Lecture 3. Experiments with a Single Factor: ANOVA Montgomery 3-1 through 3-3

Transcription

1 Lecture 3. Experiments with a Single Factor: ANOVA Montgomery 3-1 through 3-3 Page 1

2 Tensile Strength Experiment Investigate the tensile strength of a new synthetic fiber. The factor is the weight percent of cotton used in the blend of the materials for the fiber and it has five levels. weight tensile strength of cotton total average Page 2

3 Data Layout for Single-Factor Experiments treatment observations totals averages 1 y 11 y 12 y 1n y 1. ȳ 1. 2 y 21 y 22 y 2n y 2. ȳ a y a1 y a2 y an y a. ȳ a.... Page 3

4 Analysis of Variance Interested in comparing several treatments. Could do numerous two-sample t-tests but this approach does not test equality of all treatments at once. ANOVA provides method of joint inference. Statistical Model: y ij = µ + τ i + ɛ ij i =1, 2...a j =1, 2,...n i µ - grand mean; τ i - ith treatment effect; ɛ ij N(0,σ 2 ) - error a Constraint: i=1 τ i =0. Hypotheses: H 0 : τ 1 = τ 2 =... = τ a =0 vs H 1 : τ i 0for at least one i Page 4

5 Partitioning y ij Notation y i. = n i j=1 y ij y i. = y i. /n i (treatment sample mean, or row mean) y.. = y ij y.. = y.. /N (grand sample mean) Decomposition of y ij : y ij = y.. +(y i. y.. )+(y ij y i. ) Estimates for parameters: So y ij as ˆµ +ˆτ i +ˆɛ ij. ˆµ = y.. It can be verified that a i=1 n iˆτ i =0 ni j=1 ˆɛ ij =0for all i ˆτ i =(y i. y.. ) ˆɛ ij = y ij y i. ( residual ) Page 5

6 Partitioning the Sum of Squares Recall y ij y.. =ˆτ i +ˆɛ ij =(y i. y.. )+(y ij y i. ) Can show i j (y ij y.. ) 2 = i n i(y i. y.. ) 2 + i j (y ij y i. ) 2 = i n iˆτ i 2 + i j ˆɛ2 ij. Total SS = Treatment SS + Error SS SS T =SS Treatments +SS E Derive test statistics for testing H 0 : look at SS Treatments = n iˆτ i 2 = n i (y i. y.. ) 2 Small if ˆτ i s are small If large then reject H 0, but how large is large? Standardize to account for inherent variability Page 6

7 Test Statistic F 0 = SS Treatments/(a 1) SS E /(N a) Mean squares: = ni (y i. y.. ) 2 /(a 1) (yij y i. ) 2 /(N a) MS Treatments = SS Treatments /(a 1); MS E = SS E /(N a) Can show that the expected values are E(MS E )=σ 2 E(MS Treatment )=σ 2 + n i τ 2 i /(a 1) Hence, MS E is always a good estimator for σ 2, that is, ˆσ 2 =MS E. Under H 0,MS Treatment is also an estimator for σ 2. F 0 is the ratio of the mean squares, under H 0, it should be around 1. Large F 0 implies deviation from H 0. What is the sampling distribution of F 0 under H 0? Page 7

8 Sampling Distribution for F 0 under H 0 Based on model: y ij N(µ + τ i,σ 2 ) and y i. N(µ + τ i,σ 2 /n i ) j (y ij y i. ) 2 /σ 2 χ 2 n i 1 Therefore: SS E / σ 2 = i j (y ij y i. ) 2 /σ 2 χ 2 N a Under H 0 : SS Treatment /σ 2 = i n i(y i. y.. ) 2 /σ 2 χ 2 a 1 SS E and SS Treatment are independent. So F 0 = SS Treatment/σ 2 (a 1) SS E /σ 2 (N a) = χ2 (a 1)/(a 1) χ 2 (N a)/(n a) F a 1,N a F -distribution with numerator df a 1 and denominator df N a. Page 8

9 Analysis of Variance (ANOVA) 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 balanced: SS T = y 2 ij y2../n ; SS Treatment = 1 n y 2 i. y 2../N SS E =SS T -SS Treatment If unbalanced: SS T = y 2 ij y2../n ; SS Treatment = y 2 i. n i y 2../N SS E =SS T -SS Treatment Decision Rule: IfF 0 >F α,a 1,N a then reject H 0 Page 9

10 Connection to Two-Sample t-test a=2 Consider the square of the t-test statistic t 2 0 = (ȳ 1. ȳ 2. ) 2 (s pool 1/n1 +1/n 2 ) 2 = n 1 n 2 n 1 +n 2 (ȳ 1. ȳ 2. ) 2 s 2 pool = n 1 n 2 n 1 +n 2 [(ȳ 1. ȳ.. ) (ȳ 2. ȳ.. )] 2 s 2 pool = n 1 n 2 n 1 +n 2 (ȳ 1. ȳ.. ) 2 + n 1n 2 n 1 +n 2 (ȳ 2. ȳ.. ) 2 2n 1n 2 n 1 +n 2 (ȳ 1. ȳ.. )(ȳ 2. ȳ.. ) s 2 pool Page 10

11 Consider ȳ.. = 1 n 1 + n 2 ( n 1 j=1 y 1j + n 2 j=1 y 2j )= n 1 n 1 + n 2 ȳ 1. + n 2 n 1 + n 2 ȳ 2. One has t 2 0 = [n 1 (ȳ 1. ȳ.. ) 2 + n 2 (ȳ 2. ȳ.. ) 2 ]/(2 1) [ n 1 j=1 (y 1j ȳ 1. ) 2 + n 2 j=1 (y 2j ȳ 2. ) 2 ]/(n 1 + n 2 2) = MS Treatment MS E When a =2, t 2 0 = F 0 When a =2, F -test and two-sample two-sided test are equivalent. Page 11

12 Example Twelve lambs are randomly assigned to three different diets. The weight gain (in two weeks) is recorded. Is there a difference between the diets? SS T = /12 = 246 Diet Diet 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 =4and N =12 SS Treatment = (33 2 / / /4) /12 = 36 SS E = = 210 F 0 = (36/2)/(210/9) = 0.77 P-value > 0.25 Page 12

13 Using SAS (lambs.sas) option nocenter ps=65 ls=80; data lambs; input diet wtgain; cards; ; Page 13

14 symbol1 bwidth=5 i=box; axis1 offset=(5); proc gplot; plot wtgain*diet / frame haxis=axis1; proc glm; class diet; model wtgain=diet; output out=diag r=res p=pred; proc gplot; plot res*diet /frame haxis=axis1; proc sort; by pred; symbol1 v=circle i=sm50; proc gplot; plot res*pred / haxis=axis1; run; Page 14

15 SAS Output The GLM Procedure 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 Source DF Type I SS Mean Square F Value Pr > F diet Source DF Type III SS Mean Square F Value Pr > F diet Page 15

16 Side-by-Side Boxplot Page 16

17 Residual Plot Page 17

18 Tensile Strength Experiment options ls=80 ps=60 nocenter; goptions device=win target=winprtm rotate=landscape ftext=swiss hsize=8.0in vsize=6.0in htext=1.5 htitle=1.5 hpos=60 vpos=60 horigin=0.5in vorigin=0.5in; data one; infile c:\saswork\data\tensile.dat ; input percent strength time; title1 Tensile Strength Example ; 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; Page 18

19 plot strength*percent/boxstyle=skeletal pctldef=4; proc glm; 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 normal; 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; Page 19

20 Tensile Strength Example Obs percent strength time : : : : Page 20

21 Scatter Plot Page 21

22 Side-by-Side Plot Page 22

23 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 Page 23

24 Residual Plot Page 24

25 The UNIVARIATE Procedure Variable: res 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 Normality Test --Statistic p Value Shapiro-Wilk W Pr < W Kolmogorov-Smirnov D Pr > D Cramer-von Mises W-Sq Pr > W-Sq Anderson-Darling A-Sq Pr > A-Sq Page 25

26 Histogram of Residuals Page 26

27 QQ Plot Page 27

28 Time Serial Plot Page 28