij N(µ+τ i,σ 2 ) This means... y i. N(µ+τ i,σ 2 /n i ) (yij y i. ) 2 /σ 2 χ 2 n i 1 (yij y i. ) 2 /σ 2 χ 2 N a Under H 0... y i. N(µ,σ 2 /n i ) ni (y i. y.. ) 2 /σ 2 χ 2 a 1 STAT 514 Topic 4 7

8 Test Statistic Distribution II If we can show independence of numerator and denominator SS, then ratio is F distributed Cochran s Thm Since SS T with N 1 degrees of freedom is partitioned into SS Treatment and SS E and (a 1) + (N a) = N 1, then the two sum of squares are independent Therefore, use F with a 1 and N a degrees of freedom to determine the degree of evidence against H 0 Distribution approximately F when y ij non-normal STAT 514 Topic 4 8

9 Analysis of Variance Table Source of Sum of Degrees of Mean F 0 Variation Squares Freedom Square Between SS Treatment a 1 MS Treatment F 0 Within SS E N a MS E Total SS T N 1 If F 0 > F α,a 1,N a then reject H 0 (using critical value) If P(F a 1,N a > F 0 ) α then reject H 0 (using P-value) STAT 514 Topic 4 9

10 Similarity with Two-Sample t-test Consider the square of the t-statistic (n 1 = n 2 = n) ( ) 2 ( ) 2 t0 2 y = 1. y 2. (y = 1. y.. ) (y 2. y.. ) S p 2/n S p 2/n = 2((y 1. y.. ) 2 +(y 2. y.. ) 2 ) S 2 p(2/n) = MS(Between) MS(Within) = MS Treatment MS E = n((y 1. y.. ) 2 +(y 2. y.. ) 2 ) S 2 p When a = 2, t 2 o = F 0 F-test gives identical results as equal variance two-sample t-test for two-sided alternative H A : µ 1 µ 2 STAT 514 Topic 4 10

11 Example Twelve lambs are randomly assigned to three different diets. The weight gain (in two weeks) is recorded. Is there a difference in the average weight gain among the diets? Diet 1 Diet 2 Diet yij = 156 and y 2 ij = 2274 y 1. = 33, y 2. = 75, and y 3. = 48 n 1 = 3, n 2 = 5, n 3 = 4 and N = 12 STAT 514 Topic 4 11

12 Hand Calculations SS T = /12 = 246 SS Treatment = (33 2 / / /4) /12 = 36 SS E = 2274 (33 2 / / /4) = 210 As a check, see if SS T = SS Treatment + SS E F 0 = (36/2)/(210/9) = 0.77 P-value > 0.25 Do not reject - There is not strong evidence against the null hypothesis that the average weight gains are equal STAT 514 Topic 4 12

13 Using SAS (lambs.sas) data lambs; input diet wtgain cards; ; proc glm plots=all; class diet; model wtgain=diet; output out=diag r=res p=pred; symbol1 v=circle i=sm50; axis1 offset=(5); proc sort; by pred; proc gplot; plot res*pred / haxis=axis1; run; STAT 514 Topic 4 13

14 Log File (.log) 395 data lambs; 396 input diet wtgain; 397 cards; NOTE: SAS went to a new line when INPUT statement reached past the end of a line. NOTE: The data set WORK.LAMBS has 12 observations and 2 variables. NOTE: DATA statement used: real time 0.03 seconds 401 ; proc glm; 404 class diet; 405 model wtgain=diet; 406 output out=diag r=res p=pred; 407 run; NOTE: The data set WORK.DIAG has 12 observations and 4 variables. NOTE: PROCEDURE GLM used: real time 0.61 seconds STAT 514 Topic 4 14

15 The GLM Procedure Output File (.lst) Class Level Information Class Levels Values diet Number of observations 12 Dependent Variable: wtgain Sum of Source DF Squares Mean Square F Value Pr > F Model Error Corrected Total R-Square Coeff Var Root MSE wtgain Mean STAT 514 Topic 4 15

16 SAS Default Plot STAT 514 Topic 4 16

17 SAS Diagnostics Plot STAT 514 Topic 4 17

18 Residual Plot Created with Gplot STAT 514 Topic 4 18

19 Using R - to perform a permuation test - trt <- c(1,1,1,2,2,2,2,2,3,3,3,3) trt <- as.factor(trt) data <- c(8,16,9,9,16,21,11,18,15,10,17,6) bc <- cbind(trt,data) ####### One-way ANOVA ####### result <- lm(data ~ trt) summary(result) fstat <- summary(result)$fstatistic[1] ###### Permutation test ######### fstat1 <- numeric(length=1000) for(isim in 1:1000) { data1 <- sample(data) bc1 <- cbind(trt,data1) result1 <- lm(data1 ~ trt) fstat1[isim] <- summary(result1)$fstatistic[1] } hist(fstat1,nclass=20) pvalue1 <- length(fstat1[fstat1 > fstat])/1000 print(pvalue1) STAT 514 Topic 4 19

20 Histogram of fstat1 Frequency Estimated P value= fstat1 STAT 514 Topic 4 20

21 Handout Example - tensile.sas data one; infile c:\saswork\data\tensile.dat ; input percent strength time; title1 Example Code ; proc print data=one; run; symbol1 v=circle i=none; title1 Plot of Strength vs Percent Blend ; proc gplot data=one; plot strength*percent/frame; run; proc boxplot; plot strength*percent/boxstyle=skeletal pctldef=4; proc glm plots=all; class percent; model strength=percent; output out=oneres p=pred r=res; run; proc sort; by pred; symbol1 v=circle i=sm50; title1 Residual Plot ; proc gplot; plot res*pred/frame; run; proc univariate data=oneres pctldef=4; var res; qqplot res / normal (L=1 mu=est sigma=est); histogram res / normal; run; symbol1 v=circle i=none; title1 Plot of residuals vs time ; proc gplot; plot res*time / vref=0 vaxis=-6 to 6 by 1; run; STAT 514 Topic 4 21

22 Output Chapter 3 Example Obs percent strength time STAT 514 Topic 4 22

23 Plots STAT 514 Topic 4 23

24 Plots STAT 514 Topic 4 24

25 The GLM Procedure Dependent Variable: strength Sum of Source DF Squares Mean Square F Value Pr > F Model <.0001 Error Corrected Total R-Square Coeff Var Root MSE strength Mean Source DF Type I SS Mean Square F Value Pr > F percent <.0001 Source DF Type III SS Mean Square F Value Pr > F percent <.0001 STAT 514 Topic 4 25

26 SAS Diagnostics Plot STAT 514 Topic 4 26

27 STAT 514 Topic 4 27

28 The UNIVARIATE Procedure Moments N 25 Sum Weights 25 Mean 0 Sum Observations 0 Std Deviation Variance Skewness Kurtosis Uncorrected SS Corrected SS Coeff Variation. Std Error Mean Tests for Location: Mu0=0 Test -Statistic p Value Student s t t 0 Pr > t Sign M 2.5 Pr >= M Signed Rank S 0.5 Pr >= S Fitted Distribution for res Parameters for Normal Distribution Goodness-of-Fit Tests for Normal Distribution Test ---Statistic p Value----- Kolmogorov-Smirnov D Pr > D Cramer-von Mises W-Sq Pr > W-Sq Anderson-Darling A-Sq Pr > A-Sq STAT 514 Topic 4 28

29 STAT 514 Topic 4 29

30 STAT 514 Topic 4 30

31 Using R - to perform a permuation test - tensile <- read.table("tensile.dat",col.names=c("percent","strength","time")) attach(tensile) percent <- as.factor(percent) ####### One-way ANOVA ####### result <- lm(strength ~ percent) summary(result) fstat <- summary(result)$fstatistic[1] ###### Permutation test ######### fstat1 <- numeric(length=1000) for(isim in 1:1000) { data1 <- sample(strength) result1 <- lm(data1 ~ percent) fstat1[isim] <- summary(result1)$fstatistic[1] } hist(fstat1,nclass=20) pvalue1 <- length(fstat1[fstat1 > fstat])/1000 print(pvalue1) STAT 514 Topic 4 31

32 Histogram of fstat1 Frequency Estimated P value< fstat1 STAT 514 Topic 4 32

33 Background Reading Analysis of variance: Montgomery 3.2 Decomposition of Sum of Squares: Montgomery Statistical Analysis: Montgomery Estimation of model parameters STAT 514 Topic 4 33